Stepper Motor Voltage

Thread Starter

smokie

Joined Apr 15, 2009
13
Why are stepper motors rated at such odd voltages? i.e. 3.42v

Why not 5v, can a stepper rated at 3.42v be run at 5v?
 

hgmjr

Joined Jan 28, 2005
9,029
To answer this one we are going to need to consult the datasheet for the part if it can be located. Can you provide a part number and a manufacturer's name?

hgmjr
 

Papabravo

Joined Feb 24, 2006
14,872
Why are stepper motors rated at such odd voltages? i.e. 3.42v

Why not 5v, can a stepper rated at 3.42v be run at 5v?
The answer to this question is that it is not really a rating. It is the result of choosing a particular gage of wire and placing as many turns as possible inside a particular case. The amount of wire has a DC resistance and an inductance. When the stepper is "holding" the "rating" tells you a voltage and a current. Ohms law tells you the DC resistance of a winding. You can operate a stepper motor at any voltage as long as you use a series resistor to drop the difference between your supply and the voltage on the coil. As an additional benefit the L/R time constant goes down so that you can increase the speed. What you pay for that benefit is the heat dissipated in a 100W resistor, or whatever power level is required.

In my youth I ran steppers for ean engraving table rated at three volts from an 8 volt @ 25A and a 40 volt @ 4A supply. Of course I had the magic switching logic to turn the high voltage supply off as the current in the coil rose and to adjust the ducty cycle of the low voltage supply to regulate the phase current.
 

Thread Starter

smokie

Joined Apr 15, 2009
13
I've decided to run the stepper motor at 12v and use the 317T to limit the current to .88 amps with a single resistor.

The formula as I understand it is:

Iout=1.25/R

R=1.25/.88 = 1.42 ohm

1.42 ohm hard to find, use 1.5 ohm

Iout = 1.25/1.5 = .83 amp ok

Is my math ok?
does the resistor have to be rated for .88 amps (.88x12=10.56Watts)?
 

THE_RB

Joined Feb 11, 2008
5,438
The resistor will dissipate E*I or 1.25v * 0.88A or 1.1 watts.
The LM317 will dissipate approx (12-3-4.4)*0.88 = 4.0 watts.

If the motor requires 0.88A for EACH phase then as it has 2 phases those figures will be doubled (as amps rises from 0.88 to 1.76).
 
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