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.
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.