is it possible that a RLC filter works as an amplifier? If it works at resonant frequency, then the output will be greater than the input.

enter image description here

Thank for your time.

  • 3
    \$\begingroup\$ Where should the energy come from? \$\endgroup\$
    – PlasmaHH
    Jun 8, 2016 at 9:12
  • \$\begingroup\$ Notice that you don't need an LC resonator for obtaining a voltage enhancement: it can be obtained also with an RC circuit. \$\endgroup\$ Jun 8, 2016 at 9:45
  • 1
    \$\begingroup\$ @MassimoOrtolano, an RC circuit with no switches or diodes? Can you post a schematic/ link? \$\endgroup\$ Jun 8, 2016 at 13:15
  • \$\begingroup\$ @GeorgeHerold Sorry for the late reply. A paper which describes RC circuits with gain greater than one is this one. If you don't have access to IEEE journals, you can find the analysis of one of those circuits in this discussion (it's in Italian, but the circuit and the calculations should be clear; I'm the author of the linked calculations). You can find several other example circuits there, with calculations. \$\endgroup\$ Jun 13, 2016 at 19:49
  • \$\begingroup\$ @GeorgeHerold If I recall correctly -- I don't have time to check now -- in the cited IEEE paper there were a few calculation errors, but the circuits remain valid. \$\endgroup\$ Jun 13, 2016 at 19:56

1 Answer 1


It can amplify the voltage (near to the resonant frequency) at the expense of load current. In other words you can't get power amplification. The frequency response will look like this: -

enter image description here

For a high damping (R is low in value, zeta = 0.5) the frequency response will be the lower blue curve but, with a high value resistor several tens of dB voltage amplification can occur. That amplification is not constant with frequency (as you can see) but can be significant from half resonance to a little over resonance.

A problem of significance occurs if you try and get too much amplification. Consider the situation when damping is really low i.e. resistance really high. For now the resistor can be ignored and what you are left with is a series resonant LC tuned circuit to ground. As you may know, a series resonant circuit (with zero or very low damping) will act as a short circuit - if you try and apply a voltage input at the resonant frequency it will be shorted by the LC. Because the inductor will never be perfect the input impedance won't be zero but the DC resistance of the inductor (maybe an ohm).


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