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schematic

simulate this circuit – Schematic created using CircuitLab

I need to anlayse state of OPAMP for V1=1V ,10V ,15V ,40V Its not ptractical problem just want to learn what happens if a voltage source is connected to output of OPAMP.Think of OPAMP constituting from near ideal BJT D.A. Below figure is why I am biased towards saying that OPAMP will remain in linear region for any of case. enter image description here

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closed as unclear what you're asking by MCG, Dave Tweed Mar 1 at 12:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Veiwpoints so far: For V2<Vsat{15volts} OPAMP is less likely to be in saturation region as differential amplifiers output is fixed to make that bjt operate in active region For V2>Vsat OPAMP goes to saturation Please correct me. \$\endgroup\$ – Buzz bee Mar 1 at 8:55
  • \$\begingroup\$ To me, it makes not much sense to feed an external output voltage into a low-resistive output (to be honest: it sounds crazy). \$\endgroup\$ – LvW Mar 1 at 9:14
  • \$\begingroup\$ @Buzz Bee: Concerning the text: It doesn't make any sense. Please look at it and improve it. Concerning the circuit: as others have noted: connecting two voltage sources (OpAmp Output) and V1 doesn't make sense (would cause a short). \$\endgroup\$ – Curd Mar 1 at 10:29
  • \$\begingroup\$ @MCG Firstly which proffessor will ask question of little practical importance.I just wanted to understand OPAMP a bit more conceptually and see if the above will work or burn because if differential amplifiers are treated as core of OPAMP then above should work and not burn try yourself it's just changing path of biasing current from one of transistor. \$\endgroup\$ – Buzz bee Mar 1 at 12:38
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From an educational perspective, this circuit is invalid and unsolvable. The output of an ideal opamp can be considered an ideal (controlled) voltage source. This means that two ideal voltage sources are potentially applying two different voltages to the same output node which is naturally unsolvable.

how can I analyse state of OPAMP

Analyzing the shown schematic assuming ideal components is impossible.

I can venture a guess what will happen in real-life though. Typically, if you short-circuit two voltage sources, it will result in a very high current through the very low-resistive connection. This connection will heat up and burn through. It is also possible that the connection doesn't break, but that the output driver of the opamp blows up instead. Perhaps for some opamps, you will be able to keep it running if the opamp sufficiently limits the current, in which case the opamp has no control over its output and you can basically remove the opamp without changing the circuit voltages.

I'm not willing to bet my money on it that these are the only ways your circuit can fail though.

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  • \$\begingroup\$ The opamp in your schematic will try to behave like a differential amplifier, but will fail to because V1 is shorting the output. An ideal opamp is a differential amplifier with an infinite gain, input impedance and zero output impedance. \$\endgroup\$ – Sven B Mar 1 at 12:39
  • \$\begingroup\$ please look at above differential amplifier figure and if it is possible the way i see. \$\endgroup\$ – Buzz bee Mar 1 at 12:47
  • \$\begingroup\$ Because the output is at V1, regardless of the state of the op amp, there is no negative feedback, there is no guarantee that the two input nodes are equal, and it is most likely that the internal state of the output is saturated at a rail, generating a short. \$\endgroup\$ – Scott Seidman Mar 1 at 13:10

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