Q&A

How do I design a flyback converter? What are the basics I should know?

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I need to design a simple flyback converter like this: -

I want to control duty cycle from another circuit such as an MCU but, I'm unsure how to proceed. My requirements are: -

• Input voltage 125 volts
• Output voltage 500 volts
• Load resistance is 10 kohm
• Transformer turns ratio 1:1
• Primary/secondary inductance 1 mH
• Operating frequency 100 kHz
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Equivalent Circuit, 1:1 transformer

Start by simplifying the circuit and change the transformer to a single inductor. This simplification ignores the benefits of isolation but, flyback converter theory is about transferring energy and not fundamentally about isolation: -

We also don't need to worry about the output polarity being inverted because, when a transformer is used in the real circuit, isolation and correct DC polarity are delivered

The Operating mode, 1:1 transformer

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 (red), transfer (green) and hold. The length of the hold-phase accommodates load current variations: -

CCM has just two phases; charge (red) and transfer (green). Load current changes are accommodated by the whole of the inductor current waveform rising or falling: -

CCM and DCM summary

Only one mode will deliver 500 volts to the 10 kΩ 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 do this we consider the boundary condition.

Boundary condition, 1:1 transformer

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: -

$$\color{red}{\boxed{\large\text{At the boundary, is the load power sufficient to sustain V_{OUT}}}}$$

If the answer is yes, then we will operate in DCM. If the answer is no we will operate in CCM.

Boundary Power Transfer Analysis

We need to find the input output relationship and duty cycle in the boundary condition. Simple analysis tells us: -

$$\text{boundary condition }\Bigg\rvert\dfrac{V_O}{V_{IN}} = \dfrac{D}{1-D}\text{ and } D = \dfrac{V_O}{V_{IN}+V_O}\Bigg\rvert\text{ boundary condition}$$

Calculate D using the values in the question: -

$$D = \dfrac{V_O}{V_{IN}+V_O} = \dfrac{500}{125+500} = 0.8000$$

This also delivers $t_1$ (time taken to charge the inductor): -

$$t_1 = \dfrac{D}{F_{SW}} = \dfrac{0.8}{\text{100 kHz}} = \text{8 μs}$$

Calculating $I_{PK}$ in the boundary condition

Using Faraday's equation for an inductor we know: -

$$\boxed{V = L\cdot\dfrac{di}{dt}}$$

• $V = V_{IN}$ = 125 volts
• $di = I_{PK}$ (the peak current at the end of phase 1)
• $dt = t_1$ = 8 μs
• $L\ =$ inductance (1 mH in this example)

Hence:

$$V_{IN} = L\cdot\dfrac{I_{PK}}{t_1}\therefore I_{PK} = \dfrac{t_1\cdot V_{IN}}{L} = \text{ 1.00 amps}$$

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 500 μJ.

The power transfer in the boundary condition

Inductor stored energy of 500 μJ is transferred to the load 100,000 times per second ($F_{SW}$).

$$\color{red}{\boxed{\large\text{This is an equivalent continuous power transfer of 50 watts}}}$$

How much power is transferred to the actual load?

Unlike the boost converter, all the load power comes through charging the magnetics hence the load requires 500$^2$/10 kΩ = 25 watts. This is significantly less than the power that could be transferred in the boundary condition therefore: -

$$\color{red}{\boxed{\large\text{The flyback converter will operate in DCM}}}$$

What DCM duty cycle is needed?

We need to deliver a power of 25 watts to sustain 500 volts across a 10 kΩ output load. So now we reverse back from 25 watts and calculate energy transfer (W) per switching cycle.

$$W = \dfrac{\text{25 watts}}{\text{100 kHz}} = \text{250 μJ}$$

This means that the peak current in the inductor is: -

$$I_{PK} = \sqrt{\dfrac{2\cdot W}{L}} = \text{ 0.7071 amps}$$

And, given that we know $V_{IN}$ and $L$, we can calculate $t_1$ by: -

$$t_1 = \dfrac{L\cdot I_{PK}}{V_{IN}} = \text{5.657 μs}$$

$$\color{red}{\boxed{\large\text{This of course is a duty cycle of 0.5657}}}$$

Flyback voltage considerations

In the example above we used a 1:1 flyback transformer to produce 500 volts DC on the secondary. The "flyback" that creates 500 volts DC on the output also gets 1:1 reflected to the primary winding and this means that when the MOSFET switch is deactivated, the DRAIN will see the input voltage (125 volts) plus the 500 volts of flyback. This means that the MOSFET will need to be rated sparingly in excess of 625 volts to avoid damage.

Using a step-up transformer

To reduce this we can use a step-up secondary winding. For example, if a 1:2 step-up transformer is used then the secondary "flyback" of 500 volts is only 250 volts on the primary and, this means the MOSFET can be rated greater than 375 volts on the DRAIN. If a 1:4 transformer is used, the primary "flyback" is 125 volts and a MOSFET with a rating greater than 250 volts can be chosen: -

Note the "N" factor in the above relationships for $V_{SECONDARY}$ and $V_{PRIMARY}$.

Leakage Inductance considerations

There will always be energy stored in the primary that cannot be transferred to the secondary because the two coils will never be 100% coupled. The outcome of this is that when the MOSFET turns off, in addition to the fly-back voltage raising the drain higher than $V_{IN}$ there will be an extra spike due to leakage inductance. This can be dealt with using a diode-capacitor-resistor clamp or a zener clamp: -

Above picture from here.

Note that the zener diode has to allow the normal flyback voltage to occur hence its voltage rating has to be greater than the natural flyback voltage seen on the primary winding.

Micro-cap 12 simulation 1:1 transformer

• $V_{IN}$ is 125 volts
• D is set to 0.5657
• The peak $V_{OUT}$ is 498.45 volts (target 500 volts)
• The reflected primary voltage is 623.45 volts (498.45 + 125 volts)