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Series resonant circuit with parallel capacitor:

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What is effect of that parallel capacitor (C2) on the circuit? And what happens if we remove this capacitor?

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2 Answers 2

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Think of it as the same as a crystal: -

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There are two resonant conditions (fs and fp) shown in the response curve on the right. fs is series resonance and fp is parallel resonance. A more conventional look at the impedance magnitude reveals where the resistance comes into play during series resonance: -

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In terms of formulas here's the deal: -

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Look initially at just the series branch. At the L.C1 resonance, they will go short circuit. Below that frequency, you will see mainly C1, above mainly L.

The whole circuit will reduce to R in parallel with C2 at the L.C1 resonance. Let's assume R is small, and C2 is small, as this is the most common place this sort of circuit is found. With a small R across C2, the whole network will have roughly the impedance R.

As the frequency increases, the conductance of C1 increases, that of L decreases, so the series branch becomes inductive.

Now we have two branches in parallel, C2, and the series branch looking inductive. These will go into a parallel resonance at some higher frequency. A parallel resonance is open circuit, when there are no losses. The loss of R means that the resonance does not go entirely open circuit, but the impedance is still much higher than you would expect for having C2 in parallel.

The reason why this circuit is very interesting, and the open circuit behaviour due to the parallel resonance is unexpected and bad, is that this circuit is often created accidentally from decoupling capacitors.

Let's say C1 is a large 10uF electrolytic, with its corresponding residual inductance and loss shown as L and R. We put C2, say a 10nF ceramic, across C1 for RF decoupling, because C1 is 'bad' at high frequencies due to the residual L.

In this case, R may be sufficiently big that we get away with it. The parallel resonance of the 10nF with the residual inductance of the electrloytic may be so de-Q'd by the high loss, that the parallel resonance impedance doesn't go too high.

These days however, we can get 10uF ceramics. If we replace the C1 electrolytic with a ceramic, R is an order of magnitude or two smaller, and the parallel resonance between the residual L and C2 is vicious. We end up with no decoupling, just at the frequency where we thought C2 ought to be providing a very low impedance to ground.

Parallel decoupling capacitors with care, a bit of extra loss between them will de-Q any resonances.

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