# Potentiometer Resistance ?

Discussion in 'General Electronics Discussion' started by ohm, Feb 5, 2014.

1. ### ohm

7
0
Jan 16, 2014
Hey guys.

I am working on a solar electronics project just for education and fun really, and what I want to do is use a pot to control the speed of a small motor.

The solar cells turn the motor in direct sunlight but I need to regulate it and turn the speed up and down.

My solar cells are wired in parallel and putting out 5.5v and apparently they give 90ma each but I have only seen 130ma on a good day.

I see all these different resistance strengths for pots and I have looked today at it and I am still confused.

I take it I need a linear pot ? but what I am unsure of is the resistance of one given the above voltage and amp rating.

If anyone can help it would be much appreciated.

Many thanks

2. ### duke37

5,364
769
Jan 9, 2011
You have not said what current the motor takes.
You have not said how stable the speed should be.

A potentiometer is normally a small component and will fail if too much power is dissipated by it. It would be better to use the potentiometer to set a small current to the base of a transistor and use the amplified current to drive the motor.

3. ### ohm

7
0
Jan 16, 2014

Its a solar motor 1.5VDC to 12VDC, current: 84mA under load.

What you said went straight over my head, I am just starting out in electronics and am at level 1 lol.

This will better explain what I am wanting to do.

Direct light powers the motor, however too much light makes it spin faster and I want to be able to regulate it so during summer I can turn it up, and down, and the reason for this is basically to get a better understanding of how pots work so I can use them in future projects, I just dont know which value pot to get.

4. ### BobK

7,682
1,686
Jan 5, 2010
Use Ohms law.

If the motor is drawing 84mA at 5.5V then it has an effective resistance of

V = I R
R = V / I = 5.5 / 0.084 = 65.4 Ohms.

A 100 Ohm pot would be therefore cut the voltage to the motor by a factor ofL

65.4 / 165.4 = 0.395

That should be enough to slow it down a bit.

Now consider power. The equation is:

P = I^2 R

The max power dissipated in the pot will be when it is the same resistance as the motor, so 65 Ohms, and the current will be half of the 84mA. So:

P = 0.042 ^ 2 * 65.4= 0.115 W. A typical 1/2 W 100 Ohm pot should do fine.

Bob

5,164
1,081
Dec 18, 2013
You use a shunt regulator for solar panels to limit the current in bright sunlight. I think you will have problems having to keep adjusting the speed of the motor if you don't regulate it. You could then use an emitter follower current amplifier to control the voltage of the motor using the pot. This allows you to use a small pot on the input to the transistor. You might also benefit from some capacitance on the output of the regulator to give the motor a bit of a kick when you switch it on for the first time.

6. ### ohm

7
0
Jan 16, 2014
Awesome guys thanks for the info

7. ### duke37

5,364
769
Jan 9, 2011
This is a suggestion for crude motor speed control. When the slider is at the bottom there will be no voltage on the transistor and so no collector current. When the slider is at the top, there will be 5.5V across the 1k resistance so about 5mA base current. If the transistor has a gain of 100 then it could deliver about half an amp.

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5,164
1,081
Dec 18, 2013
If you need a greater level of adjustment then I would go for this one. At half pot position you will have half the supply current of the max. Although the circuits maximum current will be reduced slightly because of the VE drop.