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  1. #1
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    How to determine voltage level on Pin 2 of the comparator

    I'm trying to get a complete understanding of a schematic I've been given. On a high level, I got it. It's a portable reusable "fuse" that you can hook in-line in a circuit with the potentiometer setting up the threshold for what current level will trip the device. When the switch 2 is set to connect pins 4/5 and 2/3 it'll only notify you of high current running through input 1 & 2 by D6 turning off. When the switch is set to 5/6 and 1/2, current above the threshold will turn on the transistors which will shut the MOSFETs off and disable any current flow at the inputs.

    What I can't quite figure out though is calculating the voltage level seen at pin 2 of the op-amp. Looking at an ideal analysis, it should just be input current x .22 Ohm since no current would flow into the collector for a transistor in cut off mode or into the input of the op-amp. But that produces a voltage level too low to actually trip the device. I'm hoping someone with a lot more practical knowledge than me can help me out on this.

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  2. #2
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    Re: How to determine voltage level on Pin 2 of the comparator

    the current flowing through 0.22 ohm produces a drop which is compared with the Vref you set with R2 pot.

    Based on the circuit given a rough calculation estimates the current to have a maximum of 220 mA(approx).

    This will produce a voltage of the order of 50 millivolt(approx) which is set by R2 in max position.

    the question too low voltage does not arise here.
    It is just comparing.



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  3. #3
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    Re: How to determine voltage level on Pin 2 of the comparator

    Thanks for explaining it. Part of my confusion actually stemmed from an error in my initial assessment of thinking V_Ref started at 50mV and went up. (Nothing like basic circuits I level errors). Appreciate the 2nd set of eyes.



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