Electronic – Diode in positive feedback


I have a colleague who is off work for the next 2 weeks and has asked me to finish off one of his schematic designs. I have a list of operations it needs to be able to perform, seems simple enough.

I got round to making a start on it today, and after having a browse through what he has already done, I notice something that I haven't seen before.

Here is the basics of what it looks like:


simulate this circuit – Schematic created using CircuitLab

I have not seen an op amp circuit which uses a diode in the feedback circuit in this way. I recognise it is a window comparator, and this part of the circuit is used to detect a voltage level and turn on an LED if it goes above or below a threshold. I just can't work out what the point of the resistor and diode is in the feedback.

My go-to op amp configuration PDF is one from texas instruments (LINK) and I couldn't find one like this. So can anyone tell me what the function is of this feedback circuit?

NOTE: I have labelled things as V1, V2, OUT etc as they should be irrelevant to the circuit, V1 and V2 are measuring the input voltage, Vref is the threshold, and the output toggles LEDs

EDIT: I have updated the schematic to include the resistors that Andy aka mentioned, the values of resistance are what was on the schematic at the time, they may be incorrect, I am unsure, as the schematic is not finished.

Best Answer

It looks like the intention is to provide hysteresis. For instance (and assuming that V1 and Vref have series resistance that is not shown on the OP's diagram), if V1 drops below Vref then OUT will drop to 0 volts and R1/D2 will further enhance the effect of V1 lowering below Vref. This is called hysteresis and is used to avoid a situation where V1 is hovering close to the value of Vref and causing OUT to oscillate high and low due to noise.

That is what hysteresis does - once a comparator switches it stays switched with no ambiguity.

With a diode (D2) in series with R1 and with OUT high there is NO current passing back through D2 to V1. This means that if Vref were to increase toward V1, OUT would switch low at precisely the point Vref = V1.

It's a kind of one-sided hysteresis and this is due to the diode blocking the hysteresis effect in one direction.