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Would like to confirm current rating of a stepper motor

I took of it off Epson printer/scanner (CX5200). It's rated at 42VDC (which seems really high) and 26 Ohm. I used formula I=V/R and got 1.6 Amps, which seems awfully big current for this tiny bipolar stepper motor (it moves scanner head). Am I correct with calculations or missing something?

I could be that this motor uses a higer drive voltage to achieve the needed torque at the required speed. When the windings are first turned on, the current is limited by the winding inductance, and will increase until it reaches its steady state value. The time required for the current to rise to a desired level will depend on the drive voltage and the L/R time constant of the winding.

This is just my speculation in this case, but stepper motor are sometimes driven that way. If you do drive with that high a voltage, make sure your drive pulses are short enough that the current can rise to a level that would burn out the windings.

There is no harm in driving with a lower voltage, but you may not attain the torque you need in your application, especially at higher speed. Torque will tend to drop off as drive speed is increased, as eventually you will reach a point where the drive pulses are too short to allow the current in the winding to rise to the level needed to achieve the desired torque.
It doesn't seem rate. With the figures you give, the windings would dissipate 67 watts. Are the markings clear? Could it read 4.2V instead?

If there are any part numbers visible on the motor, try Googling them to see if you can get any specs. If only a manufacturer is listed, search their site to see if you can find the motor or something similar to get an idea of whether the "42 Volt" rating makes any sense.
bratan (author)  LargeMouthBass5 years ago
I got number from Repair Manual PDF for this scanner. It says Drive Voltage: +42VDC (maybe it means driver chip voltage?). I tried googling motor p/n but only few results showed up none of them show voltage, just Ohms value.