# Capacitance multiplier

#### MrSoftware

Joined Oct 29, 2013
1,935
Maybe this will help:

#### Irving

Joined Jan 30, 2016
1,064
That circuit is some sort of long time oscillator? I can't make it do anything useful. All it seems to do is charge c1 up to full output of amp and sit there...

#### Zeeus

Joined Apr 17, 2019
602
Maybe this will help:
Thanks.. was going to watch it later but not sure it helps for the equation

Thanks all the same.

#### WBahn

Joined Mar 31, 2012
26,295

Post #15

View attachment 212615

Done the math more than 3 times now and getting this :

$Ceq = [(R1+R3)/R3] * [C1/ (sR1C1 + 1)]$

Kindly assist

Thanks
Show the work of how you got that, but I think you are fine.

Are you sure the given equation isn't an approximation for the case R1 >> R3 and for sufficiently low frequency?

The key to understanding how it works is that the output of the opamp is (nominally) always equal to the voltage on the capacitor. If you apply a voltage at the output of the circuit that is different than the capacitor output the current that goes into changing the capacitor voltage is (Vout-V+)/R1. But the opamp draws an additional (Vout-V+)/R3 from the output node. So the total current draw to produce the same change in capacitor voltage is increased by a factor of (R1+R3)/R3.

Further, note that this circuit is not going to act like an ideal capacitor. In an ideal capacitor, if you apply a voltage that is different than the capacitor voltage you get an impulse of current to create a step change in charge on the capacitor. Here it acts as though there is a resistor in series with the capacitor. The effective series resistor is (R1||R3) but since this is in series with the multiplied capacitance, the time constant is going to work out to R1·C1.

So that handwavy analysis would seem to agree with your results.