Slowing down a fan (motor)

[Guest]

I'm not sure if this is the right place to ask, but I'm not sure where to ask, so here goes:

I have a small DVD drive motor. Mabuchi // RF-310T-11400 // D/V 5.9 // TD744720

If I connect it to a 9V battery it works perfectly, spinning relatively quickly.

Using this guide (http://www.ecawa.asn.au/home/jfuller/electronics/resistors.htm) I'm trying to slow it down.

I put a 1k-Ohm Resistor (1/4 watt * 5% tolerance) (from Radioshack, 271-1321) between the power source and the motor and it just doesn't work. I've tried multiple (identical) resistors, from all directions in relation to + and - end of battery, etc. and it just .. won't work.

IF there's anything anyone can tell me to help me that would be fantastic. This is for a group assignment and there's a little bit of time crunch. >.<

Thank you in advance,
Zack

 

[jack_attack]

It won't work, meaning at all, or won't slow it down?

 

[Guest]

no movement

 

[Guest]

http://tinypic.com/r/2py0w1y/6
my components

 

[angelobike]

It doesn't work because the Ohm resistance of the moter is about 4-6 Ohms. Putting an 1 KOhm resistor, you leave nothing but 50mV to the moter.

You must use a resistor with a value of about the same with that of the moter.

 

[Paperdoc]

follow angelobike's lead, and try a resistor with a small value. I'm skeptical about 4 to 6 ohms, though. That may be the reading across the motor's terminals with a resistance meter, but the actual impedance of the motor as it runs is higher. A 9V battery won't keep pushing much current for very long. I suspect the motor may run at about 1 amp or less current draw, which makes its impedance more like 10 Ohms, maybe more. What you want to achieve for starters to is to have the voltage across the motor's terminals when it is running about half (or 2/3) of the 9V battery's value.

Watch for the wattage rating of the resistor, too. For example, if the motor is about 10 ohms and you use a 10 ohm resistor in series with it, the voltage drop across the resistor will be roughly 4.5 volts. Now, since V=IR (Ohm's Law), then I= V/R. Then put that into P=VI, and you have P=V^2/R. At 4.5 volts across a 10 ohm resistor, the power dissipated by the resistor is about 2 watts - MUCH more than 1/4 watt.