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I greet everyone in the community,

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While trying to solve such an opamp circuit on paper, I could not find the formulation, I simulated it, but I could not find the value in the simulation.

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I find the Vout value as -13V in the simulation, but I couldn't find it in the formula on paper. How can we find the Vout value?

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I find the Vout value as -13V in the simulation, but I couldn't find it in the formula on paper. How can we find the Vout value?

You have your op-amp connected as a comparator. This means that the output will be close to the positive supply or close to the negative supply rail.

In your situation you have 3.5 volts at the inverting input and 3 volts at the non-inverting input. This means that the inverting input "wins" the battle and makes the output -13 volts (close to the negative supply rail).

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  • \$\begingroup\$ the inverting input is 3V, not 3.5V. but I don't fully understand what you are saying. How do we subtract -13V from the formula? \$\endgroup\$ Commented Jul 7 at 9:48
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    \$\begingroup\$ @smartony152 Imagine the opamp sets its output to A*(pin3-pin2). Pin3 is at 3 V. Pin 2 is at 3.5 V. Suppose A=500,000!. Then the output would be 500,000*(3-3.5)=-250 kV!!! Obviously, that's not going to happen. Reality impinges. The opamp will get as close to -250 kV as it can. But it appears that -13 V is about as close as it can manage. Opamps have very large values for A. On the order of 10^5 to 10^7. So very high values of open loop gain, A, are to be expected. The "problem" here is that there's no feedback from the output back to the inputs to set a smaller closed loop gain. \$\endgroup\$ Commented Jul 7 at 10:02
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    \$\begingroup\$ @smartony152 The non-inverting input is 3 V. The inverting input is 3.5 V because of the voltage divider formed by R19 and R20. \$\endgroup\$
    – Hearth
    Commented Jul 7 at 18:47

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