Re: Stepper Drivers

stratton@... wrote:

> From: stratton@...
>
> > From: "Matt Shaver" <mshaver@...>
> >
> > > From: stratton@...
> > > Prototype is going to be with the big chip...
> >
> > What are you going to use between the step and direction signals and
> the
> > H-bridge? Ericsson?
>
> Probably an L297 doing step/direction to phase sequencing and
> chopping.
> Why? Well I made it work with the little L298 bridge...
>
> Next question: power supply sizing. Say I run a parallel-wired
> stepper off a chopper drive set for 6.6 amps per phase with a phase
> resistance of .29 ohms, using a 48 volt nominal supply. The motor is
> consuming around I^R or 12.6 watts.
>
> According to the user's manual for a pacific scientific drive, the
> secondary of a brute force supply should be rated for 1.8 RMS amps
> times the chopper setting. In other words, they want a transformer
> rated for nearly 12 amps at 48 volts, or 756 VA to drive this little
> 12.6 watt motor.
>
> Does this make sense? I would think that the filter caps would absorb
>
> all the chopping-frequency alternating currents, leaving the
> transformer to simply deliver the overall power needs. I can't see
> where the difference between 756 VA and 13 watts is going. Up in
> heat? Where? What gets hot?

Hmm, maybe someone at Pac Sci has big investments in power transformer
manufacturers. :-)
Perhaps this was a worst case calculation assuming there was no
recirculation of the
winding current. That would probably burn up the motors, however, due
to eddy and
hysteresis losses.

There is no 756 VA going anywhere. There will be losses in the
switching transistors,
and in the freewhelling diodes when the chopper cuts that winding off
from the power
source. That could be significant with a 6.6 A, 2 V per phase stepper.
So, you might
double motor static dissipation, or try to compute the losses for the
specific transistors,
etc. used. That 12.6 W is PER PHASE, double it for both phases. So,
you would
most likely be in good shape with a power source capable of about 50
Watts per
motor.

>
> Perhaps I misunderstand the physics of chopper driven motors. In
> fact, the more I think about it I'm sure I do. Mechanical power is
> related to torque and speed, and is related to electrical power.
> Sitting still with a non-chopping driver and consuming 13 watts, the
> motor may be holding against its rated 450 oz/in, but it is not
> moving, hence it is not doing any work. The faster I want to make it
> spin (using my chopper to keep the torque up) the more work it must
> do, so the more electrical power it draws.

This is generally true of servo motors. It is NOT true of stepper
motors. They generally
draw the most power at standstill, and power in slowly declines as speed
increases.
The reason is that when stepping fast, the applied voltage can not ramp
the winding current
up to the setpoint before the phases are switched to the next step.
This is also why torque
declines with increasing speed.

> Perhaps I should estimate the torque/speed I need and convert
> backwards into units of electrical power?

Totally meaningless for steppers.

Jon