PSU math

>I thought this way, too, when I first bench-tested my first (only, so far)
>bipolar-CHOPPER Gecko-drives (G210). Mariss explained to me why the MAGIC I
>observed: I had about 30 Volts, unregulated but filtered ("smoothed" in
>English-English; We Yanks call that "filtered"), DC feeding two G210's,
>those driving two 3+ amp motors, with two 3+ volt windings, each, i.e., about
>20 watts per motor. With BOTH motors stepping simultaneously, meaning that
>neither was drawing only 1/3 current "at standstill" like the Gecko drives do
>after a second or so pause, the DC from the 30 V supply was less than two
>amps! Yet, both motors had "full rated torque". Mariss pointed out that the
>supply was supplying WATTS, as the chopper-drive still puts-out about 30
>volts with this supply to the motor-windings, but "chopped", to average the
>current down to the 3+ amps rating.

This is all true, but there's a catch.

The reason it works is the way the chopped waveform from the driver
interacts with the inductance of the motor. To keep the example simple,
suppose we have a 30 V supply and a motor rated 3 A at 3 V. The drive will
automatically adjust its pulse width to keep 3 A (average) flowing in each
winding. In this example, this will happen with a driver duty cycle of 10%.
For 10% of the time, the motor inductance "sees" 27 V applied across it (30
V supply minus 3 V resistive loss in the winding), and the current increases
at some rate that depends on inductance. For the other 90% of the time, the
inductance sees -3 V, and the current decreases. The rate of decrease is
only 1/9 as fast as the increase, but it happens for 9 times as long. The
net effect is that the average current remains 3 A, but the instantaneous
current goes above and below that in a sawtooth waveform.

So the power supply does supply 3 A of current, but only 10% of the time,
for an average current of 0.3 A, and a power of 9 W. During that 10% on
time, the voltage stuffs energy into the inductor in the circuit (which
happens to be the motor winding). The other 90% of the time, the current in
the winding is actually supplied by the stored energy in the magnetic field.
The winding has an average current of 3 A, and an average voltage of 3 V,
so it's drawing 9 W. Thus, we've "transformed" 9W from the power supply
into 9 W in the motor, without using a transformer, even though the power
supply is running at 30 V and the motor at 3 V. This is very unlike a
system using dropping resistors, where the supply would have to provide 3A
continuous at 30 V (90 W) and the resistors would have to dissipate most of
that. (There are some losses in a chopper system - switching losses in
transistors, core losses in the motor).

The "catch" is that the analysis above only applies when the motor is
stationary. Once you start turning the motor, you get back EMF developed in
the windings. This requires more voltage across the windings to maintain
the same current, which makes the PWM driver increase its "on" time. And
that draws more average current from the power supply. At some sufficiently
high RPM, the motor will need 30 V across it, the driver will be "on" all
the time, and the power supply will have to supply 3 A per winding. That's
at the maximum full-torque speed. You can run the motor even faster, but
now the current drops below 3 A and the torque goes down.

Thus, with a PWM drive, the current needed from the PSU is low at zero
speed, rises to PSU voltage times motor current at a particular speed, and
then falls above that speed. In comparison, a resistor-type drive demands
the maximum power when the motor is stationary, and current (and torque)
drops with increase in speed.

The peak power out of the PSU depends on how fast you will run the
motors, and how many you will have running at once. It also depends on how
long you'll maintain those peaks - a transformer can withstand a lot of
overload for a short period, but the diode bridge in the power supply may
have less "headroom".

Dave