Dumb question about circuits with diodes

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The discussion centers on understanding the output voltage in a circuit involving diodes and an operational amplifier (op-amp). The diodes serve as clamps to prevent the output from exceeding the positive rail or dropping below ground. The circuit is identified as a comparator, where the op-amp triggers when the voltage across a variable resistor equals the input voltage, causing the op-amp to saturate near the supply voltage. The voltage drop across the diode is specified as 0.6 V, and the circuit's behavior is linked to the concept of unity gain in op-amp configurations.

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I'm at a loss as to how there is an output voltage for this circuit when I think the diode is facing in the wrong direction. Or is it possible to get an output voltage? Can someone help me analyze it. Just assume resistance is R and current through resistor is I and voltage drop across diode is 0.6 V. Thank you!
 

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The diodes are just clamps to be sure that the output does not go above the + rail or below ground. The rest of the amp circuit makes no sense, unless it represents some fixed gain (unity gain?) amplifier. If you do not connect feedback on an opamp circuit, the output will indeed peg.
 
the rest of the amp circuit makes no sense
This kind of circuit is called comparator. The opamp will trigger when voltage over variable resistor will be equal to the input voltage -> making the opamp go into saturation i.e to ~+Vc, assuming that Vo should be less than Vc, those clamps makes sense.
 

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