Sorry if I'm being dense, but where is that 2 mA measured? It only has three pins: input, output, and reference. There's no ground pin. The reference presumably draws almost no current, as drawing significant current would affect the measurement. So the current on the input & output must be the same, and the same as what we're feeding into the electrodes. No? Does this just mean you can't use the simple circuit composed entirely of the LM317 and a single resistor between the output pin and the reference?
Sorry my mistake, I was talking without looking at the datasheet.
From the datasheet
>all quiescent operating current is
returned to the output terminal. This imposes the
requirement for a minimum load current. If the load current
is less than this minimum, the output voltage will rise.
I've used an LM317 drawing less than 2mA, so I assumed it was possible (in hindsight I must have used the solution, below).
The solution is very simple however, make a separate parallel circuit that will draw at least 2mA. This can be accomplished just by placing a resistor of appropriate value between the Vout and ground.
