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Question about power factor corrections etc in flyback converters

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powerelec

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Hi, I have a couple of questions about flyback converters. Your help in answering these is much appreciated!

1) I have read some papers in which isolated flyback operating from the AC mains, sometimes use constant on time DCM control to get high power factor. I understand why this is so. But if the output load is constant then we end up needing a very high value output capacitor and a very slow feedback loop to filter out the 60 Hz ripple. Is there any technique to keep power factor high and also deliver constant power to the load?

2) I have a problem understanding the electricity-to-magnetism-to-electicity conversion when there are two windings on the secondary side of the flyback. When the switch is off, both the windings on the secondary side have a voltage of V. So why the primary side MOSFET drain does not go to VIN+V+V. Instead it goes to VIN+V. Could you please explain this.

Thanks!
 

Hi, I have a couple of questions about flyback converters. Your help in answering these is much appreciated!

1) I have read some papers in which isolated flyback operating from the AC mains, sometimes use constant on time DCM control to get high power factor. I understand why this is so. But if the output load is constant then we end up needing a very high value output capacitor and a very slow feedback loop to filter out the 60 Hz ripple. Is there any technique to keep power factor high and also deliver constant power to the load?
By definition for PFC to work you have to draw current from the line at the line frequency (power draw is at twice the line frequency). So there must be one stage where the feedback bandwidth is low, and the output has ripple at twice the line frequency. In order to get an overall high loop bandwidth, you have to use a second DC-DC converter stage after the PFC. The output capacitors of the PFC stage must still be somewhat large. Something has to absorb that line frequency ripple in order for PFC to occur efficiently.
2) I have a problem understanding the electricity-to-magnetism-to-electicity conversion when there are two windings on the secondary side of the flyback. When the switch is off, both the windings on the secondary side have a voltage of V. So why the primary side MOSFET drain does not go to VIN+V+V. Instead it goes to VIN+V. Could you please explain this.
When the primary side FET goes off, the secondary winding clamps to Vout. That secondary voltage Vs=Vout will be seen on the primary side Vp as well (multiplied by the turns ratio). In the case of a turns ratio of one, Vp will equal Vs. And from inspection of the flyback circuit it should be obvious that the MOSFET voltage will be Vin+Vp. So for N=1, the MOSFET voltage will be Vin+Vout.
 
By definition for PFC to work you have to draw current from the line at the line frequency (power draw is at twice the line frequency). So there must be one stage where the feedback bandwidth is low, and the output has ripple at twice the line frequency. In order to get an overall high loop bandwidth, you have to use a second DC-DC converter stage after the PFC. The output capacitors of the PFC stage must still be somewhat large. Something has to absorb that line frequency ripple in order for PFC to occur efficiently.
When the primary side FET goes off, the secondary winding clamps to Vout. That secondary voltage Vs=Vout will be seen on the primary side Vp as well (multiplied by the turns ratio). In the case of a turns ratio of one, Vp will equal Vs. And from inspection of the flyback circuit it should be obvious that the MOSFET voltage will be Vin+Vp. So for N=1, the MOSFET voltage will be Vin+Vout.

Hi mtweig, thanks for your reply. It is very helpful to improve my understanding.

For the second part of the question, I understand if there is one secondary winding. But how about if there are two secondary winding Vs1 and Vs2. Vs1 is regulated (say with an opto-coupler). If this is the case, Vs2 will have the same voltage as Vs1. My question is, why is the voltage on the MOSFET drain not VIN+Vs1+Vs2. Instead it is just VIN+Vs1? Hope that is a bit clearer.
 

For the second part of the question, I understand if there is one secondary winding. But how about if there are two secondary winding Vs1 and Vs2. Vs1 is regulated (say with an opto-coupler). If this is the case, Vs2 will have the same voltage as Vs1. My question is, why is the voltage on the MOSFET drain not VIN+Vs1+Vs2. Instead it is just VIN+Vs1? Hope that is a bit clearer.
I don't see why the addition of a second winding would increase the EMF seen on the primary side.

Just go back to the basic rules for transformer operation. At any given time, each winding should have the same volt-turn product. The actual sum of voltages can be anything. The sum of currents, however should be zero (neglecting magnetizing current). So if you have another secondary winding, the primary may see additional current draw, but not additional EMF.
 
Thanks for the helpful datasheet ZekeR

- - - Updated - - -

Thanks for the additional explanation mtwieg. I will need to think about it to better understand.
 

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