Edit: I solved it, and found that it should be 30v because that causes a 24v drop across 40kohm resistor which is 6v, which is equal to the voltage at the non inverting terminal, meaning Vout will be +VSat, but I don't understand why this is correct, because my notes say that if the voltage at the non inverting terminal is greater than the voltage at the inverting terminal, then Vout will be +VSat, which is what we want to power the LED, but assuming the opposite that is, assuming Vout will be +VSat if the inverting terminal is **greater** than the non inverting terminal led to the correct answer. I do not understand why, though 1. The problem statement, all variables and given/known data For the circuit below, to what value would the 5 V source have to be changed so that the LED lights when Vin reaches 9V? Figure: http://i.imgur.com/1AIQvVy.jpg 2. Relevant equations Vout = -VSat if VRef > VSignal Vout = VSat if VRef < VSignal 3. The attempt at a solution I used voltage division to find VRef and VSignal: 40000/(40000+10000)*5=V drop across 40kΩ = 4 3000/(3000+6000)*9=V drop across 3kΩ = 3 Then, VRef = 5v-4v = 1v, and VSignal = 9v-3v = 6v As 6v > 1v, Vout = VSat, which should power on the LED, which means the 5 V source should not have to be changed. Though, this clearly circumvents what this problem is trying to show me so I think there's something fundamental I am not understanding about comparator op amps.