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I have made a PCB based planar transformer with a turns-ratio of 8:6. I am considering the T-model as shown below: Transformer T-model

In the T-model, Lp is the primary side leakage inductance, Ls' is the secondary side leakage inductance referred to the primary and Lm is the magnetizing inductance. The following tests are conducted (the values are consistent for a frequency sweep from 100kHz to 1 MHz):

  • Connect the primary side to the impedance analyzer keeping the secondary open: (Lp + Lm) = 105.6 µH
  • Connect the primary side to the impedance analyzer keeping the secondary shorted: (Lp + (Lm || Ls')) = 1.26 µH
  • Connect the secondary side to the impedance analyzer keeping the primary open: (Ls' + Lm) = 60.76 µH
  • Connect the secondary side to the impedance analyzer keeping the secondary shorted: (Ls' + (Lm || Lp)) = 0.74 µH

I have solved these equations and have determined the component values as:

  • Lp = 26 µH
  • Ls' = -19 µH
  • Lm = 80 µH

Are these values correct? The secondary side leakage (transferred to the primary) is negative. The transformer self resonant frequency is in the MHz range. Are my values correct? If I am to design a series resonant converter utilizing this transformer, can I consider 26 µH as the resonant inductor along with an external capacitor?

If I am wrong in my calculations, how should I proceed correctly?

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    \$\begingroup\$ The reason for the close vote is: needs more detail or clarity. Please specify what more is required. I will be more than happy to add it to the question. \$\endgroup\$
    – Jonathan_the_seagull
    Commented Jul 7 at 8:39

2 Answers 2

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Made with Maple ...

enter image description here

It seems that there are 4 solutions.
The solution where Lm1 and Lm2 (and positive) are "near" should be cared.
It is the first in the sequence list where Lm1=79.6 uH and Lm2=79.9 uH.
Then Lp=26.4 uH and Ls= -19.1 uH ...
So, you are "right", but I don't understand why Ls is "negative" ... (not checked).
As it is a "theoretical" background, it is possible. Confirmed by microcap v12.

enter image description here

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  • \$\begingroup\$ Thank you. Even if the secondary side leakage is negative, can I still use the primary side leakage as it is to build a series resonant circuit utilizing an external capacitor? \$\endgroup\$
    – Jonathan_the_seagull
    Commented Jul 7 at 8:40
  • \$\begingroup\$ Yes. You can use it. I have done this in microcap v12. \$\endgroup\$
    – Antonio51
    Commented Jul 7 at 8:41
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When you used the values from connecting the impedance analyzer to the secondary, did you account for the turns ratio?

You should be using Ls, Lp', and Lm' for those two values, where Lp' and Lm' are referred from Lp and Lm to the secondary. That would apply a factor of (6/8)^2 to Lp, Lm, and Ls'

I'm not at my computer to calculate exactly, but that would seem to line up with your values.

On the primary side:

Lp is around 0.7 uH

Ls' ~ 0.7 uH

Lm ~ 100 uH

On the secondary side:

Lp' is around (6/8)^2 * 0.7 uH = ~0.4 uH

Ls ~ 0.4 uH

Lm' ~ 60 uH

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