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 :

enter image description here

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 :

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


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

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 : enter image description here

Thank you very much and have a nice day !

  • 1
    \$\begingroup\$ Note, you can use the \cdot operator in MathJax to create an algebraic multiply (dot) symbol. Normally these are omitted however. \$\endgroup\$
    – rdtsc
    Commented Jul 29, 2020 at 14:34
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Jul 29, 2020 at 16:28
  • \$\begingroup\$ 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 ! \$\endgroup\$
    – Jess
    Commented Jul 29, 2020 at 18:20
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ Commented Jul 30, 2020 at 7:24
  • \$\begingroup\$ Ok I see ! Thank you :D \$\endgroup\$
    – Jess
    Commented Jul 30, 2020 at 7:38

1 Answer 1


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\$.

  • \$\begingroup\$ 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. \$\endgroup\$
    – Jess
    Commented Jul 29, 2020 at 15:03
  • 1
    \$\begingroup\$ @Jess this answer is correct. Impedance transfer between primary and secondary uses the square of the turns ratios. \$\endgroup\$
    – Andy aka
    Commented Jul 29, 2020 at 16:03
  • \$\begingroup\$ @Andyaka I ask for seeing the demonstration :D and why my calcul would be wrong ... \$\endgroup\$
    – Jess
    Commented Jul 29, 2020 at 18:21
  • \$\begingroup\$ or @skvery :D :D \$\endgroup\$
    – Jess
    Commented Jul 29, 2020 at 18:24
  • \$\begingroup\$ @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. \$\endgroup\$
    – Andy aka
    Commented Jul 29, 2020 at 18:53

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