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An on-line boost converter calculator Image from this website. The Operating Mode The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefo...
Answer
#20: Post edited
by
Andy aka
·
2020-09-21T16:51:16Z (over 4 years ago)
# The Schematic- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- - The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase increases.
- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
$$D = t_1*F_{SW}$$- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # [An on-line boost converter calculator](http://www.stades.co.uk/Boost%20converter/Boost%20calculator.html)
- 
- Image from [this website](http://www.stades.co.uk/Boost%20converter/Boost%20calculator.html).
- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- - The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase increases.
- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = t_1\cdot F_{SW}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#19: Post edited
by
Andy aka
·
2020-09-16T16:54:29Z (over 4 years ago)
- # The Schematic
- 
- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- - The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase increases.
- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
$$D = \dfrac{t_1}{F_{SW}}$$- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # The Schematic
- 
- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- - The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase increases.
- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = t_1*F_{SW}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#18: Post edited
by
Andy aka
·
2020-06-27T10:58:32Z (over 4 years ago)
- # The Schematic
- 
- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase lengthens.- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # The Schematic
- 
- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- - The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase increases.
- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#17: Post edited
by
Andy aka
·
2020-06-27T10:57:11Z (over 4 years ago)
- # The Schematic
- 
- # The Operating Mode
The circuit can operate in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
**DCM** "hold" is when the all the inductor's stored energy has transferred to the load before the next switching cycle begins. If the hold-phase becomes zero and the load current is still rising, **DCM** will "slip into" **CCM**. This can cause instability but, it's beyond the scope of this question.- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # The Schematic
- 
- # The Operating Mode
- The circuit operates in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- -----
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- - The"hold" phase begins when the all the inductor's energy is transferred to the load. If the load "demands" more energy/power, then the "hold" phase must shorten and the charge phase lengthen. If the load requires less energy, then the "hold" phase lengthens.
- - Therefore, in order to maintain the desired output voltage, the duty cycle has to be continually adjusted to accommodate load current changes.
- - The slopes of the currents (\$\frac{di}{dt}\$) are non-negotiable because they define the input and output voltages as per Faraday's law of induction.
- - If the input voltage reduced by 10% then the current charge slope (red) also must reduce by 10%.
- - If the output voltage requirement increased by 20% then the current transfer slope (green) must also increase by 20%.
- - If the input voltage changes, duty cycle must also be adjusted to maintain the correct output voltage for a given load current. This is also true for **CCM** operation.
- - If the hold-phase becomes zero and the load current demand is still rising, **DCM** will "slip into" **CCM**. This can cause output voltage instability.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#16: Post edited
by
Andy aka
·
2020-06-23T16:53:31Z (over 4 years ago)
- # The Schematic
- 
- # The Operating Mode
- The circuit can operate in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- **DCM** "hold" is when the all the inductor's stored energy has transferred to the load before the next switching cycle begins. If the hold-phase becomes zero and the load current is still rising, **DCM** will "slip into" **CCM**. This can cause instability but, it's beyond the scope of this question.
- -----
**CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # The Schematic
- 
- # The Operating Mode
- The circuit can operate in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" is when the all the inductor's stored energy has transferred to the load before the next switching cycle begins. If the hold-phase becomes zero and the load current is still rising, **DCM** will "slip into" **CCM**. This can cause instability but, it's beyond the scope of this question.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#15: Post edited
by
Andy aka
·
2020-06-23T15:23:29Z (over 4 years ago)
- # The Schematic
- 
**The Operating Mode**The circuit can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
**DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
Then solve the emerging quadratic equation to get: -- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # The Schematic
- 
- # The Operating Mode
- The circuit can operate in either of two modes but, the voltage transfer equation is different for each mode. Therefore, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" is when the all the inductor's stored energy has transferred to the load before the next switching cycle begins. If the hold-phase becomes zero and the load current is still rising, **DCM** will "slip into" **CCM**. This can cause instability but, it's beyond the scope of this question.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get this **DCM** formula: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- **Conclusions**
- - The output voltage in **DCM** is observably load dependent.
- - The output voltage in **CCM** is not load dependent.
- When dropping into **DCM** from **CCM**, this can lead to output instabilities especially if load variations cause the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#14: Post edited
by
Andy aka
·
2020-06-23T14:55:09Z (over 4 years ago)
# Background- **The Operating Mode**
- The circuit can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.- **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # The Schematic
- 
- **The Operating Mode**
- The circuit can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents. Duty cycle lower than **CCM**.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavier load currents. Duty cycle higher than **DCM**.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#13: Post edited
by
Andy aka
·
2020-06-23T14:51:58Z (over 4 years ago)
- # Background
- **The Operating Mode**
The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- 
- # Background
- **The Operating Mode**
- The circuit can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#12: Post edited
by
Andy aka
·
2020-06-23T13:09:57Z (over 4 years ago)
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
This means that a duty cycle of (say) 0.5, **DCM** won't deliver the same output voltage as **CCM** for a given load. This is an important distinction.- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- Lighter loads are serviced in **DCM**- Heavier loads are serviced in **CCM**- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that for a given load and duty cycle, **DCM** won't deliver the same output voltage as **CCM**. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are usually serviced in **DCM**
- - Heavier loads are usually serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#11: Post edited
by
Andy aka
·
2020-06-23T12:52:13Z (over 4 years ago)
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that a duty cycle of (say) 0.5, **DCM** won't deliver the same output voltage as **CCM** for a given load. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
<sub>At this point I'm assuming that the diode is ideal. In the real world we would add-on a few watts for the diode forward conduction losses. I'm going to ignore this for now because, any decent boost-converter will slightly increase its duty cycle (D) to accommodate the forward volt-drop if the diode. In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}\normalsize}}$$
- This means that a duty cycle of (say) 0.5, **DCM** won't deliver the same output voltage as **CCM** for a given load. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>In the real world we would add-on a few watts for diode forward conduction losses. I'm going to ignore this because, any decent boost-converter will raise its duty cycle (D) to accommodate diode losses. In other words, I can assume that the control loop will do what it needs to do but, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- Also recall that the **boundary** and **CCM** formula is this: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-D}}$$
- And this tells us that upon entering **DCM** from **CCM**, output voltage regulation becomes very-much load dependent and this can lead to output stability problems especially if load changes caused the mode to continually fluctuate.
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#10: Post edited
by
Andy aka
·
2020-06-23T12:12:27Z (over 4 years ago)
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
$$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}ormalsize}}$$- This means that a duty cycle of (say) 0.5, **DCM** won't deliver the same output voltage as **CCM** for a given load. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we would add-on a few watts for the diode forward conduction losses. I'm going to ignore this for now because, any decent boost-converter will slightly increase its duty cycle (D) to accommodate the forward volt-drop if the diode. In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer equations }}
- ormalsize}}$$
- This means that a duty cycle of (say) 0.5, **DCM** won't deliver the same output voltage as **CCM** for a given load. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we would add-on a few watts for the diode forward conduction losses. I'm going to ignore this for now because, any decent boost-converter will slightly increase its duty cycle (D) to accommodate the forward volt-drop if the diode. In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#9: Post edited
by
Andy aka
·
2020-06-22T14:56:13Z (over 4 years ago)
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to remember this.- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
The time taken for phase 1 to complete is: -- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
# How much power is transferred to the load?- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
<sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub><sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>- -----
The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a duty cycle of (say) 0.5, **DCM** won't deliver the same output voltage as **CCM** for a given load. This is an important distinction.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- This means that the time taken for phase 1 to complete (\$t_1\$) is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the actual load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we would add-on a few watts for the diode forward conduction losses. I'm going to ignore this for now because, any decent boost-converter will slightly increase its duty cycle (D) to accommodate the forward volt-drop if the diode. In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed \$\boxed{\text{45 watts}}\$.</sub>
- -----
- The \$\boxed{\text{45 watts}}\$ figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#8: Post edited
by
Andy aka
·
2020-06-22T12:55:21Z (over 4 years ago)
- # Background
- **The Operating Mode**
The boost circuit in the question can operate in two modes but, the voltage transfer equation is not consistent for both modes. In other words, you need to know the operating mode before you can calculate duty cycle. Operating modes: -- **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps- **CCM** (continuous conduction mode) - inductor current remains above zero amps- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
The **DCM** hold phase occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
**CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, this should be noted: -- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to recognize this.- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- The boost converter schematic in the question can operate in two modes but, the voltage transfer equation is different for each mode. In other words, you need to establish the operating mode before you can calculate duty cycle. The operating modes are: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps. Accommodates light to medium load currents.
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps. Accommodates medium to heavy load currents.
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **DCM** "hold" occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, here's a reminder: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to remember this.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#7: Post edited
by
Andy aka
·
2020-06-22T12:46:22Z (over 4 years ago)
- # Background
- **The Operating Mode**
This must be established before calculating the duty cycle because, the voltage transfer equation is not consistent for both operating modes (despite the circuit being identical). In other words, you need to know the operating mode before you can calculate duty cycle. Operating modes: -- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- The **DCM** hold phase occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to recognize this.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- The boost circuit in the question can operate in two modes but, the voltage transfer equation is not consistent for both modes. In other words, you need to know the operating mode before you can calculate duty cycle. Operating modes: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- The **DCM** hold phase occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to recognize this.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#6: Post edited
by
Andy aka
·
2020-06-22T12:13:08Z (over 4 years ago)
- # Background
**First - determine the operating mode**- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- The **DCM** hold phase occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to recognize this.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **The Operating Mode**
- This must be established before calculating the duty cycle because, the voltage transfer equation is not consistent for both operating modes (despite the circuit being identical). In other words, you need to know the operating mode before you can calculate duty cycle. Operating modes: -
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- The **DCM** hold phase occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to recognize this.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#5: Post edited
by
Andy aka
·
2020-06-22T11:50:15Z (over 4 years ago)
- # Background
- **First - determine the operating mode**
- **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps- **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.- Lighter loads are serviced in **DCM**- Heavier loads are serviced in **CCM**- **CCM** is a natural outcome when **DCM** can't service the load- **DCM** is a natural outcome when **CCM** can't service the load-----**The DCM hold phase**This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -$$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as a **CCM** duty cycle of 0.5. It's important to recognize this.The next step is to decide which mode the boost converter operates in. This is largely determined by the load power requirement.- -----
To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We need to ask this hypothetical question: -Can the **boundary condition** deliver enough power to the load to sustain the output voltage?- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **First - determine the operating mode**
- - **DCM** (discontinuous conduction mode) - inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) - inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- The **DCM** hold phase occurs when the inductor has transferred all its stored energy into the load and, is waiting for the next switching cycle to begin.
- -----
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- **CCM** doesn't have a "hold" phase because it's continually charging and transferring energy however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** at the same duty cycle. It's important to recognize this.
- -----
- **CCM and DCM summary**
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the heavier load
- - **DCM** is a natural outcome when **CCM** can't service the lighter load
- The next step is to decide which operating mode fulfills the requirements of the proposed design.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We then need to ask this hypothetical question: -
- $$\color{red}{\boxed{\large{\text{At the boundary, is the load power sufficient to sustain } V_{OUT}?}\normalsize}}$$
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- **Voltage Transfer Equation**
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#4: Post edited
by
Andy aka
·
2020-06-22T10:33:13Z (over 4 years ago)
- # Background
- **First - determine the operating mode**
- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
**CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the load
- - **DCM** is a natural outcome when **CCM** can't service the load
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as a **CCM** duty cycle of 0.5. It's important to recognize this.
- The next step is to decide which mode the boost converter operates in. This is largely determined by the load power requirement.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We need to ask this hypothetical question: -
- Can the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **First - determine the operating mode**
- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor current waveform rising or falling: -
- 
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the load
- - **DCM** is a natural outcome when **CCM** can't service the load
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}\normalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as a **CCM** duty cycle of 0.5. It's important to recognize this.
- The next step is to decide which mode the boost converter operates in. This is largely determined by the load power requirement.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We need to ask this hypothetical question: -
- Can the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}\normalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}\normalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}\normalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#3: Post edited
by
Andy aka
·
2020-06-22T09:15:08Z (over 4 years ago)
- # Background
- **First - determine the operating mode**
The two operating modes are either: -- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -
- 
Only one operating mode can deliver 16 volts to the 5 Ω load from a 12 volt power supply with \$L\$ and \$F_{SW}\$ as specified.- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
$$\color{green}{\boxed{\normalsize{\text{ CCM and DCM have different input-to-output voltage transfer functions }}ormalsize}}$$This means that (for example) a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** using a duty cycle of 0.5. It's important to recognize this.The next step is to analyse the input supply voltage (\$V_{IN}\$), the switching frequency (\$F_{SW}\$), the inductance (\$L\$) and the load resistance (\$R\$) in order to decide which mode the boost converter might operate in.- -----
<sub>It should also be noted that there are many examples where a boost converter will switch between the two modes when there is a more complex load changing scenario.</sub>-----To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We should therefore ask this question: -Will the **boundary condition** deliver enough power to the load to sustain the output voltage?- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
This formula tells us that if we were operating the boost converter on the borderline between **DCM** and **CCM**, D is revealed. So, let's calculate it using the numbers in the question: -$$D = \dfrac{16 - 12}{16} = 0.25$$Knowing D and \$F_{SW}\$, we can then determine the time (\$t_1\$) required to charge the inductor during phase 1. This will lead us to uncover inductor peak current, \$I_{PK}\$.With \$F_{SW}\$ at 100 kHz (a period of 10 μs) and D = 0.25, \$t_1\$ = 2.5 μs.- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- \$V = V_{IN}\$ (the input supply voltage)- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
$$V_{IN} = L\dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L}$$From that we find that \$I_{PK}\$ = 30 amps.- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
The inductor's stored energy of 450 μJ is transferred from the inductor to the load 100,000 times per second (\$F_{SW}\$). This is an equivalent continuous power transfer of 45 watts.-----\$\color{magenta}{\large{\text{Remember this is at the boundary condition.}}ormalsize}\$# How much power should the converter transfer?This may not be as obvious as you think. On one hand you could say that the output voltage is 16 volts and the load is 5 Ω therefore the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the charging of the inductor.- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
<sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 51.2 watts.</sub>- -----
The 51.2 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -\$\color{red}{\large{\text{The boost converter will operate in DCM.}}ormalsize}\$- # What DCM duty cycle is needed?
We need a power uplift of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
$$I_{PK} = \sqrt{\dfrac{2\dot W}{L}} = 16 \text{ amps}$$- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
\$\color{blue}{\large{\text{Given that 100 kHz has a period of 10 μs, D will be 0.1333.}}ormalsize}\$- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **First - determine the operating mode**
- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -
- 
- - Only one mode will deliver 16 volts to the 5 Ω load with \$V_{IN}\$, \$L\$ and \$F_{SW}\$ as specified.
- - Lighter loads are serviced in **DCM**
- - Heavier loads are serviced in **CCM**
- - **CCM** is a natural outcome when **DCM** can't service the load
- - **DCM** is a natural outcome when **CCM** can't service the load
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
- $$\color{red}{\boxed{\large{\text{ CCM and DCM have different voltage transfer functions }}
- ormalsize}}$$
- This means that a **DCM** duty cycle of 0.5 won't deliver the same output voltage as a **CCM** duty cycle of 0.5. It's important to recognize this.
- The next step is to decide which mode the boost converter operates in. This is largely determined by the load power requirement.
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We need to ask this hypothetical question: -
- Can the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{\text{16 volts - 12 volts }}{\text{16 volts}} = 0.25$$
- The time taken for phase 1 to complete is: -
- $$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.25}{100\text{ kHz}} = 2.5\text{ μs}$$
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage = 12 volts)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\cdot \dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{30 amps}$$
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred to the load 100,000 times per second (\$F_{SW}\$).
- $$\color{red}{\boxed{\large{\text{ This is an equivalent continuous power transfer of 45 watts }}
- ormalsize}}$$
- # How much power is transferred to the load?
- This may not be as obvious as you think; you might say that the output voltage is 16 volts and the load is 5 Ω therefore, the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the inductor's stored energy.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 45 watts.</sub>
- -----
- The 45 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- $$\color{red}{\boxed{\large{\text{ The boost converter will operate in DCM }}
- ormalsize}}$$
- # What DCM duty cycle is needed?
- We need a **power uplift** of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- This then leads us to find duty cycle because: -
- $$D = \dfrac{t_1}{F_{SW}}$$
- $$\color{red}{\boxed{\large{\text{ Given that 100 kHz has a period of 10 μs, D will be 0.1333 }}
- ormalsize}}$$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#2: Post edited
by
Andy aka
·
2020-06-22T07:37:16Z (over 4 years ago)
- # Background
**A bit about the operating mode**Start by determining the only operating mode that can be used to deliver 16 volts to a 5 Ω load from a 12 volt power supply with an inductor of 1 uH operating at 100 kHz. The operating mode options are (in case you didn't know): -- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -
- 
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
- $$\color{green}{\boxed{\normalsize{\text{ CCM and DCM have different input-to-output voltage transfer functions }}\normalsize}}$$
- This means that (for example) a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** using a duty cycle of 0.5. It's important to recognize this.
- The next step is to analyse the input supply voltage (\$V_{IN}\$), the switching frequency (\$F_{SW}\$), the inductance (\$L\$) and the load resistance (\$R\$) in order to decide which mode the boost converter might operate in.
- -----
- <sub>It should also be noted that there are many examples where a boost converter will switch between the two modes when there is a more complex load changing scenario.</sub>
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We should therefore ask this question: -
- Will the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- This formula tells us that if we were operating the boost converter on the borderline between **DCM** and **CCM**, D is revealed. So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{16 - 12}{16} = 0.25$$
- Knowing D and \$F_{SW}\$, we can then determine the time (\$t_1\$) required to charge the inductor during phase 1. This will lead us to uncover inductor peak current, \$I_{PK}\$.
- With \$F_{SW}\$ at 100 kHz (a period of 10 μs) and D = 0.25, \$t_1\$ = 2.5 μs.
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L}$$
- From that we find that \$I_{PK}\$ = 30 amps.
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred from the inductor to the load 100,000 times per second (\$F_{SW}\$). This is an equivalent continuous power transfer of 45 watts.
- -----
- \$\color{magenta}{\large{\text{Remember this is at the boundary condition.}}\normalsize}\$
- # How much power should the converter transfer?
- This may not be as obvious as you think. On one hand you could say that the output voltage is 16 volts and the load is 5 Ω therefore the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the charging of the inductor.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 51.2 watts.</sub>
- -----
- The 51.2 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- \$\color{red}{\large{\text{The boost converter will operate in DCM.}}\normalsize}\$
- # What DCM duty cycle is needed?
- We need a power uplift of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\dot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- \$\color{blue}{\large{\text{Given that 100 kHz has a period of 10 μs, D will be 0.1333.}}\normalsize}\$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **First - determine the operating mode**
- The two operating modes are either: -
- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -
- 
- Only one operating mode can deliver 16 volts to the 5 Ω load from a 12 volt power supply with \$L\$ and \$F_{SW}\$ as specified.
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
- $$\color{green}{\boxed{\normalsize{\text{ CCM and DCM have different input-to-output voltage transfer functions }}\normalsize}}$$
- This means that (for example) a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** using a duty cycle of 0.5. It's important to recognize this.
- The next step is to analyse the input supply voltage (\$V_{IN}\$), the switching frequency (\$F_{SW}\$), the inductance (\$L\$) and the load resistance (\$R\$) in order to decide which mode the boost converter might operate in.
- -----
- <sub>It should also be noted that there are many examples where a boost converter will switch between the two modes when there is a more complex load changing scenario.</sub>
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We should therefore ask this question: -
- Will the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- This formula tells us that if we were operating the boost converter on the borderline between **DCM** and **CCM**, D is revealed. So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{16 - 12}{16} = 0.25$$
- Knowing D and \$F_{SW}\$, we can then determine the time (\$t_1\$) required to charge the inductor during phase 1. This will lead us to uncover inductor peak current, \$I_{PK}\$.
- With \$F_{SW}\$ at 100 kHz (a period of 10 μs) and D = 0.25, \$t_1\$ = 2.5 μs.
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L}$$
- From that we find that \$I_{PK}\$ = 30 amps.
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred from the inductor to the load 100,000 times per second (\$F_{SW}\$). This is an equivalent continuous power transfer of 45 watts.
- -----
- \$\color{magenta}{\large{\text{Remember this is at the boundary condition.}}\normalsize}\$
- # How much power should the converter transfer?
- This may not be as obvious as you think. On one hand you could say that the output voltage is 16 volts and the load is 5 Ω therefore the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the charging of the inductor.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 51.2 watts.</sub>
- -----
- The 51.2 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- \$\color{red}{\large{\text{The boost converter will operate in DCM.}}\normalsize}\$
- # What DCM duty cycle is needed?
- We need a power uplift of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\dot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- \$\color{blue}{\large{\text{Given that 100 kHz has a period of 10 μs, D will be 0.1333.}}\normalsize}\$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
#1: Post edited
by
Andy aka
·
2020-06-21T21:20:13Z (over 4 years ago)
- # Background
**A bit about the Modes**Start by determining the only operating mode that can be used to deliver 16 volts to a 5 Ω load from a 12 volt power supply with an inductor of 1 uH operating at 100 kHz. The operating modes are (in case you didn't know): -- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- **CCM** (continuous conduction mode) i.e. inductor current always flows**DCM** means there are three distinct phases per switching cycle; charge, transfer and hold. The length of the hold phase accommodates load current variations: -- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -
- 
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
\$\color{green}{\normalsize{\text{CCM and DCM have different input-to-output voltage transfer functions}}ormalsize}\$- This means that (for example) a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** using a duty cycle of 0.5. It's important to recognize this.
- The next step is to analyse the input supply voltage (\$V_{IN}\$), the switching frequency (\$F_{SW}\$), the inductance (\$L\$) and the load resistance (\$R\$) in order to decide which mode the boost converter might operate in.
- -----
<sub>It should also be noted that there are many examples where a boost converter will switch between the two modes when there is a more complex load scenario.</sub>- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We should therefore ask this question: -
- Will the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
$$D = \dfrac{V_O - V_{IN}}{V_O}$$- This formula tells us that if we were operating the boost converter on the borderline between **DCM** and **CCM**, D is revealed. So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{16 - 12}{16} = 0.25$$
So, it's important to calculate D because knowing \$F_{SW}\$, we can determine the time (\$t_1\$) required to charge the inductor during phase 1. This will lead us to uncover inductor peak current, \$I_{PK}\$.- With \$F_{SW}\$ at 100 kHz (a period of 10 μs) and D = 0.25, \$t_1\$ = 2.5 μs.
- Knowing \$t_1\$ we then ask this question: -
Will we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
$$V = L\dfrac{di}{dt}$$- - \$V = V_{IN}\$ (the input supply voltage)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L}$$
- From that we find that \$I_{PK}\$ = 30 amps.
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
$$W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2$$- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred from the inductor to the load 100,000 times per second (\$F_{SW}\$). This is an equivalent continuous power transfer of 45 watts.
- -----
- \$\color{magenta}{\large{\text{Remember this is at the boundary condition.}}\normalsize}\$
- # How much power should the converter transfer?
- This may not be as obvious as you think. On one hand you could say that the output voltage is 16 volts and the load is 5 Ω therefore the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the charging of the inductor.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 51.2 watts.</sub>
- -----
- The 51.2 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- \$\color{red}{\large{\text{The boost converter will operate in DCM.}}\normalsize}\$
- # What DCM duty cycle is needed?
- We need a power uplift of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\dot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- \$\color{blue}{\large{\text{Given that 100 kHz has a period of 10 μs, D will be 0.1333.}}\normalsize}\$
- Summarizing the above process, we get: -
$$D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}$$This is my preferred formula but plenty of time you see this: -$$D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}$$- And, you often see this: -
$$D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}$$- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.
- # Background
- **A bit about the operating mode**
- Start by determining the only operating mode that can be used to deliver 16 volts to a 5 Ω load from a 12 volt power supply with an inductor of 1 uH operating at 100 kHz. The operating mode options are (in case you didn't know): -
- - **DCM** (discontinuous conduction mode) i.e. inductor current will fall to zero amps
- - **CCM** (continuous conduction mode) i.e. inductor current remains above zero amps
- **DCM** has three distinct phases per switching cycle; charge, transfer and hold. The length of the hold-phase accommodates load current variations: -
- 
- **CCM** has just two phases; charge and transfer. Load current changes are accommodated by the whole inductor charge/discharge waveform rising or falling: -
- 
- -----
- **The DCM hold phase**
- This is when the inductor has transferred all its stored energy into the load and is waiting for the next switching cycle to begin. **CCM** doesn't have a "hold" phase because it is continually charging and transferring however, this should be noted: -
- $$\color{green}{\boxed{\normalsize{\text{ CCM and DCM have different input-to-output voltage transfer functions }}
- ormalsize}}$$
- This means that (for example) a **DCM** duty cycle of 0.5 won't deliver the same output voltage as **CCM** using a duty cycle of 0.5. It's important to recognize this.
- The next step is to analyse the input supply voltage (\$V_{IN}\$), the switching frequency (\$F_{SW}\$), the inductance (\$L\$) and the load resistance (\$R\$) in order to decide which mode the boost converter might operate in.
- -----
- <sub>It should also be noted that there are many examples where a boost converter will switch between the two modes when there is a more complex load changing scenario.</sub>
- -----
- To perform this next step we consider the **boundary condition**. This is where **DCM** meets **CCM**; the inductor has delivered all it's energy to the output and the switching cycle restarts immediately. We should therefore ask this question: -
- Will the **boundary condition** deliver enough power to the load to sustain the output voltage?
- If the answer is yes, then we will operate in **DCM**. If the answer is no we will operate in **CCM**.
- # Boundary condition
- 
- Right at the bottom of the picture is an important formula: -
- $$\boxed{D = \dfrac{V_O - V_{IN}}{V_O}}$$
- This formula tells us that if we were operating the boost converter on the borderline between **DCM** and **CCM**, D is revealed. So, let's calculate it using the numbers in the question: -
- $$D = \dfrac{16 - 12}{16} = 0.25$$
- Knowing D and \$F_{SW}\$, we can then determine the time (\$t_1\$) required to charge the inductor during phase 1. This will lead us to uncover inductor peak current, \$I_{PK}\$.
- With \$F_{SW}\$ at 100 kHz (a period of 10 μs) and D = 0.25, \$t_1\$ = 2.5 μs.
- Knowing \$t_1\$ we then ask this question: -
- Can we acquire enough energy in the inductor to sustain the 5 Ω load at 16 volts with a charge time of 2.5 μs? To answer this we must find the peak inductor current (\$I_{PK}\$) during phase 1.
- # Calculating \$I_{PK}\$ in the boundary condition
- Using Faraday's equation for an inductor we know: -
- $$\boxed{V = L\cdot\dfrac{di}{dt}}$$
- - \$V = V_{IN}\$ (the input supply voltage)
- - \$di = I_{PK}\$ (the peak current at the end of phase 1)
- - \$dt = t_1\$ = 2.5 μs
- - \$L\$ is inductance (1 μH in this example)
- Hence:
- $$V_{IN} = L\dfrac{I_{PK}}{t_1} \therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L}$$
- From that we find that \$I_{PK}\$ = 30 amps.
- # The stored energy (W) in boundary condition
- The inductor's stored energy (W) is found using this well-known formula: -
- $$\boxed{W = \dfrac{1}{2}\cdot L\cdot I_{PK}^2}$$
- If we plug the numbers in for current and inductance we get an energy figure of 450 μJ.
- # The power transfer in the boundary condition
- The inductor's stored energy of 450 μJ is transferred from the inductor to the load 100,000 times per second (\$F_{SW}\$). This is an equivalent continuous power transfer of 45 watts.
- -----
- \$\color{magenta}{\large{\text{Remember this is at the boundary condition.}}\normalsize}\$
- # How much power should the converter transfer?
- This may not be as obvious as you think. On one hand you could say that the output voltage is 16 volts and the load is 5 Ω therefore the load power is 16\$^2\$/5 = 51.2 watts. That is true but, it isn't the power that actually needs to be transferred via the charging of the inductor.
- In fact, the inductor only needs to "uplift" the output from 12 volts to 16 volts (Δ 4 volts) hence, the power is less. So, if we calculated the load current (16/5), we get 3.2 amps. Therefore, the power-uplift required from the inductor is 4 volts x 3.2 amps = 12.8 watts.
- -----
- <sub>At this point I'm assuming that the diode is ideal. In the real world we might need to add-on a few more watts for the diode forward conduction losses. I'm going to ignore it for now because, any decent boost-converter will continually adjust the duty cycle (D) to accommodate changes in the output voltage and, the "extra" voltage due to diode forward should be accommodated.</sub>
- <sub>In other words, I can sweep this little problem away and "know" that the control loop will do what it needs to do. But, I should check that the likely power needed by the load (plus that dissipated by the diode) does not exceed 51.2 watts.</sub>
- -----
- The 51.2 watts figure is the maximum power that can be transferred at the boundary condition. Clearly 12.8 watts plus a few more watts is not going to exceed this therefore we can say: -
- \$\color{red}{\large{\text{The boost converter will operate in DCM.}}\normalsize}\$
- # What DCM duty cycle is needed?
- We need a power uplift of 12.8 watts to sustain 16 volts across a 5 Ω output load. So now we reverse back from 12.8 watts and calculate energy transfer (W) per switching cycle.
- $$W = \dfrac{12.8\text{ watts}}{100\text{ kHz}} = 128\text{μJ}$$
- This means that the peak current in the inductor is: -
- $$I_{PK} = \sqrt{\dfrac{2\dot W}{L}} = 16 \text{ amps}$$
- And, given that we know \$V_{IN}\$ and \$L\$, we can calculate \$t_1\$ by: -
- $$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = 1.333 \text{ μs}$$
- \$\color{blue}{\large{\text{Given that 100 kHz has a period of 10 μs, D will be 0.1333.}}\normalsize}\$
- Summarizing the above process, we get: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot P_{UPLIFT}}}$$
- The above is my preferred formula but, often you see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{2\cdot L\cdot F_{SW}\cdot (V_{OUT}-V_{IN})\cdot I_{LOAD}}}$$
- And, you often see this: -
- $$\boxed{D = \dfrac{1}{V_{IN}}\cdot\sqrt{\dfrac{2\cdot L\cdot F_{SW}\cdot V_{OUT}\cdot (V_{OUT}-V_{IN})}{R_{LOAD}}}}$$
- To calculate the voltage transfer equation, take the above formula and re-arrange to get this: -
- $$\dfrac{D^2\cdot R_{LOAD}\cdot V_{IN}^2}{2\cdot L\cdot F_{SW}} = V_{OUT}\cdot(V_{OUT} - V_{IN})$$
- $$= \dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}} = \dfrac{V_{OUT}^2}{V_{IN}^2} - \dfrac{V_{OUT}}{V_{IN}}$$
- Then solve the emerging quadratic equation to get: -
- $$\boxed{\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} +\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}}$$
- # Simulation
- This is the schematic: -
- 
- This is a transient analysis of the resulting waveform: -
- 
- As you can see, the output voltage (\$V_{O}\$ in blue) hits 16 volts very nicely and, as expected, the peak inductor current (\$I_{PK}\$ in green) is 16 amps.
- Transient analysis with non-ideal (UF5408) diode: -
- 
- Now, the output voltage is around 15 volts due to the forward conduction diode losses. But, as I said earlier, the control-loop that monitors \$V_{OUT}\$ will make the necessary adjustments to D to ensure that 16 volts is produced.