Re: [CAD_CAM_EDM_DRO] Re: welding machine as a power supply

On Tuesday 23 March 2004 09:33 am, caudlet wrote:

> --- 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.

Right. Now figuring that the current is ten times on the secondary side what
it'd be on the primary side for the same effect. The formula is
P=I<squared>R, so for putting the resistor on the secondary side it'll have
to handle ten times the current, or 100 times the power dissipation...

This sort of thing when considering transmission line losses, is exactly why
transformers are used to step the voltage way up in the first place.
Currents go down proportionally to the turns ratio, and power losses go down
proportionally to the square of the turns ratio.