# Equivalent input capacitance

Joined Oct 18, 2011
19
I am having difficulty with understanding how to solve the circuit with the capacitor tied to the output of the op-amp. I assume that I break the circuit down into two parts and then solve for the Thev. equivalent circuit but with no input voltage I am at a loss. Do I add a source at the output and then open circuit the capacitor and solve the circuit that way?

any help would be appreciated.

the problem asks for the equivalent input capacitance Ceq.

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#### crutschow

Joined Mar 14, 2008
33,358
The first op amp is a follower/buffer. The second op amp is an inverter with gain. So what effect would this have on the apparent capacitance caused by C at Ceq? Remember C = Q/V.

Joined Oct 18, 2011
19
So if the gain of the follower is one, and the gain of the inverter is R2/R1 that leads me to
Ceq = C(1+Av)? where Av = R2/R1?

#### Ron H

Joined Apr 14, 2005
7,014
So if the gain of the follower is one, and the gain of the inverter is R2/R1 that leads me to
Ceq = C(1+Av)? where Av = R2/R1?
Close. This is not the same as Miller capacitance.

EDIT: It is the same as Miller capacitance. A thousand pardons.

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Joined Oct 18, 2011
19
I guess I am trying to make this harder than it really is. I understand the C=Q/V but what is confusing me is the charge. I thought of using a capacitive divider, but in the end I get a negative value for Ceq because in the problem C = 6nF, R1 = 3.8k, and R2 = 24.8k.

Vo/Vin = C/(C+Ceq)

and since Vo/Vin = R2/R1 then

Ceq = C* (R1-R2)/R2

Thank you all for your help

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#### Ron H

Joined Apr 14, 2005
7,014
I am very embarrassed.
You were correct in post #3. It is the same as Miller capacitance. I'm sorry for confusing you.

Carl, I needed you to keep me honest. Where were you?

#### crutschow

Joined Mar 14, 2008
33,358
I am very embarrassed.
You were correct in post #3. It is the same as Miller capacitance. I'm sorry for confusing you.

Carl, I needed you to keep me honest. Where were you?

#### Ron H

Joined Apr 14, 2005
7,014