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I have made a simple pulse transformer circuit in LTSPICE using a pulse and two coupled inductors. L1 has a resistance of 1 Ω and L2 has a resistance of 0 Ω.

I am having trouble understanding why this circuit does what it does. I have graphed the primary coil current vs the secondary voltage output. What is very strange is that from 0.1 μs onwards, primary current is constant yet secondary voltage is nonzero.

I was under the impression that the secondary voltage was equal to mutual inductance times di/dt on the primary coil, yet di/dt=0. Can somebody help me understand why LTSPICE is is producing these results?

Primary current and secondary voltage on the same graph

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  • \$\begingroup\$ You have a voltage (L2?) plotted there. Show a plot with all the interesting variables shown together. \$\endgroup\$
    – Kevin White
    Jun 28, 2022 at 0:34
  • \$\begingroup\$ V2 / I1 ~ 9 Ohms and coupling factor is unspecified as well as turns ratio but is implied by sqrt(inductance) ratio \$\endgroup\$
    – Tony Stewart EE75
    Jun 28, 2022 at 1:51
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    \$\begingroup\$ Try simulating for a longer period of time, say 1m, or use smaller values for inductors. The magic will be gone. LTspice (note the spelling) shows the correct output for the given input (GIGO). \$\endgroup\$
    – a concerned citizen
    Jun 28, 2022 at 5:57
  • \$\begingroup\$ You don't need dI/dt for a transformer to function in simulation. It will always copy over voltage from the primary to the secondary according to turns ratio. \$\endgroup\$
    – winny
    Jun 28, 2022 at 7:03

1 Answer 1

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I simulated your circuit, and the primary inductor current is actually changing, from 8.9029 A to 8.9040 A at the end of the 4 us pulse. The output voltage drops from 80.454 V to 80.436 V. This is 583 A/s. I'm not sure how to calculate the expected dI/dt for a coupled inductor, but there is definitely a finite value which explains the output voltage. I just let LTSpice take care of the math.

Here is my simulation of the first 200 ns:

Pulse transformer simulation

The current in L1 rises from -12 A to 8.9 A in the 100 ns of the applied voltage V1. The voltage on L1 rises from zero to 3.1 V. At the same time, the output current on R1 rises from zero to 804 mA and 80.4 V. This corresponds to the 26:1 turns ratio for voltage. Considering that the input current changes by 20.9 A which also confirms the ratio.

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  • \$\begingroup\$ Since K=1, it's the same as the magnetizing inductance. Note that the voltage applied to the inductor is less, because primary ESR is considerable versus R1 reflected resistance (100 * 1/675 = 0.148Ω). This still doesn't give the measured 8.9A; evidently V1 has 0.2Ω ESR also. \$\endgroup\$
    – Tim Williams
    Jun 28, 2022 at 1:13
  • \$\begingroup\$ It has been my understanding that reflected resistance is according to the turns ratio, which in this case is about 26:1. Thus the reflected resistance would be about 4 ohms. However, this would seem to make the peak primary current 12/5 = 2.4 amps. So there must be a much higher primary voltage caused by stored energy in the inductor. A simulation with external resistances may clarify this. \$\endgroup\$
    – PStechPaul
    Jun 28, 2022 at 2:13
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    \$\begingroup\$ Impedance ratio is square of turns ratio; and impedance ratio is given by inductance ratio (impedance is proportional to inductance) :) \$\endgroup\$
    – Tim Williams
    Jun 28, 2022 at 3:57
  • \$\begingroup\$ OK, I see now. I am still trying to understand some of the voltage and current values as shown in the simulation, which I will add to my answer above. \$\endgroup\$
    – PStechPaul
    Jun 28, 2022 at 5:23

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