Average input current in forward converter

Question

I am trying to understand how to calculate the average input current of a typical forward converter. If the converter has negligible magnetizing current, a large inductance and a turns ration of n1:n2:n3, based on my understanding, the input current would be:

Ig = D*Iinductor * n3/n1

But this does not seem to be correct.

Any help/pointer will be much appreciated.

Thanks

Answer 1

In a forward converter, the primary current is the sum of the magnetizing current \$I_{mag}\$ and the reflected inductor current \$I_L\$ which, on average, is the output current \$I_{out}\$ scaled by the transformer turns ratio 1:N. This current circulates in the primary side during the on-time duration. The below drawing shows a typical waveform excerpted from a report I posted on the over-power phenomenon in the forward converter:

If we neglect the magnetizing current circulation, it is possible to approximate the triangular waveform into a square waveform and easily calculate the average input current. Average calculations are about area determination. Look at the below picture to see how to simply determine the average value by stretching the original area along the switching cycle:

Assume you have a forward converter delivering 5 V 10 A with a transformer turns ratio 1:N=1:0.16. The input voltage is 100 V and the duty ratio is 31.3% (neglecting all the drops). The below Mathcad sheet shows the input current to be, on average, 502 mA:

Answer 2

Switching converters convert power.
So if a load is 10V @ 1A that's 10Watts. Now the input side will take that same 10Watts but at a different current and voltage.

Maybe you have 120V input, then that is

\$\frac{10W}{120V}=0.0833A\$ input current.

All of this neglected efficiency. So adding that in: at 80% efficiency the input power is \$10W*1.2=12W\$.

Going back to calculate current now gives \$\frac{12W}{120V}=0.1A\$ input current.