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I am simulating this simple circuit on PSIM; a single-phase AC voltage source connected to the primary of a pair of coupled inductors, with a 12 Ohms load on the secondary side. I want to increase the voltage across the load. The PSIM manual mentioned the following formula for the voltage across these two coupled inductors:

$$V_{1}= L_{11}\tfrac{d_{i1}}{dt} + L_{12}\tfrac{d_{i2}}{dt}$$ $$V_{2}= L_{12}\tfrac{d_{i1}}{dt} + L_{22}\tfrac{d_{i2}}{dt}$$

Where V1 is the voltage across the primary inductor (righthand side) in volts. V2 is the voltage across the secondary inductor (lefthand side) in volts. L11 is the self-inductance of the primary inductor in Henry. L22 is the self-inductance of the secondary inductor in Henry. L12 is the mutual inductance between the two inductors in Henry.

So, as per the formula, I tried increasing the frequency to increase the rate of change of current, and, consequently, increase V2. Raising the frequency from 10 Hz to 40 Hz did increase the V2, but any frequency of 41 Hz and higher resulted in a decrease in the output voltage.

I also tried increasing the self-inductance of the coils (and consequently the mutual inductance at a coupling factor of 0.9) but the V2 still decreased, regardless of the frequency range...

I don't understand why my output voltage does not match this formula...What am I doing wrong? enter image description here

Edit: One of the answers suggested that the voltage drops due to the leakage inductance of the coils. However, when I want to assign the inductance parameters for two coupled coils, PSIM clearly refers to them as self-inductance, and not leakage inductance, as shown in the image below. And PSIM clearly differentiates between leakage inductance and self-inductance because if one were to use a single phase transformer (not ideal), there will be a parameter for leakage inductance as shown in the lowermost image. I assume that this coupled pair of inductors has (ideally) zero leakage reactance...

enter image description here

enter image description here

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  • \$\begingroup\$ Your question is seriously lacking details on all magnetics and drive signals. But perhaps I=V/ωL reduces with rising ω might explain it. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 26 '18 at 5:57
  • \$\begingroup\$ I heavily edited my question to make my problem more clear. Can you please read it again? I mentioned all the parameters I can think of. \$\endgroup\$ – DigiNin Gravy Sep 26 '18 at 10:53
  • \$\begingroup\$ All transformers have a frequency dependence eventually and most common is Audio transformer rarely with more than 3 decades of bandwidth. At the low end amplitude drops with lower frequency due to coupling, and upper end drops due to core losses. WHat impedance ratios do you expect in your analysis that might explain this dependance ( high sourceR/load ZL ratio? ) @Andyaka ’s answer is correct \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 27 '18 at 1:48
  • \$\begingroup\$ I fully understood what @Andyaka meant. My reasoning was that as inductance increases, the alternating magnetic field that links with the other coils increases and therefore the voltage across the secondary coil increases. I totally forgot about inductive reactance. But what confuses is: which portion of the inductance is responsible for inductive reactance and which portion is responsible for creating the magnetic field. \$\endgroup\$ – DigiNin Gravy Sep 28 '18 at 17:08
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From what I read about coupled inductors, an increase in frequency should result in an increase in the voltage across the secondary inductor.

That's only true if the RMS current in the primary remains constant and then the output voltage is proportional to frequency because the induced voltage is proportional to \$\dfrac{d\Phi}{dt}\$.

However, you have a primary winding and that winding has inductance and as frequency rises the reactance increases and the current falls with frequency. The result is that you get a constant output voltage with frequency.

Any difference from the theoretical will be down to other components affecting things. Consider your circuit broken down into three reactive components; the leakage in the primary due to imperfect coupling, the secondary leakage also due to imperfect coupling and the transformer representing the fraction of perfect coupling that remains: -

enter image description here

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  • \$\begingroup\$ but if the current in the primary remains constant then how will the primary generate an alternating magnetic field? \$\endgroup\$ – DigiNin Gravy Sep 26 '18 at 10:58
  • \$\begingroup\$ I'm talking about the RMS value of current. I shall make that clearer. \$\endgroup\$ – Andy aka Sep 26 '18 at 11:00
  • \$\begingroup\$ I edited my question and used a simpler circuit. Can you please read the question again? \$\endgroup\$ – DigiNin Gravy Sep 26 '18 at 11:04
  • \$\begingroup\$ I read it several minutes ago and I noted your amendment. What is it that you don't understand about (or that isn't explained in) my answer? \$\endgroup\$ – Andy aka Sep 26 '18 at 11:13
  • \$\begingroup\$ the new circuit I used only has a single phase voltage source and the pair of coupled inductors, so shouldn't the RMS current stay constant? Unless I am misunderstanding what you mean by constant RMS current... \$\endgroup\$ – DigiNin Gravy Sep 26 '18 at 11:21

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