Timeline for What is the amplification when both input terminals of an opamp are at same potential? [closed]
Current License: CC BY-SA 3.0
14 events
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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
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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 |