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Circuit A is a classic non-inverting amplifier with a common-mode DC voltage 1 V to both its inverting and non-inverting input. Since 1 V has to appear at the inverting input, there is no voltage drop across \$ R_i \$ and subsequently there cannot be any current through \$ R_f \$ either. With no current and no voltage drop across \$ R_f \$, the output \$ V_{out} \$ has to hold the same voltage as that appears on the inverting input: \$ V_{out} = 1\,\text{V} \$.

I will certainly measure 1 V at the output but conceptually that doesn't feel right...

If no current passes through \$ R_f \$, the feedback loop is essentially an open circuit (circuit B). If the feedback is open, there is no feedback and the circuit is simply an open loop op-amp (circuit C): if both inputs in circuit C is 1 V, there is no potential between the inputs and the output must be \$ V_{out} = 0\,\text{V} \$! After all, an op amp by itself amplifies any potential difference with infinite gain and drives itself to saturation - but there is no potential difference, output should be zero.

To sum things up, if I go with ideal op-amp analysis, I get \$ V_{out} = 1\,\text{V} \$. But if think through the circuit conceptually from step A, B to C, then I would arrive at \$ V_{out} = 0\,\text{V} \$.

There has to be something wrong in my thought process but I don't know where it is.

enter image description here

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  • \$\begingroup\$ The ideal op amp model, with a gain of infinity, is bound to fail in edge-cases such as these; consider doing the analysis with an op amp that has a high, but finite gain (e.g. 1 million). \$\endgroup\$
    – nanofarad
    Apr 29, 2021 at 16:42
  • \$\begingroup\$ @nanofarad I just edited my question. If the input potential is zero, shouldn't the output be zero too? Actually with finite gain as you pointed out, the argument would be even more compelling - finite open loop gain multiplied by zero potential must be zero! With ideal infinite gain, it's difficult to say .... as I would run into infinity multiplied by zero... \$\endgroup\$
    – KMC
    Apr 29, 2021 at 16:52
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    \$\begingroup\$ "If no current passes through Rf, the feedback loop is essentially an open circuit (circuit B)." No it isn't. Any vanishingly small deviation from 1V on the inverting input would be corrected by a change to Vout, so the loop is still closed. At the limit with gain approaching infinity the output approaches exactly 1V, but Rf is still present and necessary. \$\endgroup\$
    – John D
    Apr 29, 2021 at 16:56
  • \$\begingroup\$ Please ask a specific question \$\endgroup\$
    – Voltage Spike
    Apr 29, 2021 at 17:01
  • \$\begingroup\$ In case "A" you have an example of negative feedback in action. If Vout = 0V then the voltage at inverting input will be +0.5V. And this difference will be amplified by an op-amp. Thus, the opt amp output will be driven towards the positive voltage to "forced" V_"+" = V_"-". Because V_"+" > V_"-". As this is what negative feedback "wants". If the op-amp output overshoots the sign of a voltage difference (V_"+" - V_"-") will change (V_"-" > V_"+") so that the output will now start to decrease until equilibrium point is reached. \$\endgroup\$
    – G36
    Apr 29, 2021 at 17:07

4 Answers 4

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Step A

Circuit A is a classic non-inverting amplifier with a common-mode voltage 1VDC to both its inverting and non-inverting input.

Circuit A is more than a classic non-inverting amplifier; it is a "bad" differential ampifier with unequal gains at both inputs - a noninverting gain of Rf/Ri + 1 = 2 and an inverting gain of -Rf/Ri = -1. That is why the sum of the two partial output voltages is 1 V instead of 0 V (superposition). You can make it a perfect differential amplifier by attenuating the noninverting gain with a ratio of Rf/(Ri + Rf)... and this is the most probable scenario for its invention...

Since 1V has to appear at the inverting input...

+

the output Vout has to hold the same voltage as that appears on the inverting input: Vout=1V...

+

so there is no voltage drop across Ri and subsequently there cannot be any current through Rf either.

As you can see, I rearranged your (correct) thoughts and put together a sentence from them with a more correct causal relationship.

Step B

If no current passes through Rf, the feedback loop is essentially an open circuit (circuit B).

Your observations are very interesting (+1 for this "discovery"). Yes, this can be seen as a kind of "open circuit"... which we can call a "virtual open circuit". This circuit trick is known as "bootstrapping" and is believed to have been invented (in a non-electrical form) by Baron Munchausen several centuries ago:)

The idea is very simple and intuitive - just insert an equal but opposite voltage source in series to the input voltage source. Thus it neutralizes the input voltage and no current flows in the circuit. This creates the illusion of infinite resistance ("open circuit"). But does it mean "broken circuit" as it is shown in your figure?

Step C

However, this "virtual open circuit" (or, as they say, "bootstrapped resistor") does not mean "broken circuit". The paradox of this phenomenon is that there is a resistor "bridge" between the two voltage sources... and it can be low resistive enough... but the input voltage source has the illusion that there is no connection between the sources. Thus, your Fig. C is not correct either...

"Golden rule"

We can summarize this wisdom in another "golden rule" for stopping the current in a circuit of a voltage source and resistor in series. So, we can stop the current in a branch of circuit in three possible ways:

  • by short connecting the circuit (diverting the current),
  • by breaking the circuit (open circuit),
  • by inserting an opposing voltage source.

See also

I recommend you to visit an interesting Wikibooks circuit story that I created with my students in 2008. There we were exploring the same arrangement as yours - a resistor circuit (potentiometer) connected between two sources.

Voltage diagram of a resistor summer

Here is a movie of a computerized experiment with the same arrangement (described in the story).

As you may have guessed, this is a great idea attributed to Miller.

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Your initial understanding is correct. Ideal Op Amps assume the differential inputs draw no current and require unsaturated negative feedback to make Vin-=Vin+. Thus using your 1st diagram with 1V on both inputs and negative feedback, both inputs and output must equal 1V.

If Vin+ were 0V= Gnd, then the output would be -1V and thus Vin-=Vin+=0V.

  • by shifting Vin+ up to 1V, the output goes from -1 to +1V because the non-inverting gain Av+= (Rfb/Rin +1) = 2 while the inverting gain Av-=-1 (with Rfb=Rin)

  • so the net output Vout = 2-1=1V

  • The open loop version is just a comparator and if the output is at either rail then you know there is no gain at that point to go further and Vin- cannot be equal to Vin+. (Yes that means the gain goes from infinite or huge to zero)

    • i.e. it is no longer linear with a virtual null input or a virtual ground.
  • ground just means 0V as a reference and the virtual ground means there can be a common mode voltage with this null difference.

  • in practice, many forget to check the input common mode range, or Vcm specs on single supply in order to satisfy the bias for input circuits. Those with PNP inputs are designed to work at and slightly below Vee. Some such as CMOS inputs can go rail to rail on inputs and outputs.

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  • \$\begingroup\$ Tony, it was a real pleasure for me to read your precious thoughts. There is a lot of wisdom in them derived from a lot of experience. \$\endgroup\$ Apr 29, 2021 at 22:31
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    \$\begingroup\$ You're welcome Cyril, TY for the grammatical improvements., \$\endgroup\$ Apr 30, 2021 at 0:23
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Actually with finite gain [...] the argument would be even more compelling - finite open loop gain multiplied by zero potential must be zero!

I find it even more clear to see with finite gain. You are forcing only one input to 1V. The other depends on the voltage source and the op. amp. output for its voltage. If you break the negative feedback, you remove the mechanism that tries to "zero" the differential input voltage.

This circuit uses an op. amp. only with finite gain as a DC non ideal characteristic (in fact, a very small one, 1000, to make it even more clear):

enter image description here

There is current through the resistors in this condition.

$$\frac{1-0.99901088}{1000} = 989.1 nA$$

$$\frac{0.99901088-0.98911969}{10000} = 989.1 nA$$

If you break the feedback with this ideal model, except for the finite gain, everything works as you expected (\$(1 - 1) \times 1000 = 0\$):

enter image description here

Trying to follow a thought process while multiplying \$0 \times \infty\$ is really a problem. As the open loop gain tends to infinity, the differential input voltage tends to zero (with negative feedback). The current through the resistors will tend to zero, but non-zero current means that \$ V_{in-}\$ > Input- pin \$> V_{out}\$. The two op. amp. pins will tend to the 1 V @ the non-inverting pin, but not together if the feedback resistor is non-zero.

Nice reading here: https://math.stackexchange.com/questions/28940/why-is-infty-cdot-0-not-clearly-equal-to-0.

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Very short answer about the main point where your assumptions are wrong:

Zero feedback current does not imply open-loop.

As a strong counter example consider a opamp voltage follower. If the inverting input is connected only to the opamp output and nowhere else (voltage follower), then the ideal opamp never has any feedback current but of course is in a closed loop with the output following input. A real opamp will have some feedback current though in this scenario.

If the inverting input is also used as an input (in addition to negative feedback), then many opamp configurations (ideal or real) will have one particular set of input voltages at which feedback current changes sign. But even this zero crossing does not imply an open loop.

Open-loop means that there is no causal dependence of any one input on the opamp output.

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