# Miller's Theorem - Input Capacitance

Please see attached image. I've become confused if the answer to this question, which uses Miller's Theorem is correct as it should not be a minus, rather it should be a plus. Please see my working out.

Thanks

In general case the situations look like this

Now let as try to find a input resistance.

Rin = Vin/Iin

In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

Rin = Vin/Iin = R/(1 - A)

As you can see we have a "minus" sign. We get the "plus" sign only when our amplifier gain is negative (inverting amplify)

Rin = R/(1 - (-A)) = R/(1+|A|)

• Okay, so when the gain is negative the resistance would be plus, but when the gain is positive the resistance would be minus? If that made sense Commented May 15, 2016 at 19:33
• If you have a inverting amplifier we will have a "plus" sign and our Rin resistance is (1 + |A|) smaller then R or our capacitor is (1 + |A|) larger then C en.wikipedia.org/wiki/Miller_effect. But if our amplifier is "voltage follower"(non-inverting) and your circuit is a source follower (voltage gain from 0 to 1). Now Rin will be 1/(1 -A) times larger. And sometimes we call this a "bootstrap". But for A = 1 we will have Rin = oo. And if A > 1 we have a special case. Rin becomes negative. We create a negative resistor. For example if A = 2 and R = 10 Ohm ---->Rin = -10 Ohm
– G36
Commented May 16, 2016 at 16:50
• And this is how Negative impedance converter work. en.wikipedia.org/wiki/… have a non-inverting amplifier with gain of 2 and a feedback resistor (R3 in wiki) between the input and output terminals.
– G36
Commented May 16, 2016 at 16:52