Opamp - preceded by a unity buffer

[kimandrew20]

Homework Statement

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.

Homework Equations

wb = 1/CRf

The Attempt at a Solution

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

 

[berkeman]

Homework Statement

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.

Homework Equations

wb = 1/CRf

The Attempt at a Solution

1/CRf = 2 * 1khz
Rf/Ri = 1 so Rf = Ri which means Rf = 27kΩ
C = 1/(2∏ * 1khz * 27kΩF)View attachment 58405

Looks like you have done it correctly. The unity gain buffer just let's 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).

 

[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?

 

[berkeman]

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?

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.

 

[rezeew]

I am not sure what you mean by "preceded by a unity buffer," as a unity buffer does not have any effect on the circuit's frequency response. However, assuming that the unity buffer is simply a voltage follower, the circuit can be analyzed as if it were not present.

To find the component values for the desired frequency response, you can use the standard formula for a low-pass filter: wb = 1/CRf, where wb is the desired cut-off frequency. Since the low frequency magnitude gain is 1, the ratio of Rf to Ri (the input resistor) must also be 1, which means Rf = Ri. Using this information, you can solve for the component values as you have done in your attempt at a solution.

However, if the unity buffer is not a simple voltage follower and has an effect on the circuit's frequency response, then the analysis would need to be adjusted accordingly. It would be helpful to have more information about the unity buffer and its purpose in the circuit to provide a more specific solution.