Re: It's more VOLTAGE needed to overcome back emf in the coil

--- In CAD_CAM_EDM_DRO@yahoogroups.com, R Rogers <rogersmach@y...>
wrote:

> I understand it to be when AC is rectified to DC the voltage

increases by a factor of 1.414 and the current/amperage decreases by
a factor of 1.8.

>
> Ron
>

Ron,

You have the factors applied correctly, but remember:

The "1.414 times AC voltage = DC voltage" describes the maximum
voltage that will be output. This is indeed what you must worry about
to keep from blowing up motor drivers, but it is not the whole story.
If you load the power supply to its full rated current the output
voltage will drop down to approximately 1.0 times the AC voltage.
Most of us operate with the supply lightly loaded and will see the DC
voltage ranging from 1.25 to 1.414 times the AC input. If you want to
include the diode voltage drop, it gets applied after the factor. For
example 10 volts AC input, 14.14 volts before rectifier, and 14.14
minus 1.5 (typical drop across full wave bridge lighty loaded) equals
12.64 volts DC output.

The "AC amps divided by 1.8 = DC amps" is for derating based on heat
buildup in the transformer with 100% duty cycle. Most CNC setups have
much less than 100% duty cycle at full power. I prefer to figure out
average DC current drawn with the duty cycle figured in, then apply
the 1.8 factor. The thermal time constant of a transformer is in
terms of many minutes or even hours (for large transformers) so short
bursts are safely integrated into average duty cycle.

Regards,
Steve Stallings
www.PMDX.com