Potentiometer question.

Thread Starter

mpierich

Joined Nov 26, 2023
2
Hi. I'm a hobbyist so please excuse the ignorance.

I'm trying to control a 12v, 10,000 rpm motor with a 10k pot. The pot tests ok and varies from 0 to 10k.
But when I wire it to a battery and the motor, no current passes through it. At all.

Any suggestions? Thanks.
 

dl324

Joined Mar 30, 2015
16,688
Welcome to AAC!
The pot tests ok and varies from 0 to 10k.
But when I wire it to a battery and the motor, no current passes through it. At all.
It's probably too low for you to measure. Assuming 0 ohms resistance for the motor, you'd have 1.2mA through the pot (at maximum resistance).

How are you trying to measure the current? Voltage across the pot? Or an ammeter?

If you set the resistance of the pot low enough, you probably destroyed it.

If you're trying to vary the speed of the motor, a pot isn't the way to do it. Google PWM.
 

Papabravo

Joined Feb 24, 2006
21,003
Hi. I'm a hobbyist so please excuse the ignorance.

I'm trying to control a 12v, 10,000 rpm motor with a 10k pot. The pot tests ok and varies from 0 to 10k.
But when I wire it to a battery and the motor, no current passes through it. At all.

Any suggestions? Thanks.
Draw a diagram of how you wired it. No current means no pathway for current to flow.
 

Thread Starter

mpierich

Joined Nov 26, 2023
2
Welcome to AAC!
It's probably too low for you to measure. Assuming 0 ohms resistance for the motor, you'd have 1.2mA through the pot (at maximum resistance).

How are you trying to measure the current? Voltage across the pot? Or an ammeter?

If you set the resistance of the pot low enough, you probably destroyed it.

If you're trying to vary the speed of the motor, a pot isn't the way to do it. Google PWM.
Ok, thanks. I'll look into that.
 

MrChips

Joined Oct 2, 2009
30,488
It is very unlikely that you can control the motor with a 10kΩ potentiometer.
The first thing you need to do is to measure the current the motor draws with 12V supply, without any potentiometer.
 

Jon Chandler

Joined Jun 12, 2008
987
How much current does the motor draw? Chances are, the pot is rated for nowhere near that current draw.

Even if the pot could handle the current, the resistance value is way too high.
 

MisterBill2

Joined Jan 23, 2018
17,814
Hi. I'm a hobbyist so please excuse the ignorance.

I'm trying to control a 12v, 10,000 rpm motor with a 10k pot. The pot tests ok and varies from 0 to 10k.
But when I wire it to a battery and the motor, no current passes through it. At all.

Any suggestions? Thanks.
That is not the way to control the speed of a motor. At the higher resistance values it will not pass enough current to run the motor, and as it gets into a resistance range that will allow enough current it will not be able to handle the power dissipated within the resistance element. That problem will not appear when using a SPICE simulation because simulation parts do not burn up.
No need to show a circuit diagram, that potentiometer will not provide speed control with any connection arrangement.
To operate that motor you will need either a variable voltage power source able to deliver the required current for the motor..
We do not have any information about the motor that would help us specify a control scheme,
What we do need to know is the specified voltage and current for running the motor at the claimed 10,000RPM speed. If you bought it through amazon they do not have that information.
 

MrChips

Joined Oct 2, 2009
30,488
We don't known how much current your motor takes to run at 12VDC.
Let us suppose that the motor draws 100mA at a given load. That means that the effective resistance is 12V/0.1A = 120Ω. The power consumed is 12V x 0.1A = 1.2W.

If you want to insert a resistor in series with the motor to control the motor speed, you want to try something like 1000Ω 1W wire-wound variable resistor. A 10kΩ pot is going to burn out the carbon track very quickly when turned up to full speed.
 

MisterBill2

Joined Jan 23, 2018
17,814
understand that the wattage rating for a variable resistor is for the whole device, because it is based on power dissipation in the whole device. At least most of the time. Some are also rated in amps, and that is a different situation.
 

WBahn

Joined Mar 31, 2012
29,865
Hi. I'm a hobbyist so please excuse the ignorance.

I'm trying to control a 12v, 10,000 rpm motor with a 10k pot. The pot tests ok and varies from 0 to 10k.
But when I wire it to a battery and the motor, no current passes through it. At all.

Any suggestions? Thanks.
What is the claim that no current passes through it at all? Is it because the motor doesn't rotate, or are you actually measuring the current in the motor windings?

The basic problem is that you are dropping almost all of the voltage across the pot and hardly any is getting to the motor.

Pots make reasonable adjustable voltage references, but generally make lousy variable power supplies.

You need a completely different approach. There are several options. One is to use the pot to get a reference voltage and then use an H-bridge to drive the motor with that average voltage. This is normally done using some kind of pulse-width modulation so as to minimum power losses in the HBridge transistors.
 

dl324

Joined Mar 30, 2015
16,688
Here's a single comparator PWM. The trade-off for not using more components is that the frequency changes, but you can still control duty cycle.

1701061139298.png
EDIT: In the circuit I breadboarded, I inserted 2 signal diodes above R7 to avoid violating the input voltage range of the comparator. C1 was 10nF, and the frequency of oscillation was 800-1200Hz.

Hysteresis from R4 is easier to calculate when R2-4 are the same value. I reduced R5 to 5.1k.
 
Last edited:

crutschow

Joined Mar 14, 2008
34,047
Welcome to the results of not using Ohm's law.
Understand that basic electrical law, along with the current your motor requires, then you will see why your pot won't work to control the motor.
 
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MisterBill2

Joined Jan 23, 2018
17,814
It is not likely that the TS is a master at constructing electronic circuits, especially the one with the power connections missing. And it is not suitable for a 12 volt application.
 

Tonyr1084

Joined Sep 24, 2015
7,782
At the higher resistance values it will not pass enough current to run the motor, and as it gets into a resistance range that will allow enough current it will not be able to handle the power dissipated within the resistance element.
Was going to say the same thing.
A 10kΩ pot is going to burn out the carbon track very quickly when turned up to full speed.
This is another important point. You'll get virtually no response on the motor until you're at 99.9% of the throw (rotation). When you hit that last tiny portion you'll have massive amounts of current passing through a very tiny carbon ribbon and it will burn out faster than you could ever hope to react. Yes, as stated, you will burn up the pot.
when I wire it to a battery and the motor, no current passes through it. At all.
That's either because you already burned out the pot OR you've wired it wrong. There are other possibilities such as an undersized battery, undersized wiring, which could burn out or you have a dead motor. Those are on the outside of the realm of possibilities. Likely when you connect the motor directly to the battery everything works just fine. That eliminates the questions of whether you have a sufficient battery, correct size wiring and whether the motor works or not. However, we don't know all the testing you may have done. We tend to make assumptions you already know a little about it. Hence, the reason why you're here asking for help. Which we will gladly help.

So how big is the motor?
How much current does it draw?
What is the capacity of your battery in volts and amp hours?
What gauge wiring are you using?
How are you wiring it up? (wire schematic needed)

Give us that and we can help.
 

MisterBill2

Joined Jan 23, 2018
17,814
Was going to say the same thing.
This is another important point. You'll get virtually no response on the motor until you're at 99.9% of the throw (rotation). When you hit that last tiny portion you'll have massive amounts of current passing through a very tiny carbon ribbon and it will burn out faster than you could ever hope to react. Yes, as stated, you will burn up the pot.
That's either because you already burned out the pot OR you've wired it wrong. There are other possibilities such as an undersized battery, undersized wiring, which could burn out or you have a dead motor. Those are on the outside of the realm of possibilities. Likely when you connect the motor directly to the battery everything works just fine. That eliminates the questions of whether you have a sufficient battery, correct size wiring and whether the motor works or not. However, we don't know all the testing you may have done. We tend to make assumptions you already know a little about it. Hence, the reason why you're here asking for help. Which we will gladly help.

So how big is the motor?
How much current does it draw?
What is the capacity of your battery in volts and amp hours?
What gauge wiring are you using?
How are you wiring it up? (wire schematic needed)

Give us that and we can help.
It is also entirely possible that the current is not great enough to provide an indication on the meter. That happens quite a bit. I have a power supply with digital meters that only read amps, tenths of amps, and hundredths of amps. A 5 milliamp draw will not register at all. But it will light an LED connected to the supply.
 

Ya’akov

Joined Jan 27, 2019
8,973
The way that a potentiometer could control a motors speed is by limiting current. It would do this by dropping the voltage which has the effect of dropping the current, as Ohm’s Law says:

\[ \large{\mathsf{I = {V\over R}}} \small{ \space \space \space \text{ where I is current in amps (A), V is voltage in volts, and R is resistance in ohms} \space ( \Omega )}\]
You can see that we are taking advantage of the fact that if we reduce the voltage the amperage must follow in direct proportion. We are doing this by changing R using the pot since the larger we make R the less current will flow. But it doesn’t change the amount of power that is available to the circuit, it just steals some from the load and turns it into heat.

This means the pot has to dissipate any power we want to keep from getting to the motor by heating up with the equivalent power. This is very inefficient and will require a hefty potentiometer. Rather than the fractional Watt rating of your current pot, you will need something in whole Watts.

Additionally, a little math sill show that the resistance value of the pot will have to be quite low. If you use the current version of Ohm’s Law, above, you can get an idea of just how low R would be for a motor. Since the lower resistance gets the current, the pot will have to spend its time operating in a range where its resistance is lower than the motors. This means your current pot is probably about three orders of magnitude too resistive.

As has been recommended, a PWM solution is probably best. If you don’t want to build one yourself, you are in luck—there are many inexpensive PWM motor controllers available on sites like Amazon, AliExpress, and Temu. Just be sure to buy one in the right rage of voltage and power.

 
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