If we have a circuit that - given some input voltage \$U\$ - produces an output voltage \$f(U)\$ (where \$f\$ is continuous, monotonic descending), we can construct a new circuit that behaves like the (negated) inverse \$f^{-1}\$ of \$f\$ (meaning the function-inverse) as follows:


simulate this circuit – Schematic created using CircuitLab

Now I've tried applying that with a funtion \$f\$ that roughly looks like a \$\arctan\$-function (up to a scaling factor) where \$f\$ is implemented using following circuit. I have tested this circuit and it works fine.


simulate this circuit

To check if that worked I built this on a breadboard with a NE5532 op-amp (with bypass cap with +-12V supply) used a 20Hz saw-wave and hooked up an oscilloscope. It seemed to work fine but the output was quite noisy. This was especially noticeable in the flat part (so I'm not sure if it's actually particularly strong there or it is just particularly noticeable to the flat slope).

Now I was wondering, where does this noise come from, and how can I reduce it? Below is the circuit I actually used (can be simulated, see "DC Sweep")


simulate this circuit

Here's the simulation output that works as desired:


  • \$\begingroup\$ You made an attenuator , the inverse of your feedback gain = - 10 until the diode compresses the feedback and thus amplifies the input signal into a quadratic expander with less phase margin and more noise., with unity gain where the diode voltage/current= impedance = 10k What results did you get? What did you expect? \$\endgroup\$ Feb 20, 2021 at 16:40
  • \$\begingroup\$ I'm not sure I understand: Do you say that reducing the gain (by setting R2 to e.g. 10k) would decrease the noise? I have played around with various values of but none seemed to alleviate the issue. \$\endgroup\$
    – flawr
    Feb 20, 2021 at 16:53
  • \$\begingroup\$ Does my answer make sense yet? \$\endgroup\$ Feb 20, 2021 at 19:55
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I don't know enough yet to understand all the details. But your answer helped me see the issue when thinking about the situation of a constant output. Thanks a lot! \$\endgroup\$
    – flawr
    Feb 20, 2021 at 21:04

1 Answer 1


Your overall gain 0.1 , which by design is the inverse of the feedback gain = 10 as OA1 reduces its output so that 10 times its output matches the input of the OA1 as an "error amplifier".

So that rather than a resistor attenuator for feedback to get gain, you have gain feedback to make an attenuator.

Each is internally compensated to get a typical phase margin at a unity gain of 60 deg. (IDK what your IC specs are but I'll look later) Now cascading in closed-loop two amplifiers with some phase margin each overall reduces the phase margin.

Now understand if the output doesn't change in a linear amplifier due to diode limiting or saturation, all single-stage amplifiers and comparators alike the loop gain drops to zero, Av=Δout/Δin=0. If that happens to your x10 feedback amplifier, it becomes a closed-loop noise oscillator amplifier with rail-to-rail saturation noise feeding while operating non-linearly from no gain at one rail then racing back towards the other rail while amplifying the very low input thermal noise by 1e5 or more.

This configuration is inherently unstable at some point whenever the initial condition or transient condition, the output of OA2 happens to saturate, causing OA1 to oscillate with its input noise amplified by its open-loop gain and the open-loop gain of OA2.

By reducing the input signal with a sinewave to say 100mV max and restarting, it might be barely stable but as you increase the level where the phase margin at high frequency drops to <=0 you have by Barhausen criteria, a closed-loop feedback oscillator of input noise.


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