What configuration is this op-ampworking in?

Hello everyone. I was working on some circuit and I came across this op-amp configuration and I am confused in what configuration is this working in and what is the purpose of those diodes ?. So the purpose of this op-amp in the circuit is that it gives different voltages output depending on the input signal the output voltages varies from 0-10.6V.

Analysing op amp circuits is very easy if you remember the characteristics of an ideal op amp;

The feedback circuit acts to maintain a zero difference between + and - inputs.
If the input voltage is less than v(pin12) pin 14 will go more positive and D4 will conduct - so the gain is set by the 4k7 and 56k resistors
If the input voltage is more than v(pin12) pin 14 will go less positive and D18 will conduct - so the gain is set by the 4k7 and 5k6 resistors.

1 Like

@johnerrington it would be good of you to say a bit more; I'm no op-amp expert but have thrown them around here and there in a cookbook fashion. Mostly getting what I need them to do done.

So I wonder about the capacitor, which is in there like we'd see in an integrator.

And presumably those are real diodes, which means… well rather than use the classic internet technique of saying the wrong thing to most rapidly get the right answer, Imma ask what effect they might have.



Some sort of fuzz or distortion circuit or maybe for a zero-crossing detector. This is a basic inverting configuration.

However the diodes mean that when the output is positive (input negative) there will be a diode drop added to the voltage. Similarly when negative a diode-drop is subtracted from the voltage, so that a small signal (millivolts) will produce at least 0.6V peak-to-peak output.

The capacitor is just for high frequency stability and will be something like 10pF. This is very common in opamp circuits.

In the above "positive" and "negative" are relative to the virtual ground (which is 5.5V as applied to the non-inverting input of the opamp).

One odd feature is the gain is different for each polarity, perhaps this is conditioning some sort of pulse waveform and negative-going inputs are the pulses and get much larger gain.

The back-to-back diodes are classic fuzz box, yes. Never seen a sfx circuit with the diodes and a series resistor though.



What circuit?
We need some context as to what the bigger picture is?
What does the "circuit" do?

Is it an audio, RF, power supply , DC measurement, AC measurement system?

Thanks.. Tom... :smiley: :+1: :coffee: :australia:

I don't know what this is for, but it is easy enough to see what it does. The basic formula is

Vout = β * (V12 - v13)    

where β is the gain of the opamp.

Here V12 = 5.5, and for V13 we have two cases:

<if Vout > 0)
   V13 =  (5.6 Vin + 4.7 Vout) / 10.2
<if Vout < 0)
   V13 = (56 Vin + 4.7 Vout) / 60.7

Plugging these into the first equation and solving for Vout gives

<if Vout > 0>
  Vout = β * (56.2 - 5.6 Vin) / 10.2    / (1 + β * 5.6 / 10.2)
<if Vin < 0>
 Vout = β * (370.35 - 56 Vin)/60.7 / (1 + β * 4.7 / 60.7)

But now, if Vin > (56.2 / 5.6) ~ 10 then the formula for Vout > 0, makes Vout negative. Thus if Vin > 10 then Vout will turn negative.

Similarly, if Vin < 370.35 / 56 ~ 6.6 the formula for Vout < 0 makes Vout positive. Thus if Vin < 6.6 then Vout will turn positive.

To sum up:
if Vin < 6.6, Vout > 0
if 6.6 < Vin < 10, then polarity will be retained.
if Vin > 10, then Vout < 0

1 Like

really nice explanation, thanks alot sir

this portion of the circuit is just driving VCO to generate sound on the basis of input signal if the signal is strong the loud sound will be generated if signal is weak the sound will be really low

capacitor is being used to minimize the noise it is making a low pass filter, if i remove this capacitor there is serious noise in the circuit (I have done it in practical)

1 Like

No it is not.

What is the value?
It is more likely that it is acting an an integrator and reducing the rise and fall time of the edges, thereby removing high frequency components to the output.

That might sound like a filter but it is not.


That noise is ultrasonic oscillation intermodulating down into the audible range - am I right it was about 10 to 30pF in value?

The problem with diodes in the feedback path is that they are open-circuit around 0V, so that the circuit becomes completely unstable and parasitics dominate the behaviour. A small negative feedback capacitor restores stability by swamping any parasitics and ensureing negative feedback (rather than positive).

1 Like

It seems the circuit is taking a slowly changing voltage and "mangling" it to produce a different but related voltage that will be used to control a VCO.

I dont see the reason for the assymmetry though.

I'd suppose the capacitor (220nF) is to limit the transient response of thecircuit. As @Grumpy_Mike says, an integrator; and also as @MarkT says, to prevent oscillation.

Common error @mitchell2090 - equations that work for an "ideal" op amp dont take the limitations of power supplies into account; but a nice analysis nonetheless.

1 Like

What is the actual purpose of this nonlinear amplifier circuit?

Thank you, @johnerrington, for the complement. I just want to note that I don't regard my omission of "the the limitations of power supplies" as an error. I was just trying to answer the original poster's question. (In fact, I couldn't have taken the limitations of the power supply into account, as I had no information about the power supply.)

I'd also like to thank @MarkT for his explanation of how the capacitor improves the stability.


"sound will be really low" is that volume or frequency?
Is the output of this circuit controlling the frequency of the VCO or its output amplitude?

Thanks.. Tom... :smiley: :+1: :coffee: :australia: