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I was trying to solve the below question. As per my understanding at t=0+ the difference voltage between (Vnoninvering -Vinverting)=-2.5v so output will change to -5. After then as Vin increases it will behave like normal noninverting amplifier with gain 2. The output will be distorted to give 5v.

Edit: The question is to draw output waveform Vo over time where time varies from time=0 to time= infinity. enter image description here

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  • \$\begingroup\$ I am still not clear what you want to know? My guess is the transient nature of opamp (at t=0+)!! \$\endgroup\$
    – Mayank
    Sep 21, 2017 at 6:09
  • \$\begingroup\$ As per my understanding at t=0+ the difference voltage between (Vnoninvering -Vinverting)=-2.5v Explain why you make that (wrong) assumption. An opamp in with negative feedback will always try to make the input voltage difference zero. Only when it cannot do that you should assume otherwise. \$\endgroup\$
    – Bimpelrekkie
    Sep 21, 2017 at 6:36
  • \$\begingroup\$ @Bimpelrekkie: He's not assuming it; it's stated in the image. It must be a start-up transient. \$\endgroup\$
    – Transistor
    Sep 21, 2017 at 7:13
  • \$\begingroup\$ @Transistor OK, you conclude that from intitial Vo = +5V so - input = 2.5 V but +in put is 0 V. That makes it an inconsistent question as no information is given about the transient behavior. Also, the situation Vo = 5V while Vin = 0 V is not a proper solution for this circuit. I think Vo = 5V initially is a typo and should say: Vo = 0 V initially. \$\endgroup\$
    – Bimpelrekkie
    Sep 21, 2017 at 7:31

1 Answer 1

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  • The opamp is wired in an inverting configuration.
  • The output will attempt to drive the inverting input to match the non-inverting input so that the voltage difference between them is zero.
  • At t = 0 \$ V_+ = 0 \$ so \$ V_- = 0 \$. This means that \$ V_{OUT} = 0 \$ too.
  • If the question is how does the opamp recover from saturation at power-up then we don't have enough information. As an approximation we can say that it immediately drops to zero.
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  • \$\begingroup\$ perhaps a better approximation would be that it 'quickly' drops to zero, as fast as it can but limited by unspecified time constants. \$\endgroup\$
    – Neil_UK
    Sep 21, 2017 at 7:57

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