Re: What size capacitor should I use? resistors....

> I understand the desire to limit the starting inrush current as

well

> as the need for the smallest time constant.
>
> would it not be prudent to charge the cap slowly up to some value,
> and at that point, switch over to the raw voltage directly ?
>
> It would seem a simple electromechanical relay would be a

possibliity

> to eleminate the hum of initial power on.
>
> Dave

You are exactly right dave...And for a example Jon Elson used to have
on his site a schematic of what you just described (not sure if it is
still there). He used a time delay relay to slowly charge the caps...

This method also helps save the bridge rectifier from the huge
inrush...I usally spec the bridge rectifier current rating 50 to 100%
larger than desired current because of this huge inrush...


Wally



--- In CAD_CAM_EDM_DRO@yahoogroups.com, "turbulatordude"
<dave_mucha@y...> wrote:

>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "cnc_4_me" <cnc4me@g...>
> wrote:
> >
> > Attn CNC power supply people...Please DO NOT put a resistor

between

> > your bridge rectifer and your capacitor...Because of the time
> > constant increase (T=RxC)x5, you will render the recovery of
> > capacitor voltage to be to long in time to be of any use to PWM
> > drives...
> >
> > T = time in seconds
> > R = resistance in ohms
> > C = capacitance in farads
> > X 5 = 6 time constants for a full charge
> >
> > example... Power supply has 30,000Uf and you add a .5 ohm

resistor.

> > C = 30,000Uf is 10,000 to the minus 6 farad.
> > R = .5 ohm
> > T = C x R
> > T = .03farad x .5 ohms = .015 seconds for 1 time constant.
> > It takes 5 time constants to charge a capacitor...So..
> > 5 x .015 seconds = .075 seconds for capacitor to recover...
> > .075 seconds = 75mS = 75000uS
> >
> > We are PWMing around 22khz or 22,000 cps..At this frequency each
> > chopped wave can be as small as 1/22,000 = .0000454 seconds...Or..
> > .0454mS....Or 45.4uS
> >
> > With resistor in circuit it takes 75mS to recover and we need a
> fully
> > charged cap in .045mS....Adding a .5 ohm resistor makes the system
> >
> > 75mS/.0454mS = 1670 times to slow.
> >
> > And now you can see why many power supplys use buss bar or very
> heavy
> > wire....
> >
> >
> > Wally
>
> I understand the desire to limit the starting inrush current as

well

> as the need for the smallest time constant.
>
> would it not be prudent to charge the cap slowly up to some value,
> and at that point, switch over to the raw voltage directly ?
>
> It would seem a simple electromechanical relay would be a

possibliity

> to eleminate the hum of initial power on.
>
> Dave