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There is a lot of information about RLC circuits and resonance, and many times it is mentioned that when a series (parallel) circuit resonates, a really high voltage (current) can be found across the inductor and the capacitor and this can lead to faults of components.

But why? I mean, it's clear that if I take a component and put it under 1kV, it's going to break (unless designed for sustaining high voltages...), but I see the case of resonance a bit different: while it's true that there is a high voltage/current across the components, the net voltage/current is zero if you consider both the capacitor and the inductor. Let's take an RLC series: the voltage across the series of the capacitor and the inductor is zero, at the resonance. The power supply or the load are not interested in whatever high voltage is there across the capacitor or the inductor. So why should it lead to problems?

Please let's assume that the capacitor and the inductor have no problem in sustaining the voltage across themselves and they don't break because of it. The issue is how this can lead to problems for other components in the circuit.

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  • \$\begingroup\$ Isn't the voltage for any circuit zero if you count the way you do? That's almost the definition for a circuit. \$\endgroup\$ – pipe Jul 7 at 17:45
  • \$\begingroup\$ It's not exactly that case, across the LC you have zero voltage even though it's not a closed loop, and that's what really makes the difference as I see it. It's like putting two voltage supply in series, one of 5V and one of -5V and the result is as if there were no voltage supplies. \$\endgroup\$ – Elia Jul 7 at 18:42
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A fairly common example of when RLC circuits can produce an excessive voltage is a 2nd order low pass filter used in conjunction with a voltage regulator. I've fallen foul of this myself so I know it can be a problem.

I designed a 2nd order filter to remove noise from the input feed to a voltage regulator. The voltage regulator was nominally fed about 20 volts at the input and the filter removed a lot of noise that was present thus, the output from the voltage regulator was also cleaner. One of the tests of the circuit was to check that a short on the output didn't damage the design but it did. The problem was that an excessive current was taken through the regulator and also through the inductor (in the filter) that fed the regulator.

Upon removing the short on the output, the RLC filter did the normal thing and produced an under-damped ringing waveform whose peak voltage exceeded the input voltage rating for the regulator. The regulator fried. It took a little while to figure out because it was initially assumed that the original short circuit on the output did the damage (but it didn't). So, what was it all about: -

enter image description here

The circuit above was the type of filter used and the simulation shows the problem. In the bode plot there is very high gain at 15.9 kHz (40 dB) and this translates to a voltage multiplication of 100:1. In the transient response clearly a 1 volt step produces nearly a doubling of the 1 volt at the output.

This is just one specific example.

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  • \$\begingroup\$ Yeah, I didn't think that it could be used the voltage across capacitance/inductor as the output! Anyway, now I'm curious, how did you solve the problem with the voltage regulator? \$\endgroup\$ – Elia Jul 9 at 18:58
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    \$\begingroup\$ @Elia in the end I bit the bullet and got rid of the inductor. I did think about using a zener clamp but it all worked out fine without the inductor i.e. my worries about noise were not realistic. \$\endgroup\$ – Andy aka Jul 10 at 8:51
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Consider that the circuit board itself and/or any insulated wiring has a breakdown voltage. After breakdown, any nearby components might be adversely affected (exceed specified maximums on random pins, etc.)

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The high Q of Series RLC circuits amplify current according to f and Q= 2pifL/(DCR+ESR+Rs) ratio.

Parallel RLC circuits amplify voltage at fo, according to shunt R for high Q=R/(2pi*fL) at fo.

Damage depends on component ratings for ripple current induced power dissipation or BDV.

Any questions?

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  • \$\begingroup\$ So what you're saying is that it's not the high voltage across L or C that can lead to problems, but the - possible - high current through the circuit? Anyway, I think the current is even easier to control than the voltage across L or C. The first one depends only on the Rs value, the other two by Rs, and the impedance of L and C at the resonant frequency. It still doesn't seem a big issue to me... Am I missing something else? \$\endgroup\$ – Elia Jul 7 at 18:49
  • \$\begingroup\$ Yes read again... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 7 at 18:50

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