Re: Convert an AC to a DC Welder


"GregS" <zekfrivo@xxxxxxxxxxxxxxxx> wrote in message
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In article <fnvmkj$r24$1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
zekfrivo@xxxxxxxxxxxxxxxx (GregS) wrote:
In article <%DIoj.64$Jh5.47@xxxxxxxxxxxxxxxxxxxx>, "Arfa Daily"
<arfa.daily@xxxxxxxxxxxx> wrote:

"GregS" <zekfrivo@xxxxxxxxxxxxxxxx> wrote in message
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In article <fnsiav$5fg$2@xxxxxxxxxxxxxxxxxxxxxxxxx>,
zekfrivo@xxxxxxxxxxxxxxxx (GregS) wrote:
In article <fnrpah$ab5$1@xxxxxxxx>, someguy@xxxxxxxxxxx wrote:

On Wed, 30 Jan 2008 09:34:24 GMT, "Arfa Daily"
<arfa.daily@xxxxxxxxxxxx> wrote:


<someguy@xxxxxxxxxxx> wrote in message news:fnpbr9$319$1@xxxxxxxxxxx

I recently read that an AC welder can be converted to a DC welder,
and
all thats needed are either two or four very high amp diodes. The
welder puts out up to 240amps. I never measured the voltage, but I
think it's around 25 to 30. (I'll have to check). Two diodes will
make a half wave rectifier, but if my memory is right, I'll only
get
half the voltage, whereas using 4 diodes as a bridge rect. I will
get
the full voltage output.

Now, my question is this. How do I figure the amperage of the
diodes
that I would need (for both methods), so I have at least 250amp
capacity. I'd guess 300 would be better to cope with heating. Of
all
the years I puttered with electronics, I was never good at math.
My
guess is that each diode should be 150A to achieve 300A. But I
might
be wrong.
Can someone please help.

Also, where could I get something like this?

Thanks


I made a converter using a bridge made up of diodes of 1/2 the required
amperage,
and a series coil made out of a 2KW variac core with a cut going all
the
way
through the torroid. I used parallel winding for the output through the
indictor.

greg

Before I did the mod, I just asked around like you did. The slotted
inductor makes
the current more constant. I don't imagine it would work well without
it.
It seems like you have the same basic figuring as I in making a bridge.
If a bridge is outputting 150 amps, then each diode average current is
75
amps.

greg


If what I learnt nearly 40 years ago is correct, then that isn't true.
With
a single untapped winding, at any given instant, one diode from the upper
arm of the bridge is conducting, and one from the lower. When the phase
at
the winding ends reverses, during the following half cycle, the other two
diodes will conuct instead. The two diodes are in series in the circuit,
not
parallel, so it's winding end - diode - load - diode - other winding end.
Thus any current in the load will be the same current as in either of the
diodes. If there is 150 amps in the load, then there will be 150 amps in
both the diodes, not 75 amps in each.

Arfa


If the input to the bridge is AC each set of two diodes conduct for 1/2
cycle.
On the other half they are NOT conducting and cooling down.
If in that 1/2 cycle two diodes carry 100 amps, then the average
of 8.4 msec. zero amps, and 8.4 msecs 100 amps, I calculate 50.

How else do you do it ?


On an AC line driven Xmas light st, the LED is on for 50% of the time
If the peak current is 30 ma. then the average is 15 ma. !! ??

greg

OK Greg, I see what you're saying, but I think that really, you are playing
with words and math here, rather than exploring a concept that's relevant to
speccing the diodes. Average diode current is not a good way to be choosing
the device, because it relies on the limiting factor being power related,
rather than a possible physical limitation of the construction. Any diode
that's going to be correctly rated for this job, must be capable of handling
the peak instantaneous current. How long it's passing that current for,
followed by how long it's then got as a 'rest' period, is neither here nor
there. So, if we were discussing a power rating, rather than a current
rating, then your reasoning would have some validity.

If, in your christmas tree light example, the LEDs were specced for a max
forward current of say 20mA, then by your reasoning, with an average current
through the string of just 15mA, we should be OK. I would venture to suggest
that this would not in fact be the case, as the peak current of 30mA through
the string would over-run the diodes by some 50% at the peak of the cycle.
If the reason for the 20mA max figure was one of power dissipation in the
LED die, you might get away with a15mA average current figure, but if the
limitation is on say the maximum current handling capability of the
connections to the chip, then the likely outcome would be a blown LED. I
think we're probably talking the difference between a guaranteed reliable
setup, and a 'getting away with it' setup.

Arfa


.

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