I am going to use the SM42HT47-1684A stepper motor, its datasheet is following: http://p.globalsources.com/IMAGES/PDT/SPEC/001/K1052328001.pdf

In the datasheet the rated voltage of the motor is 2.8 V.

I found a drive chip, which can achieve the required resolution, it is the following: http://www.st.com/web/en/resource/technical/document/datasheet/CD00255075.pdf

The minimum motor supply voltage in the drive chip is 8V. Does this mean, that the motor can't be directly driven from this chip? I have chosen this chip, because I'm going to use microstepping with 1/32 or 1/64 resolution. Is it recommended to look for an other drive with lower motor supply voltage?

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    \$\begingroup\$ Stepper motors are normally driven with a chopping current regulator operating at several-to-many times their steady-state supply voltage rating (but not exceeding their insulation rating, of course). \$\endgroup\$ Commented Mar 21, 2014 at 18:45

1 Answer 1


The 2.8v rating of your motor is just the voltage for which the rated coil resistance (1.65 Ohm) draws the rated current (1.68A).
In other words \$ V = I \times R = 1.65 \times 1.68 = 2.77V \$

Stepper motors like this (low resistance models) are intended for constant current operation rather than constant voltage, so the voltage range of the power supply for L6470 (the driver) is not a limiting factor and it doesn't prevent the use of the particular motor. On the contrary driving the motor with higher voltage increases the torque of the motor as the speed increases.
As long as the driver programmed current is below the max spec of the motor you are fine.

Running a motor at higher voltages leads to a faster rise in the current through the windings when they are turned on, and this, in turn, leads to a higher cutoff speed for the motor and higher torques at speeds above the cutoff.

Please read more http://homepage.cs.uiowa.edu/~jones/step/current.html


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