I want to analyse a parallel LC circuit in terms of inputs and outputs as I am studying them in the context of using them as a tuned antenna.

LC circuit

However I am unsure of where to identify the inputs and outputs in order to find a transfer function between them.

Thinking of the mass on a spring analogy the input may be the applied force and the output is the displacement which becomes very large when the applied force nears the resonant frequency.

Is there some kind of equivalent for this circuit?

  • \$\begingroup\$ Why would you need a mechanical analogy? This only adds complication. \$\endgroup\$ – Chu Oct 5 '18 at 15:00
  • \$\begingroup\$ I do not require a mechanical analogy, I'd just like to know what the input and output of this resonant circuit are, in the same way the input to the mass on a spring is the force and the output is the displacement \$\endgroup\$ – EEvan Oct 5 '18 at 15:09

Your circuit is a bit incomplete in the sense that it only shows the inductor and capacitor in parallel. This circuit is often called an "LC tank" and it has an impedance that is frequency dependent.

There are two ways to determine an impedance:

  • A: apply a voltage and measure the current
  • B: apply a current and measure the voltage

Like so:


simulate this circuit – Schematic created using CircuitLab

Now the "special" property of the LC circuit is that in some situation the energy travels back and forth between inductor and capacitor. That's called "resonance" and an important property of you want to use this as an antenna.

Can you spot the issue with circuit A? It is related to this resonance.

The issue is the following: suppose capacitor C1 has a certain charge (and therefore some energy) and since we want the circuit to resonate the energy has to go into the inductor. But can this happen? No it can't because the voltage source V1 dictates the voltage across C1 so even if C1 had a certain charge that would immediately be overruled by the voltage set by V1. So circuit A isn't useful.

Is circuit B better?

Yes it is, I2 is a current source so it will not enforce any voltage. But what about the current? Indeed I2 does force the current but that is the current through both L2 and C2. So C2 can still push out extra charge (current) into L2. This means the energy can be exchanged between L2 and C2 so we can have resonance.

So the input signal for your circuit is the current and the output signal will be the voltage that develops across the LC circuit as a result.

I'm not going into the analysis of that and it is not needed as it is all explained on the LC circuit article on Wikipedia

Also, if you want a voltage input but and current output that's also possible but then you have to use a series LC circuit.

  • \$\begingroup\$ Thanks for the help! So for when the circuit is used as an antenna, the EM waves carrying the radio signal induce a current in the inductor? I would have thought the law governing this was Faraday's law of induction which states that the changing magnetic field induces a voltage not a current so I am unsure how to proceed. \$\endgroup\$ – EEvan Oct 5 '18 at 13:56
  • \$\begingroup\$ Voltage and current are closely related, it depends on the circuit connected to the inductor what the induced EMF will result in. If the inductor has nothing connected to it (open circuit) there will be a voltage. Current cannot flow as there is no loop. If you short the inductor then there can be no voltage (due to the short) but there will be a current. When an impedance is connected to the inductor (that can be a capacitor) then a bit of both happens. You get voltage and current. \$\endgroup\$ – Bimpelrekkie Oct 5 '18 at 15:12
  • \$\begingroup\$ Say the resonance circuit is connected to a buffer which then goes to the rest of the circuit e.g. a demodulation circuit. If there is an induced EMF across the inductor then surely this is the voltage seen at the input to the buffer and so I am not sure where the whole resonance concept comes into this which is what confused me to begin with. I thought the input would be the induced EMF then the resonance would somehow 'amplify' it giving a higher output voltage, but clearly in this circuit what I had initially thought of as input and output voltages are one and the same. \$\endgroup\$ – EEvan Oct 5 '18 at 15:44
  • \$\begingroup\$ A buffer is a voltage sensing circuit, it takes no current from the LC tank and therefore will not influence it. Indeed a buffer (or an amplifier with a high input impedance) are often used for making a receiver. The resonance is essential at the resonance frequency, the least amount of energy is needed to get a certain voltage. So at resonance the LC circuit is most sensitive. At much higher and lower frequencies you will receive almost nothing. \$\endgroup\$ – Bimpelrekkie Oct 5 '18 at 15:59
  • \$\begingroup\$ Is there a way to mathematically represent this 'sensitivity' that the circuit gains near resonance in terms of some input and some output? This is essentially what I am looking for. \$\endgroup\$ – EEvan Oct 5 '18 at 16:05

Given that the capacitor is less likely to be the antenna than the inductor you would put the induced voltage source in series with the inductor. That is what drives the circuit. The output would be across the capacitor forming a resonant (but also a low pass filter). There would also be losses that are modelled by the resistor below: -

enter image description here

Interactive calculator.

Is there some kind of equivalent for this circuit?

Yes it can certainly be modelled as a spring and mass with damping but the tool above allows you to enter values and further info on the website allows you to understand the math.


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