Re: [CAD_CAM_EDM_DRO] Stepper current in partial-step positions

"Kevin P. Martin" wrote:

> This is partly a question to Mariss and Dan Mauch... It concerns how much
> current stepper drivers apply to the motor windings during partial steps.
> This started with a conversation in the yeagerautomation group, where a person
> mentioned running 3 Camtronics 5A drivers in half-step mode off an 8A power
> supply, and the question arose as to what was the max current draw on the power
> supply, 15A, 21.2A, or 30A.
>
> When the motor is at a full-step position, of course, only one winding is
> energized, and it should be getting the full rated current of the motor (Imax).
> This current causes a power dissipation within the motor of Imax*Rwinding, and
> it is the motor's ongoing ability to dissipate this power (heat) which
> determines the motor's max current rating in the first place.
>
> Suppose instead the motor is at a half-step position. In this case, both coils
> are getting the same current input, producing a magnetic field at the half-step
> orientation.
>
> The question is: How much current should each coil get?
>
> If the magnetic fields of the two windings just add up as vectors, then to get
> the same net field (and thus the same torque) as at the full-step positions,
> each coil should be getting 0.707*Imax (0.707 being approx. the square root of
> 0.5).
> However, if the driver does this, each winding is dissipating
> 0.707*Imax*Rwinding, for a total power (heat) production of 1.414*Imax*Rwinding,
> which exceeds the motor's rated ability to shed heat.

Not quite right! Power is I ^ 2 * R (I squared times R). Note that .707 ^ 2 is
.5 ! So, it does NOT overheat the motor. Knowledge of RMS relationships would
tell you that right away, though. So, 2 windings at .707 A each will dissipate the

same power as one winding at 1 amp, assuming the resistance is linear. Actually,
depending on the way the motor is constructed, the 2 * .707 may dissipate just
a microscopic hair less, as the 2 windings will probably run a few degrees cooler
than one winding taking full current. The cooler windings have less resistance.

If you don't understand the squared bit, it is because power is Volts x Amps, and
if you cut the current to .707 A, the voltage dropped by the winding will reduce
to .707 times what is was at full current. So, both voltage and current drop
by the same amount, and the power is reduced by .707 times .707.

>
> To avoid overheating the motor, the total current in both windings should never
> exceed Imax, so in the half-step positions, each coil would get 0.5*Imax, and
> the half-step positions would only have 70% of the holding torque that the
> full-step positions do.
>
> A similar analysis can be made for the 1/10th-step positions in a 10-step
> microstepper driver.

All based on the wrong calculation of power, therefore, wrong.

Jon