0
\$\begingroup\$

I've got a 3V DC toy motor that I'd like to convert to variable speed using a PWM circuit. I found an $5 PWM controller that would work. This expects a 100 K\$\Omega\$ potentiometer as its speed input.

I want to set the speed using a mechanical trigger (like on a variable-speed drill), but that means I have an input that travels only 5-10 mm. The potentiometers I've found are rotary (270° travel) or linear/slider (20 mm travel). If I were to use these, I'd only be able to use part of the 0-100 k\$\Omega\$ range.

How should I approach this?

  • create a mechanical system to convert the trigger travel into the full range of the pot,
  • is there a way to get the full rpm range out of the PWM circuit using only part of the potmeter range?
  • or should I build a custom PWM circuit instead?
\$\endgroup\$
12
  • 2
    \$\begingroup\$ What range does your PWM controller expect? 0-100k? then buy a 0-200k and use only half of it. \$\endgroup\$
    – PlasmaHH
    Dec 6, 2016 at 10:28
  • \$\begingroup\$ If you want "variable speed" then you need a "once per rev" sensor. "I have an input that travels only 5-10 mm" - I don't understand the relevance given you want "variable speed". Your question is confusing. \$\endgroup\$
    – Andy aka
    Dec 6, 2016 at 10:33
  • 1
    \$\begingroup\$ @Andyaka: I think he means that he wants a mechanical trigger with 5-10mm travel that he uses to control the PWM with his fingers. \$\endgroup\$
    – PlasmaHH
    Dec 6, 2016 at 10:35
  • \$\begingroup\$ @PlasmaHH: correct. \$\endgroup\$
    – Hobbes
    Dec 6, 2016 at 10:36
  • 1
    \$\begingroup\$ Actually you could do what they put on Wha-Wha pedals, its a rotary pot on the inside with a cog, then a rack with teeth that turns the cog. Think of it like a rack and pinion style. Actually just google rack and pinion potentiometer and you'll see what I mean. \$\endgroup\$
    – crowie
    Dec 6, 2016 at 14:53

1 Answer 1

1
\$\begingroup\$

I would use a slider something like this; Slider potentiometer. I think it would be easier to make a linear mechanism than a rotary one.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.