Re: [CAD_CAM_EDM_DRO] Can someone please explain this.......

Greg Jackson wrote:

>Jon,
>Ok, I get it on the power thing: the motors are sucking up power as VI with
>the voltage drop being the small voltage drop of the motor, not the 60 some
>volts of the DC bus. I also get the fact that the drive, because it's an
>on/off device, does not represent the remaining power discrepancy. In
>other words, just because the bus is 60 VDC and the motor delta V is 2.5
>volts, the drive is not heating up like 57.5 volts would. With PWM the
>drops do not have to add up. Never the less, 7 amps is 7 amps. This
>represents a certain number of electrons that are going through the motor.
>If those electrons come through the transformer then its amperage is
>overloading.
>

No. The reason is TIME! The switching transistor is only on a SMALL
fraction of the time.
Ignoring edge effects, if the supply is 60 V, and the idle drop in the
winding is 2 V, then the
transistor should be on approximately 2/60 = 3.33% of the time. So, it
will be drawing 7 A
from the power supply, in pulses, only 3.33% of the time. The AVERAGE
current draw
will be 233 mA (per winding) or 466 mA for both windings (when both
windings are at
full current, such as in a full-step drive). If the current draw was
still 7 A, and the power
supply was 60 V, it would be drawing 420 W, not the 28 W it actually
draws. And, the
driver would HAVE to, by the laws of physics, dissipate the difference.

>It seems possible that, with some inductance, the current through the motors
>does not have to add up to the current out of the bus. If the inductance of
>the motor keeps things going in circles through free wheeling diodes, then
>the electrons can just go in circles between the drive and the motor with a
>bit of extra current tossed in every time the PWM circuit is open. This is
>the only way I can figure that the electrons out of the DC bus are less than
>the electrons flowing through the motor. Is this what is going on?
>
>

Yes, absolutely. The motor inductance is the entire key to how this works.
Making the current recirculate until it needs to be bumped up again allows
a "free ride" by recycling the current.

Jon