Re: PSU math

Wow Dave,

great explanation.

Very simple and it puts all the pieces in order.

Dave




--- In CAD_CAM_EDM_DRO@yahoogroups.com, Dave Martindale
<dave.martindale@s...> wrote:

>
> >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