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What is the -3dB bandwidth of an inverting op amp

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simbaliya

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In the following figure, if the two resisters are equal, what is its -3dB bandwidth? Compare its stability with that of a source follower.
Untitled.jpg

This is one of my interview question today, still can not figure it out when I am back home. I only know how to solve for non-inverting op amp case.
 

FvM

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I guess you want to compare the G=-1 amplifier circuit with a G=+1 non-inverting buffer (not a "source follower").

The difference is the so-called "feedback factor". In the present circuit, only half of the output voltage is feed back to the inverting input, the resistors act as a voltage divider. Thus the loop gain is halved. Assuming a single pole open loop gain characteristic, the -3 dB bandwidth will be halved, too.
 

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I guess you want to compare the G=-1 amplifier circuit with a G=+1 non-inverting buffer (not a "source follower").

The difference is the so-called "feedback factor". In the present circuit, only half of the output voltage is feed back to the inverting input, the resistors act as a voltage divider. Thus the loop gain is halved. Assuming a single pole open loop gain characteristic, the -3 dB bandwidth will be halved, too.

FvM, I suppose you mean half of the opamp´s transit frequency, correct?
(Transit frequency identical to the gain-bandwidth product).
The -3dB bandwidth always is identical to the loop gain`s cross-over frequency (which is the frequency where the loop gain is 0 dB).
 
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FvM

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Thanks for clarification. I primarly meaned half of the unity gain buffer bandwidth, the latter is also identical to the OP GBW.
 

simbaliya

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FvM, I suppose you mean half of the opamp´s transit frequency, correct?
(Transit frequency identical to the gain-bandwidth product).
The -3dB bandwidth always is identical to the loop gain`s cross-over frequency (which is the frequency where the loop gain is 0 dB).

Hi LvW, I do not understand why "The -3dB bandwidth always is identical to the loop gain`s cross-over frequency", can you explain with details?
 

LvW

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simbaliya, I was not able to insert my answer (formulas) into this reply with copy and paste.
Therefore, please find my response in the pdf attachement.
 

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  • 3_dB_corner.pdf
    18.7 KB · Views: 79

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When you think about the frequency response of an opamp you also must consider its slew rate at the output level you need.
For example, an LM358 dual or LM324 quad opamp with a 30V supply has a gain of about 200 at 10kHz. But if its output level is 20V p-p then its output cannot swing higher than only about 2kHz.
 

simbaliya

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simbaliya, I was not able to insert my answer (formulas) into this reply with copy and paste.
Therefore, please find my response in the pdf attachement.

Hi LvW,

Thank you very much for your attachment, I almost understand the back theory except for one small part, you said that the open loop gain of op amp around unity gain frequency can be approximated as wt/jw, I wonder why the j is in the denominator?
 

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Hi LvW,
Thank you very much for your attachment, I almost understand the back theory except for one small part, you said that the open loop gain of op amp around unity gain frequency can be approximated as wt/jw, I wonder why the j is in the denominator?

All unity-gain compensated opamps have a frequency response which can be approximated (in the active region with Aol>1) by a first-order lowpass function

Aol=Adc(1+jw/wo).

(with wo: 3-dB corner frequency of Aol).
For frequencies w>>wo we have

Aol=Adc*wo/jw=wT/jw.

Explanation: The imaginary sign "j" indicates a PHASE shift of -90 deg.
 

simbaliya

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All unity-gain compensated opamps have a frequency response which can be approximated (in the active region with Aol>1) by a first-order lowpass function

Aol=Adc(1+jw/wo).

(with wo: 3-dB corner frequency of Aol).
For frequencies w>>wo we have

Aol=Adc*wo/jw=wT/jw.

Explanation: The imaginary sign "j" indicates a PHASE shift of -90 deg.

You cleared all my doubts, a million thanks!
 

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