I don't understand why when we design a proportional amplifier, the feedback has to be negative. When I run the simulation in simulink I get the same result for both circuits.

positive feedback

negative feedback

  • \$\begingroup\$ Where are you taking your output voltage from? \$\endgroup\$ – DKNguyen Mar 27 '19 at 0:16
  • \$\begingroup\$ @Toor between the output of the op amp and ground \$\endgroup\$ – Pedro Mar 27 '19 at 0:17
  • \$\begingroup\$ Adding info OP put into a now deleted response: "I used a 1v DC supply and the resistors are 2k and 4k. The output is -2V in both." You can't get a -2V output if your op-amp only has 1V supply (the power supply for the opamp). Or do you mean your input signal is 1V? What is the opamp supply? Or does your simulator not require a supply for the opamp? Try this test: If you delete the feedback resistor for the top circuit, what do you get for the output? \$\endgroup\$ – DKNguyen Mar 27 '19 at 2:32

The top one is a non-inverting Schmitt trigger, the bottom one is an inverting amplifier. The output of the top one will always saturate, the output of the bottom one will only saturate if the gain is large enough.

  • \$\begingroup\$ This answer actually makes sense and has practical value. Upvote. \$\endgroup\$ – Elliot Alderson Mar 27 '19 at 15:12

The thing here is that you are looking (and very likely your simulator) at those two opamps circuits from a 'static' point view.

If you actually go through the math, you indeed find that the gain for both circuit (even the positive feedback one) is the same.

You start out with:

$$V_o = A_{ol}(V^+-V^-) \tag1$$

For the one with positive feedback, you get a gain (static, no dynamics involved) of

$$G=-\dfrac{R_2}{R_1}\tag2 $$

Which is the same as in the negative feedback case. Where is the catch?

I don't understand why when we design a Proportional amplifier, the feedback has to be negative. When I run the simulation in simulink I get the same result for both circuits.

Think of it this way: If for whatever reason, the output voltage drifts, even the slightest, that perturbation is in phase with the input in the positive feedback circuit, that means, that the output will keep rising until it hits one of the rails (saturates). Which rail is going to hit (either +VCC or -Vss) will depend on the sign of the perturbation.

Another way to think about it is by looking at equation (1). Remember that in the positive feedback circuit, you are feeding the output back into the \$V^+\$ terminal. So if \$V_o\$ rises a bit, the right hand side of (1) will tend to either grow positively or negatively, again, depending on the sign of the perturbation.

You don't have this problem in the negative feedback circuit. If \$V_o\$ rises, that pertubation will affect \$V^-\$ in equation (1). So if there is a rise in \$V_o\$, the right hand side in (1) will go down. Once it goes down, it forces the output back down. You can see this happening back and forth until you reach an equilibrium point. That is why negative feedback is used and its importance.

In an ideal world, you don't have the perturbations I've been telling you about, but in the real world, those will be present. The negative feedback will counteract it. It's probably the case that the simulator is not taking those into account.

I'll link these two resources, they answer your exact question, and there is a more rigorous explanation in them:


youtube video


As per one of the comments in my answer. Equation (1), as you said it is the open loop gain. Then equation (2) is the closed loop gain for the inverting configuration under negative feedback. I do a agree that the concept of 'gain' under positive feedback does not make sense for the reasons I have explained (the output will hit one of the rails) but if you start out with the open loop gain (1) for the positive feedback configuration you will arrive at (2). That is why I also said that the 'gain' under positive feedback includes no dynamics and that is the reason (likely) his simulator gives the same answer for both configurations.

Now, how would you get the same mathematical expression? Start out with the open loop equation and notice that:

$$ V^+ = V_i\dfrac{R_2}{R_1+R_2} + V_o\dfrac{R_1}{R_1+R_2}$$

Now plug that into equation (1), keeping in mind that \$V^-\$ is zero (grounded) in the positive feedback case:

$$V_o = A_{ol}\bigg(V_i\dfrac{R_2}{R_1+R_2} + V_o\dfrac{R_1}{R_1+R_2}\bigg)$$

Since \$A_{ol}\$ is extremely large, for a finite value of \$V_o\$, the left hand side of the previous equation is negligible so:

$$ -V_i\dfrac{R_2}{R_1+R_2} = V_o\dfrac{R_1}{R_1+R_2}$$

And that will give you the 'static gain', \$G=-\frac{R_2}{R_1}\$ which again I agree with the assessment that it does not make sense since any perturbation will steer the opamp out of the linear region into saturation, but from a mathematical (ideal) point of view, they do show the same mathematical expressions.

This is very likely the reason why the simulator gave the OP the same answer for both configurations (ideal case, opamps, etc )and that is what I tried to answer.

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    \$\begingroup\$ You talk about the gain of the circuit but then give the equation for the open-loop gain of the op amp itself. How can you say that the gain with positive feedback is the same as the gain for negative feedback? With positive feedback the circuit is non-linear and the notion of "gain" no longer makes sense. \$\endgroup\$ – Elliot Alderson Mar 27 '19 at 12:43
  • \$\begingroup\$ @ElliotAlderson please see my edit. I agree 100% percent that it does not make sense to talk about 'gain', in the ordinary sense, for the positive feedback case since any perturbation will steer the opamp out of the linear region into saturation. However, from a mathematical point of view (ideal case), it does the turn out expression for the output voltage (closed loop) is the same for both. Could you please read the first page of the link I posted? It provides more info (and the same conclusion I have posted here). \$\endgroup\$ – Big6 Mar 27 '19 at 14:41
  • \$\begingroup\$ @ElliotAlderson Also, I did reference the open loop gain but that is used as a starting point to try and get the closed loop gain. You can always start from there and work your way to a closed loop equation. \$\endgroup\$ – Big6 Mar 27 '19 at 14:44
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    \$\begingroup\$ Two comments: (1) Interestingly, for ideal opamp models even the simulation programs reach a stable operating point and give the gain (-R2/R1) for the version with positive feedback. (2) There is an indication that "something" must be wrong with the pos. feedback version: The signal is injected into the non-inv. input, but the gain turns out to be negative....a point to humble... \$\endgroup\$ – LvW Mar 27 '19 at 15:01
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    \$\begingroup\$ @LvW Yes, I believe that the conclusion provided in the linked lecture notes is strictly an academic exercise in mathematics and has no practical value. It predicts a single metastable value for Vout but the equation given is not describing the relationship between Vin and Vout, as it does for the case with negative feedback. The slightest perturbation of the input due to thermal noise will send the output banging into the rails. The conclusion shown in the lecture notes is, in my opinion, so misleading that it is harmful to students. \$\endgroup\$ – Elliot Alderson Mar 27 '19 at 15:10

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