2 switch flyback

[wurring]

Hello,

I am requesting your assistance if i may as i must make a DCDC converter to have V(in) = 400 Volts and V(out) = 12 Volts.

The high V(in) makes me want to use a 2 switch flyback, as in the following article......

http://www.edn.com/contents/images/47173.pdf

...the two switches, and the clamping to V(in) mean less voltage on the MOSFETs.

...However, i am uneasy as when the secondary is conducting, it will refer back to the primary a voltage to the primary of V(sec) * Np/Ns....

...However, the primary at this (off) time, is clamped to V(in).....so i am wondering , which voltage wins out and ends up across the primary when the secondary is conducting?......by Faraday's Law the primary (should)experience a voltage of N*d(phi)/dt .....the d(phi)/dt being same for both primary and secondary.

So i wonder how the primary can be clamped to V(in) under these circumstances?

 

[electricuwe]

The 2 diodes will clamp against the supply. This is very advantageous regarding limiting voltage coming from the transformers stray inductance, but compared to a sinlge swicth flyback, it restricts you in a way that you have to keep the reflected voltage below minimum input volatge.

 

[wurring]

I thankyou for this reply, and i am still concerned since as you rightly point out, there is a reflected voltage on the primary from the secondary which happens when the switch is off.

The thing is, the primary can only have one voltage across it at any one time......not two different ones at the same time.

By Faraday's Law, we must believe that during secondary conduction, the voltage on the primary coil is given by...

Np/Ns * V(secondary output)

However, by Kirchoff's Law, during secondary conduction, we can see that the voltage across the primary coil is
V(in).

These are most likely not the same voltages..

The mystery is, which voltage does appear across the primary during secondary conduction?...or....who wins out of messr's Kirchoff and Faraday in this case?

 

[electricuwe]

The diodes take care of that:

calculated reflected voltage smaller than Vin --> Vout as desired

calculated reflected voltage larger than Vin --> this will limit vout below the desired value

 

[wurring]

If for example as a 2 switch flyback…..

Np = 100
Ns = 10

Vin = 400V
Vout = 20V
…………………………………………..
When mosfets go off…..

Secondary starts conducting.

Vs / Ns = d(phi) / dt (1)

At the same time the primary is conducting through the diodes.

Vp / Np = d(phi) / dt (2)

d(phi) / dt must be the same for primary and secondary as they are on the same core.

From (1) and (2) we can say that during the mosfets OFF time…..

Vs / Ns = Vp / Np (3)

Substituting into (3)…..

20 / 10 = 400 / 100 (4)

…..(4) is not correct.

This is proof that the 2 switch Flyback converter has disobeyed Faraday’s Law.

Does any reader know how it works?

 

[Madcow]

Don't forget the polarity of the transformer
-200 volts should be reflected back to the primary.
(Ignoring parasitics).
ie. 400 volts is across the primary during the on time and
-40v on the secondary (no conduction due to rectifiers
being reverse biased). During the off time the secondary
conducts +20 and reflects -200 on the primary (no
conduction on the primary because the fets are off and the
body diodes are reverse biased).


Hope this helps

 

[wurring]

Hi


Madcow, i agree with you that -200v is reflected back to the primary.

However, if diodes D1 and D2 of 2nd page, fig 5 of following

http://www.edn.com/contents/images/47173.pdf

...were on during the off-time of the Mosfet switch's then this would not be possible -since it would be 400V across the primary and not 200V.

I think that the only way that figure 5 of the above link can work is if diodes D1 and D2 only conduct for say less than 200ns after the MOSFETs are switched OFF ?

-This means that the primary leakage inductance must be small enough that it can discharge its current in this short space of time ?

If diodes D1 and D2 conduct throughout the off time of the two mosfets, then this converter with....

Np = 100
Ns = 10

Vin = 400V
Vout = 20V

....would be only able to give an output voltage of 40 Volts ?

Any thoughts appreciated.

I have finished the PFC stage now, and i hope to switch the 400V with an IR2110 (hi and lo gate driver IC) and hopefully a two switch flyback, if i can understand how it works.

I believe its advantageous for me as instead of the 2 switch forward converter it doesnt need the output inductor.

 

[DHambley]

If the output voltage is 20V and your turns ratio is 10:1, when the MOSFET is off, the transformer primary will have (-)200V across it for the whole duration that the secondary current is falling due to the magnetizing inductance. The diodes won't be conducting if the transformer has less than (-)400V. The only time that you will see the diodes conduct (more than -400V on the primary) is during the spike after turn-off. This will last for typically 100 nanoseconds or so. This spike is due to the leakage inductance (bad coupling between windings) and NOT the magnetizing inductance.

 

[gabisuciu]

you can design the flyback circuit for reaching the critical
discontinuous mode at a lower voltage (e.g Vin(min)=200V;D.C=0.4); ypu will rich by calcul. a N1:N2~10, trafo ratio.
So, the reflected input voltage will be:10*12V=120V<<Vin=400V. No current feedback will occur during OFF periodes.500V mosfets will suits the application (for more safety: 600V).