Timeline for OpAmp with common-mode voltages and zero feedback current
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2021 at 7:39 | vote | accept | KMC | ||
| Apr 30, 2021 at 11:49 | history | edited | Elliot Alderson | CC BY-SA 4.0 | Fixed SI units |
| Apr 30, 2021 at 6:00 | history | tweeted | twitter.com/StackElectronix/status/1388010259323703296 | ||
| Apr 30, 2021 at 2:17 | answer | added | tobalt | timeline score: 1 | |
| Apr 30, 2021 at 0:37 | history | became hot network question | |||
| Apr 29, 2021 at 22:44 | history | edited | Circuit fantasist | Added tag | |
| Apr 29, 2021 at 22:41 | comment | added | Circuit fantasist | @qrk, No current flows in this circuit... | |
| Apr 29, 2021 at 21:26 | answer | added | Circuit fantasist | timeline score: 2 | |
| Apr 29, 2021 at 20:39 | answer | added | Tony Stewart EE since 1975 | timeline score: 5 | |
| Apr 29, 2021 at 19:20 | answer | added | devnull | timeline score: 5 | |
| Apr 29, 2021 at 17:23 | comment | added | qrk | You show floating 1V sources which will cause your output to slam to a rail due to the floating positive input wrt DC. The sources need to be 2-terminal devices. If the 1V sources are referenced to ground, then you have current flow through your feedback resistors and the circuit will give you a 1V output as expected. If you try to connect a single source between the inputs, you'll get a clipped output since the circuit gain is the open loop gain of the op-amp. | |
| Apr 29, 2021 at 17:07 | comment | added | G36 | In case "A" you have an example of negative feedback in action. If Vout = 0V then the voltage at inverting input will be +0.5V. And this difference will be amplified by an op-amp. Thus, the opt amp output will be driven towards the positive voltage to "forced" V_"+" = V_"-". Because V_"+" > V_"-". As this is what negative feedback "wants". If the op-amp output overshoots the sign of a voltage difference (V_"+" - V_"-") will change (V_"-" > V_"+") so that the output will now start to decrease until equilibrium point is reached. | |
| Apr 29, 2021 at 17:01 | comment | added | Voltage Spike♦ | Please ask a specific question | |
| Apr 29, 2021 at 16:56 | comment | added | John D | "If no current passes through Rf, the feedback loop is essentially an open circuit (circuit B)." No it isn't. Any vanishingly small deviation from 1V on the inverting input would be corrected by a change to Vout, so the loop is still closed. At the limit with gain approaching infinity the output approaches exactly 1V, but Rf is still present and necessary. | |
| Apr 29, 2021 at 16:52 | comment | added | KMC | @nanofarad I just edited my question. If the input potential is zero, shouldn't the output be zero too? Actually with finite gain as you pointed out, the argument would be even more compelling - finite open loop gain multiplied by zero potential must be zero! With ideal infinite gain, it's difficult to say .... as I would run into infinity multiplied by zero... | |
| Apr 29, 2021 at 16:47 | history | edited | KMC | CC BY-SA 4.0 | added 192 characters in body |
| Apr 29, 2021 at 16:42 | comment | added | nanofarad | The ideal op amp model, with a gain of infinity, is bound to fail in edge-cases such as these; consider doing the analysis with an op amp that has a high, but finite gain (e.g. 1 million). | |
| Apr 29, 2021 at 16:33 | history | asked | KMC | CC BY-SA 4.0 |