# OP-AMP frequency response of feedback network? help!

#### Dennis Atwood

Joined May 26, 2007
8
I'm having trouble calculating the -3db freq of a non-inverting op amp circuit.

basically, i'm looking at:

in ---|\
| \_____ out
| / |
---|/ R1
|______|
|
R2
|
C
|
GND

(hopefully my ascii art makes sense)

i've calculated that at high frequencies, the impedance of C is negligible, so the gain is equal to 1 + R1/R2 - no problem there.

now, according to the textbook i'm referencing, the -3db point is the freq at which the impedance of C equals R2

however, when i do the calculations with real values, it doesn't make sense to me. i'm using R1=18k, R2=2k, and C=4.7uF, for a high freq gain of 10. this is the example my textbook gives

at lower frequencies, i'm figuring the gain is equal to 1 + impedance(R1)/(impedance(R2)+impedance(C)). according to the book, the -3db point is at 17Hz. however, at this frequency i'm calculating a gain of 5.5 whereas i would expect a gain of 7.07 (1/sqrt(2)*10)

what am i missing?????

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#### hgmjr

Joined Jan 28, 2005
9,027

#### Dennis Atwood

Joined May 26, 2007
8
i was calculating that the magnitude of the denominator is:

sqrt((2000 -2000j)(2000 + 2000j)) =

okay, wait. you're right. i just realized i was taking the magnitude after doing complex division and i must have made a mistake along the way.

...now i'm just wondering how you knew exactly what my mistake was so quickly...

(btw, thanks!! i've been beating my head against the table on that one)

#### hgmjr

Joined Jan 28, 2005
9,027
...now i'm just wondering how you knew exactly what my mistake was so quickly...

(btw, thanks!! i've been beating my head against the table on that one)
Very simple. I have made the same mistake myself. That's how.

hgmjr