Opamp Question

Thread Starter

circuit_solver

Joined May 1, 2013
2
Hi guys, I have to figure out how to design this circuit (i.e. determine component values) to give this circuit a cut-off frequency of 1khz and a low frequency magnitude gain of 1

Basically, I know how to do it if i'm ignoring the unity buffer that precedes the opamp (use Rf/Ri = 1), but im not sure how to modify my process to include the buffer.

Does anyone have any ideas?
 

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WBahn

Joined Mar 31, 2012
30,082
How would you do it is the buffer weren't there?

Why do you think the buffer would require you to do anything different?
 

Thread Starter

circuit_solver

Joined May 1, 2013
2
Thanks for the responses guys.

My working so far is (ignoring the unity buffer):

1/CRf = 2 * 1khz
Rf/Ri = 1 so Rf = Ri which means Rf = 27kΩ
C = 1/(2∏ * 1khz * 27kΩF)

What I've read about the unity buffer is that it has a gain of 1, but I'm not sure if I have to change my working above to incorporate the unity buffer.
 

WBahn

Joined Mar 31, 2012
30,082
Thanks for the responses guys.

My working so far is (ignoring the unity buffer):

1/CRf = 2 * 1khz
Check the units on the above. They don't work out, so you know it is wrong.

Rf/Ri = 1 so Rf = Ri which means Rf = 27kΩ
C = 1/(2∏ * 1khz * 27kΩF)

What I've read about the unity buffer is that it has a gain of 1, but I'm not sure if I have to change my working above to incorporate the unity buffer.
If the buiffer weren't there, you would be connected to a voltage source that, as near as I can tell, is nominally an ideal source and thus has no output impedance. But, nominally, that's what a voltage buffer is doing is taking a voltage signal from a source that has a significant output impedance and buffering it so that the circuit after the buffer thinks it is hooked up to a source with very low (ideally zero) output impedance.
 
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