# Question about using a micro-motor and voltages

Discussion in 'Sensors and Actuators' started by NickGuy, Oct 2, 2012.

1. ### NickGuy

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0
Oct 2, 2012
So I'm building a small device that will be powered by a 8.4v power source.

The problem is I want to add in a micro-vibrating motor but the motor I want to use runs off of 0.9v -1.6V. And I would like to run the motor slow. I believe the motor runs at 65ma typically.

Here is the motor I will be using:
http://www.elept.com/4mm-x-8mm-vibration-pager-vibrating-vibrator-motor_p2532.html

Space is extremely limited so I can only use a single resister or diode to drop the voltage/current.

I would appreciate any help that anyone could give me on this matter.

2. ### GreenGiant

842
7
Feb 9, 2012
depending on the final voltage that you want I would say use one of the 1N4000 series diodes (1N4007 is a good one)

3. ### NickGuy

4
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Oct 2, 2012
Well I would need about 6-7 of those in series in order to drop the voltage enough right? (at 1.1v drop)

4. ### GreenGiant

842
7
Feb 9, 2012
nope, you use its forward voltage (1V)

5. ### BobK

7,682
1,689
Jan 5, 2010
Actually more than that. At 60ma the forward voltage is only about 0.7V, so more like 9 or 10 to drop the voltage to 1.6. A single resistor of about 120 Ohms, 1 Watt should work.

Bob

6. ### NickGuy

4
0
Oct 2, 2012
That definitely sounds more like it, but I can't fit that many diodes in the space I have to work with...

So would just a resister work?

At my desired running speed the motor would run at: 35ma/0.8v which would be 28mW.

Could I put a ~2500ohm resister in series with the motor so that the total power getting to the motor would be 28mW?

Or would the voltage still be high when it got to the motor and screw things up?

7. ### BobK

7,682
1,689
Jan 5, 2010
If you want 35ma at 0.8V, all the rest of the voltage goes to the drop in the resistor. Use Ohm's law:

E = I R

(8.4-0.8) = 0.035 R

217 = R

Then calculate the power requirement for the resistor:

P = V I = (8.4-0.8) * 0.035 = 0.266 Watts. So a 1/2 Watt resistor will give you a good safety margin.

Bob

8. ### NickGuy

4
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Oct 2, 2012
Well I did that calculation earlier but for some reason I was thinking that would give me 35ma at 8.4v.