This video lecture (start at 32:01), the professor showed a demo about the effects of feedback on noise and non-linearities.

The graph below is a system with input signal which is applied directly to an operational amplifier(the amplifier models the linear portion of the forward path gain), followed by a nonlinear element (has dead zone and two different degrees of compression of linear region). A potentiometer is put between the output and input of nonlinear element to moderate the effects on the non-linearity.

What I am confused is why the feedback path (potentiometer) is connected between the input and output of the non-linearity element as the figure 1 instead of the one as I modified in figure 2.

Can I use the feedback path with the voltage divider as figure 2?

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Figure 1: system from the lecture

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Figure 2: modified system

  • \$\begingroup\$ +1 for the video link! James Roberge was one of the analog masters. (I'll have to watch more.) \$\endgroup\$ Dec 9, 2016 at 14:56

2 Answers 2


The professor in that video is trying to demonstrate the effect of feed back on noise and other non-linearities.

Lets start by assuming the amplifier is ideal: infinite gain, infinite bandwidth and capable of driving any load. I know these do not exist but it makes it easier for us to do a thought experiment.

Taking the circuit of Figure 1 then the system always has unity gain but we have a source of noise and non-linearity. The amplifier is forcing the voltage at the wiper of the potentiometer to match \$ v_i \$. If we imagine the wiper fully on the right hand side as drawn then the amplifier sees the effects of noise and non-linearity and corrects for it totally, With the wiper fully on the left hand side the amplifier does not see any of these effects so there is no correction and we see the full effects of non-linearity and noise. With the wiper somewhere in-between there is partial correction.

To clarify the point regarding always having unity gain I mean unity gain with some superimposed distortion resulting from noise and non-linearity. The position of the potentiometer controls the amount of distortion seen.

With your circuit of figure two the amplifier sees all the effects of non-linearity and noise so these are totally corrected for. However this circuit also introduces gain. The larger \$ R_1 \$ or the smaller \$ R_2 \$ the more gain and.

$$v_o = v_i \left(1+\frac{R_1}{R_2}\right)$$

  • \$\begingroup\$ I have some problems relating to your answer. 1. "Taking the circuit of Figure 1 then the system always has unity gain but we have a source of noise and non-linearity." - Why unity gain here as the wiper of the potentiometer is not fully on the right hand? 2. "With the wiper fully on the right hand side the amplifier does not see any of these effects so there is no correction and we see the full effects of non-linearity and noise." - Did you mean on the left hand side instead? \$\endgroup\$
    – emnha
    Dec 10, 2016 at 4:59
  • \$\begingroup\$ Sorry you are correct the second "right" should have been "left". I've edited my answer to correct this and also add a note to clarify what I mean by unity gain. \$\endgroup\$ Dec 10, 2016 at 14:48
  • \$\begingroup\$ Thank you. But I still have problem with non-linear element. Assuming that it has a gain a2, then how do you calculate the output voltage? Also, where does the noise signal enter into the loop, at the input of non-linear element or somewhere? \$\endgroup\$
    – emnha
    Dec 10, 2016 at 14:54
  • \$\begingroup\$ The noise signal should not exist at the input to the non-linear element since that is also the output of the amplifier (which we are assuming is ideal) so the noise must either be being injected into the non-linear element or created with it. If we assume the non-linear element has gain the gain will be a2 with the pot wound fully to the left increasing or reducing to unit as we turn it fully to the right. \$\endgroup\$ Dec 10, 2016 at 15:15
  • \$\begingroup\$ Can you calculate the output voltage Vo in terms of Vi, the gain of two amplifiers, a and a2 and resistors? \$\endgroup\$
    – emnha
    Dec 10, 2016 at 15:22

Treat the non-linearity+noise as a composite noise signal, \$v_n\$, then the output will be: \$v_o=v'+v_n\$, where \$v'\$ is the op-amp output signal. Also assume, for simplicity, that the potentiometer is in the middle of its travel.

The signal fed back to the inverting input, and consequently equal to the input voltage, is then: \$v_i=v_o-\frac{v_n}{2}\$, giving: $$v_o=v_i+\frac{v_n}{2}$$ Thus the output noise is halved.

If the feedback is entirely from the \$v_o\$ end of the potentiometer, the noise is removed completely; if it's from the op-amp output, all the noise appears at \$v_o\$

  • \$\begingroup\$ If in the figure 2, R1 is a variable resistor, then I can vary its resistance from 0 to infinity to get the same effect as figure 1? So the two figure gives the same result? \$\endgroup\$
    – emnha
    Dec 10, 2016 at 5:04
  • \$\begingroup\$ Fig 2 gives a gain dependent on R1 and R2. Fig 1 has unity gain for all settings of the potentiometer. Hence the two circuits are not equivalent. \$\endgroup\$
    – Chu
    Dec 10, 2016 at 9:55
  • \$\begingroup\$ Can you explain why it is unity gain? If unity gain, should we connect output directly to the inverting input instead of through potentiometer? \$\endgroup\$
    – emnha
    Dec 10, 2016 at 10:22
  • \$\begingroup\$ Fig 2 is the standard non-inverting amplifier configuration, so you set the gain by R1 and R2. Fig 1 is a non-inverting buffer, and gives \$v_o = v_i+k.v_n\$, where \$0<k<1\$ is the potentiometer gain. The two circuits are quite different in characteristics. \$\endgroup\$
    – Chu
    Dec 10, 2016 at 11:10
  • \$\begingroup\$ ... I can't see what you're trying to do. Why are you looking for an equivalent to Fig 1? It seems rather arbitrary. \$\endgroup\$
    – Chu
    Dec 10, 2016 at 11:31

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