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For negative feedback op-amp circuits, we know that V+ = V-. However, does that also hold for positive feedback circuits? According to my simplistic SPICE simulations, this seems to be the case, but I want to confirm.

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  • \$\begingroup\$ +ve feedback causes hysteresis since the output has limited \$\endgroup\$ Dec 13, 2020 at 6:52
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    \$\begingroup\$ In any op-amp circuit if V+ doesn't approximately equal V- then the OP amp is maxed out positive or negative. Therefore for any desired output voltage between those extremes v+ must equal v-. \$\endgroup\$
    – Drew
    Dec 13, 2020 at 7:01
  • \$\begingroup\$ whenever the differential inputs are NOT near zero, the output will be saturated at one rail according the the polarity and Vcm input requirements. we call this non-linear mode , and with positive feedback ratio x Output swing = input Hysteresis \$\endgroup\$ Dec 14, 2020 at 23:23

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No: even though the op amp input terminal \$V_{IN-}\$ is driven to nearly the same voltage as \$V_{IN+}\$ when configured for negative feedback, when configured for positive feedback \$V_{IN+}\$ is not driven to the same voltage as \$V_{IN-}\$, and in fact it's driven away from the \$V_{IN-}\$ voltage.

Negative Feedback Configuration is Convergent

In negative feedback configuration, \$V_{IN+}\$ is the reference, and the feedback network drives \$V_{IN-}\$ such that \$V_{IN-} = V_{IN+} - (V_{OUT} / A_{v})\$, where \$A_{v}\$ is open-loop gain. If the output \$V_{OUT}\$ is "too big" compared to the target reference value, then the feedback \$V_{IN-}\$ is also "too big", and since \$V_{IN-}\$ > \$V_{IN+}\$ then \$V_{OUT}\$ is driven to a lower voltage, until \$V_{IN-}\$ reaches nearly the same voltage as \$V_{IN+}\$. Also, \$V_{OUT}\$ will be driven to whatever value the feedback network requires it to be to make \$V_{IN-}\$ match the reference \$V_{IN+}\$.

Positive Feedback Configuration is Divergent

In positive feedback configuration, \$V_{IN-}\$ is the reference, and the feedback network drives \$V_{IN+}\$ such that \$V_{IN+} = (V_{OUT} / A_{v}) + V_{IN-}\$, where \$A_{v}\$ is open-loop gain. If the output \$V_{OUT}\$ is "too big" compared to the target reference value, then the feedback \$V_{IN+}\$ is also "too big", and since \$V_{IN+}\$ > \$V_{IN-}\$ then \$V_{OUT}\$ is driven to an even higher voltage, until \$V_{OUT}\$ reaches the practical limit of how close the output can get to the power supply voltage.

Positive feedback can be useful when designing a memory cell, or when using hysteresis to help a comparator avoid responding to certain types of noise sources. But for DC amplifiers this effect is unwanted.

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  • \$\begingroup\$ Positive feedback (frequency-dependent) is widely used in active filter circuits and is always present in oscillating circuits. \$\endgroup\$
    – LvW
    Dec 13, 2020 at 14:01
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Your simulation program and the result is corrrect - there is no error. However: The result is heavily unrealistic. What does this mean?

  • Under IDEAL conditions (ideal opamp and no power switch-on transient) the simulator can find a fixed DC bias point and, hence, can multiply the input signal with a gain factor - determined by the positive feedback path. This situation can be compared with two balls riding on each other. This would also be possible, when there is absolutely no disturbing effect (which in reality, however, never will happen)

  • But the simulation result will be realistic (Vin- NOT equal to Vin+) when you either will use a real opamp model (with signal delay) or when you are switching-on (at least) one of the two supply voltage sources at t=0 (simulation start). In this case, the simulation result will be the same as in practice: Positive feedback causes instability - and the output will be latched at one of the supply voltages.

  • Summary: The simulator did not make any errror - however, the user should know if the simulation set-up is a realistic one. In most (if not in all) cases, the mistake is on the users side (misinterpretation)

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The simple answer is "no." In general when an op-amp has positive feedback only, the inverting and non-inverting inputs cannot be assumed to be the same. That assumption is only valid when there is negative feedback.

Probably what is happening in your simulation is that you have a condition occurring that is not possible in the real world, but only possible in the noise-free realm of the simulator. Try applying a tiny signal to one of the input terminals of the op-amp and see what happens.

Positive feedback is sometimes useful for a variety of purposes. But the assumption that V- = V+ is only valid when you have negative feedback.

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For negative feedback op-amp circuits, we know that V+ = V-. However, does that also hold for positive feedback circuits?

Yes, it is possible if only the input voltage (V2) is zero... the op-amp is ideal... and there are no other disturbances. In this situation, the output voltage (VOUT) will be zero... and the voltage of the non-inverting input (V+) will be zero as well... so all voltages will be zero...

If some input voltage (disturbance) is applied, it will affect V(-)... that will be amplified... and VOUT will affect V(-) even more through the positive feedback network (R1-R2). Now there are two possible situations.

If the loop gain <= 1 (this will be the case if your op-amp had a fixed gain of 7 or less), the circuit will still reach a equilibrium. But if the loop gain > 1, the amp will reinforce itself finally reaching the supply rail. This arrangement is used in latches and Schmitt triggers.

It is interesting that even in the latter situation, there is a (short) time interval when V(-) = V(+). This is because, when the circuit changes its state, the input and output voltage "move" against each other... and at some point they "meet" (become equal) for a while.

Finally, there are circuits with positive feedback where V(-) = V(+)... but they have also an additional negative feedback that dominates over the positive feedback. Such circuits are, for example, negative impedance converters (NIC).

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