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Gain Bandwidth (really get confused)

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imijoon

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Hi all,

I want to measure my op-amp GBP(fT) . I have a non inverting op-amp amplifier circuit with the gain of 10 . I increase the frequency until the gain becomes 7.07 . Now the frequency is 85khz (-3db frequency) . what is the formula for GBP. according to the bottom picture it should be (( GBP=7.07*85=600.95khz )) but according to application notes it should be ((GBP=10*85=850khz)).

which one is correct ? (GBP=A0*fc) or (GBP=A*fc) of (GBP= (any A in slope)*(it's frequency))

main-qimg-e50942c69985f5228303fd416d48a8e9.jpg
 

imijoon

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Thanks chuckey but I didn't get my answer clearly.
which one is correct ? (GBP=A0*fc) or (GBP=A*fc) of (GBP= (any A in slope)*(it's frequency))
 

KlausST

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Hi,

you know that GBW is open loop gain. But since you work with gain of 10 you work with feedback...

***
to your problem. You have a very nice picture.
* There is a horizontal line giving DC gain
* then there is an arc. This is where the phase shifts from 0 to 90°
* Then there is a line with constant 20db/decade fall rate. Only here GBW is constant. Also here are three examples that all show a GBW of 10MHz.

To measure at cutoff frequency (-3dB, = 0.707) is possible. Extend the horizontal line a bit to the right, and the constant falling line to the left until they meet. (This should be the -3dB frequency)
Take frequency and gain form the meeting point. (in your case this is A = 10 and frequency = 85kHz, giving a GBW of 850kHz)

Hope this helps.

Klaus
 
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