# Equivalent transformer model AN1679/D

I found the following transformer model in the AN1679/D written by ON SEMICONDUCTOR. The document is really interesting and very well explained.

Nevertheless, I do not understand this point, how the two model are equivalent :

If I try to put Ll2 which is at the secondary to the primary, I do not find the relation given :( Here are my calculs :

$$\mathrm{ L_{l2}*\frac{dI_{0}}{dt} = \frac{N_{p}}{N_{s}}(V_{p}-V_{L1}) -V_{0} }$$

and

$$\mathrm{ L_{l2'}*\frac{N_{s}*dI_{0}}{N_{p}*dt} = V_{p}-V_{L1} -V_{0} }$$

We then express each relation equal to

$$\mathrm{ \frac{dI_{0}}{dt} = \frac{dI_{0}}{dt} }$$

For express Ll2 in function of Ll2', and we get the correct relation only if Ns = 1 and Np = 1...

Did I do an error ? I m really interested by this subject !

For informations :

Thank you very much and have a nice day !

• Note, you can use the \cdot operator in MathJax to create an algebraic multiply (dot) symbol. Normally these are omitted however. Commented Jul 29, 2020 at 14:34
• I was in short trousers when I wrote this AN during the glorious MOT days : ) One little correction, the primary ohmic loss is in series with the winding not with the inductance which, alone, sits in the primary. Commented Jul 29, 2020 at 16:28
• Hi, @VerbalKint I did not get what you said :( Is it right or wrong what is written into your AN ? There is no primary ohmic loss in the model ? Which inductance are you talking about ? I didn't even get your expression "I was in short trousers". What does it mean in french ? ^^ Thank you very much for your help !
– Jess
Commented Jul 29, 2020 at 18:20
• Hello Jess, I simply meant that I wrote this application note long time ago. In figure 8, resistor Rp should not be in series with Lm: Lm should be alone and Rp in series with the upper input terminal as Rs1 in figure 11. That's all. Commented Jul 30, 2020 at 7:24
• Ok I see ! Thank you :D
– Jess
Commented Jul 30, 2020 at 7:38

We are referring the secondary inductance to the primary to simplify calculations.

For an impedance $$Z = \frac{V}{I}$$.

If any impedance $$\Z_s\$$ on the secondary is referred to the primary the equation is:

$$Z_p = \frac{V_p/V_s}{I_p/I_s}Z_s = \frac{N_p/N_s}{N_s/N_p}Z_s = \left(\frac{N_p}{N_s}\right)^2Z_s.$$

This can also be checked by calculating a short-circuit test $$\(R_\text{load}=0.)\$$ The same ratios apply to the inductance $$\L\$$.

• Hi, thank you for your anwer. Your result is correct for a transformer model which is more simple than the one I showed I think.
– Jess
Commented Jul 29, 2020 at 15:03
• @Jess this answer is correct. Impedance transfer between primary and secondary uses the square of the turns ratios. Commented Jul 29, 2020 at 16:03
• @Andyaka I ask for seeing the demonstration :D and why my calcul would be wrong ...
– Jess
Commented Jul 29, 2020 at 18:21
• or @skvery :D :D
– Jess
Commented Jul 29, 2020 at 18:24
• @Jess your solution is over-complicating things - L1 has got nothing to do with the L2 transformation from primary to secondary - it's just turns ratio squared. Commented Jul 29, 2020 at 18:53