# Opamp - preceded by a unity buffer

1. May 1, 2013

### kimandrew20

1. The problem statement, all variables and given/known data

Hi, I'm trying to figure out the component values(Rf and C) for the circuit given. The circuit must be designed so that it has a cut-off frequency of 1khz and a low frequency magnitude gain of 1. I know how to do it if i ignore the unity buffer in front but I'm not sure how to incorporate the unity buffer into my working.

2. Relevant equations

wb = 1/CRf

3. The attempt at a solution
1/CRf = 2 * 1khz
Rf/Ri = 1 so Rf = Ri which means Rf = 27kΩ
C = 1/(2∏ * 1khz * 27kΩF)

2. May 1, 2013

### Staff: Mentor

Looks like you have done it correctly. The unity gain buffer just lets you ignore the output impedance of the signal source, since the buffer's output impedance is low compared to the 50 Ohms of a typical signal generator (or much higher impedance of other signal sources like microphones).

3. May 1, 2013

### kimandrew20

Thanks for the response Berkeman!

So does this mean that the presence of the buffer doesn't really change any calculations for the behavior of the opamp? Like, if I were to draw a bode plot for the second opamp, it wouldn't be affected by the presence of the unity buffer?

4. May 1, 2013

### Staff: Mentor

Let's say that you were modeling this with SPICE as part of a product design. You would then have a different impedance to add in series between an ideal voltage source -- either the output impedance of the buffer or the output impedance of the real signal source. Since those impedances add with the input resistor of the 2nd stage, they will give you different errors in the Bode plot. The unity gain buffer version probably will give you negligible errors in this case.