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How do I go about calculating the voltage output for an operational amplifier when the only connection to the the voltage source is through the operational amplifier itself?

For example, the diagram below has a 1.8 V source and 7 kohm resistor connected to Vp and a 1.2 V source and 10 kohm resistor connected to Vn of the operational amplifier. These sources are not bridged by an additional resistor or wire connecting either source to Vo.

Op Amp

I understand that Kirchoff's laws can be used to determine the output when the voltage source connects to it via a node at Vp or Vn, but I don't see how to determine the output voltage without such a connection.

Edit: The answer should be a numerical value. The hint that was given is that the op amp will saturate to +Vcc if V+ > V- or will saturate to -Vcc if V- > V+

Given this, my understanding is that this is an instance of positive saturation since 1.8 V > 1.2 V. According to Fundamentals of Electric Circuits, Vout = Vcc for positive saturation and -Vcc for negative saturation, where Vcc is the supply voltage.

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  • \$\begingroup\$ It will be unpredictable and nonlinear at best, and destructive at worst. Most op-amps have input protection diodes, which will be forced into conduction. \$\endgroup\$ – Caleb Reister Oct 16 at 21:38
  • \$\begingroup\$ Where did you find the diagram? Are you sure the amplifier symbol is meant to represent an op-amp? \$\endgroup\$ – The Photon Oct 17 at 2:53
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    \$\begingroup\$ @CalebReister It won't be destructive, you won't get enough power dissipation to harm anything with less than 0.5mA current into a part, that won't harm any diode \$\endgroup\$ – Voltage Spike Oct 17 at 3:31
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    \$\begingroup\$ A diagram very similar to this was posed by a Circuit Theory professor with the expectation that the answer will have a numerical value. Other students seem to be struggling with this as well, so we were recently given the hint that the op amp will saturate to +Vcc if V+>V- or will saturate to -Vcc if V->V+. I understand the gist of how saturation works, but I don't understand how this is helpful when only the inverting and non-inverting inputs are connected. I'll edit my post and update it with this hint. \$\endgroup\$ – Octavius Oct 17 at 3:38
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    \$\begingroup\$ Do you have numerical values for the voltages of the power supplies? \$\endgroup\$ – The Photon Oct 17 at 5:23
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The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated. In the case of a real opamp (with rails, that are not shown in the model above) then it can't source more voltage than it has available and the output will be limited by what the rails can source.

This circuit is simply a comparator. The open loop equation above applies to both opamps and comparators for ideal simple amplifiers, however with rails either amplifier will saturate.

enter image description here Source: https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html

In the early days of amplifiers, people would try and run them open loop, with little success because of variations in amplifiers. It was then that someone realized that negative feedback can take out the inconsistencies

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    \$\begingroup\$ With 600 mV between the inputs the formula you gave is almost certainly not applicable. \$\endgroup\$ – The Photon Oct 17 at 2:53
  • \$\begingroup\$ This is the equation for a basic operational amplifier (with input impedance and open loop gain), remember that basic operational amplifiers only have input impedance and open loop gain, no noise, no common mode, no rails. There is even a model in Lt spice that does this (.lib opamp.sub). In a perfect world (acedemic, simulation) you can run models like this and they handle open loop conditions just fine. You can do this with regular opamps to some extent, but the open loop gain is so high, the most likely thing is for you to hit the rails.The other problem is input bias tolerances. \$\endgroup\$ – Voltage Spike Oct 17 at 3:27
  • \$\begingroup\$ I'm fairly certain that this is to be treated as an ideal op amp, although that isn't stated explicitly. It's helpful knowing that this exercise is more academic than practical. My understanding was that an open loop gain in an ideal op amp is treated as infinite, and so I was looking for a work-around that doesn't involve A(v+ - V-). Am I mistaken? \$\endgroup\$ – Octavius Oct 17 at 3:45
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    \$\begingroup\$ @VoltageSpike Sometimes "V+ = V-" is taken as one of the ideal op amp equations, and that one is only true with negative feedback, but the equation used here does apply at all times regardless of feedback. \$\endgroup\$ – Hearth Oct 17 at 15:15
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    \$\begingroup\$ As it turns out, we were to "assume that each op amp is powered by +/- 10 volt power supplies", but the professor hadn't made this clear in the initial problem, hence my ongoing confusion. As @VoltageSpike has said, the op amp here saturate. So in the example I included, positive saturation will occur since V+ > V-, and thus Vo = +Vcc, which in this case is + 10 V. \$\endgroup\$ – Octavius Oct 17 at 22:33

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