# Infinite impedance at resonance in parallel LC circuit ? does that mean it is not usable?

In a parallel (tank) LC circuit, this means infinite impedance at resonance. In a series LC circuit, it means zero impedance at resonance: What does that mean?

Does this infinite impedance at resonance in the parallel LC circuit means it is not feasible to use at resonance frequency ? and vice versa for the serial circuit, does it's performance better than serial circuit at resonance ?

## 3 Answers

Both series and parallel resonances are useful One use of these circuits would be a band pass filter. So for instance if I put a parallel resonance after the resistor of a Thevenin AC source, the frequencies well below resonance would be shorted by the inductor. Above resonance the capacitor would short the signal. At resonance the the high impedance of the circuit would let the signal through. This same thing could be done with series resonance followed by the load resistor. In the first case we short unwanted signals to ground. In the second the unwanted frequencies are blocked by high reactances. A use of this would be accepting a radio signal but blocking all others.

Neither is really better than the other. Picking which one to use depends on the problem being solved.

Both circuits can be used at their resonant frequencies.

In practice because all components have parasitic resistance it is not possible to make a perfect resonator. Well not at room temperature. Some pretty cool things can probably be done at temperatures near absolute zero.

OK, as an example this filter is centered near 1MHz and with 3db frequencies of 900kHz and 1.1MHz. For a stop band, 600kHz and 1.4MHz with a minimum of 24db was chosen. The first thing to do is to find filter attenuation curves for the type of filter that works best for you. Here is an example of a curve for 1db ripple Chebyshev filter; Using this curve this filter has an f/fc of 4. To get 24db this chart shows n=2. Now look up the low pass prototype. This filter uses 150 Ohm source and 50 Ohm load so it comes up to be the first line of the table. To convert the low pass to a bandpass use the formulas in the schematic. Here is a quick simulation to show the result;

• Thanks. First thing the load single handedly the biggest (apart from LC) manipulator in the output behaviour to the input. So in case I put a serial resonator circuit at the transmit end and parallel resonator circuit at the load, it still acts as the band pass filter. But how does these two resonances interact with each other ? can you please comment a bit on that. Commented Nov 7, 2016 at 18:22
• It is not possible to describe how the elements of a band pass filter are calculated in a paragraph. But if your goal is to build a band pass filter you don't have to be a filter expert to do it. There is a simple process that will get the filter you want. The way to do this to use a prototype LP filter and then scale it to your bandpass filter termination and frequency requirements. If you goal is to design a LC band pass filter, say so I will post a how to do that here. If you just want to understand, the book I learned from is; "Intro to Circuit Synthesis and Design" Temes and LaPatra Commented Nov 7, 2016 at 19:56
• Thanks. Yeah it would be a lot of help if you can show me how you go about the design of LC band pass filter. And thanks again for the book, I will take a look now. Commented Nov 7, 2016 at 21:50
• I updated the answer to provide and example. Commented Nov 8, 2016 at 1:52
• Awesome. Thanks. final question, May I know why there seems to be two small peaks in the frequency sweep plot ? Commented Nov 8, 2016 at 3:12

Think of a inductor and capacitor in parallel. At DC, it's a short due to the inductor. At infinite frequency it's a short due to the capacitor. You'd therefore expect the impedance to be high in the middle.

It is, but there's a special case. Inductors and capacitors aren't just frequency-variable resistors. For one thing, the current is shifted by 90° phase from the voltage, although this phase shift is in opposite directions for inductors and capacitors.

Now think again what happens to a parallel inductor and capacitor as the frequency changes. We already said at DC the inductor is a short, so the combination is a short too. As the frequency starts to go up, the inductor still draws large current, and the capacitor starts to draw a small current. The net result is that the inductance dominates, and the current lags the voltage by nearly 90°.

At the other end, with high but finite frequency, the capacitor dominates. The current leads the voltage by nearly 90°.

There is a frequency somewhere such that the current magnitude thru the capacitor and thru the inductor are equal. At that special frequency, the inductor current with its 90° phase lag and the capacitor current with its 90° phase lead combine to cancel each other out. We have no current when a voltage of this frequency is applied. That's infinite impedance. A little off this special frequency, and the current phase is no longer 0°, and the above doesn't apply.

Real parts aren't ideal. The capacitor and especially the inductor each have a little inevitable resistance in series with it. This prevents the overall impedance to become truly infinite at the special frequency. With good quality parts, it can come fairly close.

The result is a peak in the graph of impedance magnitude as a function of frequency. The frequency of the peak is the resonant frequency. The more ideal the parts, the sharper the peak, meaning it is tighter in frequency.

This relatively sharp peak can be exploited to filter in or out just the resonant frequency (or frequencies close to it).

A series L-C combination works similarly, except that the impedance at 0 and infinite frequency is infinite. It has lower impedance to in-between frequencies. At resonance, it's impedance is ideally 0.

• An ideal series LC circuit exhibits infinite impedance at DC and at infinite frequency, but at resonance it exhibits zero impedance. Commented Nov 7, 2016 at 17:24
• @EMFi: Doh! That's what happens when you get lots of interrupts while writing a answer. Fixed. Commented Nov 7, 2016 at 17:40
• "See what they made me do?" ;) Commented Nov 7, 2016 at 18:12

With ideal (aka perfect) components, yes, you get zero or infinite impedance at resonance.

However no component is free from losses, and this makes the impedance of any resonant circuit non-zero, or non-infinite.

In practice, any resonant circuit is coupled to some other circuit, and it is often the losses from the other circuit that control how non-zero or non-infinite the overall impedance at resonance becomes.