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Convert an AC to a DC Welder

G

GregS

Jan 1, 1970
0
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
 
J

James Sweet

Jan 1, 1970
0
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

The average current is less than the peak current, but when talking current
ratings of semiconductors, it's the peak current that matters. 100A
rectifier needs 100A diodes, because they still have to carry 100A,
regardless of how much time they spend actually doing it. It isn't just a
matter of heating.
 
G

GregS

Jan 1, 1970
0
Yes and Yes... but diode capacity must be calculated base on PEAK
current, not the average current.

http://www.answers.com/topic/diode-bridge?cat=3Dtechnology

This might help explain as well. Each diode is carrying the entire
current, not half the current.

FOR HALF THE TIME.

I have said nothing of peak current. A diode has an average
rating, a peak rating, a voltage rating, etc. EVERYTHING must
be selected as needed. There should also be a plot of current vs time
allowed for peak current. I can't find that loking at a 1N4004. A 30 amp
non repetative peak and 1 amp average, but average IS average.

greg
 
G

GregS

Jan 1, 1970
0
The average current is less than the peak current, but when talking current
ratings of semiconductors, it's the peak current that matters. 100A
rectifier needs 100A diodes, because they still have to carry 100A,
regardless of how much time they spend actually doing it. It isn't just a
matter of heating.

When you say you want a 3 amp bridge. You get a 3 amp bridge.
Nothing was said about peak current. Very rarely do I worry about peak current.
In most circuits the current is limited by other resistances or Z's.
OK the 3 amp bride has a 200 amp peak.

greg
 
D

default

Jan 1, 1970
0
Yes and Yes... but diode capacity must be calculated base on PEAK
current, not the average current.

http://www.answers.com/topic/diode-bridge?cat=technology

This might help explain as well. Each diode is carrying the entire
current, not half the current.

Look at some diode datasheets. When they say a one amp diode that is
one amp AVERAGE current - or having one amp of DC flow 24/7. If you
have a power supply that puts out one amp of DC at the output
terminals - the diodes are seeing peaks of 6 X that depending on type
of filtering - a capacitive input filter sucks a lot of current only
when the diode voltage equals the charge voltage on the capacitor and
then the cap looks more like a short circuit.

Now, how does that relate to welders? I don't know. My feeling is
that the current output on the knob on the front is only a guide to
give you some relative reference.

Realistically - My arc welder puts out about 50 volts on the highest
setting (225 amps) The primary circuit is fused at the recommended 30
amps at 240 volts or 7,200 watts. 50 volts at 225 amps is more like
11,250 watts - and if you keep the output electrode shorted for more
than 5 seconds it pops the breaker. So I see the 225 amps on the
front but assume that must be the short circuit current or just when
you strike the arc.

My feeling is a 4 diode bridge with 100 amp diodes will likely work
well enough. Welding 1/8" steel I've never gone over the 120 amp
setting and 75 is more normal.
--
 
J

James Sweet

Jan 1, 1970
0
Realistically - My arc welder puts out about 50 volts on the highest
setting (225 amps) The primary circuit is fused at the recommended 30
amps at 240 volts or 7,200 watts. 50 volts at 225 amps is more like
11,250 watts - and if you keep the output electrode shorted for more
than 5 seconds it pops the breaker. So I see the 225 amps on the
front but assume that must be the short circuit current or just when
you strike the arc.


Welders have a high leakage transformer in them, it works like a ballast for
a discharge lamp. When you load down the output, the voltage drops
dramatically and the current remains at more or less the output on the dial.
In a nutshell, it's a constant-current source, not the constant-voltage
source you get from the wall.
 
D

default

Jan 1, 1970
0
Welders have a high leakage transformer in them, it works like a ballast for
a discharge lamp. When you load down the output, the voltage drops
dramatically and the current remains at more or less the output on the dial.
In a nutshell, it's a constant-current source, not the constant-voltage
source you get from the wall.

Then I guess the next question would be how sharp is the "knee" when
it reaches current limiting? Its been a long time since I had the
welder apart but the transformer looked like a pretty normal IE
lamination but with more empty space than most transformers and taller
than wider.

I don't remember seeing a magnetic shunt - but it has been awhile. I
don't remember seeing a gap either. Primary on the bottom in one
winding and secondary also wound close to the core, above it, with
about an inch space between the two. I would think it would limit
current some, but wouldn't be a very sharp limit.

--
 
M

mike

Jan 1, 1970
0
default said:
Then I guess the next question would be how sharp is the "knee" when
it reaches current limiting?

It doesn't limit in the sense that it has a knee.
The leakage inductance has a reactance
at the line frequency that limits the current. Yes, all kinds
of transient stuff happen when the arc strikes, but you get the idea.

There IS a different kind of KNEE where the core saturates. Depending
on the construction
of the inductor, this shouldn't happen, and would blow the line fuse
if it did.

But when you start modifying the welder, you can no longer count on
the safety features built into the original design. You kinda have
to know what you're doing and have a fire extinguisher handy.

One more thing. The AC characteristics of the core demand that the
current be symmetrical about zero. If you try to use a half-wave
rectifier, you wind the core off the end of the B-H curve and
(probably) blow the line fuse...i suspect there are many different
core designs, but an AC welder engineer doesn't need to worry about
core saturation with a half-wave rectifier 'cause he ain't using one.

When you calculate the $avings of making your own DC welder instead
of buying one, you should consider that you'll probably blow up at
least one set of diodes in your initial experiments. People who know
how to do it right the first time, don't have to ask about half-wave
rectifiers on a repair newsgroup. ;-)
The oscilloscope I'd use to verify the diode voltage/current
costs 50X the cost of the project.
mike


Its been a long time since I had the
 
A

Arfa Daily

Jan 1, 1970
0
GregS said:
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
 
M

mike

Jan 1, 1970
0
Arfa said:
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
Average current is what heats up the diode package.
Peak current is what melts the silicon in the junction.
You need to consider both. So quit arguing about it.

The welder diode specs don't care how many Xmas lights you have.
There is no need for speculation...it's written in the spec.

Just look up the specs and design with both numbers in mind.
mike
 
A

Arfa Daily

Jan 1, 1970
0
Average current is what heats up the diode package.
Peak current is what melts the silicon in the junction.
You need to consider both. So quit arguing about it.

Isn't that what I just said ... ? It's not me that's arguing the point.

Arfa
 
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