Need Help Understanding How HP E3610a Can Output Above +5 Volts

[Efe_114]

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

[pqass]

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?

[Efe_114]

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

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

OH everything makes sense now. So the +5 is actually Vout+5 and the formula for the voltage divider ratio is Vout/(Vout + 5)? Did not check the resistor values but im assuming it goes down to 0 volts up to a point decided by the value of R15? and the diodes must be there to improve transient as the difference between S+ and the noninverting input will be close to 0 unless there is an overshoot or an undershoot going on. Thank you so much