Re: welding machine as a power supply

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Roy J. Tellason"
<rtellason@b...> wrote:

> On Monday 22 March 2004 01:14 pm, Jon Elson wrote:
> > Roy J. Tellason wrote:
> > >On Monday 22 March 2004 12:37 am, Jon Elson wrote:
> > >>ibewgypsie wrote:
> > >>>Probably, the best thing would be 3 smaller individual

transformers. I

> > >>>got one large 2kva 220/24+24=48(dual tap) transformer and it

weighs

> > >>>about 50 pounds. It cost about $200 Local. Problems with one

large

> > >>>transformer is the bridge, you want to use a 35amp 600 volt

cube for

> > >>>simplicity. I have lost 3 of them in a year. When the

capacitors are

> > >>>discharged and you hit the go relay and throws power to the

drives,

> > >>>charges the capcitors (looks like a short for a second) it

fries the

> > >>>cube one time outa a hundred. I tried dual switches that

would charge

> > >>>the capacitor bank half wave but quickly tired of it.
> > >>
> > >>You can connect a couple Ohm resistor between the bridge and

the

> > >>capacitors. Have a time-delay relay short the resistor after

one second.

> > >>This will reduce the power-on surge by a factor of 100 or so.
> > >
> > >Yeah, but since power dissipation in the resistor is current-

squared

> > > times the resistance, putting it on the primary side is

probably a

> > > *much* better idea...
> >
> > Nope. You get no benefit that way, as then you need a resistor

that is

> > N times larger, where N is the winding ratio of the

transformer. The

> > dissipation will be identical on either side of the

transformer. That's one

> > of the corollaries of using a transformer.
>
> Huh? I don't see how you figure that...
>
> This doesn't make any sense to me.

Watts is watts. Power dissipation is measured in watts. It's
simple Ohms law. Take a transformer with a given ratio. Primary
current is a reflection of secondary current divided by the turns
ratio (ignoring transformer losses and winding resistance). Let's
use some simple numbers. Say we want to limit current in the
secondary to 10A. To calculate the resistor we need to know the
applied voltage, min load impedence (close to zero for this
calculation). Given: transformer ratio 10:1. Lets say to get 10A we
work out that we need to drop 10V across the resistor and that say
we have to use a .3 Ohm series resitor in the secondary. If we
place a resistor in the primary we find we need a 30 ohm resistor to
limit it to 1A. (1/10 the secondary) In the first circumstance you
get 30W. In the second you get 30 Watts.....Where you place the
current limit does not matter since the value has to change. Watts
is watts. The numbers above are for a circuit with 100V in 10V out
and a 10A secondary limit need. Obviously the numbers will change
with different voltages and currents but the end result is the same.

The actual size in watts can be a lot less because of the
exponential change in the load impedence over time. The smaller the
cap the faster the change. By calculating the charge time and
selecting the time to bypass the resistor at some point along the
charge curve the average wattage can be determined.