How will the OPA4227 sense the current when it cannot sense up to V+? You'd need a Jfet op-amp there no?
V+ will always be about 4-5V higher than any of the inputs since it's supplied from the tracking pre-regulator's output. I have read the OPA4277 datasheet a dozen times before choosing it. However, I've only tested the PSU with the LM324 because I had half a dozen lying around.
Because of the voltage over the adjuster, the common mode signal on the current sense amplifier will never swing to the opamp's positive rail.
Jfet-opamps, due to their inability to swing close to either V+ or V-, are not a good choice here.
And even if the output is shorted the inputs will be 3.3V above V-.
OK, I am officially confused. I will explain why I made this comment. On the schematic you show the op-amp being powered by "VCC". Through a 10R resistor which might drop say 50mV. Now the next component is the pass transistor, looks like a darlington pair. Obviously to open this transistor up fully we need 1.2V over the rail. That should be one consideration. Of course I have no idea what Vcc actually is and what the max output requirements are. Now assuming we need max voltage on a small load, there will be no voltage drop on the pass transistor and then we hit the current sense resistors. We know the op-amp cannot "sense" that high, it needs a few volts off Vcc at its best. Else it may, probably will, invert. So the question is how sure are we that the pass transistor, that actually controls the output voltage, always, but always (under any settings and any condition, drops at least 2 V - at worst ?
Another thing : in my transistorised PSU design I used a complimentary pair - a MJE15028 bypassed by a MJL1302A. The pair is driven via a 2.2K resistor straight from the op-amp, nothing else, no other transistors or anything. And for current control like you I use a BC337 to short the base of the MJE15028, where the BC328 is driven by the current sense op-amp. I only drop 0.6V at the pass transistors. My op-amps are powered by almost double voltage (two diodes two caps off of the main transformer). But I do not need negative voltages anywhere.
As an improvement, since the PSU should be floating, we could actually place the sense resistors at the V- lead. That way sensing the voltage can be done by any "single supply" op-amp (eg MC33072) without "extra" Vcc and does not require a differential circuit, since we are already referenced to "ground".
I would also suggest placing a strong diode at the output to prevent damage to the PSU or the components (imagine you plug in another PSU or a battery for example). With a power Schottky you'd be sacrificing Iout * 0.5 Watts. As you are sensing output voltage anywhere you like, even right at the load, it is not going to cause regulation issues.
I would also suggest a relay to take the load "off load" when power is lost and at the press of a button, so that you can adjust the voltage as you wish while the load is off, and if you ever have multiple PSUs in the same enclosure/supply, you would not need to switch the whole damn thing off just to take one of the loads "off load".
Finally, a very useful feature which I do not have, but you have MCUs so I presume you can do what you like, is the "slave" mode, where one PSU "follows" the settings of another so for example when you have symmetrical supplies, you only adjust one button.