My existing setup:

  • Rae Corporation gear motor (90vdc, 4.5amps, 120rpm, 150 in/lb torque, Part number 6040042). I can't find a datasheet, but it looks a lot like this
  • Dart Controls motor controller 15DV1A, (datasheet) which takes in AC main power and runs the motor. Pots on the board control max and min rpm as well as implementing IR compensation.
  • the main speedpot shown attached to the 15DV1A relocated to the end of a long cable with connecting it back to the board. I've measured the voltage across this speedpot - it's about 6.5V from Pot-Low to Pot-High, with the Pot-wiper being somewhere in between depending on the pot position. The site says it's a 5k ohm pot.

Here's the question: I would like to control the motor with a raspberry pi - how?

I see a couple of approaches:

#1 Replace the pot:

  • there are variable resistor ICs, but reading around it seems like they would not handle the current that passes through when the resistance is low.

  • I've also read that mosfets can be used to create variable voltages, but they are very sensitive and only operate this way over narrow range. Also, it seems like hooking this up is complicated.

  • digital potentiometer, like this or this. Would this be able to handle the current as a replacement for the pot? If so, this seems like the best approach but how can I wire it so the 6.5v motor control signal is isolated from the RPi?

#2 Control the existing pot:

  • I've seen folks use a stepper motor to control a knob, sort of like this. While I appreciate that approach wouldn't modify the existing arrangement at all, I want something smaller and more direct ideally.

#3 Replace the entire Dart Board

  • I really don't want to do this, since the board provides some nice features. But if this would be cheaper than either of the above two, I'm listening.
  • \$\begingroup\$ Did you read this: dartcontrols.com/wp-content/uploads/2011/04/… \$\endgroup\$
    – user80875
    Dec 15, 2015 at 18:30
  • \$\begingroup\$ I had not seen that - thank you. So it looks like my suggestion of using a mpc4725 DAC + LM124 op-amp would require an additional isolation board between the LM124 outputs and the LOW and WIPER (leaving high open and disconnected) (and hence also probably another power supply to for the high end of the LM124 and the end of the isolation board NOT attached to the pot inputs... sound right? \$\endgroup\$
    – bhoward
    Dec 20, 2015 at 2:29
  • \$\begingroup\$ I think that is right. \$\endgroup\$
    – user80875
    Dec 20, 2015 at 3:38

2 Answers 2


I suspect that injecting a voltage into the wiper wire from a DAC and a buffer will work. What won't work is a digipot of the type you linked - they are only rated for 5.5V max despite the digital inputs being capable of 12V+.

This also needs to be an isolated solution but you may find that there is enough power from the outer two pot wires to power the DAC and op-amp. This just leaves digital control of the DAC via a digital isolation chip like an ADuM3401: -

enter image description here

It will need a little bit of testing first to ensure that adding say 1k across the pot doesn't significantly reduce the 6.5 volts. Then try injecting a dc control signal from a power supply between most negative pot position and wiper. If this all seems to work you should be OK.

  • \$\begingroup\$ Thanks for the suggestion - I'll try adding the resistor and checking for voltage drop, and then injecting maybe a 3v signal into the wiper. \$\endgroup\$
    – bhoward
    Dec 20, 2015 at 2:11
  • \$\begingroup\$ It looks like I could build a circuit to do this by hooking up a mcp4725 to something like a LM124 opamp. The MCP can run at 3.3v or 5v (I can use an arduino to talk to it). I can take advantage of the isolation in the LM124, using the MCP to power it on one side, but it looks like the LM124 can't reach the full voltage on the other side and may need some extra power supply? Is this the right direction? \$\endgroup\$
    – bhoward
    Dec 20, 2015 at 2:19
  • \$\begingroup\$ An opamp doesn't provide isolation but you may get it working sufficiently to prove the concept. That should be your first step. \$\endgroup\$
    – Andy aka
    Dec 20, 2015 at 10:01

Normally the control pots don't run any particularly large currents.

Have you measured the control pot current? Per your numbers it should be all of 1.3 mA (end to end) and the wiper current in most controls is virtually none (since a large wiper current makes the voltage on the wiper non-proportional to position.)

  • \$\begingroup\$ I have not measured the current across the pot, in part because my multimeter does not read current directly (only voltage and resistance). As I mentioned, the voltage is a constant 6.5v, but the resistance can vary - wouldn't that mean that the current varies as well? How would I go about measuring the current? \$\endgroup\$
    – bhoward
    Dec 17, 2015 at 19:07
  • \$\begingroup\$ The resistance across the pot is quite utterly fixed. The only change is where the wiper is on the pot. The reason a pot is called a pot is that it's short for "potentiometer" meaning that when the ends of the pot are connected to a voltage and ground, the wiper is of variable potential (old-speak for voltage), since it makes the pot into a voltage divider which varies as the wiper moves - how true that is is inversely proportional to how much current is drawn from the wiper, (no current = very true) but that's normally very little when set up this way, to keep it accurate. \$\endgroup\$
    – Ecnerwal
    Dec 17, 2015 at 19:50

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