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My textbook states that the voltage across the input terminals of an ideal op amp is equal to zero. From what I understand, this is only true if there is a feedback loop connecting the output and input terminals. The examples in my textbook all use the ideal op amp in a feedback loop. But when the ideal op amp is used in a circuit without a feedback loop, does the assumption that the voltage difference between the input terminals is zero still hold? In other words, does the drawing of an ideal op amp, which looks identical to a non ideal op amp implicitly include a feedback loop?

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  • \$\begingroup\$ An ideal op amp does what it does and the operation what the whole circuit does depends on the circuit. If there is no feedback the output can't affect the inputs so an ideal op-amp output can go to +/- infinity if the output has no effect on input. \$\endgroup\$
    – Justme
    Oct 15, 2022 at 19:04

3 Answers 3

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The voltage difference between the input terminals of an ideal op-amp that is used in a circuit that employs overall negative feedback is zero.

When the ideal op amp is used in a circuit without a feedback loop, does the assumption that the voltage difference between the input terminals is zero still hold?

Without negative feedback to close the op-amp loop, the input voltage difference can be anything i.e. it is subject to the vagaries of the circuit.

From what I understand, this is only true if there is a feedback loop connecting the output and input terminals.

Usually true but, without the feedback, the circuit could force the inputs to be the same value such as by connecting them together.

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    \$\begingroup\$ Also, feedback doesn't necessarily imply that the terminals are at the same voltage. Even for an ideal op amp with unlimited output voltage, positive feedback will never make the inputs equal. Negative feedback specifically is required for that. (though you can have a stable system with both positive and negative feedback). \$\endgroup\$
    – Hearth
    Oct 15, 2022 at 18:54
  • \$\begingroup\$ @Hearth I decided to add the word "overall" to the first sentence. \$\endgroup\$
    – Andy aka
    Oct 15, 2022 at 19:12
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My textbook states that the voltage across the input terminals of an ideal op amp is equal to zero.

The specification that allows this claim is that, "An ideal op-amp has infinite gain \$A_{\text{VOL}}\$." This discussion requires that the amplifier is not in a saturated condition, that it is operating within its "continuous" range. I suppose that ideally the amplifier has a linear range, but is not always necessary.

The input-output relationship is: $$V_{\text{out}}=A_{\text{VOL}}V_{\text{D}}$$, where \$V_{\text{D}}=V_{\text{noninv}}-V_{\text{inv}}\$.

As \$A_{\text{VOL}}\rightarrow \infty\$, \$V_{\text{D}}\rightarrow 0\$ for a finite \$V_{\text{out}}\$ within its continuous range.

This makes operating an "ideal" op-amp in open loop mode useless. Zero volts in corresponds to any output voltage.

Negative feedback allows the use of infinite gain by adjusting the output so that: $$V_{\text{D}}\rightarrow \frac{V_{\text{out}}}{A_{\text{VOL}}}$$ Which is zero for an ideal op-amp.

So the ideal \$V_{\text{D}}=0\$ is not a closed loop requirement but an infinite open loop gain requirement. However it is really useful only in closed loop for continuous (linear) operation.

Update: The input differential pair is the mechanism that subtracts the two input signals, open loop or closed loop. So really it is the applied inputs that make \$V_{\text{D}}\$ have any voltage including zero.

In closed loop, the application of voltage to the inverting input is automated by applying a portion of the output to it. When a change is applied to the non-inverting input of a real amplifier, there is a small delay before the amplifier stabilizes. During this time \$V_{\text{D}}\$ has a significant voltage that settles at "steady state" to \$\frac{V_{\text{out}}}{A_{VOL}}\$.

when the ideal op amp is used in a circuit without a feedback loop, does the assumption that the voltage difference between the input terminals is zero still hold?

Yes, with the qualification that for a finite output the voltage difference between the input terminals must be zero. That it is not an assumption. It is a requirement of infinite \$A_{\text{VOL}}\$. The input voltages must be adjusted to meet this requirement.

In other words, does the drawing of an ideal op amp, which looks identical to a non ideal op amp implicitly include a feedback loop?

No.

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  • \$\begingroup\$ Mathematically, I understand why Vd will have to be zero when AVOL is infinite and Vout finite, but what is the actual mechanism making VD zero if there is no feedback? \$\endgroup\$
    – nova
    Oct 16, 2022 at 20:38
  • \$\begingroup\$ Please see my update \$\endgroup\$
    – RussellH
    Oct 17, 2022 at 0:05
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When the opamp is required to work in its linear region (amplification) it must have negative feedback.

For any real opamp with finite open-loop gain Aol the result of negative feedback is a very small differential voltage Vd across both opamp input terminals in the (upper) µV range.

For most applications this differential voltage Vd is much smaller than all other voltages in the circuit and can, therefore, be neglected. This is a reasonable approach with respect to other uncertainties and simplifications (neglected input and output resistance, parts tolerances,...).

As a consequence, we simply set (assume) Vd=0 during all calculations.

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