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An op-amp is defined by the following equation

$$ V_o = A(V_p - V_n) $$ Rearranging,

$$ V_p-V_n = V_o/A $$

A real op-amp will have a gain around 100 and the output will be limited by the supply voltage say 1.8V. Let's put this op-amp in an open-loop configuration.

$$ V_p-V_n = 1.8/100 $$ This is 1.8 mV difference between the two inputs. I don't understand how this is possible. There is no connection from output to input, so how this equation predicting 1.8mV difference between two inputs

In summary, why does op-amp virtual ground not hold in open-loop and positive feedback cases?

Null
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AlfroJang80
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    Virtual ground is a circuit property, not an op-amp property. – DKNguyen Aug 13 '22 at 19:26
  • @DKNguyen I see. But the virtual ground is derived with use of the equation for an op-amp. – AlfroJang80 Aug 13 '22 at 19:27
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    No, it's derived using the circuit which makes it a combination of the op-amp open-loop gain equation, the wiring, and other components. Not just the op-amp open-loop gain equation. – DKNguyen Aug 13 '22 at 19:40
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    "*A real op-amp will have a gain around 100 ...*" They're much better than that! Real op-amps have open-loop gain of 100,000 to > 1 million. – Transistor Aug 13 '22 at 19:47
  • It is the input signal swing multiplied by the open loop gain which gives the resulting output swing. The combination of the input signal and the open loop gain determines the output signal. The input is not caused by the output but if you know the open loop gain then you can use that equation to determine either the input given the output or the output given the input. –  Aug 22 '22 at 17:01
  • The open-loop gain of an opamp is often greater thano 100 dB (factor of 100000), not 100 times. Hence, Vp - Vn in your example will be 18 uV if the open-loop gain is 100 dB, not 18 mV. As others stated, the virtual ground (should be taught as a zero voltage between input terminals) concept is with feedback, not open-loop. The zero-voltage concept greatly simplifies analysis of feedback networks. – qrk Aug 22 '22 at 19:38

2 Answers2

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  1. Virtual ground is a circuit property, not an op-amp property. It's derived using the circuit which makes it a combination of the op-amp open-loop gain equation, the wiring, and other components. Not just the op-amp open-loop gain equation.

    Therefore it naturally follows that if you change the circuit so that the conditions for virtual ground are no longer met, such as an open-loop or positive feedback circuit, then virtual ground no longer holds.

    Virtual ground is not mathematically rigourous. It is an approximation used as a mental short cut and can't just be blindly applied. To use it, you need to already know enough about what the circuit is doing to verify it can be treated as virtual ground. If you do not know then you cannot assume.

    If you assume nothing about virtual ground and forget about it and blindly do an analysis on circuit with the open-loop gain equation (such a simple buffer circuit) you will find virtual ground doesn't really exist. What does exist is a node with an equation for its potential that approximates the ground potential for very large open-loop gains.

    And this simple linear analysis will fail in circuits where the behaviour enters the non-linear region such as open-loop, or positive feedback circuits that enter saturation, or wiring up the op-amp as a comparator. You may get a result but it won't be correct because you incorrectly assumed linearity. Which leads us to...

  2. Your error of 18mV rather than 1.8mV error aside, you cannot use the open-loop gain equation that way because saturation is not a linear phenomeon, therefore the linear gain equation is inapplicable.

    Only when your input difference is very small, your open-loop gain very low, or your power rails very large (all relative to each other of course) such that the output does not saturate then the open-loop gain equation would be valid.

DKNguyen
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  • Check out this question which has the nonlinear problem. Note all the comments. Last I checked no one was able to answer how you would analyze it without knowing ahead of time whether it was linear or nonlinear : https://electronics.stackexchange.com/questions/516790/why-is-this-derivation-of-schmitt-trigger-circuits-output-signal-incorrect – DKNguyen Aug 13 '22 at 20:03
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An op-amp is defined by the following equation $$V_o = A(V_p - V_n)$$

That formula only applies universally when the op-amp is operating in closed loop and the output is not saturated. And even then, the formula you wrote is missing an important detail: $$V_o = A_{\boxed{OL}}(V_p-V_n).$$

\$\boxed{OL}\$ stands for open loop. It's the open loop gain that applies here, even though the amplifier must be configured with a closed loop for the formula to apply.

There is no connection from output to input

For the formula above to hold in all conditions, there must be a connection from output to input, and it this connection must form net negative feedback.

In summary, why does op-amp virtual ground not hold in open-loop and positive feedback cases?

Because it is the closed-loop negative feedback that establishes the virtual ground condition. Without that feedback network, there is no way for the output of the op-amp to control the inputs. Op-amp inputs are high impedance. The only way an op-amp can control any of its inputs is via the feedback network. It is the action of this feedback network that allows the op-amp to "create" a virtual ground at one of its inputs.

A real op-amp will have a gain around 100

Closed-loop gain - sure. Open loop gain - no. Not at DC. Any practical op-amp will have an open-loop DC gain of 10V/mV or more. Even the venerable LM741 has typical open-loop gain of 200V/mV: that's 200,000.

Table 6.5, Electrical Characteristics, from the LM741 Datasheet

Image source: Texas Instruments LM741 datasheet

Assuming an op-amp without an offset voltage, the closed-loop voltage between the inputs will be impossible to measure with cheap multimeters. 1mV of error (without offset) is a lot for any modern low-voltage op-amp.

Thus, a more realistic relationship would be $$V_p-V_n = 1.8/(50\cdot10^3) = 1.8/(5\cdot10^4) \approx \frac{2}{5}10^{-4} = 40{\,\rm \mu V}.$$

You'll need a decent multimeter to measure that. Or a preamplifier, with the offset trimmed out. Even precision and "zero offset" op-amps may have offset too high to amplify such a small voltage accurately without offset nulling.

SamGibson
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Kuba hasn't forgotten Monica
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