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Figure 1

In figure 1, is the feedback regenerative in nature? If so, can I apply a virtual short to solve it?

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  • \$\begingroup\$ What are you trying to solve? \$\endgroup\$
    – winny
    Feb 27, 2019 at 18:04
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    \$\begingroup\$ Regenerative feedback is positive. Degenerative feedback is negative. It's just one of the traps native English speakers like to leave lying about for the unwary. \$\endgroup\$
    – TimWescott
    Feb 27, 2019 at 19:46

5 Answers 5

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Negative feedback means: Within the feedback loop there must be for DC as well as for low frequencies one (in very rare cases: three) phase inversion (one single minus sign).

In most cases, for this purpose the inverting input is used. However, it is also possible - as in your example - to perform the necessary phase inversion (180 deg) within the loop. In this case, you must use the non-inv. terminal for feedback, because otherwise you would have two phase inversions (360 deg equivalent to 0 deg), which results in pos. feedback for DC - no stable operating point.

However, as outlined by Sunnyskyguy, the two opamps in one common loop will create unwanted phaseshift for rising frequencies which may lead to unwanted oscillations. This is true, in particular, if both opamps in the loop have similar frequency characteristics. Therefore, in practical circuits, the voltage divider R3-R4 is designed for a much smaller divider ratio (in you case: 0.5) - resulting in a larger overall gain.

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  • \$\begingroup\$ Can I use virtual short to find the output as sergio franco's book says that only inverting input tracks non inverting input and not the vice versa; and that it is necessary to have a feedback path connected to inverting input. \$\endgroup\$ Feb 28, 2019 at 1:25
  • \$\begingroup\$ Yes - you can use ther virtual short concept. As a general rule - the voltage DIFFERENCE between both inputs disapears (nearly) - and is set to zero for idealized amplifiers. In most cases (exception: Your example) we have feedback to the inverting node - and this was assumed in Sergio Franco`s book. \$\endgroup\$
    – LvW
    Feb 28, 2019 at 9:34
  • \$\begingroup\$ The differential input is a virtual gnd which means 0V due to input different / open loop gain. I disagree that output only tracks the +input. If the output is in the linear range ( not saturated) it tracks the difference of BOTH inputs as long as the Vcm is also in the operating range.(even though those inputs are almost 0. So you can always use the logic if - input is lower than + or + is great than -in then the output >0V (leaving input mismatch offset out of this) \$\endgroup\$ Mar 1, 2019 at 1:38
  • \$\begingroup\$ (1)...."The diff. input is a virtual ground"..? I rather think, the diff. voltage disappears... (2) The output "...tracks the difference of BOTH inputs..." ? Sounds a bit unclear to me.. \$\endgroup\$
    – LvW
    Mar 1, 2019 at 9:55
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Yes the logic of feedback polarity in the loop must be negative applies to linear feedback. The same rule is used in analyzing stages of inversion for transistors inside the Op Amp.

i.e. yes in your case, you swap the inputs

But as frequency increases and bandwidth limits there exists a phase lag which accumulates to 90 deg per 1st order effect and must be compensated internally to prevent positive feedback oscillations at the unity gain max frequency. Since OA's typically are compensated to only 60 deg margin, unity gain stability is not possible without filter 1/s compensation.

For stability criteria, you should not expect unity gain stability in other words unless compensation is understood. (Barkhausen criteria) But you will learn this later.

Also understand that OA's have current limits so observe minimum R feedback .. > 300 or > 10k for CMOS types.

enter image description here

Swapping inputs with inverted feedback to Vin+ makes this a negative feedback circuit again.

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  • \$\begingroup\$ But the feedback here red OPAMP part is inverting configuration and it is applied to positive terminal so it should be negative feedback.?? \$\endgroup\$ Feb 27, 2019 at 18:20
  • \$\begingroup\$ That is correct .. but with unity gain , it is not stable due to phase margin \$\endgroup\$ Feb 27, 2019 at 18:24
  • \$\begingroup\$ But for Ideal , it is OK \$\endgroup\$ Feb 27, 2019 at 18:30
  • \$\begingroup\$ Thanks . Can I use virtual short with it as franco says otherwise. \$\endgroup\$ Feb 27, 2019 at 18:33
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    \$\begingroup\$ SERGIO FRANCO is Emeritus Professor of Electrical Engineering at San Francisco State University \$\endgroup\$ Mar 1, 2019 at 0:13
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Further analysis show that the input nodes of OA1 can be swapped without changing the closed loop gain, which will be also -1.

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  • \$\begingroup\$ Welcome to EE! Could you shed some more light on your idea? Written so briefly it looks more like a comment. It would also be good to illustrate it with a CircuitLab schematic (simulation). Thanks! \$\endgroup\$ Nov 3, 2023 at 12:48
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Yes, sometimes you require feedback to be applied to the non-inverting input of the error amplifier, as I discussed here.

In the classic amplifier setup, feedback from the overall output is applied to the inverting input of the op-amp, but this assumes that the signal fed back is in phase with the op-amp output. In both circuits below, this is indeed the case:

schematic

simulate this circuit – Schematic created using CircuitLab

In both circuits, a rise in overall output potential at OUT1/OUT2 causes a rise at the op-amp's inverting input, resulting in negative feedback.

If however, the overall output is inverted with respect to the error amplifier's own output, which would happen if an additional inverting stage is inserted in that path, then to still obtain negative feedback, the behaviour of the error amplifier itself must be inverted too, by applying feedback to the non-inverting input instead:

schematic

simulate this circuit

In my above circuit, A1 represents your own inverting amplifier section consisting of OA2, R1 and R2. A2 represents the potential divider formed by your R3 and R4.

The combined gain of A1 and A2 is \$(-1) \times (+0.5) = -0.5\$, with the negative sign implying an inversion. This inversion means you must apply feedback from the overall output OUT to the op-amp's non-inverting input, to ensure that feedback remains negative.

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What is virtual short?

... can I apply virtual short to solve it?

It is a big challenge to uncover what lies behind verbal clichés that have been used mechanically for many years... so much so that they have lost their meaning like in the popular in the past children's telephone game. Such famous names are virtual ground and virtual short.

For example, it is very common to say that the inputs of an op-amp with negative feedback are "virtually short-circuited". But where is this "short circuit"? Between the two inputs inside the op-amp? Many people imagine it that way. Or is this connection external? But then where is it? Few people can see it.

Basic idea

Actually (see the schematic 1.1 below), the real ground is a voltage source V1 with zero or some voltage, the virtual ground represents its copy source V2 = V1, and the virtual short represents the network of the two sources connected oppositely in series (V1 - V2 = 0). I have outlined this idea below step by step using CircuitLab conceptual experiments.

Short-circuiting elements

Real and virtual ground

To implement this concept, first we get an "original" voltage source with zero voltage V1 = 0 V (terminal A). Then get another ("copy") variable voltage source and adjust its voltage V2 (terminal B) equal to V1. For this purpose, in the schematic below, I have used the so-called "behavioral voltage source" V2 whose voltage V(A) is a copy of V1 (point A). Thus point A is a real ground, point B is a virtual ground and the section A-B between them is a virtual short. This arrangement is typical for the op-amp inverting circuits.

schematic

simulate this circuit – Schematic created using CircuitLab

Virtual short

In op-amp non-inverting circuits, the input voltage V1 applied to the non-inverting input, widely varies; so we can consider point A as a "shifted real ground". Accordingly, the "copy" voltage V2 follows V1; so point B acts as a "shifted virtual ground". What is the voltage between point A and B? It is always zero since two equal but opposite voltage sources are connected in series.

schematic

simulate this circuit

Virtual conductor

If we draw the string from the two sources as a line...

schematic

simulate this circuit

Real conductor

... we can think of it as a "piece of wire".

schematic

simulate this circuit

Shorted circuits

Since this is a piece of wire, let's check this by measuring its resistance with a DIY "ohmmeter" - another (constant) voltage source Vin, a limiting 1 kΩ resistor R and an ammeter Iin. Note that the voltage source is "floating" since the other two voltage sources are grounded.

Virtual circuit

The result is amazing - Iin = (Vin + V1 - V2)/R = Vin/R = 1 mA (KVL and Ohm's law), as though point A and B are joined. So, we conclude, the "virtual conductor" really has zero resistance...

schematic

simulate this circuit

To be fully convinced of this, let's see the current curve by sweeping Vin in the range 0 ÷ 10 V.

STEP 2.1

Real circuit

... as if it were a real conductor! As you can see, the result is the same - 1 V, 1 kΩ and 1 mA (according to Ohm's law).

schematic

simulate this circuit

The current curve is the same.

STEP 2.2

Op-amp applications

Let's finally see this phenomenon in the famous circuit of an op-amp non-inverting amplifier.

Op-amp virtual short

Here the op-amp (and the R1-R2 voltage divider) acts as the clone voltage source V2 from the conceptual circuits above; it makes the voltage of the non-inverting input (almost) equal to the voltage of the inverting input.

schematic

simulate this circuit

Op-amp virtual circuit

Now we will perform a sophisticated experiment not seen in the textbooks - we will pass a current by our ohmmeter from above through the "virtual wire" that connects the op-amp inputs:-) The result is amazing as above - the current flows unimpeded between the two inputs and is still 1 mA. And here arises the most interesting question, "Where does it come from?" Think for a moment, and if you find it difficult to answer, follow the current path indicated by the arrows. It is also good to explain why...

schematic

simulate this circuit

The current curve is the same as above.

STEP 3.2.1

It is really not that easy, especially with this "ideal" op-amp without power pins; okay, let's do it together then...

The current exits the positive Vin terminal, passes through the resistor R, enters the positive V1 terminal, and goes to ground. From there, it enters the negative terminal of the positive power supply Vcc (not shown in the schematic), exits through its positive terminal, and enters the op-amp positive supply pin. It then exits the op-amp output, passes through the resistor R2, exits the output of the voltage divider R1-R2, passes through the ammeter Iin, and finally enters the input source Vin through its negative terminal. So it is not just a "short" between the inputs; it is a rather complex circuit involving various devices. Very interesting - the extremely high differential input resistance of the op-amp is as if shunted with a piece of wire!

Op-amp reaction

There is also another interesting phenomenon here that is extremely important for circuits (systems) with negative feedback - the voltage at the op-amp output rises from 2 to 3 V (hover the mouse over this part of the circuit). This is its reaction to our intervention at its inputs, a basic property of these circuits. In more detail, by drawing current from the output of the voltage divider, the input source Vin lowers the voltage at the inverting input. The op-amp senses this and raises its output voltage to compensate this disturbance.

STEP 3.2.2

Conclusions

In op-amp non-inverting circuits, the op-amp subtracts in a series manner an equivalent voltage from the input voltage, and the zero difference is applied between its inputs. Thus, they are virtually short-circuited (shunted) by a network of two opposite voltage sources of equal voltage that behaves like a zero resistance.

In op-amp inverting circuits, the op-amp does the same thing but in a slightly more complicated way.

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  • \$\begingroup\$ I invite the anonymous downvoter to provide some explanation for their downvote from two days ago; this would benefit everyone. \$\endgroup\$ Nov 5, 2023 at 12:50

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