Non linear inductor in LTspice

Question

I am trying to construct a non linear inductor in LTspice and have a question regarding the two plots provided below. In the first image current is swept from -30A to 30A in steps. The result yields the required behavior where the initial inductance is 500uH and with 27A peak current the left over inductance is around 40 uH. However, when using the same flux function measuring inductance across L1 the obtained result shows that with zero current the inital inductance is around 15mH, this is not correct. I wonder if there is something I may do not understand here. Spice netlist is provided below:

L1 ind1 0 Flux=0.007*tanh(0.072*x)
I1 0 ind1 PWL(0 0 1 30)
I2 0 ind2 {I_DC} AC {1/(2*pi)}
L2 ind2 0 Flux=0.007*tanh(0.072*x)
.tran 1
.step param I_DC -30 30 0.1
;.ac list 1
.backanno
.end

Answer 1

Your second analysis is in the frequency domain, so the results that you get are a function of frequency, first and foremost. You are using the card .ac list 1, which means you are evaluating the frequency response at 1 Hz, for a variation in the source's DC value.

If you'll change the analysis to be .tran 10 and set the current source to be sin 0 1 1 (unity amplitude, 1 Hz), you'll see that the voltage across the inductor will be ~3.16 mV peak. If you now divide this by 2*pi it will show ~504 uV peak, which is the same as what you got there.

Don't forget that, if you want a .DC analysis it will fail since that only evaluates the operating point, without any dynamics involved, and the inductor is considered a short. If you want a .TRAN analysis then you will get the dynamic variation, provided that it varies much slower than the variation, itself. And an .AC analysis will give you a response as a function of frequency, and if you perform one for 0 Hz (only works with .ac list 0) you will get exactly what a .DC analysis will get you: an operating point with shorted inductors and opened capacitors; otherwise you'll need to take frequency into account.