I have a question about the current ratio in this type of circuit. I know that the voltage ratio is simply, since it's based on a buck converter:

\(\frac{Vout}{Vin}= \frac{Nsec}{Npri}\cdot D\)

Where D = duty cycle, and the N's is the transformer ratio.

My question: what about current? If I assume that Pout=Pin (meaning no losses), then the above should simply be reversed.

\(\frac{Iout}{Iin}= \frac{Npri}{Nsec}\cdot \frac{1}{D}\)

Is this correct?

But, what about the demagnetizing winding N3? How would it affect the current in this stage

Next question: how about a voltage drop on the input side - how would this affect the input current if the output current is kept constant? (really wierd but my guess is that it shouldn't at all, since the circuit is drawing the same current).

Very thankful for any pointers in the right direction!

 

Schematic!
I went to the effort to google images of forward converters and did not find N3 in the first 3 tries. We REALLY need schematics to do anything helpful.

 

#12

You're correct. A picture tells a thousand words.

Source: wiki.

So the question is about current, see first post.