LC circuit voltage amplification

Thread Starter

Virium

Joined Jan 8, 2016
4
Hello everybody, small question here, could someone give me an explanation or some material to read, as to why the VOLTAGE between the conductor and capacitor (red circle, relative to ground) is amplified ? Specially when we reach resonant frequency. I always had the impression that the voltage should remain the same or slightly drop there.
Thank you for reading.
 

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mcasale

Joined Jul 18, 2011
210
If you write the equations for XC and XL in a series circuit, you will see that the impedances cancel each other out at the resonance point. So, you have a direct connection to the load resistor. There is really no amplification going on. It may look like it because XC and XL are battling each other over the frequency range. They are opposite in phase regardless of the frequency.

If you Google "resonant frequency" I'm sure you'll get more info than you want.
 

#12

Joined Nov 30, 2010
18,210
I watch in anticipation because I am not clear on this point. I think running the math might provide some insight, but I'm not sure.
I'd really like to see somebody point out the equation where voltage approaches infinity.
 

mcasale

Joined Jul 18, 2011
210
Okay, I'll skip the one about infinite voltage. Sounds like a "black hole" :(

If you look at the circuit as a voltage divider, you get
Vout/Vin = [XC + R]/[XC +XL +R]
at resonance XC + XL = 0 (in an ideal world). Therefore,
Vout/Vin = [XC + R]/R = 1 + XC/R (whew!)
So in a hypothetical case where XC >> R at resonance, it looks like a bigger voltage, but the phase is 90 deg out.

Maybe try doing the calculations for some real examples and see what you get.
 
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