Re: Stepper current in partial-step positions

Hi,

Actually both coils carry the motor's rated current in full-step
operation. You can think of it as a single-bit approximation of a
sine and cosine. Microstepping is just a progressively finer
approximation.

Motor heat dissipation is independent of partial step angle (sine
squared + cosine squared = 1). The exception is a common half-step
technique where both windings, then one winding carries current.

Based on power dissipation, one would assume the current can be set
to 1.41 times the unipolar rating when microstepping. Doing this can
have unwanted effects on motor smoothness however. The motor windings
do not share a common flux path; you can prove that by trying to use
the motor as a transformer. This means the iron may saturate at 1.41
times rated current and cause the motor to become non-linear.

I recommend using the motor's unipolar current rating for the
microstepping peak winding current for best all around performance.
High speed torque is not affected by the set current because at high
speeds inductive reactance limits motor current, not the drive.

Power supply current (and watts delivered) is a combination of motor
heating and mechanical power generated. This means power supply
current increases as applied motor load increases.

As an interesting aside, you can calculate the motor's torque at any
speed using just a multimeter. Here's how:

1) Measure power supply voltage.
2) Set multimeter to DC Amps to measure power supply current.
3) Accelerate motor to the speed you want to measure torque at.
4) Measure the power supply current at no motor load.
5) Gradually load the motor while monitoring current.
6) Note the supply current at the moment the motor stalls.
7) Subtract (4) from (6), then multiply with (1).
8) Multiply (7) by 4506, then divide result by (3) full steps/sec.
9) The answer will be in in-oz

in-oz = (4506 * V supply * (I stall - I noload)) / (full-steps/sec)

Mariss

--- In CAD_CAM_EDM_DRO@y..., "Kevin P. Martin" <kpmartin@t...> 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.
>
> 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.
>
> So what fraction of the limit current do drivers supply at the
partial-step
> positions?
> -Kevin Martin