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Consider this circuit for instance.

schematic

simulate this circuit – Schematic created using CircuitLab

If I calculate the closed loop gain of this circuit using Kirchhoff's law it turns out to be

$${V_o\over V_i} = {-R_2 \over R_1}\left( \frac{R_4}{R_3} + \frac{R_4}{R_2} + 1 \right)$$

Since one end of \$R_3\$ is at 0 potential and \$V_1\$ is also at zero potential, \$R_2\$ and \$R_3\$ can be considered to be connected in parallel connection, which is then in series with \$R_4\$. This way the closed loop gain turns out to be

$${V_o\over V_i} = {-1 \over R_1}\left( \frac{R_2 R_3}{R_2 + R_3} + R_4 \right)$$

This value isn't equal to the above calculated value, hence there must be something wrong with this method. The only mistake I can think is short circuiting \$R_3\$'s one terminal with \$V_1\$ junction. But why can't I do that?

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    \$\begingroup\$ "V1 is also at zero potential ..." Almost, but not. Remember that the \$ V_O = (V_{IN+} - V_{IN-})A \$ so if there is no difference between the inputs then the output is zero. I like to think of it as "virtually" zero. \$\endgroup\$ – Transistor Jun 2 at 11:41
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You can't do that because even though V1 sits at the same potential as ground, it is a different node in the circuit and as such has an isolated current path. Sharing the same potential is not a sufficient condition to consider components to be in parallel; they must share the same nodes. If you connect anything else to that point (such as R3) you disrupt the operation of the circuit by allowing current to flow through a new path.

In this circuit, the current into the op amp input pins is zero, so the currents through R1 and R2 must be identical. However, if you connect R3 in parallel with R2, then the current through R1 will be split between R2 and R3 instead of flowing exclusively through R2. As a result, the circuit will behave differently.

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  • \$\begingroup\$ " Sharing the same potential is not a sufficient condition to consider components to be in parallel" - This is an excellent point, and it why I emphasize that two parallel connected circuit elements have identical voltage across rather than the same voltage across \$\endgroup\$ – Alfred Centauri Jun 2 at 12:29
  • \$\begingroup\$ i.stack.imgur.com/LHFra.png @alex.forencich in the above schematics (R2 and R3) and (R4 and R5) though not sharing the same node are in parallel. \$\endgroup\$ – Akash Karnatak Jun 3 at 6:02
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    \$\begingroup\$ They are not parallel as they don't share the same nodes. However, it is possible that assuming they are parallel still enables you to get the correct answer due to symmetries in the circuit. You have to be very careful when you make that kind of assumption - as you can see here, we have one case where it works and one where it fails. \$\endgroup\$ – alex.forencich Jun 3 at 6:15
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    \$\begingroup\$ @AkashKarnatak, if it takes two volmeters to measure the voltage across two circuit elements, they're not parallel connected even if the voltmeters read the same voltage. If the voltage across two circuit elements can be measured with one voltmeter, they have identical voltage across. In this case, the voltmeter across R2 may read the same as the voltmeter across R3, but the voltmeters are not connected across the same circuit nodes. \$\endgroup\$ – Alfred Centauri Jun 3 at 11:31
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    \$\begingroup\$ @AlfredCentauri thank you, that make it clear \$\endgroup\$ – Akash Karnatak Jun 3 at 11:54
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In effect, you are asking why these two circuits are not equivalent: -

enter image description here

And the obvious reason is that R3's current instead of going directly to ground, would pass through R1.

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  • \$\begingroup\$ I think the example might be improved if R2 and R3 had the same infinitesimal value. In the second circuit, as the value of R2 and R3 approaches zero, the effect on circuit operation would decrease to nothing. In the first circuit, however, their presence would radically increase the gain; the effect would approach a factor-of-two increase as the resistors approached zero. \$\endgroup\$ – supercat Jun 2 at 17:41
  • \$\begingroup\$ How is this relevant to the question? What is your 100R resistor supposed to represent? As far as I can see, neither circuit is even slightly similar to anything which would give the OP more information. \$\endgroup\$ – Graham Jun 3 at 10:23
  • \$\begingroup\$ @Graham the 100R resistor is the one shown in the OP's original post (and as reproduced in the top diagram of my answer). The 2nd circuit in my answer reproduces what happens when you use the 2nd formula in the OP's question. I don't see your objection so please think again about your reason for downvoting. \$\endgroup\$ – Andy aka Jun 3 at 10:27
  • \$\begingroup\$ My mistake on the 100R - sorry. I still don't see how it helps the OP answer his question though. \$\endgroup\$ – Graham Jun 3 at 10:31
  • \$\begingroup\$ Well, if you remove the downvote I will try and explain. \$\endgroup\$ – Andy aka Jun 3 at 10:33
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The essential problem here, beyond what other answers (at this time) have pointed out, is that the node voltage at the lower end of R3 is zero volts by definition (the circuit common is the zero volt reference).

But the action of the (ideal) op-amp in this circuit is to keep the node voltage \$V_1\$ equal to \$V_2\$. That is, the output voltage \$V_o\$ will be whatever it must be to maintain the condition \$V_1 = V_2\$.

So, while it is tempting to think that one can connect the inverting input (in this circuit) to the circuit common since \$V_1\$ is (ideally) zero volts, doing so would 'disconnect' the feedback from the output terminal, i.e., the output would 'loose control' of the voltage \$V_1\$

If it helps to see this, try driving the non-inverting input with a voltage source \$V_{i2}\$ and then redo the analysis.

Also, it might be helpful to back off from the ideal op-amp approximation in your circuit and redo the analysis from scratch. That is, set the open loop gain of the op-amp to \$A\$ and, making no assumption that \$V_1 = V_2 = 0\$, solve the circuit. You'll find that only in the limit \$A\rightarrow\infty\$ does the voltage \$V_1 \rightarrow 0\$

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    \$\begingroup\$ It would be very useful for understanding to carry out the experiment at the end of your answer by a real amplifier with variable gain (e.g., from 1 to 100)... and to observe the difference V1 - V2. The input voltage should be constant. \$\endgroup\$ – Circuit fantasist Jun 4 at 17:24
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There are three points to nail down here.

The first is that I strongly suspect you haven't applied Kirchoff correctly. I would need to spend some time working out the mistake, but I strongly suspect it's related to currents in R3, because this is a tricky circuit with two voltage sources.

The second point is the concept of a "virtual ground". It's not actually 0V, it's just a very, very low voltage which is small enough that for practical calculations we can assume it's zero. The output can't go past the power supply rails, so knowing the gain of the op-amp, we can work out how big the difference between V1 and V2 can be if we're running the op-amp without clipping. The datasheet for the TL071 op-amp, for instance, specifies a typical gain of 200V per mV difference, which is a gain of 200,000. If you've got a 10V power supply to your TL071 then, that means the voltage on V1 cannot be any greater than 50uV. In a practical circuit this is probably below the noise level of your measurements, and that's why you can assume it's basically zero for your calculations.

Relating to this, it's worth mentioning that the fact you do need a voltage between V1 and V2, however small, is why you can't swap the inverting and non-inverting inputs. That small voltage on the inputs needs to be going in the right direction, or else your output won't go in the right direction. The results will depend on the circuit, but often this will result in the circuit being unstable and turning into a kind of oscillator.

And the third important point is that the resistance between the op-amp inputs is very, very high. Looking at the TL071 again, the datasheet says 1012R - yes, that's 1 teraohm. This is why your op-amp theory says that all the current going through R1 is also going through R2, because the current through the op-amp input is below the noise level of your measurements. (Although again, it doesn't mean the current is actually zero!)

To cover the second two points, it's useful to think about the "equivalent circuit" for the op-amp. Op-amp equivalent circuit

This is taken from a good tutorial about op-amps which you can find here.

To get a feel for the effects of the gain and input impedance, it's very useful to start with a simple circuit (a basic inverting amplifier, for instance) and replace the op-amp with a resistor between the inputs and a voltage source on the output. (You can probably ignore the output impedance for now.) Pick some resistor values for your circuit (say, a 1K and 10K to get a gain of 10), then pick an input resistance and a gain and do the sums. Start with maybe an input resistance of 100K and a gain of 100, and see how it works out. Then increase the input resistance to 1M and the gain to 1000, and try again. You should see that you get a small error on the output so that it isn't exactly a gain of 10, but the more you increase the input resistance and gain, the smaller that error is. Then finally try with 1 teraohm and a gain of 200,000 to get the numbers for a real circuit.

I find this is actually a really important exercise to do. Generally it's covered a little further on in your electronics course, but the problem there is that you start off having to assume some "magic rules" for your op-amp which don't really make sense. Less able students can often just accept the rules and plod along, but more able students start asking all the awkward questions about why it works that way. If you start off with the basic concept of how the op-amp behaves though, you can get that intuitive grasp of why those "magic rules" work.

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  • \$\begingroup\$ I really like such experiments with equivalent circuit diagrams as described in the last part of your answer; I often use such both real and imaginary experiments where I replace active elements with variable resistors... op-amps with voltage sources, etc. I also share your observations on the students' understanding of circuit concepts; I have the same impressions. Judging by the specific way you think and present your ideas, I guess you are a teacher. \$\endgroup\$ – Circuit fantasist Jun 3 at 18:09
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An input current of 1mA thru R1 will flow thru R2 and this forces a current of 10mA thru R3 assuming R2 is 10xR3. The input source does not supply the current for R3, this is sourced from the opamp's o/p as the opamp action aims to maintain both of its inputs at virtually the same potential (0V). The sum of IR2 and IR3 flows thru R4 effectively amplifying the feedback current.

By placing R3 in parallel with R2 the input current of 1mA splits between R2 and. R3 so IR4 remains at 1mA.

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  • \$\begingroup\$ Interesting observations... Really, R3 acts as a current-to-voltage converter like Re in a common-emitter stage (the so-called "emitter regeneration" in the past). If a (floating) load is put in the place of R4, the circuit will act as a true current amplifier with a gain of 10. I suppose the name of this application is "current multiplier". In its usual application, OP's circuit can be thought as of a "disturbed inverting amplifier" where the voltage divider R3-R4 is the disturbance. \$\endgroup\$ – Circuit fantasist Jun 5 at 6:01
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enter image description here

first we calculate Z1 to find the equivalent impedance of the feedback enter image description here

witch give us V/I =R and and then we cane find Vo/Vin

1.from the results it is clear that the output is in negative saturation 2.if R2 and R4 are >> than R3 it is very big impedance and need to use this circuit only for low current or have sum DC voltage in "+" of the opamp. 3.R2 + R4 in this case are negligible 4.this circuit usually us to detect smoke or dust in the air enter image description here

if we put values for example in the privies circuit R2 =100K R4=100K and R3=100 we gate 100MEG resistor!!!

so if we have current of 30nA and our positive saturation is 3v we need in the feedback resistance of 100MEG To decrease the band width of frequency response add all the capacitors These are the calculations enter image description here

and the results are: enter image description here

for this values: enter image description here

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  • \$\begingroup\$ for the opamp I use OPA320 opamp \$\endgroup\$ – Koren Reuben Jun 8 at 9:20
  • \$\begingroup\$ Interesting... I1 in the first picture exits the ground. Really, T network (R2 + R3 + R4) can be considered as a "resistor" with high resistance. For example, I have used it instead of 1 Gohm resistor in parallel to the capacitor of an op-amp inverting integrator. \$\endgroup\$ – Circuit fantasist Jun 8 at 10:28
  • \$\begingroup\$ Can you explain exactly why "it is clear that the output is in negative saturation"? I don't see that. \$\endgroup\$ – Elliot Alderson Jun 8 at 12:46
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I am glad to see this question here because, 7 years ago, I joined the electronics forum of ResearchGate with the same question, "As the virtual ground is a kind of ground, can we connect it to the real ground?" ... and I received 38 answers (a good achievement for a newcomer). A few years later, after asking more 123 questions, I moved to SE EE… and now I am answering the same question here. But I realize that after all these years, I should say something better and more intriguing.

Although the virtual ground evokes a sense of mystery, I will try to show that it is a simple concept. The paradox is that I have to use a lot of words for this purpose... but I do not see other way to convince you...

Virtual short

The main task of the op-amp in negative feedback circuits is to maintain (almost) zero voltage between its two inputs. This creates the illusion of a short circuit between the inputs even though no current flows between them. Since this is not a real short circuit, we call it "virtual short circuit" or simply "virtual short" (in circuit theory, it is named "nullator"). What is important to understand for beginners is that the short circuit is not internal between the inputs but external (by the network of two elements in series - R2 and VOUT). The op-amp can maintain the equality between the two input voltages in several ways:

1. Changing the voltage of the inverting input (an example are inverting circuits with negative feedback)

2. Changing the voltage of the non-inverting input (non-inverting circuits with negative feedback)

3. Simultaneously changing both voltages (negative impedance converters - NIC, having both negative and positive feedback)

Virtual ground

While in the last two cases above the "virtual short" is floating, in the first case its non-inverting end is grounded and we say that the inverting end is a "virtual ground". Or the virtual ground is a node that is virtual short connected with the real ground. Thus the virtual ground is just a copy of the real ground... a node whose voltage copies (follows) the voltage of the real ground… a clone of the real ground. So it is implemented by the output of a voltage follower whose input is connected to the real ground. But what is this follower in the OP's circuit?

My answer is simple and maybe surprising: The whole circuit of four resistors and op-amp is a follower… but a "disturbed follower". I will explain what I mean by showing the evolution of the circuit from an undisturbed to a disturbed follower.

1. Undisturbed follower. If we connect the op-amp inverting input to its output, and the non-inverting input to ground, we obtain the classic op-amp voltage follower - Fig. 1.

Undisturbed follower

Fig. 1. Classic op-amp voltage follower with zero input voltage (zero voltage stabilizer)

The op-amp will do whatever to keep (almost) zero voltage between its inputs (the H&H's "golden rule"). So the voltage of the inverting input is a copy of the ground voltage (almost zero)... and the circuit acts as a zero voltage stabilizer with negative feedback. This node serves as another but artificial ("virtual") ground.

2. VIN+R disturbed follower. Now we decide to see if this stabilizer is reliable… and, for this purpose, we intend to "provoke" it with another voltage source. But let's not be so brutal and connect it through a resistor to this node - Fig. 2.

VIN+R disturbed follower

Fig. 2. Disturbed but not reacting op-amp follower with zero input voltage (disturbed zero voltage stabilizer)

The (input) voltage source "pulls down" the virtual ground... but the op-amp, behaving as a perfect voltage source, does not allow it to *move". So, the op-amp does not react noticeably to this additive disturbance. We have to do something else...

3. VIN+R1+R2 disturbed follower. The problem is that the two sources try to set the voltage of the same node but they are not under the same conditions - VOA does it directly while VIN through a resistor. Then let's make it difficult for VOA by connecting another resistor R2 in series - Fig. 3.

VIN+R1+R2 disturbed follower

Fig. 3. Once-disturbed op-amp follower (inverting amplifier)

The resistors R1 and R2 constitute a voltage divider acting as a multiplicative disturbance (beta1) for the op-amp. To compensate it, the op-amp increases R2/R1 times its output voltage… thus acting as an inverting amplifier (if we take this voltage as an output VOUT).

Very interesting - we managed to explain in a logical way the need for both resistors R1 and R2 in an inverting amplifier! But note the voltage follower is still here and it continues keeping steady zero voltage at the virtual ground node.

4. VIN+R1+R2+R3+R4 disturbed follower. If we want a higher gain, the ratio R2/R1 becomes too high... and we decide to put another multiplicative disturbance beta2 (voltage divider R3-R4) in the feedback loop - Fig. 4.

VIN+R1+R2+R3+R4 disturbed follower

Fig. 4. Twice-disturbed op-amp follower (inverting amplifier with T feedback network)

The op-amp reacts to this intervention by increasing more (1 + R4/R3) its output voltage. So, this "twice-disturbed follower" acts as an amplifier with a higher gain. Note that the previous voltages remain and there are in total three circuit outputs - the undisturbed VOUT1, once-disturbed VOUT2 and twice-disturbed VOUT3... but here we are interested only in the first.

The conclusion is that when disturbing more and more the op-amp, it overcomes any disturbances (here, one additive and two multiplicative) in the name of maintaining zero voltage at its inverting input (of course, its output voltage must not reach the supply rails). In its quest to achieve this, the follower becomes an amplifier.

It is so interesting conclusion: An amplifier with negative feedback is actually a disturbed zero voltage follower! And this is the recipe how to convert a follower into an amplifier - by disturbing it.

We can see this technique everywhere around us... and even here, in SE EE:) Let's consider this intriguing psychological phenomenon.

As you know, there are people who regularly downvote our answers; thus they regularly "disturb" us. But since we behave (like the op-amp above) as a negative feedback system with a goal, we overcome these "disturbances" and write even better answers.

This was my incredible story about the virtual ground… particularly, in this circuit solution aka "inverting amplifier with T feedback network". I hope that now you will better understand why you should not connect the virtual ground to real ground...

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  • \$\begingroup\$ This is indeed an "incredible" story, but I'm not sure I would call it intriguing. \$\endgroup\$ – Elliot Alderson Jun 7 at 17:15
  • \$\begingroup\$ @Elliot Alderson, and what can be more intriguing? Maybe the definition of "virtual ground"? Or, maybe the R2/R1 "explanation" of the inverting amplifier? Or (1 + R4/R3).R2/R1 of the T inverting amplifier? Or to show the error in the last expression due to the R2 loading of the voltage divider? Intriguing is to see something new in the well-known old... and to generalize the specific circuit solutions in a common basic idea... principle... and then to use this principle to explain unknown circuits... and why not to invent completely new circuits? Thanks for the response. \$\endgroup\$ – Circuit fantasist Jun 7 at 18:06

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