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I have a microcentrifuge which has a DC motor rated 24V, 10 Watt and 7000 rpm.

I have attached one 5W, 500 Ohm potentiometer with it to control the speed and it works fine without heating up.

This microcentrifuge is supposed to be used with a light weight part. But as I need to use it with some large tubes I have 3D printed some parts and attached on the rotor part.

This time with this extra weight my potentiometer becomes very warm within 1 minute.

Is it because of the extra mechanical load or it is something wrong with my potentiometer rating?

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  • \$\begingroup\$ Define hot. 50°C will burn your skin, but not the potentiometer. You can measure and calculate if the 5W rating is exceeded. \$\endgroup\$ – Huisman Mar 18 '19 at 13:37
  • \$\begingroup\$ It is a bad idea to be putting 3D printed parts into a microcentrifuge - even cheap microcentrifuges can produce accelerations of 20,000 x g. Now two grams of liquid feels like 40 kg. \$\endgroup\$ – D Duck Mar 19 '19 at 8:55
  • \$\begingroup\$ @DDuck the one I am using it has the maximum RCF 2,680g \$\endgroup\$ – Rahat Mar 19 '19 at 10:02
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My guess: When you want to use the extra tubes, you are probably changing the potentiometer setting.

If you have 500 ohm and 24 V, nothing happens. If you reduce the value to 50 ohm, though, Joule losses become 10 times as large and at 10 ohm 50 times:

$$ \begin{equation} P(R) = \frac{V^2}{R} \end{equation} $$

If you can, you should measure the potentiometer voltage and current.

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    \$\begingroup\$ Probably "joulean" after James Prescott Joule. SI units named after a person are lowercase when spelled out. Same with "ohm" but I don't know about "joulean". Anyway, not bad. I wish my Deutsch was as good as your English. \$\endgroup\$ – Transistor Mar 18 '19 at 14:25
  • \$\begingroup\$ @Transistor: Right, I've simply got too used to "oo" by writing "bool" or "boolean". I've had to read up about the capitalization of units: All the style manuals recommend to not capitalize units named after a person. Strange, I never observed this difference. \$\endgroup\$ – Frank from Frankfurt Mar 18 '19 at 15:03
  • \$\begingroup\$ See the last sentence of the first paragraph of List of scientists whose names are used as SI units. \$\endgroup\$ – Transistor Mar 18 '19 at 15:55
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Controlling a 10W motor with a 5W pot isn't great in the first place. At certain speeds, you could already be getting close to 5W, and that won't be spread evenly across the whole length of the pot.

The more mechanical load you place on a motor, the more current it draws. More current means more heat in the potentiometer. If you are turning the pot towards one end to increase the speed to compensate for the extra load, then the heat is now being dissipated in an even smaller part of the pot's track.

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  • \$\begingroup\$ Thanks for your reply. The problem is 10 watt potentiometer isn's available from our supplier. In this case do you think aDC Motor Speed Control PWM HHO RC Controller will serve the purpose? like (ebay.co.uk/itm/…) this one? thanks \$\endgroup\$ – Rahat Mar 18 '19 at 15:18
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Theoretically, additional weight should only effect the load during acceleration. However you stated "I have 3D printed some parts and attached on the rotor part." The additional load may be due to the aerodynamic drag created by that modification. Also, the "large tubes" would cause additional drag if they are not entirely enclosed by centrifuge rotor compartments.

Another factor could be the operating speed. If you are operating at a different speed than your initial test speed, the heat dissipated in the pot would be different. Also, 500 ohms is not the optimum value for a 10 watt motor. it should be closer to the effective resistance of the motor, 24V^2 / 10W = 58 ohms.

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    \$\begingroup\$ @Huisman Nice to have someone who is awake helping out. \$\endgroup\$ – Charles Cowie Mar 18 '19 at 14:08
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You could just go buy a bigger wirewound pot. You could buy a DC motor speed control (a few dollars from China). You could make a PWM controller using a 555 and a power MOSFET. If you go the latter way you'll also need a voltage regulator such as LM7812 because the MOSFET gate and the 555 won't like 24V.

Here is a rough idea, diagram was originally on twovolt.com but that site seems to be down so no link.

enter image description here

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  • \$\begingroup\$ Thanks for your reply and help. It makes sense. But the problem is I don't have an Electrical background. I only know these stuff a little bit. So these MOSFET understanding will be difficult for me. I am feeling like a dump student right now. Do you think if I buy these (ebay.co.uk/itm/…) then it will serve the purpose? \$\endgroup\$ – Rahat Mar 18 '19 at 15:22
  • \$\begingroup\$ Looks okay to me. \$\endgroup\$ – Spehro Pefhany Mar 18 '19 at 15:24

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