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Mar 18, 2017 at 17:21 history closed Dmitry Grigoryev
Voltage Spike
pipe
Andrew
ThreePhaseEel
Needs details or clarity
Mar 15, 2017 at 17:45 review Close votes
Mar 18, 2017 at 17:21
Mar 15, 2017 at 17:35 comment added brhans @skvery - only for an ideal opamp with negative feedback.
Mar 15, 2017 at 17:14 comment added skvery They are always at the same potential for a linear op-amp.
Mar 15, 2017 at 17:13 answer added Adam Haun timeline score: 3
Mar 15, 2017 at 17:09 answer added analogsystemsrf timeline score: 0
Mar 15, 2017 at 17:06 comment added user3219492 If Vd is differential gain, then I can't come up with a meaning for A=Vo/Vd. You mean gain = volatge/gain ?
Mar 15, 2017 at 17:05 comment added Bodhisattya Dutta Vd is differential gain. Thanks for pointing it out. I have edited it in.
Mar 15, 2017 at 17:04 comment added dannyf The amplification of an (ideal) opamp is always infinite. How its closed loop amplification is depends on how the feedback is arranged.
Mar 15, 2017 at 17:03 history edited Bodhisattya Dutta CC BY-SA 3.0
added 25 characters in body
Mar 15, 2017 at 17:03 comment added brhans I think you're tying yourself in a knot unnecessarily. If V1=V2, then Vo is 0 irrespective of what the gain is. Vo=A x (V1-V2) and A x 0 = 0.
Mar 15, 2017 at 17:02 review First posts
Mar 15, 2017 at 17:11
Mar 15, 2017 at 17:01 comment added user3219492 What is Vd in your equations? In this Vd=Vo/(V1-V2), you are treating vd as differential gain (has no unit). Whereas in this A=Vo/Vd, you are treating Vd as voltage(has unit volt). Tell what it is
Mar 15, 2017 at 16:55 history asked Bodhisattya Dutta CC BY-SA 3.0