0
\$\begingroup\$

I am trying to control the power of a vacuum pump using a 10k potentiometer with 12v as its input source. I've tried wiring it (with the know facing you)with the first lead as ground, second one being the output of the pot and the last one being the 12v input but when I go to measure the voltage of the output it reads 0 volts, is there a different way I need to wire this, I've tried on 2 individual pots so I doubt it is a defect.

\$\endgroup\$
  • \$\begingroup\$ Can you measure the resistance between the first and third terminals? Between the center terminal and either end? \$\endgroup\$ – Elliot Alderson Sep 16 '18 at 20:32
  • 3
    \$\begingroup\$ You can use a pot to control a power supply but not pass power thru it. It will burn out. \$\endgroup\$ – Sunnyskyguy EE75 Sep 16 '18 at 20:32
  • \$\begingroup\$ What do you mean by, "control the power of a vacuum pump"? Are you powering it with the output of your potentiometer? What is the second lead (output) of the pot connected to? Please add a schematic/diagram? \$\endgroup\$ – Daniel Sep 16 '18 at 20:33
  • \$\begingroup\$ @Daniel I have the second pin going to the positive end of the vacuum pump \$\endgroup\$ – Pagaley 12 Sep 16 '18 at 20:38
  • \$\begingroup\$ In that case, refer to @TonyEErocketscientist 's comment, and answers from questions such as this. \$\endgroup\$ – Daniel Sep 16 '18 at 20:43
2
\$\begingroup\$

I am trying to control the power of a vacuum pump using a 10k potentiometer with 12v as its input source.

Let's say, for ease of mathematics, that your vacuum pump is 24 W so it will draw \$ I = \frac {P}{V} = \frac {24}{12} = 2 \ \text A \$ when connected directly to 12 V.

The current through a resistor is given by \$ I = \frac {V}{R} \$ and at 10k you will get \$ I = \frac {12}{10k} = 0.0012\ \text A \$. The motor will do nothing.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Your circuit.

... but when I go to measure the voltage of the output it reads 0 volts ...

The resistance of the motor is much, much lower than that of the pot so it appears as a short circuit. The pot can't supply enough current.

Is there a different way I need to wire this?

Yes. You need a speed controller of some sort.

I've tried on 2 individual pots so I doubt it is a defect.

You may have burnt out part of the track nearest the 12 V pin if you turned them all the way up. Most pots are rated at about 1/8 W or so and are not designed for high current applications. The defect, I'm afraid, is in the use of the pots.

\$\endgroup\$
  • \$\begingroup\$ I have a spare dc motor controller for up to 24V, is it safe to use that as a replacement? \$\endgroup\$ – Pagaley 12 Sep 16 '18 at 20:46
  • 2
    \$\begingroup\$ How could I know? You have provided no details of the motor in your question and no details of your controller in the comment. Ideally you would provide a datasheet link for each and we could point out to you the relevant details that would help answer the question. Thanks for accepting my answer but I recommend that you unaccept for a day or two to encourage other answers which may give you further insights. \$\endgroup\$ – Transistor Sep 16 '18 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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