Purpose of these diodes in inverting opamp feedback

Hi, I was reverse engineering a circuit for learning purposes and I came across this strange configuration of op-amp. I want to ask, what is the purpose of these diodes? are they being used to stop feedback current/voltage ? I was looking at log opamp but it seems to me this is something different.


Do You know the function of this circuit?
If it is correct, C1 has only influence, if the inner resistors of V2 and V3 are big, what I would not expect here.
What You have here, is a kind of comparator. As long as V1 is less V2, the output of the OP is high (voltage depends on Type of OP).
If V1 is greater V2, the output is low (voltage depends on R2, R3 and the diodes, which are bypassing R3 when output goes below V1 (when switching).
So You can influence the switching with R2 (R3 const.) and the low level with the diodes.

For understanding such circuits, I suggest to use LTspice (search for it).

That's an easy one. Do you understand the principle of negative feedback?

yes, feeding some of the output to input to reduce noise and distortion

Ok, so "no".

then ?

I believe there is an op amp tutorial posted at the top of this or the electronics forum... spend some time there and then come back and ask again.

ok, I'll check it

The circuit is a "clipper" or voltage limiter...

i will look into it.

I was using multi-sim for simulation but I think I was not getting accurate results. Will try ltspice now

You don't need spice. This is basic circuit theory. Do the math. In the inverting configuration, there is a "virtual ground" at the inverting input, and the current entering the node that is attached to it, is equal to the current leaving it.

That is all you really need to know to understand this circuit, once you follow those current paths.

yes, I did that, the only thing I am confused in is will the current pass through diodes or resistor R3.

Consider first, the V/I characteristics of the combination of resistor and diode string.

If V1 is near V2 (5V), then You have (without the diodes) a simple inverting amplifier (amplification -10 = - R3/R2).
So if V1 is less then V3/10 near to V2 (4,9 V to 5 V), output is following V1 (inverted) from V3 to GND).
But if V1 = V2, output will be V3/2 and with a little bit more the difference of V1 - output will get above the forward voltage of the diodes (about 1,4 V = 2*0,7V).
Because R3 is rather high and the diodes have an exponential characteristic, it is not a real edge.

For a "clipper", the diodes should be anti-parallel, so this is more a one sided clipper.
Because the input voltage V1 is given with 0-10V, the behavior is more that of a comparator with adjustable (R2,R3) low output.
For the behavior of the amplifier (and one sided clipper), only the input voltage from 4,9 V to 5,1 V is of interest.

There is a use case for one sided clippers, e.g. in a distortion amplifier (for guitars), one sided clippers have a more soft sound than both sided clippers (with only odd harmonics).

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I am confused in is will the current pass through diodes or resistor R3.

Both, of course.

Thank You @jremington, your simulation with LTspice is a perfect completion for my explanation.
I only want to add, that the output high level depends on the OP.
If you would use e.g. a rail-to-rail type, we should see, that the high level is at V3 (your V1) and the low level is not depending on the OP, but on R2 and R3.
If we make R2 >= R3 (your R1 and R2), we should more get to the one side clipper and if we make R2 << R3 we should see more the comparator effect.

Noise cannot be reduced by feedback. That is a reason, why very low noise amplifiers are developed with discrete elements (J-Fet) and drain-bias-current is optimized for the generator (e.g. microphone) resistor.

I don't see any comparator, or any even remotely comparator-like effect in that circuit. I'm calling foul.

Think again.


Thanks alot, you explained it real nice, I did simulation myself on ltspice and starting to understand it better. Will do more research on it.