I'd like to (attempt to) clarify the DIY-SMU / K236 output stage. It's complicated, kind of mind-twisting. The amplifier board itself, AMPIN to OCOM out (without local feedback) is voltage input and current output, so is a Transconductance amplifier. This type of amplifier configurations is pretty rare, but used occasionally. If it helps, think of an LM13700.
Why current output? The output current path is the common of the +/- 170V supplies, back through the FET drains (cascode amplifiers, common gate) to the bipolar transistor (Q3/Q4) collectors, which appear as a current output. The open-loop transconductance gain is set by R9/R26, 25 ohms, so -25A / V gain. This is high gain, considering that the max output current is 100mA, so the input V is -25V/A * 0.1A = -0.25V The common emitter transistors invert the gain.
Then local feedback is applied to reduce the amplifier gain to x-20. The feedback circuit is on the main board, buffer opamp U12B, R33 (10K) and R25 (200K). The effect of this is to convert the Transconductance amplifier into a voltage amplifier with fixed voltage gain.
I'm not 100% sure why it was built this way, but I suspect that this clever configuration provides a high voltage amplifier with only a couple of transistors and a low voltage opamp, and helps stabilize the circuit for all types of load impedances. Electrometers, current sources, SMUs and other DC instruments use similar configurations, so I think it is useful to understand it.
Clear as mud, right? Hope this helps.
Cheers, Dave