> From:         PicDude[SMTP:picdude2@NARWANI.ORG]
> Sent:         Monday, May 31, 2004 1:20 PM
> To:   PICLIST@MITVMA.MIT.EDU
> Subject:      [EE:]  Choosing stepper drive voltage

> Hi,

> I'm designing a stepper motor driver based on a PIC, but trying to figure out
> the best drive voltage to use for the steppers.  My goal is to build a PS
> that's will be optimized for near-peak performance, yet not cause any damage,
> and trying to get a handle on if for example 10x vs 20x the motor's static
> voltage is going to make much of a difference.

> The motor is bipolar, with a rating of .23 ohms and 5.6 A in bipolar parallel
> mode.  This gives a static voltage of 1.288 V.  The driver circuit will be a
> chopper-type.

> I've seen recommendations of using anywhere from 4x to 25x the motor's static
> voltage, but trying to understand where these magic numbers come from.  Also,
> this motor has a label with "Vs(dc): 35V" printed on it.  Is this some
> indication of what voltage I should use?

> I expect the *best* voltage would be based partially on the frequency of the
> chopper pulses, and the rpm of the motor, but what else?  Is there a more
> scientific way to choose my ideal voltage w/o just guessing?

> Thanks,
> -Neil.

When the motor is stationary, the holding torque is determined by the current in
each phase. As voltage is switched into each motor phase, the current will have
to build up with a time constant of L / R (L = inductance of the motor winding, R =
total series resistance). The larger R, the faster the current will build and the
faster the motor can step. The smaller R, the more current is available to produce
torque. The ideal is to use a current source set for the maximum current which
each phase can handle. During switching, the voltage compliance of the source
will allow a higher voltage on the coil, speeding up the transition to full current.
Once rated current is achieved, the voltage will drop to I * R. (Of course, the
excess voltage is now dropped in the current source pass transistor.) I have
seen series resistors used in place of the current sources. The great amount
of heat produced in these resistors would have been dumped in transistors
instead, so this isn't that bad.

The most efficient way to control dissipation is to use a bilevel power supply
which can be switched between a low voltage for holding current and a higher
running voltage. If the voltage source is a switcher, this could be done without
a dissipation penalty.

John Power

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