help with voltage comparator

Audioguru

Joined Dec 20, 2007
11,249
OK in short we divide the voltage across the non inverting terminal right?
so that the photodiode itself will form a resistor and this photodiode will vary the resistance depending on received IR light.
The (-) input of the comparator is biased at a reference voltage with two resistors making a voltage divider. When light shines on the reverse-biased photodiode then it leaks a current that causes the (+) input's voltage to rise. When the voltage rises higher than the (-) input voltage then the output of the comparator will oscillate, or go high.
 

Kermit2

Joined Feb 5, 2010
4,162
You give the positive input some of the output through a large value resistor. This creates a 'new' trigger voltage level at the negative input that is lower than it previously was. In essence it creates 'hysteresis' for the input toggle points. The opamp output is either feeding 0 volts back to the input, or 5 volts back to the input. The two different voltage levels 'move' the trigger point on the negative input. this means that the op amp can't oscillate at the crossover point, because as soon as it switches(goes high) the crossover point changes. No more oscillations
 
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