How to create a linear current response from a passive pot circuit?

Question

I am wanting to linearise the current response of a circuit which currently has the resistance change from a linear pot.

I have a current input that I want to test using a test jig/simulator. The limits of the current inputs are 4mA and 20mA. I want to be able to drive the current above and below these limits to ensure appropriate warnings are being triggered. The voltage output is supplied from the control as well as the return reference.

The existing test jig is shown below. A linear 10K pot was used with a 470ohm resistor too.

^{simulate this circuit – Schematic created using CircuitLab}

The current response of this circuit is in orange, while the ideal linear is in blue (point 1 is 0% twist while point 21 is 100% twisted):

This obviously does not have a linear current response to the twisting of the pot knob, only utilising around half/third of the pots range in the desired current range. I have attempted a few different layouts to try and linearise the response slightly, but none were significantly better than simply choosing a 4k pot and 750ohm (E24) resistor.

Was there a resistor layout to linearise this response so it was closer to the ideal?

Answer 1

Use the pot to drive an opamp current regulator.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

You could try something like this:

^{simulate this circuit – Schematic created using CircuitLab}

Of course this wastes more power in the pot, but it will have a more linear relationship between wiper rotation and current through R2. The source impedance of the pot is 125 ohms +/-125 ohms, which is quite a bit less than the 125+250+470 = 845 ohms total resistance.

It would be better to use an active circuit to do this, and that is what I would do, but this will work.