Is the formula X\$_L\$ = 2\$\pi\$fL still true for rectified (but unfiltered) AC (in which case the frequency f, and therefore the reactance, would be doubled in the case of full-wave rectification)?

enter image description here

I forgot to add a parasitic resistor in series with the inductor. The inductor is not ideal as depicted in the image above.

  • \$\begingroup\$ Comments Removed \$\endgroup\$ – W5VO Feb 26 '18 at 7:16

The answer to your question is NO. With such a waveform for voltage (or current) the reactance is not defined by the same formula used with sinusoidal steady state inputs and outputs (with or without the modification in the factor 2 for the frequency) because the concepts of reactance, impedance and phasors only apply to sinusoidal steady state.

Applicability of the concept of impedance

Sinusoids, cosinusoids and their complex relatives, exponentials, have the very special property that they retain their waveform shape in linear time invariant circuits. The reason for this all boils down to the self-similarity of the exponential function, but you can think of a more 'real' explanation considering that the derivative of a sine is a cosine (another sinusoidal function, just shifted) and likewise, the derivative of a cosine is a sine (ok, with a sign change, it can still register as a phase shift). And the constitutive relation of (linear, time-invariant) inductors and capacitors are linear relationship involving derivatives. So, basically: sinusoidal voltage or current IN ---> sinusoidal current or voltage OUT.

The only effect a circuit with R, L and C can have on a sinusoid is to attenuate it and to phase shift it. One can describe this effect with a mathematical quantity that include these two pieces of information. And guess what, a complex number does just that.

The impedance is described by this complex number. You have a sinusoidal stimulus, and a sinusoidal response. When described by phasors, their ratio is just a complex number - the impedance, or the admittance depending on how you like to see it - describing how much the response has been attenuated and shifted in phase.

Inapplicability of the concept of impedance

BUT all of this simplified machinery can work only if you have sinusoidal IN and sinusoidal OUT. It does not work with other waveform shapes because they get 'distorted' by derivatives (and integrals). This means that When you feed a R-L-C linear time invariant circuit with a nonsinusoidal input, the concept of impedance can no longer be used because it would be meaningless.

We can see that by solving the differential equations governing the circuit or... by simply using a simulator :-) I ran a couple of LTSpice simulations feeding an inductor with a full-wave rectified sinusoidal voltage and current generators controlled by this voltage:

enter image description here

I had to use voltage controlled voltage and current generators to make sure the L circuit did not load the rectifier (which it does, and a lot). The results are strikingly different.

When a voltage V(out2) with that shape is forced across an inductor, we get a current that builds up indefinitely, as shown by the purple waveform I(L2). This is not surprising, since to get the current we need to integrate the voltage over time and since V(out2) never go negative, we can only add, and add, and add...

enter image description here

But if a current I(L1) with that shape is forced into an inductor, we get a periodic distorted triangular-like voltage V(out) across it. The reason for this strikingly different behavior is that now to get the shape of the voltage we have to take the derivative of the current.

enter image description here

It worth noting that the concept of impedance requires that the signals be both sinusoidal and steady state. The above example has used a piecewise sinusoidal stimulus and although in each period the derivative and integral are still sinusoidal in shape, the overall waveform shape is not. When the derivative is involved we have discontinuities (in the above simulation they are softened because the input signal was, since I have used real diodes in my full-wave rectifier); when the integral is involved, we have a build up due to the value of the integration constant set by the boundary conditions.

In either case, since derivatives and integrals of functions that are not exponentials, sines or cosines return in general functions with a different shape, you can no longer describe the effect the inductor has on the stimulus waveform as a simple attenuation and phase shift. The bottom line is that you can kiss the concept of impedance goodbye.

Fourier analysis to the rescue

You can still use the useful impedance concept, though, if you apply it within its limits. If you decompose the non-sinusoidal input signal into a sum of sinusoids (even a series, or an integral if it is not periodic) of different frequencies, you can use the concept of impedance on each single sinusoidal component to find the sinusoidal components of the output signal and then reconstruct the resulting waveform.

  • \$\begingroup\$ You got my vote too! \$\endgroup\$ – jonk Feb 26 '18 at 3:50

Sure, the formula is still the same. And yes, you're correct that the fundamental frequency is doubled from the original sine wave.

What's different is how the formula is used. This reactance formula is a single frequency representation of the time dependent properties of an inductor. A pure sinusoid is only composed of a single frequency, so you can easily calculate the reactance at that frequency.

A rectified sinusoid is composed of an infinite sum of pure sinusoids at every integer multiple of the fundamental frequency. So, the original equation is accurate... but only for one frequency component at a time. Technically, you could solve the circuit by calculating the reactance at each of the (infinite) frequency components, finding the voltage or current of interest, and summing the results over all frequencies, but depending on what your circuit looks like and what information you actually need, you may want to select a different approach to solving the problem.

For more information, i recommend you research the Fourier transform, including what the Fourier transform of a rectified sinusoid is.

  • 1
    \$\begingroup\$ I would clarify the "...and summing the results" making it explicit that it is referred to waveforms (voltage or current), not reactances, at different frequencies. As long as the circuit is linear (and a circuit with only linear ideal resistance and inductance is linear), this is justifiable, no matter how nonlinear is the circuit that produced the rectified waveform used as its 'input'. (provided it is not loaded by the RL circuit) \$\endgroup\$ – Sredni Vashtar Feb 25 '18 at 23:27
  • \$\begingroup\$ @Sredni Vashtar Good catch, clarified. \$\endgroup\$ – Selvek Feb 26 '18 at 0:11

Phew! The input voltage you've drawn can be approximated quite well with a few terms of a fourier series.

From this page at RFCafe, a rectified 50 Hz sine wave with a peak amplitude of V, has these components:

  • DC of 0.63 V
  • 100 Hz of 0.42 V
  • 200 Hz of 0.08 V
  • 300 Hz of 0.04 V

That's probably enough for your purposes.

The total current that flows in an RL load connected to this source is then:

Itot = 0.63*V/R + 0.42*V/(sqrt(R^2+(2pi*100*L)^2)) + ...etc

With no R of course the current is infinite because of the DC component.

  • 1
    \$\begingroup\$ Yeah. That's what I said at the outset, too. (Except for the DC part which of course occurs because of the offset.) +1 for wading in, too! \$\endgroup\$ – jonk Feb 26 '18 at 3:21
  • \$\begingroup\$ Thanks! I just accepted Sredni's answer, but this is very useful too! \$\endgroup\$ – user1247 Feb 26 '18 at 3:49

After a schematic was shown I decided to do what Chris advised you to do. Numerically simulate it.

So I chose my favorite simulator, CircuitJS, and tried to set up the same schematic as in your question.

This is my attempt:

enter image description here

Link to simulation so you can interact with it.

  • The green line is the voltage across the inductor
  • The yellow line is the current through the inductor
  • The white line is the reactive power of the inductor

The leftmost op-amp has the following mathematical expression \$V_{out}=\text{abs($V_{in}$)}\$, the input is a sine wave with an amplitude of 5 V. Then this voltage is buffered with a second op-amp. Both op-amps are ideal so they can supply, in theory, infinite current.

Then the rightmost op-amp simply acts as an ideal voltage source.

As you can see, the current through the inductor is simply the integral of the voltage across it, which will tend to infinity. The reactive power will likewise also tend to infinity.

When the reactive power is known, the reactance can be calculated as following:

\$Q = \frac{V^2}{X}\$

and we want X

\$X = \frac{V^2}{Q}\$

So \$V^2\$ is periodic with an amplitude that will never change. Never forever.

This means that \$X = \frac{V^2}{Q}\$ will tend to 0 Ω, because as you can see, Q grows larger and larger to infinity (seen in white).

Adding a simple 1 ohm resistor in series with the inductor, which obviously should be there. Duh! Stupid me.

Then you get something that looks like this:

enter image description here

Here's the link to this schematic if you want to interact with this one.

The graph's are identical to the ones above.

Derp, too tired to edit this question correctly. If anyone feel like, edit it. If not, don't. Adjö boys.

  • 2
    \$\begingroup\$ Comments Removed \$\endgroup\$ – W5VO Feb 26 '18 at 7:18

No, because that formula (\$X=2\pi f L\$) relies on the properties of sine waves, and the rectified AC is not a sine wave.

The definition of inductance, L, is:

$$V = L\frac{dI}{dt}$$

where I is the current through the inductor and V is the induced voltage across it (with I and V functions of time, t).

If the current is sinusoidal, $$I = \sin(2\pi f t)$$ with f the frequency. Differentiating gives: $$V = 2\pi f L \cos(2\pi f t)$$

so the voltage waveform is also sinusoidal but with a 90 degree phase shift (sin to cos) and a multiplicative factor of \$2\pi f L\$, which is the magnitude of the reactance you state.

If V and/or I are not sine waves, this relationship does not hold.


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