I couldnt figure out how the output can go above +5V. Isnt the op-amp keeping the output and the voltage at the non-inverting input (which cant go above +5V) the same which means the output cant go above +5V? Please help me figure this out
I looked into this schematic to fix my E3611A. Yeah, this one hurt my brain too.
The thing to understand first is that +S is attached to the +OUTPUT, positive binding post. The +S is the common for the bipolar op amp supply on the other transformer secondary (not shown in your screenshot; labeled "Reference and Bias Supply"). That means that the op amps can go above or below the +OUTPUT by 12V (there is also a +5V available too). It may be easier to imagine this PS is a negative supply where the +OUTPUT=0V, and -OUTPUT is a negative output.
The series pass transistor (NPN darlington pair in your model) emitter pins are on the other side of the shunt from the +OUTPUT and their bases are held to +12V via R3; thus are defaulted to fully-on. The "Voltage Error" op amp's job via CR5 diode is introduce a lower voltage (than +12V) to Q2 which shuts off the darlingtons.
The -ve pin on the op amp is tied to +OUTPUT, whereas the +ve pin is the center of a voltage divider between R15 and R37 (voltage set pot); between +5V (relative to +OUTPUT) and the -OUTPUT binding post (CR7 via J2 just clamps center point to +0.7V, J1 is open). So varying R37 allows the -OUTPUT to be lowered or raised such that the +ve op amp pin strives to be zero (with the -ve pin).
On power up, the darlingtons are fully-on, the voltage difference between +OUTPUT and -OUTPUT is ramping up but lower than demanded by the voltage set pot, therefore, the center of the R15/R37 divider will be positive WRT -ve op amp pin, thus op amp is outputting +rail. As soon as the voltage difference between +OUTPUT and -OUTPUT overshoots what is demanded by the voltage set pot, then the center of the R15/R37 divider will be negative WRT -ve op amp pin, thus turning off the darlingtons.
Clear as mud?