Re: [CAD_CAM_EDM_DRO] Re: LARGE DC MOTOR speed control question

Mariss Freimanis wrote:

> One thing you might want to take into account is the lousy power
> factor on the 120 VAC line with this large of a capacitor. All the
> current drawn would occur very near the top of each line half cycle
> and could easily exceed 100 Amps per pulse.

That's why I suggested one about 1/5 the first value.

> My understanding is electric meters "charge" you for a poor power
> factor load, which in this case would look like a 100 Amp draw
> instead of a 17 Amp one.

Standard home meters do NOT do that. they read real power draw, by
magnetically multiplying the instantaneous voltage times the
instantaneous
current. So, poor power factor, either due to reactive current or
non-linear
loads, do not affect power meters ability to read real power only.

Now, big energy users DO get charged by power factor and peak usage.
Our University is charged for an entire year based on the highest usage
of any 15 minute period within the year. They have a VERY fancy
electric
meter, which makes 2 cassette tapes, which are in a sealed and locked
vault, and are opened every so-and-so period when the tapes are changed,

under the observation of a Univerity official and a power company
official.
One tape is for us, one for the power company. There is a correction
for
power factor, but I don't know exactly how that is billed.

>
> A 10% ripple voltage is more reasonable from a power factor point of
> view. Use this equation for calculating capacitor size for 10% ripple
> at rated current:
>
> C = ( I * 80,000 ) / V
>
> Where C is in microfarads, I is the current you want and V is your
> power supply voltage. For your case the result would be 8,000 uF.
> Use 200 VDC rated capacitors minimum.

So, my 5,000 uF wasn't too far off.

Jon