Re: EDM home brew units ?

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Graham Stabler" <eexgs@n...>
wrote:

>
>
> > Langolis offers a couple circuts that use a FET to open the circuit
> > beteen the power supply and the cap.
> >
> > This he says, improves the finish, but he doesn't say it imporves the
> > speed.
> >
>
> When I say a pulsed circuit I refer to one without a capacitor, DC is
> simply applied to the gap via a current limiting resister for the on
> time. The limiting resister and the gap resistance form a potential
> divider, the servo tries to control the gap resistance to create the
> wanted gap voltage.

I have not seen a working non-capacitor EDM circuit. Since the readily
available circuits use capacitors, I am still thinking capacitors.

That would mean your formula is not relavant to the R/C type machines,
but is really for a non-capacitor circuit.

do you have a schematic for such a circuit ?

> > In my tests, a smaller cap delivered faster cutting speeds.
>
> Smaller than what? If you go to pF does it get even faster? ;) With
> an RC circuit (and all others) there is probably a sweet spot, a
> balance between the charging time and the power delivered per
> discharge and the flushing of course. Poor flushing and you spend
> more time going backwards than forwards.

Langolis used 100uF to 400uF (switch selectable) as his spark cap
range. I started with a camera flash capacitor and I droped to below
50 and saw improvements.

I agree that there is a sweet spot, but that has not been offered as a
known formula that I am aware of. I think there is a long list of
things that would effect operatoin, good fluid being in the two three
if not the single biggest item.

>
> > In the above formula, the energy delivered would increase greatly by
> > the isolation circuit. Take a 15 ohm resistor out of the loop and
> >you get 15 times more energy per spark.
>
> Well no, because the above formula is not for a capacitor based
> system, besides the current limit is not in the discharge path.

OK, now I'm confused. if there is no capacitor and the resistor is
not in the discharge path, but it has an effect on the energy, then it
must be in the active circuit. No ?


Dave