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.


  • \$\begingroup\$ Above a few hundred Hz the solenoid will have eddy current losses that will make any such calculations moot. The losses will 'look like' a resistance in parallel with the coil. \$\endgroup\$ – Spehro Pefhany Feb 13 '15 at 21:48
  • \$\begingroup\$ If I power this with a constant current supply wouldn't the voltage change to keep the current constant? If so, I would need to know what that voltage will be to choose a proper power supply. \$\endgroup\$ – user104385 Feb 13 '15 at 21:53
  • \$\begingroup\$ Is that really what you want? The induced current from the core losses will partially cancel out the field due to the coil, so the external field will be much less at high frequencies. \$\endgroup\$ – Spehro Pefhany Feb 13 '15 at 21:58
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    \$\begingroup\$ So Sorry, I forgot to mention that I have an air core. I would not have this effect. \$\endgroup\$ – user104385 Feb 13 '15 at 22:02
  • \$\begingroup\$ What's the coil resistance? For an air coil there is the L/R, time constant. \$\endgroup\$ – George Herold Feb 14 '15 at 2:13

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}\$

enter image description here

  • \$\begingroup\$ Thanks! I will go ahead and try that. I have an inductance of about 2.5x10^-5 H. I'm not sure if you can tell me if that is reasonable or not? \$\endgroup\$ – user104385 Feb 13 '15 at 22:17
  • \$\begingroup\$ Yes, you can select a switching current source. But the simple method above should work for you too, so long as you know the rise and fall times you require. \$\endgroup\$ – Spehro Pefhany Feb 13 '15 at 22:18
  • \$\begingroup\$ So I would really only need to get up to my desired current just by the time it needs to switch the other way. My rise time would then be (1/20000). So my voltage would be equal to (2(30)(2.5*10^-5))/(1/20000) = 30 V. This seems great but I'm having a hard time believing it for some reason. \$\endgroup\$ – user104385 Feb 13 '15 at 22:36
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    \$\begingroup\$ Simulating this myself I got the same result. Thanks for helping me out! \$\endgroup\$ – user104385 Feb 14 '15 at 1:21
  • \$\begingroup\$ It seems to me you are missing the coil resistance. \$\endgroup\$ – George Herold Feb 14 '15 at 2:19

The equation for inductive reactance is as you have shown. What you are most likely concerned with is the fundamental frequency of your square wave.

You are absolutely right in that the inductive reactance is higher for the harmonics. In fact, the inductor acts as a low-pass filter and attenuates those higher-order harmonics.

Regarding your question about current rise time: the rise time is a directly affected by the applied voltage and the inductance. Higher voltage gives you faster rise time, higher inductance gives you slower rise time. These effects are NOT trivial and you must not disregard them. Instead, calculate the rise time that you are going to get with your particular inductor and the voltage applied to it. Then you can decide if that rise time is going to affect your desired results or not.

  • \$\begingroup\$ The time constant would be equal to L/R. Two of these would provide me with 75% of my current. Doing the calculations, with my relatively low inductance, I could achieve my desired field. \$\endgroup\$ – user104385 Feb 13 '15 at 22:07

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