# How to find the secondary inductance value in flyback transformer?

I try to design smps converter with flyback topology. Parameters are below:

Vin = 100V(dc) (constant)
Vout = 50V(dc)
D(on) = 50% (So, D(off) = 50%)
fsw = 70 kHz
Np/Ns = 2
According to my calculations inductance of the primary Lp = 236µH.

What is the value of inductance of the secondary? How to calculate it?

In my opinion: Ls = Lp x (Ns/Np)^2 Ls = 236µH /4 = 59µH (I'm not sure if this calculation is true for flyback topology)

Normally, Current changes (ΔI) on primary and secondary must be equal. Because of the D(on) = D(off), slope values of current ramps must be equal.

BUT !

(Sp: slope of primary current; Ss : slope of secondary current)

Sp = Vin/Lp = 100V / 236µH = 423729 Ampere/seconds
Ss = Vout / Ls = 50 / 59µH = 847458 Ampere/seconds

Why are they not equal? I don't know my mistake

[UPDATE - 27.07.2020 - 13:28]

This image is at page 3 in AN4137 app note by OnSemi
If current changes is not equal, how can I calculate current change for secondary?

## 1 Answer

The square of the ratio of the turns ratio works irrespective of how you drive the transformer. So, if primary is 236 µH and secondary winding has half the turns of the primary then the secondary inductance is 59 µH.

The primary and secondary current slopes will only be equal when using a 1:1 transformer and input and output DC voltages are identical. The slope determines the voltage i.e.: -

$$V_{OUT} = L\cdot \dfrac{di}{dt} = 59 \text{ µH} \times 847458 \text{ A/s} = 50 \text{ volts}$$

And

$$V_{IN} = L\cdot\dfrac{di}{dt} = 236 \text{ µH} \times 423729 \text{ A/s} = 100 \text{ volts}$$

Normally, Current changes (ΔI) on primary and secondary must be equal.

That is true for a 1:1 transformer but not an N:1 transformer where N isn't 1.

If current changes is not equal, how can I calculate current change for secondary?

If the transformer steps down 2:1 then, because ampere-turns have to be equal on primary and secondary, the secondary current must be twice the primary current. Ampere-turns are transferred from primary to secondary at the moment the primary is disconnected so this is a valid approach.