Hello,

I'm trying to figure out what the input impedance to this op-amp circuit is. I have 2 ways I've been thinking about this, and I'm not sure which one is right, or if they're both wrong. I'm looking for the input impedance as seen by the AC voltage source, V1.




Way 1: It's the impedances of C1, C2, and R3 in series. Done

Way 2: Thinking about a virtual short, it's C1, C2, R3, and R2 in series. Done

Basically, I'm wondering if R2 plays a role in the input impedance (as defined above) in this circuit.

much thanks!

 

I answered your question a few minutes ago. Maybe on another website.
It is like an echo, echo, echo, .....

Why do you think that R2 is in series with the input parts??

 

Yes you did, thanks! It was on another website

I thought it might be in series because of what I was thinking was a "virtual short" but as I was corrected on the other website, it's a virtual ground not a virtual short.

I was seeing the two inputs at the same voltage potential... and taking them to be electrically the same point. This is where I was getting mixed up.

 

The negative feedback makes a DC gain of 1. Which means that the DC output voltage is the same voltage as the (+) input voltage.

If there is a dual polarity supply then the (+) input is biased at 0V so that the output can swing symmetrically up and down equally.
if the supply has a single polarity then the (+) input is usually biased at half the supply voltage so that the output can swing symmetrically up and down equally.

 

I say it's C1, C2 and R3 in series. The op amp simulates a low-impedance node at its negative input terminal, so none of the other components count. At least, that's true as long as the amp remains in its linear region (not exceeding power supply voltage or frequency limits).

 

Thanks for the responses, I have a follow up question

why is it that the feedback resistor and the load resistor don't play a role in the input impedance?

 

why is it that the feedback resistor and the load resistor don't play a role in the input impedance?

Because the impedance at the (-) input pin of the inverting opamp is zero ohms. Anything that is in parallel with zero ohms still makes zero ohms.