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 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
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
May 16, 2016 at 16:52