# What is the purpose of the positive feedback in this opamp circuit?

Consider:

Attached schematic is from a broken sampling scope that I have in the lab.

During the reverse engineering the module, I noticed that it uses positive feedback around U2. I understand that positive feedback is often used to implement a negative impedance (and thereby extending the bandwidth), but I don't see any advantage in the AC response in an LTspice simulation.

Green is without 4.2 kΩ and 1.87 kΩ in U2, and Blue is with these positive feedback networks. As shown in the figure, positive feedback reduces both the bandwidth and AC gain.

Edit: parts # for U2~U4 are OP37G.

• Here are some answers that might help. electronics.stackexchange.com/questions/99502/…. electronics.stackexchange.com/questions/98354/…. Commented Aug 4, 2023 at 5:05
• Commented Aug 4, 2023 at 5:17
• Is this in the high frequency signal path or elsewhere? Those “opamps” are probably nothing common. I imagine they would be high-bandwidth parts that may have non-standard pin outs. Please edit the question to include a picture of the part of the module that is shown on the schematic. Commented Aug 4, 2023 at 12:54
• @Kubahasn'tforgottenMonica I just updated the post. Parts # for U2~U4 are OP37G. Commented Aug 4, 2023 at 15:41

The advantage of such a "mixed" feedback is that the resulting total gain can be chosen independently of the feedback factor. This means that the loop gain (and, thus, the phase margin) can also be freely selected, which even allows the use of a non-compensated amplifier. However, it must be noted that of course the negative feedback must prevail over the positive feedback.

Simple example:

• Positive feedback with R4-R3 (input signal at R3).
• Negative feedback with R2-R1 (R1 to ground).

Gain: [R4/(R3+R4)] / [R1/(R1+R2)]-[R3/(R3+R4)].

EDIT: This result is correct - even when the forum member periblepsis thinks it would be "simply wrong". Here is the calculation:

V-=VoutR1/(R1+R2) and V+=VoutR3/(R3+R4) + VinR4/(R3+R4).

Setting V+=V- we arrive at the given gain Vout/Vin.

• ah this makes sense. thanks a lot for your response. One question: I used R1=536 Ω, R2=3.34 kΩ, R3=1.87 kΩ, and R4=4.2 kΩ, but cannot get the same gain as what @periblepsis derived. Could you also comment on how you get the gain equation please? Commented Aug 4, 2023 at 16:47
• @Emm386 Gain equation for the first stage only?
– LvW
Commented Aug 4, 2023 at 16:55
• more specifically, I was curious how you get to the equation at the bottom of your response: Gain: [R4/(R3+R4)] / [R1/(R1+R2)]-[R3/(R3+R4)] Commented Aug 4, 2023 at 17:08
• @Emm386 I have used, of course, the classical way for finding the gain expression: V+=V- (that means Vdiff=0). For finding V+ you must use superposition because two voltage sources are involved (Vout and Vin).
– LvW
Commented Aug 4, 2023 at 18:40
• I see what you are doing in the gain calculation. I also missed your assumption. Again, thanks for sharing your knowledge and answering my question! Commented Aug 5, 2023 at 17:32

## Reduction for Analysis

I've reduced the first stage of your schematic to a more behavioral one for analysis:

simulate this circuit – Schematic created using CircuitLab

From the above, we know:

\begin{align*} V_{_\text{O}}\cdot\frac{1+\frac{R_2}{R_3}}{1+\frac{R_6}{R_7}}&=\frac{\frac{V_{_\text{O}}}{1+\frac{R_4}{R_5}}\,\cdot\,R_1+V_{_\text{I}}\,\cdot\,R_{_\text{TH}}}{R_1+R_{_\text{TH}}+\frac{R_1\,\cdot\,R_{_\text{TH}}}{R_2+R_3}} \end{align*}

## Loop Gain and Closed Loop Gain

From that, find:

\begin{align*} V_{_\text{O}}\cdot\left(1-\frac{\frac{1}{1+\frac{R_4}{R_5}}\,\cdot\,R_1}{R_1+R_{_\text{TH}}+\frac{R_1\,\cdot\,R_{_\text{TH}}}{R_2+R_3}}\cdot\frac{1+\frac{R_6}{R_7}}{1+\frac{R_2}{R_3}}\right)&=V_{_\text{I}}\,\cdot\,\frac{R_{_\text{TH}}}{R_1+R_{_\text{TH}}+\frac{R_1\,\cdot\,R_{_\text{TH}}}{R_2+R_3}}\cdot\frac{1+\frac{R_6}{R_7}}{1+\frac{R_2}{R_3}} \end{align*}

That's in the form of $$\V_{_\text{O}}\cdot\left(1-B\cdot A\right)=V_{_\text{I}}\cdot A\$$ with $$\A=\frac{R_{_\text{TH}}}{R_1+R_{_\text{TH}}+\frac{R_1\,\cdot\,R_{_\text{TH}}}{R_2+R_3}}\cdot\frac{1+\frac{R_6}{R_7}}{1+\frac{R_2}{R_3}}\approx 0.877\$$ and therefore $$\B=\frac{R_1}{R_{_\text{TH}}}\cdot \frac{1}{1+\frac{R_4}{R_5}}\approx 0.419\$$.

It's also in the positive/regenerative feedback form, as you already identified. So $$\B\$$ is positive feedback. The loop gain is $$\A\cdot B\approx 0.368\$$ and since this is less than 1 the system converges on the overall closed loop gain of $$\\frac{V_{_\text{O}}}{V_{_\text{I}}}=\frac{A}{1-B\,\cdot\, A} \approx 1.39\$$.

So, expanding $$\\frac{A}{1-B\,\cdot\, A}\$$, the final closed loop gain equation is:

$$A_v=\frac{1+\frac{R_6}{R_7}}{\frac{R_1}{R_3}+\left(1+\frac{R_2}{R_3}\right)\,\cdot\,\left(1+\frac{R_1}{R_4}+\frac{R_1}{R_5}\right)-\frac{R_1}{R_4}\,\cdot\,\left(1+\frac{R_6}{R_7}\right)}$$

Let's test both:

def par(a,b): return a*b/(a+b)         # a and b taken in parallel
a = par(r4,r5)/(r1+par(r4,r5)+r1*par(r4,r5)/(r2+r3))*(1+r6/r7)/(1+r2/r3)
b = r1/par(r4,r5)/(1+r4/r5)
(a/(1-b*a)).subs( { r1:1.76e3, r2:3.64e3, r3:1.76e3, r4:4.2e3
, r5:1.87e3, r6:3.34e3, r7:536 } )
1.38762233798678                       # computed using A/(1-B*A)
(r6/r7 + 1) / (r1/r3 + (r2/r3 + 1)*(1 + r1/r5 + r1/r4) - r1/r4*(r6/r7 + 1)).subs(
{ r1:1.76e3, r2:3.64e3, r3:1.76e3, r4:4.2e3
, r5:1.87e3, r6:3.34e3, r7:536 } )
1.38762233798678                       # computed using expanded version


Note that they are both equivalent to each other. Just two different ways of saying the same thing. So the voltage gain is $$\20\cdot\log_{10}\left(1.38762233798678\right)\approx 2.84542565\:\text{dB}\$$

I've added a test using LTspice (performed a day after the answer was written):

Seems that LTspice agrees.

## Input Impedance

To get this, I simply derived it by first taking the voltage difference across $$\R_1\$$ and dividing that by $$\R_1\$$ to get the current in $$\R_1\$$. This current was then divided into the input voltage in order to work out the impedance. It's straight-forward.

The input impedance thus seen by the source will be:

$$R_{_\text{I}}=R_1\cdot\frac{1}{1-\frac{V_{_\text{O}}}{V_{_\text{I}}}\cdot\frac{1+\frac{R_2}{R_3}}{1+\frac{R_6}{R_7}}}\approx 2.432\times R_1\approx 4.28\:\text{k}\Omega$$

## Schematic Notes

The following shows my notations for the schematic used to derive the above:

• @Emm386 Let me know if there are further questions. Commented Aug 4, 2023 at 15:59
• I really want to accept this as an answer for my question, but I think response from LvW is more close to what I was looking for: the reasoning of positive feedback. But again, I really appreciate you for providing all the analysis and detailed explanation. I would greatly appreciate if you can elaborate on the input impedance calculation too!! Commented Aug 4, 2023 at 20:17
• @Emm386 That's very nice of you to say. But please don't worry. Just saying what you said is all I care about. So thanks! Very much appreciate the thought. And be happy! (I don't do any of this for points.) Just keep in mind that LvW's gain equation is ... not correct. Given your resistor values it comes out with 1.867. Which is much too high. Check it out. Commented Aug 5, 2023 at 1:08
• ah I see. yeah I was being stupid. As what you said exactly, this was a straight-forward computation. again, thanks a lot for detailed comments and everything! I learned a lot from you. Commented Aug 5, 2023 at 4:54
• @Emm386 I've added a run from LTspice to the answer and updated a few details. I am quite flummoxed by LvW's gain expression. It's simply wrong. Commented Aug 5, 2023 at 5:57