I had a pair of old Braun Oral B electric toothbrushes lying around that I decided to gut. Among other things I found a DC motor I was hoping to use with my arduino. But from what I've read it's possible to damage the arduino with currents bigger than 20 mA, DC motors being mentioned as having larger currents.
So I decided to measure the current of my salvaged DC motor. I hooked the motor up to a 3V and 1.5V battery and my multimeter in series. What I got was 0.5 A. For me this seems good I just want to know if I did it right.
One way you can measure current is by putting your meter in series with the device. It'll work well with a motor of the size you are talking about. If you want to measure voltage, you put the meter across the device you are measuring.
FYI:
The equation, V=I*R will get you all kinds information of things needed in electrical circuits and you should know it well. Otherwise, everything is just going to be a mystery.
Keep in mind when measuring motor current that it will vary depending on the mechanical load on the motor. Unless you know the max torque value that your application will require it's hard to state what worst case current draw to design for. Locked rotor current draw is one worst case specification one might be interested in if designing a drive circuit for the motor. Probably good manufacture's specification data sheet would be the best aid.
From what you've said, yes, that was a good measurement of the current. Was that 0.5A with the motor running free or with a mechanical load? You should always allow a little extra in your design, so maybe aim for a 1A driver circuit.
Actually I put one load of 1.5V and one load of 3.0V and got about 0.5A of current on both.
This was without any mechanical load what so ever. I realize now that I will have to measure again with a known load and perhaps give a litte extra room for mechanical overload.
One more question, lets say i connect a 9V battery to the DC motor, could I damage it in some way or is it just going to run faster with a higher torque. As if more power = more power?
Actually I put one load of 1.5V and one load of 3.0V and got about 0.5A of current on both.
If you got the same current from both it looks like the 0.5A was being defined by the internal resistance of the battery rather than the motor itself. Therefore, your motor could potentially draw more. Knowing the way stuff is engineered down to a price it could be that the designers were relaying on this limitation to stop the motor melting. Therefore you might have to look at some other form of current limiting if you switch to a more capable supply. Like a series resistor or constant current drive. It depends on what you are doing with it.
The toothbrush in question uses a rechargeable Ni-Cad battery that is able to supply more current than an ordinary dry cell, I have just pulled an old one apart to have a look and there appears to be 2 cells in one package giving about 2.4 volts (not sure because it is discharged).
Remember also that the initial current drawn on power up will be much higher than its free running current.
I've done some additional measuring and have the following data:
0.5A @ 3V with original load but the motor fails to rotate
1.02A @ 9V with original gear load
0.7A @ 9V without load
3V comes from two 1.5V C type batteries in series and 9V from a single 9V battery.
@brum
Could it be our tooth brushes differ somewhat since my motor wont even spin when in the original geared load? Or is it so that a Ni-Cad battery can provide more current?