How can I decrease the speed of my 12V motor without changing the battery?

[9988776655]

I have a motor (8N.m motor torque) that is attached to a 12V battery. The problem is the motor is way too fast.
I worked put that I only need 1.5N.m of motor torque to move my load. I want the motor to go as slow as possible without stalling.
I bought the expensive motor without doing too many calculations and I can't swap it for a smaller one.

What I want to do is decrease the power going to the motor. However, I am unsure of whether to decrease the voltage or decrease the current.
If I am to decrease the current, I would need to use a resistor right? But I am unsure of what resistor I need.

I can't change the battery to say a 9v because the new battery will only last like 10mins. So I have to use a rechargable 12v battery.

Thanks for the assistance

 

[Hurkyl]

Can't you just add another gear?

 

[9988776655]

Its actually a school project. I am not allowed to use gears.
The motor's operating voltage is 6-12V. I don't know if that has any effect on anything

 

[fleem]

You cannot control the voltage and current separately. They are completely related. For example, if you used a current source to set the current to 812 mA, and the voltage becomes 7.3 V, then if you used a voltage source to set the voltage to 7.3V, the current will be 812mA.

On top of that, a motor with a load has negative resistance (varying with its speed) and even unreal resistance (time delayed because of the intertia), adding to your problems of trying to regulate the speed. That's why motor control circuits must sense the speed--you can't know the speed simply by measuring or controlling the voltage or current.

So what all this means is that you've got problems.

For efficiency's sake, you ALWAYS want a motor to be running at a high speed, then gear it down if necessary. Specialty low-speed motors typically have a large diameter and many poles (take apart a disk-drive motor and you'll see).

But since you have a project to complete let's cross our fingers and try some things:

First, I wouldn't bother with a resistor. That negative resistance in the motor will have a much greater effect when there's a resistor in series with the power supply. The best value you could find for the resistor would still make it so the motor won't start up unless you spin it up by hand, then once its started it'll be going too fast. I think your best bet is reduce the voltage. Find out the voltage you need by using a variable power supply or mounting different numbers of cells in series. Also, remember to use identical cells of identical charge (i.e. new alkalines or fully-charged rechargeables of identical type and identical mAh rating).

BTW, note that a motor's start-up current is often quite high (back to that negative resistance thing). So a power supply needs to be able to handle that. Most batteries should do it (except those whimpy 9-volt batteries might not)

EDIT: One more thing. Static friction adds tremendously to the negative resistance problem. If you are, for example, simply driving fan blades (no static friction, and fortunately a dynamic friction of effectively zero at low speed) then its not nearly as bad and a resistor might actually work--although less efficient.

 

[9988776655]

My motor actually turns an arm. On the tail end of the arm is a load which is catapulted through the air. So The motor only needs a quarter turn of its shaft. If I am to try to vary the voltage, could I just use a potentiometer instead of getting new batteries? I don't need it to be too precise, all I need is to slow the motor down about a quarter of its speed.

 

[phlegmy]

i don't know a whole lot about this but, if your worried that the 9v battery will only last 10 mins, could you connect a couple in PARALELL, 6 should last an hour!

 

[NoTime]

Since you only want to go a quarter turn. You could set up a switch contact on the shaft that cuts off the power after a certain amount of rotation.

 

[9988776655]

Actually, I already have the switch implemented. I already bought the 12v battery and I don't want to buy any more. Is the resistor idea really that bad?
Is there some kind of electrical component that can half the voltage?

 

[NoTime]

You could use one. Probably a small value. Try a few.

You could also shorten the on time of your shaft switch unless it is the initial acceleration that is the problem.

 

[9988776655]

I found a universal voltage regulator:

http://www.dse.com.au/cgi-bin/dse.storefront/471998d9000a804a2740c0a87f9c06df/Product/View/K3592

 

[NoTime]

I found a universal voltage regulator:

http://www.dse.com.au/cgi-bin/dse.storefront/471998d9000a804a2740c0a87f9c06df/Product/View/K3592

I was under the impression that your motor was bigger than the 1.5A limit of the regulator in the link.
If not, I would think 4 or 6 "D" cell batteries might be cheaper and would last a long time given the duty cycle.
Using a pulse width controler tends to be better for motors if stalling is an issue.
Another option is to add a flywheel.
The resistor isn't out of the question either.

As an enginerring project it's your job to evaluate different possibilities and trade offs for a solution.

 

[Paulanddiw]

Actually, I already have the switch implemented. I already bought the 12v battery and I don't want to buy any more. Is the resistor idea really that bad?
Is there some kind of electrical component that can half the voltage?

You could put a 6 volt battery in series, but put it in backward,i.e.,+ to + and - to -

 

[budala]

You could put a 6 volt battery in series, but put it in backward,i.e.,+ to + and - to -

how do you put in backward? isn't it then in parallel? + and + =parallel.

 

[NoTime]

I would say Paulanddiw intended this as series even though the wording could be clearer.

Voltage (potential) is additive.
If you put two 1.5v flashlight batteries in series like -B+-B+ then the voltage is 3v rather than 1.5v. If you hook the two 1.5v batteries like -B++B- then the voltage will be zero. With a 12v and 6v battery the end result will be 6v.


It is not a good idea to hook real batteries up this way.
This is because the 6v battery will be in charge mode.
If the external circuit draws enough current then the 6v battery could explode.

 

[rbj]

can you control the field current in the stator and rotor separately? if so, increase the stator current to slow the motor down. increasing the stator current will increase the back e.m.f. in the rotor if the speed was constant. but the back emf cannot be more than the applied voltage to the rotor and the back emf is proportional to both the stator field (which is proportional to the stator current) as well as to the motor speed. if the back emf is limited (which it is), increasing the stator current must decrease the motor speed.

if you're dealing with a high power DC motor and have the ability to disconnect the stator current (by pulling the plug on it) while the motor is running, be sure to have your will and affairs in order. lotsa excitement.

 

[NoTime]

if you're dealing with a high power DC motor and have the ability to disconnect the stator current (by pulling the plug on it) while the motor is running, be sure to have your will and affairs in order. lotsa excitement.

Ouch! Did that once by accident. Fortunately the circuit breaker shut off before things got too exciting.

 

[circuit man]

TURN THE VOLTAGE DOWN!

sorry about the CAPS, need the emphasis

 

[rbj]

TURN THE VOLTAGE DOWN!

sorry about the CAPS, need the emphasis

so how does the OP do that? reach for the knob on his 12 volt battery that turns down the voltage?

(CAPS give your reply emphasis, but does not make it apt.)