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I'm doing an LTSpice simulation of a basic inverting op-amp. The output of the op-amp is driving a 53uH inductor. The ESR for the inductor is very small. My problem is when I try to measure the current going through the inductor, the current is sinusoidal but it's only negative. If the op-amp is in the non-inverting configuration, the current is positive only. I also noticed that increasing the value of the series resistance does solve the problem and gets rid of this offset. Is that normal? The frequency I'm using is 85kHz. enter image description here

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Welcome to the world of inductors.

If you abruptly apply a sinewave to a pure inductor, the current rises to a peak value at 180°. It then drops back to zero amps at the end of the first cycle. This is how inductors work; they integrate the voltage into current. This produces a totally asymmetrical current waveform that is wholly positive.

Because you have an inverter, the current waveform will be wholly negative until things have settled down (due to inductor resistance eating the DC losses).

If you abruptly apply a cosine wave, you will get a bipolar current (which it sounds like you were expecting) that is symmetrical.

An inductor with resistive losses will eventually produce a symmetrical bipolar current from a sinewave but a cosine wave will immediately produce a symmetrical current: -

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  • \$\begingroup\$ Sorry, I didn't know someone answered my question. Thank you for taking the time to answer my question. What you are saying does makes sense, but why does this happens though? Isn't the integral of a sine is just a negative cosine waveform? Why not the result be just a symmetrical cosine wave? \$\endgroup\$
    – Analog
    Commented Dec 3, 2022 at 17:18
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    \$\begingroup\$ For an abruptly applied sinewave not to produce the initial offset of current in the inductor, the inductor current must abruptly begin at a negative peak. Neither the process of mathematical integration nor an ideal (or non-ideal) inductor can possibly do this. There is no problem with an abruptly applied cosine because it's just a voltage that needs to be positive maximum at t=0. That then causes inductor current to rise exactly how it should and how you thought it should @Analog \$\endgroup\$
    – Andy aka
    Commented Dec 3, 2022 at 18:30
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    \$\begingroup\$ Ok, I understand now. Thank you! \$\endgroup\$
    – Analog
    Commented Dec 3, 2022 at 20:37

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