LED Power Board
Takes in +15V and +48V and uses that to provide eight very stable ~20mA constant current sources. The exact current output doesn't matter, just its long term stability.
We apply ~48V to a string of LEDs, including an indicator LED. Included in that string is a MOSFET operated in its linear region, and finally a high precision, temperature stable resistor, of approximately 100 ohms.
We apply to the inverting terminal of an op amp, the voltage from the top of the resistor, which is equal to the resistance (~100 ohm) times the current, which we hope to be 20mA. That equals 2.0V. The non-inverting terminal of the op amp is fed 2.0V from a precision voltage reference. The output of the op amp is applied to the gate of the MOSFET. This produces a feedback loop which will hold the voltage across the resistor at 2.0V, guaranteeing that 20mA does indeed flow through it.
There's no expectation of high frequency noise being generated by the op amp, but as a matter of caution, we place a small decoupling capacitor across the power rails on the op amp. Additionally, a cap is placed across the string of LEDs. In simulation it had a significant beneficial effect on clearing up high frequency current noise in the string.
Finally, and with significant trepidation, we pull off the feedback voltage, and run it through a 47kOhm resistor, and into a digital input on the MCP23017, so that we can detect string failures. (If an open circuit occurs in the LED string, the feedback voltage will drop to 0V, whereas normally it will sit at 2V, well within the logical 1 range of the MCP23017. Unfortunately, logic inputs have a tiny amount of leakage. This is specified as "max ±1uA". Probably the value is significantly less, but the amount, and any variation in it with time & temperature, represent drift in our constant current source.
The next rev of the board replaces the MCP23017 with a microcontroller, and instead of feeding the feedback nets to logic inputs, they're going to ADC inputs. Those should be higher impedance, more of the time, so we'll have less impact on the current control loop.
That rev will also generate the op amp positive supply voltage on-board, instead of needing it externally produced.