OPAMP as a comparator in positive feedback

[Kannan26nov88]

Hi,

Currently am working on the attached comparator circuit. I took that circuit from online. The circuit is Non-Inverting Comparator(Schmitt trigger) with reference voltage and am using it for monitoring the pressure. My doubt is in addition to feedback resistor, why 1000pF capacitor is added in the feedback loop of the circuit?

Thanks,
kannan

 

[ravindragudi]

As the sensor is being used, the capacitor will help in smoothing out the signal. This will function like a low pass filter.

 

[chuckey]

I disagree, the capacitor couples the output back to the input, so when the input goes the smallest amount positive, this is amplified by the opamp by a large amount and will make the input even more positive, so until the capacitor discharges, the output stays positive.
The circuit as it stands is poor as its characteristics are effected by the impedance of the sensor.
Frank

 

[arouse1973]

Shouldn't there be another resistor on the + input? I think the capacitor in this case will function as a speed up capacitor charging and discharging the internal capacitance of the op-amps + input thus increasing the switching times.
Adam

 

[Audioguru]

The impedance of the sensor is the "resistor" in series with the opamp's +input. The sensor should use the reference voltage for its common. Then the reference voltage should have a capacitor filter to ground.
The positive feedback capacitor speeds-up the opamp's switching times.

 

[ravindragudi]

The impedance of the sensor is the "resistor" in series with the opamp's +input. The sensor should use the reference voltage for its common. Then the reference voltage should have a capacitor filter to ground.
The positive feedback capacitor speeds-up the opamp's switching times.

I guess thats correct. Initially I was thinking on the same line but later thought it could be doing the filtering. Thanks !!

 

[arouse1973]

Hello Audioguru.

I don't know what sensor is being used but I guess he wants to trigger when a certain pressure has been reached. If you connect the common of the sensor to the reference pin then won't the output be high all the time as the sensor is not likely to have a zero output is it? Then to switch the output in the other direction the sensor output will need to go bellow the reference. Will it do this?

 

[Audioguru]

If you connect the common of the sensor to the reference pin then won't the output be high all the time as the sensor is not likely to have a zero output is it? Then to switch the output in the other direction the sensor output will need to go bellow the reference. Will it do this?

When the input voltage rises above the reference voltage plus the hysteresis voltage then the output goes high.
When the input drops below the reference voltage minus the hysteresis voltage then the output goes low.
But we do not know what input switching voltage is required.

If the pressure sensor in the schematic has its common connected to 0V then the opamp will switch when the sensor voltage rises above or drops below about 2.43V.

 

[arouse1973]

Yes I agree, we need to know what the output of the sensor is. For this to work this won't the output of the sensor need to be able to go negative as it's common point is now 2.5 Volts. Also I would buffer the reference voltage for this use as I suspect the pressure sensor may draw several mA's and as such a 4.7 K resistor is going to be way too high a value. Lets have a look at the sensor details if we can. Maybe we can find out a bit more about this application.
Adam