Inductor Reactance

coilfrequencyinductancesolenoid

I have a few questions about the reactance of inductors at certain frequencies.

I have been working on a project that involves flipping the field in a solenoid with an air core at ~10-20 kHz with a square wave. However, I have encountered a few problems.

What would the equation for inductive reactance be with a square wave? The general equation is \$X=2\pi fL\$ but I am pretty sure this only works with sinusoidal functions. Could breaking down the square wave and solving for the reactance of each harmonic work? This would give a huge reactance because those frequencies get pretty high.

I will probably end up stepping down the voltage and using an oscilloscope but it would be nice to know beforehand.

Also, what might be some other implications of switching a magnetic field of ~30 Amps and ~200 Gauss. I know that there is some delay between the voltage being applied and the current rising but that is small enough for me to disregard.

Thanks!

Best Answer

You can analyze the input signal as a sum of sinusoids. Maybe analyze up to the 5th or 7th harmonic. But there is an easier way..

To actually get a square wave current into a pure inductance is impossible- the voltage would have to be infinitely high to get the current to change instantaneously at the edges.

If you have some idea of the inductance (better yet, a fairly good model of the solenoid in terms of series resistance, parallel capacitance and eddy current losses), the easiest approach is to do a SPICE simulation (LTSpice is free) with an ideal current source to see where the voltages go, and what happens if the voltages are limited to some reasonable value. That will quickly give you a feel for the problem (if your model is reasonably realistic).

Edit: Below I've simulated a +/-100mA ideal current source into an ideal 100uH inductor flipped at 15kHz. The only thing non-ideal is the rise and fall times of the edges, which I've set to 1usec each in order to limit the voltage. As you can see, the voltage required to give you a rise time of 1usec for current change of 200mA (from -100 to +100mA) is 20 volts.

You can see that the voltage required to get the current to change from -I to +I or back in time tr = tf as |V+| = |V-| = \$ \frac {2 I L}{tr}\$

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