I need some explanations about Linearity, Negative Feedback and Virtual Short for an ideal Operational Amplifier.

Precisely, I have been always told that an operational amplifier with negative feedback works in linear region, i.e. that with the diagonal straight line in the following picture:

enter image description here

Now I always saw this reasoning: since the negative feedback forces the op-Amp in the linear region we may write that A = Vout/(V+-V-). Now, since A is very high, ideally infinite, we get that (V+ - V-) = Vout/A = 0. This property is called virtual short between op-Amp input terminals.

My question does not regards this last step (which I saw also in many topics here) but the sentence "the negative feedback forces the op-Amp in the linear region": why? Why does the connection of the output terminal to the inverting input terminal determine this condition? I think that probably I should know the internal structure of an op-Amp to understand it correctly, but can you give me a brief explanation?

  • \$\begingroup\$ There are many straight lines in your picture. \$\endgroup\$ – Andy aka Oct 16 '19 at 13:57
  • \$\begingroup\$ Sorry, I will correct the sentence \$\endgroup\$ – Kinka-Byo Oct 16 '19 at 14:00
  • \$\begingroup\$ Instead of an op-amp, regard it as a simple motor control system \$\endgroup\$ – Andy aka Oct 16 '19 at 14:05
  • \$\begingroup\$ Try read this electronics.stackexchange.com/questions/441184/… did that help you? \$\endgroup\$ – G36 Oct 16 '19 at 14:15
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    \$\begingroup\$ Virtual ground is an easier concept. I can draw a voltage reference at the appropriate node, label it "virtual ground" and everybody knows what's going on. Short, to many, implies a direct connection, and there is no direct connection. A ground is a voltage, and a short is a connection. \$\endgroup\$ – Scott Seidman Oct 16 '19 at 19:41

The most simple way for explaining the negative feedback effect is to split the whole procedure in several steps:

Example: Non-inverting opamp stage with two equal resistors in the negative feedback loop. Supply voltages +/- 10volts. Open-loop gain Aol.

Step 1: Apply a DC input of +1V at the non-inv. terminal.

Step 2: Because each amplifier has a certain signal delay the feedback is not yet active at t=0 and the output will "jump" to the supply rail (+10V).

Step 3: Now we have 10/2=5 volts at the inv. terminal (and still +1V at the non-inv. terminal). Hence, the voltage difference Vd between both opamp input terminals now is negative (-4.5 volts).

Step 4: Now, the opamp output "wants" to go to the negative rail (-10V). However, on the way from +10V to -10V the output voltage crosses the linear transfer region and finds one value which - together with the corresponding feedback signal - exactly fulfills the equilibrium point which is defined as Vout=Aol*Vd.

Step 5: This point of equilibrium within the linear transfer region ist stable (it is the DC operating point) because the negative feedback causes a kind of correction as described as follows:

When (during the mentioned crossing effect) the output voltage gets a bit too large (overshoot) , the feedback voltage at the inv. input also is too large and the difference Vd becomes smaller - thereby reducing the ouput voltage again (and vice versa).

Example: Aol=100. Feedback factor k=0.5. The real closed-loop gain is


and Vout=+1V*1.9608=1.9608V.

Voltage difference Vd=1-1.9608*0.5 and Vout=Vd*Aol=100(1-1.9608*0.5)=1.9608 volts (q.e.d.)

Comment: From the calculation it is evident that for larger open-loop gains Aol the voltage difference Vd becomes much smaller - and thus can nearly fulfill the common approximation Vd=0 (virtual short).

  • \$\begingroup\$ Thank you very much, it is now more clear \$\endgroup\$ – Kinka-Byo Oct 17 '19 at 6:27
  • \$\begingroup\$ Another question: what if there are both positive and negative feedback (I mean if there are both a connection of the output to the inv input and to the non inv input)? I thought this situation for instance when I saw that in this circuit (images.slideplayer.com/25/7797116/slides/slide_7.jpg) people apply the principle of virtual short. In this case output is connected, through several impedances, to both + and - inputs \$\endgroup\$ – Kinka-Byo Oct 18 '19 at 10:14
  • \$\begingroup\$ In this case, there is no new situation...for stability reasons, the negative feedback must always dominate over the positive one. A pure negative feedback with kn= - 0.1 is identical to the case with kp=+0.2 and kn= - 0.3. \$\endgroup\$ – LvW Oct 18 '19 at 11:58
  • \$\begingroup\$ So does it depend on the positive feedback and negative feedback transfer functions (which is the highest)? Or is it always true that negative feedback dominates positive feedback (and so it is sufficient to see negative feedback to apply the virtual short rule)? \$\endgroup\$ – Kinka-Byo Oct 18 '19 at 17:11
  • \$\begingroup\$ Of course, it depends on the feedback return ratio. For kp>kn there will be no stable amplifier operation. Special case: kp=kn (for one single frequency only): Oscillator (self excitement). \$\endgroup\$ – LvW Oct 19 '19 at 8:17

If \$V_+ = V_-\$ then, by definition, \$V_D = 0\$. If you look at the transfer curve you included in your answer, the region of operation where \$V_D =0 \$ is at the center of the graph. In this region, the output voltage is a linear function of \$V_D\$. Specifically, \$V_O = A_{OL}\times V_D\$.

EDIT: To clarify based on the comments, the output voltage of the op amp must be maintained within its specified linear range to maintain negative feedback. If the op amp saturates then it is no longer linear. Negative feedback does not guarantee linear operation, but the assumption that \$V_+ = V_-\$ is predicated on the presence of negative feedback.

  • \$\begingroup\$ This reasoning expresses the fact that VD = 0 means linear region, and it is visible from the graph. But my doubt is about how the negative feedback guarantees that the Op - Amp will work in the linear region (and not in saturation, where it is not true that Vo = A × VD) \$\endgroup\$ – Kinka-Byo Oct 16 '19 at 14:09
  • \$\begingroup\$ who says an op amp with negative feedback can't saturate? I promise you it can, and you can experimentally demonstrate that for yourself with a simple setup. Perhaps the idea you are conflating is that a necessary (?) condition for stability in a control system is negative feedback. \$\endgroup\$ – vicatcu Oct 16 '19 at 15:22
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    \$\begingroup\$ @vicatcu, Once the op-amp saturates, then it no longer provides negative feedback. So (OP) it's not that negative feedback forces the op-amp to operate in the linear region, it's that you must keep it in the linear region to have negative feedback. \$\endgroup\$ – The Photon Oct 16 '19 at 16:01
  • \$\begingroup\$ I have always seen that people apply the virtual short rule each time there was a connection between output and inverting input pin,, so is it not necessarily true? \$\endgroup\$ – Kinka-Byo Oct 16 '19 at 16:10
  • \$\begingroup\$ the "golden rules" of op-amps, one of which is what the OP is referring to as virtual short, are only applicable under conditions of negative feedback. It is "necessary but not sufficient" to have negative feedback for it to apply. \$\endgroup\$ – vicatcu Oct 16 '19 at 16:39

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