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Will the gain of the following IDEAL Op-Amp be zero?

My reasoning is that the positive (non-inverting) terminal voltage is zero because it is connected to ground. Since the voltage for the negative (inverting) terminal is supposed to be equal to the non-inverting terminal voltage for an ideal op-amp and that the Vin of the op-amp is equal to the non-inverting voltage minus the inverting voltage, the Vin will also be zero.

Therefore. the gain will also be zero, and the Vs will also be zero.

enter image description here

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  • \$\begingroup\$ I think that it's not that the gain is 0, it's acting as a comparator rather than an op amp. \$\endgroup\$ – Daniel Jan 19 '17 at 22:48
  • \$\begingroup\$ what is \$v_i\$ in the input-output equation? \$\endgroup\$ – robert bristow-johnson Jan 19 '17 at 23:56
  • \$\begingroup\$ Nah, its not a comparator, it still has gain \$\endgroup\$ – Voltage Spike Jan 19 '17 at 23:57
  • \$\begingroup\$ Since Vs blocks all feedback, it behaves like a comparator. \$\endgroup\$ – Sunnyskyguy EE75 Jan 20 '17 at 0:21
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Most of the answers are correct.

Trying to condense them: the feedback doesn't count. Whatever the 6 resistors and Vo try to contribute to V-, all it ever sees is Vs. Similarly, the 25k load doesn't matter. So, all the op amp sees is Vs which has it banging its head against the op amp's rails, whatever they are, with the rail bangee (invented a new term) depending on whether Vs is more or less than 0.

All in all, it looks like a mistake of some sort in the statement of the problem, or a trick question.

You're correct that the inputs to an op amp are equal for most useful connections. This is not one of them. The inputs usually are equal because the negative feedback makes them so. There's no feedback at all in the problem you show.

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  • \$\begingroup\$ Maybe yes, maybe no. Depends on other course material. Technically, g = -Aol. Of course, that won't work for a real op amp, but the couirse may not be concerned with that. \$\endgroup\$ – WhatRoughBeast Jan 20 '17 at 2:39
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Assuming Vs is a conventional 0\$\Omega\$ source impedance independent voltage source, the gain is -\$\infty\$.

All the resistors can be ignored since they are connected between 'stiff' (0\$\Omega\$ impedance) voltage sources.

Imagine the input Vs is a very tiny voltage (say +1uV or whatever you consider to be inconsequential). The differential voltage the op-amp sees is -1uV. It will multiply that differential input voltage by an enormous gain giving a huge negative voltage (perhaps limited to the supply rails depending on how 'ideal' the imaginary op-amp is). Same thing if Vs = -1uV except it will result in a huge positive output voltage.

There might be some kind of argument that it's just a huge number and not infinity since the resistors could pass enough current to overcome the zero source impedances, but I'll leave that kind of thinking to philosophers and pure mathematicians.

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  • \$\begingroup\$ It's an ideal op amp, so we are in pure math land, eh? \$\endgroup\$ – Voltage Spike Jan 20 '17 at 0:43
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Most of your reasoning is correct, but not all.

enter image description here

If you look at the diagram bellow, and pick values for R1=1kohm and R2=2kohm. And an input of 1V. Assuming that we do not get an output that is too high or too low.

Then with your correct assumption that the op-amp is trying to get the +ve input to be the same as the -ve. What output will produce this?

What will the gain be? ( I will leave this for an exercise for the reader.)

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Follow the guideline that if the inverting and non-inverting inputs are not equal, then the opamp is in saturation. That gets you through most elementary analysis. If you want to get a bit more particular, you could parallel the feedback and input resistor and tie the result to ground at the nonnverting side to balance the offset,

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The circuit is a mess:

  • If the output impedance if Vs is lower that that of the rest of the circuit it will just drive the -ve input where ever it wants.
  • The 25Kohm can be ignored.
  • The 80k and 20k can be combined (16k ohm) etc etc until you have one resistor.
  • Then you realise that you can ignore that. See answer of @SpehroPefhany gain is -∞ (or close to it).
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  • \$\begingroup\$ But, what about my analysis? is it correct or are you saying that their are two ways of solving this problem? \$\endgroup\$ – vaderrider Jan 19 '17 at 23:05
  • \$\begingroup\$ - input of opamp can not drive Vin. You have a perfect voltage source connected to a perfect watcher (I will look but will not change). The virtual ground of - input, is only true for circuits like that of my other answer. The opamp does not output through its inputs. \$\endgroup\$ – ctrl-alt-delor Jan 19 '17 at 23:46

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