Potentiometer to control motor speed

Discussion in 'Sensors and Actuators' started by sgibsley, Jun 20, 2014.

1. sgibsley

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Jun 20, 2014
Hi, I've been trying to vary a motor's speed with just a potentiometer but have so far had little luck. What I've tried is putting the full battery voltage across the entire potentiometer, then attaching my motor from the wiper pin to ground (see attached). It seems that no matter what the battery voltage, potentiometer value, or motor that I use are, I get no motion for most of the range of the pot, then a very steep ascent to full motion at the end. I've also tried it with a small incandescent bulb and gotten the same thing. The problem shouldn't be that the battery is maxing out on current draw, since I've tried reasonably large (10k, 50k, 100k) potentiometers (and gotten the same result for all of them).

One time I tried it with a 9V battery and an LED (I know that's not a safe voltage for most LEDs, but I wanted to see what would happen) and I did get the desired effect.

My physics knowledge tells me that voltage across and hence power delivered to the motor should be linear with my linear pot. So what gives? I know I could use an IC solution, but this has gotten me wondering what's going on...

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2. Harald KappModeratorModerator

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Nov 17, 2011
Your potentiometer would have to have a very low resistance and high power rating to allow the motor's current to flow. When you use a "potentiometer" you normally connect it as a rheostat and the rheostat's resistance is comparatively low. Have a look at classic slot racing car controllers.

3. davennModerator

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Sep 5, 2009
yes the voltage will change linearly, but the current is going to be limited by the pot's resistance and the motor will not get the current it needs to operate properly

The best way to control the speed of a motor is with a PWM (Pulse Width Modulation) circuit

eBay has lots of these for small prices ....
here's one example ( I'm not specifically recommending it ... just an example of what is available)
http://www.ebay.com.au/itm/FA804-DC-MOTOR-SPEED-CONTROL-12V-2A-20W-HHO-PWM-CIRCUIT-KIT-/250993391164

cheers
Dave

4. sgibsley

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Jun 20, 2014
To both commenters--thank you for the reply, but if you look at the circuit you'll see that I'm using a parallel, not series combination; in other words, I'm not using the potentiometer as a rheostat/variable resistor but as a voltage splitter. Kirchoff's loop rule and Ohm's law show that the voltage across the motor, not across the potentiometer, varies from 0 to the voltage of the battery, and if the motor is Ohmic then the current across it should vary with the same curve.

5. Harald KappModeratorModerator

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Nov 17, 2011
True only if the current through the load is negligible compared to the current through the pot. Otherwise you have to take into account the motor's resistance in parallel with the lower part of the pot's resistance. Using Kirchhoff's and Ohm's laws you will then see how the voltage breaks down due to the low resistance of the motor.

6. davennModerator

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Sep 5, 2009

Ahhh but the fact that the motor (or light globe) isn't responding as you expected, confirms what Harald and I are explaining to you

The amount of current you are going to get out of the wiper terminal of the pot is going to be very
small after going through a significant portion of resistance in the pot

Dave

7. davennModerator

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Sep 5, 2009
That's mainly because the current required by the LED is very small ... 10 - 20 mA
compared to the motor which may be a couple of 100 mA's

8. sgibsley

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Jun 20, 2014
I'm not sure I follow. If, for example, the pot is turned halfway, then the voltage across the lower resistor in my diagram is one-half of the battery voltage V. This resistor and the motor form a closed loop, so the voltage across the motor must also be V/2, dropping in the opposite direction. Using series & parallel resistance rules to find an equivalent resistance for the entire circuit doesn't seem useful here, since it's the current through the motor and not through the battery that is relevant, and for a 10k pot and motor with internal resistance of ~40 ohms, these currents will be significantly different.
---
Edit: there were some new replies. I'm not sure it's possible for the current flowing through the motor to not satisfy v=ir. If the voltage is known then I thought the current was determined.

9. Harald KappModeratorModerator

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Nov 17, 2011
No, it is not.

Assume a pot with 1kOhm total resistance and a motor with 50 Ohm.
When the pot is at 1/2, the lower resistance of the pot is 500 Ohm, as is the upper resistance. The lower resistance of the pot is in parallel with the motor's 50Ohm, totalling to 500 Ohm || 50 Ohm = 45 Ohm. The resulting divider is 45 Ohm (lower) vs. 500 Ohm (upper) which gives 0.08*Vin, not 0.5*Vin as you expect.

10. KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Say you have a 10k linear pot set to mid-position. The two resistors in your diagram will be about 5k each. The motor will be in parallel with the bottom one.

Say the motor draws 0.5A when connected directly across the battery. (You can measure this with a multimeter.) And let's pretend that the motor draws a steady current and behaves roughly like a resistor; in other words, the current it draws is proportional to the voltage you apply to it, the way it is with a resistor.

So you can calculate the motor's "resistance" as R = V / I where V is the battery voltage and I is 0.5A. Say the battery is 6V. The motor's "resistance" will be 6 / 0.5 which is 12 ohms.

Now put a 12 ohm resistor in parallel with the bottom 5k resistor in your diagram. You now have a circuit with a 5k resistor at the top, and an 11.97 ohm resistor at the bottom. From the voltage divider formula, you can calculate the voltage across the motor as VB = VIN / ((A/B) +1) = 6 / ((5000/11.97) +1) = 6 / 418.7 = 14 mV. So that's why the motor doesn't turn.

Also consider what happens when the pot is near maximum. Say the resistance of the top part of the track is 12 ohms, so the motor sees 3V. (I'm assuming a 6V battery since you didn't give a voltage.) Current will be 0.25A and power dissipation in the tiny section of potentiometer track will be 0.75W. This will burn out the little section of track and create either a complete break in the potentiometer track, or a "dead spot".

For a simple fix, add an emitter follower between the wiper of the pot and the load. That is an NPN transistor, rated for a few watts, with a high gain (use a Darlington if you can tolerate the extra voltage drop). Connect the wiper to the base, connect the collector to the battery positive, and connect the motor between the emitter and battery negative. Use a heatsink on the transistor.

As Dave says, PWM is the proper way to control a brushed DC motor's speed.

11. Harald KappModeratorModerator

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Nov 17, 2011
Right, but you have to take into account that the current has to flow through the potentiometer's resistance and will drop the voltage there according to Ohm's law. This in turn will reduce the voltage to the motor, see above calculation.

12. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Set the potentiometer to have V/2 between the slider and ground, then connect the motor and read the voltage again.

If the voltage changes, you have a problem, and Harald/Dave have explained why.

The current through the motor will vary with voltage, yes, but the voltage between the wiper of the pot and ground will vary non-linearly once your load becomes significant.

In addition to that, the current does not necessarily determine the speed, and the starting current may be many times the running current, so it may not start. In more addition, the behaviour under load will change (it will get worse) because the motor cannot get more current at the same voltage.

davenn likes this.
13. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
three replies in under a minute -- boy is he getting service!

14. davennModerator

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Sep 5, 2009
do some basic Ohms law calcs

initially just consider a motor of 40 Ohms across the 9V battery
9V / 40 Ohms = 0.225 A (225 mA)

now consider a 10k resistor across the 9V battery
9V / 10,000 Ohms = 0.0009 A ( 0.9 mA)

see where this is going ?

now lets replace the 10 k resistor with a 5k resistor and the 40 Ohm motor in series

work out the total current flow in the circuit

I = V/R = 9V / 5040 Ohms = 0.0018 A ( 1.8 mA)

That's the max current that can be flowing through the motor
it really needs 200+ mA to run properly

Dave

15. sgibsley

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Jun 20, 2014
Oh God you're right. Those two words made me instantly realize my dumb assumption; for some reason I was thinking of a voltage divider as a sort of black box that could be abstracted to a "variable battery." That's embarrassing.

Okay, thank you everyone. I'm incredibly relieved that Ohm's law still works---I was seriously worried there for a second.

Really, though, that's embarrassing. Sorry about that.

16. davennModerator

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Sep 5, 2009
Note with the first part ... the motor .... I'm ignoring losses and other thing going on in the motor as Steve hinted at
for the sake of simplicity, I'm just looking at the motor as a constant resistive load
(in reality, it ISNT, not by a long shot)

17. davennModerator

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Sep 5, 2009
no probs, we all live and learn

D