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

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.
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?
Thanks,
G. Jackson

Quoted text below........
No, this is not true. These PWM (Pulse Width Modulation) drives are
essentially
power in = power out + losses. The internal losses in these drives is
in the tens
of Watts, maximum. The voltage dropped in the motor windings at idle is
very
low, around 2 V maximum. With 7 A rated per coil at 2 V, that is 14 * 2
=28 W.
So, the drives will need something like 38 W per axis. 38 x 3 axes is
114 W,
not 1500.