# How is voltage difference of input terminals zero in open loop ideal op amp?

An ideal op-amp has infinite open loop gain. This means that the input voltage difference approaches zero. What causes the difference to go to zero in the absence of negative feedback?

• There is nothing that makes the inputs of an ideal (or real) op-amps equal in the absence of feedback. Commented Oct 26, 2022 at 23:21

The input voltage between the + and - inputs is approximately zero volts (1st order approximation) for a closed-loop configuration (negative feedback) as long as the opamp is not saturated.

In an open-loop configuration, there is nothing stopping you from inputting a non-zero value. The input will be limited by the input protection diodes going to the power supply rails. Using an opamp in an open-loop configuration is normally not a useful application for opamps. Some like to use opamps in an open-loop configuration as comparators, but the speed performance is not very good compared to a dedicated comparator.

As mentioned in the comments, "Nothing makes the inputs zero in open loop."

The infinite open loop gain part of the ideal model for an op-amp is not very useful. The discussions around the so called requirement that the input difference must be zero is a distraction. A finite gain leads to better understanding. The equation: $$V_{\text{out}}=A_{\text{VOL}}V_{\text{D}}$$

has meaning only if $$\A_{\text{VOL}}\$$ is finite.

simulate this circuit – Schematic created using CircuitLab

The diagram shows a zero-order approximation of an op-amp. It shows two parts: a device that subtracts the inputs and a device that amplifies the difference, plain and simple. There is no mechanism or magic that requires or makes the inputs zero.

However there are some limits set by the power supplies (VCC and VEE). $$\frac{V_{\text{CC}}}{A_{\text{VOL}}}>V_{\text{D}}>\frac{V_{\text{EE}}}{A_{\text{VOL}}}$$

Whether open loop or closed loop, the equation holds for analog operation between the limits. But there is nothing restricting $$\V_{\text{D}}\$$'s range necessarily.

I think you may be under the misapprehension that infinite open loop gain means that the inputs can never be unequal. That is wrong. There's no rule that constrains the input potentials of an op-amp. There's nothing that says those two inputs can't be very different.

Negative feedback simply means that one of the op-amp inputs is now under control of the op-amp itself, from its own output. Initially, the two input potentials may in fact be very different, but the op-amp quickly responds to bring the difference in input potentials towards zero, and settle when they eventually become equal.

With negative feedback, some fraction of the output potential is fed back to the inverting input, which has interesting consequences. If the op-amp's output results in the two input potentials being different, the op-amp responds by changing its output in the direction required to reduce that difference. It continues to adjust its output until the two inputs are identical, but this does not mean that the inputs can never be unequal. All it means is that the op-amp tries to make them equal, to the best of its ability.

With no negative feedback (the loop is open), there's no such self-adjustment taking place. Both inputs are now under your control, not the op-amp, and you may set those to inputs to whatever potentials you desire. The op-amp will simply multiply the difference in inputs by some very large value (its open loop gain). Of course, if the op-amp were ideal, this would mean infinite output potentials, and the universe still exists only because there is thankfully no such thing as an ideal op-amp.