Post History
How is it possible to generate 500V on a trafo secondary given the 1:1 ratio? In a DCM flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary...
Answer
#5: Post edited
by
Andy aka
·
2022-02-09T15:38:50Z (almost 3 years ago)
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- This is because CCM is constrained to have a continual rising and falling primary-secondary current waveform with no hold or dwell period (unlike DCM): -
- 
This means that the slopes of the rising and falling currents totally dictate (or are dictated by) the output voltage and the input voltage.- $$$$
- But for DCM it's this: -
- $$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
- 
- - \$D\$ is the duty cycle
- - \$N_{\text{P:S}}\$ is the primary-to-secondary turns ratio
- - \$R_L\$ is the output load resistance
- - \$L_P\$ is the primary inductance
- - \$F_{SW}\$ is the switching frequency
- The devil is in the mathematical details of course. Images from [my crappy website](http://www.stades.co.uk/Flyback/flyback%20DCM2.html).
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- This is because CCM is constrained to have a continual rising and falling primary-secondary current waveform with no hold or dwell period (unlike DCM): -
- 
- This means that the slopes of the rising and falling currents are totally dictated by the input voltage and the output voltage respectively. It cannot be in CCM if this were not the case.
- $$$$
- But for DCM it's this: -
- $$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
- 
- - \$D\$ is the duty cycle
- - \$N_{\text{P:S}}\$ is the primary-to-secondary turns ratio
- - \$R_L\$ is the output load resistance
- - \$L_P\$ is the primary inductance
- - \$F_{SW}\$ is the switching frequency
- The devil is in the mathematical details of course. Images from [my crappy website](http://www.stades.co.uk/Flyback/flyback%20DCM2.html).
#4: Post edited
by
Andy aka
·
2022-02-09T13:27:08Z (almost 3 years ago)
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- But for DCM it's this: -
- $$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
The devil is in the mathematical details of course.
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- This is because CCM is constrained to have a continual rising and falling primary-secondary current waveform with no hold or dwell period (unlike DCM): -
- 
- This means that the slopes of the rising and falling currents totally dictate (or are dictated by) the output voltage and the input voltage.
- $$$$
- But for DCM it's this: -
- $$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
- 
- - \$D\$ is the duty cycle
- - \$N_{\text{P:S}}\$ is the primary-to-secondary turns ratio
- - \$R_L\$ is the output load resistance
- - \$L_P\$ is the primary inductance
- - \$F_{SW}\$ is the switching frequency
- The devil is in the mathematical details of course. Images from [my crappy website](http://www.stades.co.uk/Flyback/flyback%20DCM2.html).
#3: Post edited
by
Andy aka
·
2022-02-09T12:24:56Z (almost 3 years ago)
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- But for DCM it's this: -
$$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- But for DCM it's this: -
- $$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
- The devil is in the mathematical details of course.
#2: Post edited
by
Andy aka
·
2022-02-09T12:24:15Z (almost 3 years ago)
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
In a flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- > _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
- In a **DCM** flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
- Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
- So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
- So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).
- It's slightly different for **CCM**; in CCM the turns ratio does affect the output voltage as per this formula: -
- $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{D}{N_{\text{P:S}}}\cdot\dfrac{1}{1-D}$$
- But for DCM it's this: -
- $$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
#1: Initial revision
by
Andy aka
·
2022-02-08T09:22:41Z (almost 3 years ago)
> _How is it possible to generate 500V on a trafo secondary given the 1:1 ratio?_
In a flyback transformer circuit, the secondary winding is being used as a pure inductor. So, once the primary is disconnected, the rate at which the core magnetic field collapses, dictates the secondary terminal voltage. If you want a bigger secondary voltage, you allow the core magnetic field to collapse at a higher rate.
Mathematically for an inductor, \$V = L\dfrac{di}{dt}\$
So, if the load is chosen to be a high value we know that \$\dfrac{di}{dt}\$ will be also high and, the output voltage will be higher than if the load was a low value.
So, providing that the secondary current (\$i\$) is of a high enough initial value to sustain \$\dfrac{di}{dt}\$ over the period in which the primary is disconnected, then we can generate whatever secondary voltage we want (within reason of course).