# Potentiometer Power Handling

I have a circuit with a voltage of about 150 volts but with just about 1.75 mA current draw. Will I be alright with a 1K, 3K or 5K 1/2 watt or a 1 watt potentiometer to adjust and dial in the voltage? This is sometimes nice for a microphone in order to get the desired sound. I would of course rather use a 5 watt potentiometer but I will need to order one.

What kind of power rating do I need for a potentiometer if I run 150V across it?

• Can you draw a circuit for that? Something sounds fishy.. – Trevor_G Mar 24 '17 at 18:23
• I don't see why you can't calculate the power dissipation in the potentiometer the same way as any other resistor. If the current in the potentiometer really is 1.75mA, then I would think a half watt will be fine for 5k. Using I^2 * R, the dissipation would only be 15mW. There could possibly be a shock hazard associated with turning the knob on the poteniometer. But I guess that depends on the details. – mkeith Mar 24 '17 at 18:38
• @mkeith, calculating it that way yes, but if that 5K pot is across 150V it's 4.5W... assuming whatever is supplying it can deliver the 30mA. Like I said.... something isn't right here. – Trevor_G Mar 24 '17 at 18:43
• That is why I said "if the current... really is 1.75 mA". I think the idea is that there is a main divider dissipating most of the power, and the pot is just to trim the voltage. But, with no schematic, I am just guessing. – mkeith Mar 24 '17 at 18:57
• One other piece of information for the OP. There is actually a built-in tool for drawing schematics in questions. You can edit your question by clicking on "edit". After you do that, there is an icon that looks like schematic symbols. You can click on the icon to draw a schematic. – mkeith Mar 26 '17 at 8:15

## 1 Answer

If you put V volts across a potentiometer of R ohms, and draw no current from the wiper, then you will be dissipating $\frac{V^2}{R}$ watts. So with 150V and a 5k pot, you need a 5 watt pot.

If you do draw a current from the wiper, this will increase the current in the top part of the pot, which will increase the track dissipation, ideally needing a higher power pot. A 5 watt 5k pot would have a maximum permitted track current of 31.6mA. When driven from 150v, the end to end current is 150/5k = 30mA, leaving very little in reserve for the wiper output. A 10 watt 5k pot would be able to tolerate a current of 44.7mA.

• Thank you for your answers. So if can locate a 2.5k 5 watt pot can I assume that it will that be acceptable also in the same situation also? – JSlick Mar 24 '17 at 23:30
• I don't know what your situation is. Faced with the inconsistency of your question, I'm guessing just like Trevor and mkeith above. But from your response, I guess that I've guessed more correctly. A 2.5k 5watt pot will tolerate a wiper current of 44.7mA, but an unloaded end to end voltage of only 111v. When we say 'draw a schematic' in comments, we mean it. You may not have the imagination to understand how many different ways the components you mention could be connected, but when faced with your inconsistent word picture, we do. Your current, voltage, resistance and power are not consistent – Neil_UK Mar 25 '17 at 4:29
• If the pot is in circuit by itself, with 150V across it, then the total power dissipation in the pot is at least 150 * 150 / R. This is the formula P = V^/R. So in this scenario, smaller R means higher power dissipation, and 2.5k would dissipate twice as much power as 5k. – mkeith Mar 26 '17 at 8:07