simbaliya
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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).
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?
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.