Re: Power Supply for 4-axis CNC stepper driver

Dave,

The function of the chopper is to limit the Current, not the Power,
flowing into the motor. When it turns on it will indeed apply the
full supply voltage to the winding until the current rises to the set
point of the chopper. This is not always going to result in a PWM
duty cycle that would produce the equivalent of the rated current
times the rated resistance. The increased power comes from the fact
that if the motor is moving, it has a higher back EMF and it will
take longer for the applied voltage to cause the current to increase
to the chopping setpoint during each cycle. In theory if the motor
has enough back EMF (or winding inductance, though that is another
issue) the chopper may never switch the current off. The current and
resistance ratings of the motor only describe the power that is lost
as heat due to the winding resistance. The mechanical output power
has to come from somewhere. Many people describe stepper motors as
constant power devices and explain that as mechanical power increased
the resistive losses decrease. This is only true if the driver is not
able to force the rated current into the motor despite the fact that
it is turning and has back EMF. With the 25x overvoltage factor
favored by many on this list, the driver will do a fairly good job of
forcing full rated current through the motor windings at reasonable
motor speeds.

Regards,
Steve Stallings
www.PMDX.com

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "turbulatordude"
<dave_mucha@y...> wrote:

> > BUT, it you did extract full
> > mechanical power from the motor, you would find that it would be
> > drawing the rated phase currents at near the power supply

voltage,

> > not at the voltage implied by resistance times current. This

would be

> > due to the back EMF and the inductance effects of the motor

windings.

> > Remember that power is RPM times torque, so a stepper motor at

rest

> > is producting no mechanical power. To get full mechanical power

you

> > must be spinning the motor fast and loading it at high torque,
>
> But the whole idea of a chopper is that the motor gets slices of
> power, PWM and all that.
>
> so if a motor is 2.4 volts and 6 amps, then powered by a chopper, it
> will get 60 volts at a pulse. as soon at it gets it's 14.4 watts,

the

> chopper cuts back or chops the power. this is repeated constantly.
>
> Even if you double the ampacity (I just like that word) of the power
> supply, the motor can only take so much. all the rest is wasted.

Not

> debating if some magical 37.5296 % is used because no one ever runs
> all 4 axes at full, but at some much lower value. Assume full

power

> for this discussion.
>
> Double it for the second phase or double it for emf or double it for
> iron losses or whatever, the total required current appears to be

much

> lower than what might otherwise be supplied.
>
> Even if we consider that each pulse will deliver 50% more current

than

> the motor rating, that is still only 10% of the amps regardless of
> voltage being supplied.
>
> What I'm missing is why if we run a 2.4 volt motor at 12 volts, we

use

> the nameplate 6 amps at 12 volts, or if we run it at 60 volts, why

do

> we need to supply so much more current ?
>
> The limitation is in the motor. The chopper will only ever deliver
> 2.4 volts at 6 amps. If the 60 volts is delivered, then 6 amps is
> reached 10 times faster and the chopper does it's thing and only has
> to PWM 1/10 the time (or some such) so, the average current is equal
> to the motor rating of 2.4 volts and 6 amps.
>
> If you consider that each phase is separate and at speed, the back

emf

> on one phase is really power for the other phase, the power supply
> does not need to be oversized.
>
> I still think I'm missing something in all this, so I could be dead

on

> my chance or dead wrong and not know it. However, my use of power
> supplies has imperically be working. (namepalate V x nameplate A X

2

> phases X # of motors)
>
> Dave