# Opamp Question

Discussion in 'Homework Help' started by circuit_solver, May 1, 2013.

1. ### circuit_solver Thread Starter New Member

May 1, 2013
2
0
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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2. ### nigelwright7557 Senior Member

May 10, 2008
488
71
f=1/(2 pi R C)

3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
How would you do it is the buffer weren't there?

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

4. ### circuit_solver Thread Starter New Member

May 1, 2013
2
0
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.

5. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Check the units on the above. They don't work out, so you know it is wrong.

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.