## Can you help me to understand this circuit it seems to be a comparator integrator?

 I don't understand the way it's working, can you show me a formula for output voltage? by georgedulica February 05, 2023 The output is Vcc if the input is less than Vref and Vee if the input is more than Vref. An OpAmp always try to supply the opposite to the (amplified) difference between the two inputs pins. When the input is 0 volt, the difference is in favor of Vref at the (-) input mode, so the output is positive, to the max possible (since it cannot provide an infinite voltage), Vcc. When the input is 5 volt, at the (+) input pin, the difference is now in favor of that (+) pin, and the output switch to Vee (the lowest voltage applied to the OpAmp). Vcc and Vee are respectively the alimentation voltages supplied to the OpAmp ( positive for Vcc, negative or ground or otherwise for Vee ). If the OpAmp is not a rail-to-rail, remove 1 volt (+/-). It is not clear what are these voltages in your case. The capacitor has an effect only when the input changes (from 0 to 5, ascending front, or from 5 to 0, descending front). If the OpAmp is NOT a Rail-To-Rail OpAmp, you have to subtract (with the proper sign) about 1 volt from the supplied voltages to the OpAmp as limit for its output. (As example, if Vee == ground, your output would NOT be able to reach 0 volt, in such a case.) You can use the simulator for futher details with that modified version: by vanderghast February 06, 2023

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