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In all the texts I read, including online tutorials, "reactance" is defined as the resistance to alternating current and they always base the equation off the assumption that the voltage is continuously changing. The equation is X(L) = 2piFL

But as far as I can tell, this doesn't seem to hold up right when you're talking about a coil's reaction to applied DC where you apply a steady ON voltage and then wait for the current to reach maximum amplitude.

As far as I can tell the inductive resistance to a 50% duty cycle square wave is actually just 2x the circuit's DC resistance if the pulses are timed to exactly 1/4 the inductor's natural frequency. The current will rise and fall steadily at the start and end of every pulse. During the transition period, it will always move from zero Amps to maximum Amps assuming you are always cutting the voltage input right when the maximum current amplitude is reached. So the result is you're getting approximately half the current you'd get if it were just a continuous flow DC circuit.

Can anyone clear this up if I'm missing something here. And if I DO understand it right, then is there some special term for what I'm describing?

Here is a diagram of what I'm talking about: enter image description here

It appears to me that the average current is just a matter of computing the average amplitude of a parabola, which I don't know how to do. LOL I can go research that, but I'm just trying to see what some of the other folks have to say about this.

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  • \$\begingroup\$ I'm not clear at all what you're getting at in the 3rd paragraph. Can you add a waveform diagram showing what behavior you're trying to understand? \$\endgroup\$ – The Photon May 28 '13 at 0:52
  • \$\begingroup\$ Also, it's not clear what you mean by the inductor's "natural frequency"...If you mean the self-resonant frequency that's related to parasitic capacitance, not resistance, and wouldn't normally come into play in the square wave scenario you're talking about. \$\endgroup\$ – The Photon May 28 '13 at 0:58
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they always base the equation off the assumption that the voltage is continuously changing. The equation is X(L) = 2piFL

That's actually not quite right. The correct assumption is that the voltage is sinusoidal. There are an infinity of voltage waveforms that are continuously changing but are not sinusoidal.

More generally, for an ideal inductor, the voltage across the inductor is proportional the time rate of change of the current through the inductor. If you place a constant voltage across an inductor, the current changes at a constant rate.

So, generally, the voltage across an inductor does not have the same form as the current through the inductor; the waveforms "look" different.

However, if the current is a sine wave, the voltage will be a sine wave though there will be a difference in phase.

Since both the current and voltage have the same form, we can take the ratio of the magnitudes of the voltage and current sine waves and call that ratio the reactance. Since higher frequency sine waves change more rapidly, the reactance increases with frequency.

For analyzing non-sinusoidal waveforms, like a square wave, one can use calculus to find the average current in the time domain. If your voltage source and inductor are ideal, the current waveform for a voltage square wave would be triangle wave:

enter image description here

not a parabolic wave. If there is significant internal resistance, the current looks more like:

enter image description here

So, your parabolic current waveform isn't the result of applying a voltage square wave across an inductor.

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  • \$\begingroup\$ Ok. Now I see. Part of the reason there's no term for what I was asking about is because the shape of the current wave varies tremendously based on the Q of the inductor. It's not just some standard reaction like reactance. It's more complex. \$\endgroup\$ – JamesHoux May 28 '13 at 2:26
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I think the word you are looking for is inductance.

This is the main property of an indcutor.

It is normally given the symbol L and it's defined in the time domain by the equation

\$ \dfrac{\mathrm{d}i}{\mathrm{d}t} = \dfrac{V}{L} \$

If L is higher, i changes more slowly in response to an applied voltage. if L is lower, i changes more quickly in response to applied voltage. This seems to fit what you describe as "resistance to change in current."

a coil's reaction to applied DC where you apply a steady ON voltage and then wait for the current to reach maximum amplitude.

I'm not sure what you're getting at here. If you apply a steady voltage to an ideal inductor, current will keep increasing continuously, and will never reach a maximum. In a real coil, the current will only be limited by the parasitic resistance of the coil, which is often very small.

Edit, replying to your comment

The maximum current is limited by the parasitic resistance. But the actual current is varying continuously. So how do you calculate the average current?

Let's draw your circuit, including the parasitic resistance of the inductor:

schematic

simulate this circuit – Schematic created using CircuitLab

Just like an RC circuit, an RL circuit has a time constant, in this case given by L/R.

If i is initially 0 and V jumps from 0 to some value V1 at t=0, then i will start to increase with an exponential (not inverse-square) curve. The time constant of the exponential is L/R.

The time it takes the curve to reach 95% of its final value is related to this time constant, R/L, not to the self-resonant frequency of the inductor.

If you apply a square wave voltage at the input, you will get repeated rising and falling edges, all following the exponential curve.

enter image description here

Notice the curve is always steepest (rising or falling) immediately after the input voltage changes, unlike in the curve given in the question.

To calculate the average current, you need to integrate:

\$ \bar{i} = \dfrac{1}{T}\int_0^T i(t) \mathrm{d}t\$

Take the integral over one cycle of the square wave.

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  • \$\begingroup\$ 'I'm not sure what you're getting at here. If you apply a steady voltage to an ideal inductor, current will keep increasing continuously, and will never reach a maximum.' That's only true mathematically -- so to speak. You're talking about an inverse square decay in the amplitude rise. In reality, the current will increase very quickly to 95% and get closer and closer to 100% but never quite make it. So for all intensive purposes, the inductance limits the current's reaching of 95% saturation by some time period. If you cut the power after that time period, current will steadily decrease. \$\endgroup\$ – JamesHoux May 28 '13 at 1:03
  • \$\begingroup\$ My point is that the inductance of an inductor limits the current to some amount if you keep switching the input voltage on and off at a repeated frequency. If you set that frequency to match 1/4 (or 1/2 depending on how you want to think of it) of the natural frequency of the coil, you will essentially cause the current to rise and fall from 0 to 95% saturation (almost 100% of DC current based on Ohm's Law), over and over and over. The result is that you're getting somewhere around half the current vs straight DC. I'm asking about the calculation of that current. Reactance doesn't work. \$\endgroup\$ – JamesHoux May 28 '13 at 1:05
  • \$\begingroup\$ @Jim, as I said, in a real inductor, the current is limited by the parasitic resistance of the coil. That resistance value can be very different for two different inductors with the same inductance value. \$\endgroup\$ – The Photon May 28 '13 at 1:06
  • \$\begingroup\$ @Jim, what do you mean by the "natural frequency" of the inductor? \$\endgroup\$ – The Photon May 28 '13 at 1:07
  • \$\begingroup\$ The maximum current is limited by the parasitic resistance. But the actual current is varying continuously. So how do you calculate the average current? \$\endgroup\$ – JamesHoux May 28 '13 at 1:07

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