ThermalRunaway

Hi everyone,

I came across this circuit earlier today (see attachment). It's actually part of a much larger circuit, but this part in particular was causing me some confusion. It's basically a comparator, and the output is used to switch an LED on or off depending on the voltage output from the POT. In practice there is some hysteresis involved, and I'm assuming that's the reason for the feedback resistor and the diode although I am completely confused with regard to how it works. Usually you operate a comparator in open-loop mode of course, but it can be triggered by noise and is very unstable so I understand why the comparator is being designed with hysteresis in mind. How does this particular circuit work though? In effect it is being used in open loop mode because the diode stops the output from being fed back to the input. How does it create the hysteresis? Who knows!!! Maybe someone here can shed some light on it for me.

I can't guarantee the resistor values because I'm trying to remember it from earlier (I didn't bring the circuit home with me)

Any advice appreciated. It's to satisfy my own curiosity, nothing more.

Brian

Attached Images

Circuit.gif (5.2 KB, 27 views)

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Electronics Test Development Engineer
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Styx

yes this comparator cct has hysteresis and it is needed to reduce its suseptablity to noise (resulting in multiply triggerings on the comparator)

The diode in the feedback is a goodthing (tm) since it simplifies the hysteresis calculations (removes the output voltage of the comparator from teh resistor chain)

The easest way to deal with comparators is to break it into two halfs
1) output high (in this case the diode and feeback resistor not in cct) and work out the potential at teh comparator input via resistor divider
2) output low (in this case you have a 0.6V voltsource & resistor) but again work out the potential at the comparator input via resistor

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JimB

OK, here is my best guess.

Assume that Vref is connected to the inverting input of the comparator.
Initially the voltage at the pot slider is less than Vref, so Vout is low.
A small current will flow through the 300k resistor and diode (assume Vslider - Vout is greater than the turn on voltage of the diode), this load current will reduce the voltage at the slider by a small amount.

When Vslider exceeds Vref, Vout goes high (greater than Vslider), there is now no current in the 300k resistor and Vslider rises by a small amount so becoming sufficiently higher than Vref to prevent noise from "chattering" the output.

On a falling voltage at Vslider, when Vout goes low, Vslider is again loaded below Vref.

Thus there is hysterysis in this circuit.

JimB

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ThermalRunaway

Yep thanks for that both. I made a slight error with my circuit - the output actually comes off the anode end of the diode, not the cathode end (which is effectively just the output of the op-amp).

I've simulated this in multisim and can see it all working so I'm pretty happy with what going on now!

Thanks,

Brian

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Electronics Test Development Engineer
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