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I'm unable to come to terms with something I think is a paradoxical situation relating to the virtual ground of an Operational Amplifier.Please pardon me if this is a really stupid question.

When the 'Negative Feedback' in an Op-Amp (Ideal) makes the difference between its input terminals equal to 'Zero'. Shouldn't the output become zero too because the Op-Amp is fundamentally a Differential Amplifier and according to the equation:

Vo = (Open loop gain)*(Differential voltage b/w the inputs)

Virtual ground Illustration

The Explanations I've come up with so far are:-

1) The Op-Amp Output is indeed zero and it is the External Circuitry (consisting of resistors Rf and Rin) that create the voltage, which adds up to the Op-Amp output voltage (in this case Zero) at point B to create the actual output of the system.

2) The virtual ground is not perfect and there exists a very very small differential voltage at the input which gets multiplied by the vary high gain and produces the output.

I'm fundamentally unable to understand how the actual definition of Op-Amp behavior is consistent with the virtual ground phenomenon without making the output zero. Please Help!

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    \$\begingroup\$ If it were exactly 0 volts then it would be 0 volts except, it's virtually 0 volts. \$\endgroup\$ – Andy aka Jun 28 '17 at 19:20
  • \$\begingroup\$ It's virtual because it an active feedback to create a 0V differential rather than an absolute 0V Reference which is th definition of any local gound. There is no Paradox. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 '17 at 3:19
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    \$\begingroup\$ This difference is exactly 0 for an ideal op-amp with an infinite gain, and \$\infty*0\$ is not necessarily 0. \$\endgroup\$ – Dmitry Grigoryev Jun 29 '17 at 11:46
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  • \$\begingroup\$ The concept of "virtual ground" is only used for explaining the opamp operation to students without confusing them. What really happens is explained in Scott Seidman's answer. I think it must be the accepted one. \$\endgroup\$ – hkBattousai Jul 3 '17 at 8:04

11 Answers 11

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It's #2. For a "perfect" theoretical opamp, the open-loop gain is infinite, and this makes the difference at the inputs zero. When introducing opamp circuits, or when working out how things are supposed to work, people normally think about the "perfect" opamp.

When thinking about the performance of a circuit, we usually have to start thinking about the imperfections of a real opamp. For a real opamp, the open-loop gain is not infinite, and there is some difference between the inputs. To take the example of an LM324, the open loop gain is about 115dB. That's a little less than a million volts/volt, so if there is a 1V DC output, then the inputs are different by about 1uV. Most of the time you can ignore that.

It gets more complicated for AC. At higher frequencies, the gain drops. For the LM324, it goes to 0dB, i.e. 1V/V at about 1MHz. At that point, the inputs certainly will have a large difference. Practically speaking, the amplifier just doesn't work any more. For frequencies in-between, the gain of the amplifier (inc. feedback) will vary. The term "Gain Bandwidth Product" is used to describe what gain you can have at what frequency for a given opamp.

This is just one of many imperfections a real opamp has. Another very relevant one is input offset voltage. This is the difference in inputs which results in a zero output, and it's not always exactly 0. This might be more important than the limited gain in many cases. Other imperfections you might want to consider are saturation/clipping, input current, PSRR, CMRR, nonzero output impedance and many more.

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  • \$\begingroup\$ So can we say that mathematically, this explanation cannot be extended to perfectly Ideal Op-Amps? Thanks for the great explanation! the first explanation I had come up with was soo convincing at first that I'd have been totally mislead. \$\endgroup\$ – Sumanth Jun 29 '17 at 11:00
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The problem is that you mix-up two different models of the op-amp.

A real, but somewhat idealized op-amp, is a differential amplifier whose output depends on the inputs as follows (neglecting saturation):

$$ V_{out} = A_{Vol} \cdot (V^+ - V^-) $$

Using this simplified model (simplified because it neglects saturation, offset voltage, bias currents, bandwidth and other real-world effects) and the fact that \$A_{Vol}\$ (open loop gain) is huge, you can prove that, when the op-amp is connected in a negative feedback circuit, then the virtual short circuit holds, but only when you approximate \$A_{Vol}\$ as infinite.

With this drastic approximation you can have a zero differential input AND still a finite output, since the open loop gain is assumed infinite.

In reality the open loop gain isn't infinite and your finite output is due to a very small differential input (in the μV range, usually). Multiply that small differential input by the actual open loop gain and you have your finite output.

Using the virtual short circuit, is much simpler, though. Once you realize that an op-amp circuit has negative feedback, you can use the virtual short circuit idealization (\$V^+ = V^-\$) to analyze how the circuit works, without bothering with the actual value of the differential input, which becomes irrelevant (unless you need the finer details), as long as you avoid output saturation.

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Let's just do the WHOLE shebang, start to finish, instead of doing this piecemeal. Let's start with the definition for the op amp.

$$ V_{out}= A_{OL}(V_+ - V_-)$$

As has been pointed out, \$A_{OL}\$ is a very big number, but let's leave it in place for the time being.

Just converting this into the notation in the original figure, $$ V_{B}= A_{OL}(0 - V_A)$$ $$ V_B=-V_AA_{OL}$$

Now, we can start applying Kirchoff's Current Law.

$$\frac{V_{in}-V_A}{R_{in}}= \frac{V_A-V_B}{R_f}$$

$$\frac{R_f}{R_{in}}(V_{in}-V_A)=V_A-V_B$$

$$ V_B=V_A - \frac{R_f}{R_{in}}(V_{in}-V_A)$$

$$ V_B = V_A \left( 1 + \frac{R_f}{R_{in}} \right)- \frac{R_f}{R_{in}}V_{in}$$

Now, we can substitute in for \$V_A\$, based on the definition of the op amp $$ V_B = -\frac{V_B}{A_{OL}} \left( 1 + \frac{R_f}{R_{in}} \right)- \frac{R_f}{R_{in}}V_{in}$$

Lastly, now we can apply \$A_{OL}\to\infty \$ , which makes the first term go to zero.

$$\lim_{A_{OL}\to\infty} V_B = - \frac{R_f}{R_{in}}V_{in}$$

This is your standard inverting amplifier equation. Also, note that \$V_A=-\frac{V_B}{A_{OL}}=0\$, leaving us with a "virtual ground" at the inverting input. Thus, there is no paradox. The virtual ground concept is entirely consistent with an infinite open loop gain op amp in a negative feedback arrangement. For giggles, try the same exercise in positive feedback, and watch it explode.

Carrying these things through without throwing out terms due to assumptions also shows you where errors are likely to crop up. For example, you can see from the equation before taking the limit that if you're asking for obscene gain, and \$R_f\$ is many orders of magnitude larger than \$R_{in}\$ that things may not work out so well.

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Math-wise, you can think of it like this: 0 * infinity (which is the ideal op-amp assumption) isn't 0, it's an indeterminate form. To be fully rigorous, you'd be taking the limit as the gain approaches infinity (and the input difference approaches zero). If you went to the trouble of doing all that (it's a pain so in practice nobody bothers, except maybe when a prof is introducing the idea), you'd see the value is determined by the surrounding circuity.

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When the 'Negative Feedback' in an Op-Amp (Ideal) makes the difference between its input terminals equal to 'Zero'. Shouldn't the output become zero too

Imagine the op-amp had an open loop gain of only 100. Negative feedback causes some fraction of the output signal to be fed back to the input and this "restricts" that output signal.

So, what would be the final steady state with equal value resistors and 1 volt at the input? What value of output voltage would satisfy the situation?

You can derive a two simple formulas for the "unknown" voltages: -

\$V_A\times 100 = -V_{OUT}\$

\$V_A = \dfrac{V_{IN}+V_{OUT}}{2}\$

This means that \$V_{OUT} = \dfrac{V_{IN}}{1+\frac{1}{50}}\$

Or, put more generally, for equal value resistors,

\$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{-1}{1+\frac{2}{A_{OL}}}\$ where \$A_{OL}\$ is the open loop gain.

This means that \$V_{OUT}\$ would be -0.9804 for 1 volt inputted.

It also means that the voltage at the inverting input is 9.804 mV.

Now that isn't a virtual ground (or zero volts) but it isn't far off. If the open-loop gain (\$A_{OL}\$) became 1000 then \$V_{OUT}\$ is now -0.998004 and the voltage at the input is fractionally under a milli volt and, by most folks practical standards, it's a virtual ground.

So, if you take this to extremes you can see that the voltage at the inverting input is "virtually" ground.

Here's a way of looking at it from a control system point of view this time using the non-inverting op-amp configuration.

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I am not sure what your question exactly is but your 2nd explanation is OK and can be applied to any op-amp circuit as long as you treat the op-amp ideal (infinite gain, infinite input impedance, zero output impedance).

You can also imagine why this operating point is the only stable one : if the voltage difference between the terminals was larger ever so slightly the op-amp would immediatelly saturate its output voltage to the opposite terminal voltage and the voltage difference would swing back and forth until the stable point (voltage difference almost zero) is reached.

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  • \$\begingroup\$ What you say in your first paragraph is incorrect and misleading: if you treat the opamp as having infinite gain, then the 2nd point of the OP cannot hold, since the input differential voltage would be exactly 0. As I explained in my answer, the confusion of the OP arises because he mixed up two different models: the one in which Avol is "simply" huge, and the one where you take the limit for Avol going to infinity. In your answer you seem to do the same mistake. \$\endgroup\$ – Lorenzo Donati supports Monica Jun 29 '17 at 5:56
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The way I think about it is, if the output voltage of an opamp in its linear region is:

$$V_o=A_{ol}(V^+-V^-) $$

You could rewrite this as:

$$V^+-V^-=\dfrac{V_o}{A_{ol}} $$

Then if \$V_o\$ is finite, and ideally \$A_{ol}\$ is infinite, then the differential input has to approach zero, \$V^+-V^-\to0\$. Even if \$A_{ol}\$ wasn't infinite, as it truly is, this number could be on the other of \$10^6\$, so the approximation is still valid.

This means there is still a small difference between the inputs, but it is convenient to assume that \$V^+=V^-\$ because it makes the analysis simpler.

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The apparent paradox arises because in one case you are dealing with a real (or at least more realistic model of) op-amp and in the other case you are dealing with an idealised abstraction that is useful for quick static (DC) analysis of the circuit.

In the real case, you do have some small differential voltage at the inputs, this is what drives the output.

If you 'let the gain go to \$\infty\$' then the small differential voltage disappears and you end up with a nullator/norator model which gives rise to the 'virtual ground'.

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rev B

A "virtual ground" means it is effectively 0V between, no matter what the common mode voltage is ( as long as the output is not saturated) The inputs are high impedance so no current between these points, but (Vin-) must be tracking the Vin+ if possible , so it always has ~ 0V between them.

This occurs due to negative feedback in the Op Amp and very high gain. This comparison is fedback via negative feedback to make it ~0V difference , yet it may be a Vcc/2 reference , then it goes to Vcc/2 but still ~0V difference.

e.g. the V in offset = Vout/k

  • where k is the open loop gain* feedback ratio .

    • if Av(ol)=1e6 and Rf/Rin gain = 100 then the feedback ratio is 1e2/1e6=1e-4 so the input voltage difference is very small. e.g. 5V/1e4= 0.5mV
  • a virtual ground may be high impedance but at DC it must be near 0V in order for the output with high gain to be in the linear region with negative feedback. Generally we try to keep the impedances balanced on each input port to match bias current Voltage drop and common mode noise from becoming a differential noise problem.

    This low voltage difference it essentially 0V so we call this difference a virtual ground on the inputs. Another circuit which uses this method is called Active Guarding, where as in EEG probes the common mode signal is buffered and drives the shield of the signals to reduce the voltage difference to ~0V with low impedance so stray noise is suppressed and capacitance is eliminated by the dv/dt reduction to 0. The same is done around high Z, or low phase noise circuits to reduce EMI from stray coupling by "gaurding" it with the common mode buffered signal around the inputs or sensor.

A floating ground means it is a 0V reference for that circuit but galvanically isolated from earth up to a limited breakdown voltage, with mandatory HIPOT tests for AC units when made. It blocks DC and AC low f but not RF. THis is good to remember when you get EMI. An RF cap to ground can reduce RF noise on floating grounds.

An earth ground is 0V reference but also tied to earth via the AC receptacle and ground path to earth for safety reasons. Even earth ground has a relative impedance. Why? becuase all grounds are 0V by definition as a point of reference and another reference point may have resistance, inductance and current flowing between will create that voltage difference. But for safety Power Line grounds may be as high as 100 Ohms or more in dry areas.

A logic ground is (again) a 0V reference for logic chips and may be noisy.

An Analog ground is (again) a 0V local reference for analog signals so that the return path is not shared with noisy loads or sources to keep ohmic loss voltages to a minimum.

So in electronics, ground ALWAYS implies 0V reference point somewhere ( by design) and the adjective in front may be implied or explicit to reference to special characteristics such as above.

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Lets talk about distortion. With 0.1volt pp output from the opamp, which has openloop gain of 1Million, and UGBW of 1Mhz. With bipolar diffpair input devices, and no resistive linearization/degeneration. The 2nd and 3rd order input-referred intercepts are, for any bipolar, approximately 0.1voltpp.

At 1Hz, the virtual-ground input will be 0.1v / 1e6 = 100 nanovolts. This differential input, across the bases of the diffpair, is 100nV/0.1v = 1millionth of the distortion intercepts, and the 2nd and 3rd order products will be -120dBc or more.

At 1MHz, the openloop gain is ONE. The virtual-ground input will be 0.1v/ONE = 0.1volt. The opamp will produce heavy distortion.

Now for some interesting results.

At 1KHz, the openloop gain is 1,000x (60db). The virtual-ground input will be 0.1v/1,000 = 100 microVolts. This 100microVolts across bases of the input diffpair is -60dB; the 2nd order distortion will be -60dBc. The 3rd order distortion will be -120dBc.

Additionally, if you reduce the input by 10dB, the 2nd order Harmonic Distortion drops by 10dB. The 3rd order drops by 20dB. Life can be very good.

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You can see an OpAmp as a P-only controller.

It will have allways some offset error if outut is non-zero.
The offset is, however, very small if open loop gain is high. It is virually zero.

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