Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated." - Trefor Thomas

Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Calculating isn't easy. Much easier to add some capacitance and check
current, add some more, keep testing until you find a minimum idle
current.

I believe that for optimal performance you would want SQRT(2*PI/LC) = 60 Hz.

You know everything in that equation except the C. That assumes you know
the effective inductance "looking into" the primary of the transformer.

I think that an LC network looks resistive at the resonant frequency, and
that the resonant frequency is SQRT(2*PI/LC). That's why if you add just
the right amount of capacitance for the inductance of your welder, then
the resonant frequency is the same as that of the power grid and you get
zero imaginary current; i.e. all of your current will be real current.

Real current and imaginary current are electrical engineering terms. Imaginary
current can burn up a wire just like real current, it just doesn't show up on
your power meter.

I used to know all this stuff ..

Grant

On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin
wrote:

Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Calculating isn't easy. Much easier to add some capacitance and check
current, add some more, keep testing until you find a minimum idle
current.

Add it where, and how much do I start with? My power bill from
Pacific Greed and Extortion (PG&E) was a $100 higher this month from
what I suspect is the welders being used a fair amount..most of which
was simply idling.

I believe that for optimal performance you would want SQRT(2*PI/LC) = 60 Hz.

You know everything in that equation except the C. That assumes you know
the effective inductance "looking into" the primary of the transformer.

I think that an LC network looks resistive at the resonant frequency, and
that the resonant frequency is SQRT(2*PI/LC). That's why if you add just
the right amount of capacitance for the inductance of your welder, then
the resonant frequency is the same as that of the power grid and you get
zero imaginary current; i.e. all of your current will be real current.

Real current and imaginary current are electrical engineering terms. Imaginary
current can burn up a wire just like real current, it just doesn't show up on
your power meter.

I used to know all this stuff ..


I never knew any of that stuff...shrug.

Grant

Gunnner

"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated." - Trefor Thomas

In article ,
"Tim Williams" wrote:

Yep, the C and L cancel. But if you're blowing breakers, you've probably
already got trouble simply turning it on. That power-on spike is none too
fun for the circuit with *just* the inductance, let alone the caps (which
look like a short to such transients)!

Would it be better to switch in the caps AFTER turn-on?

--
Free men own guns, slaves don't
www.geocities.com/CapitolHill/5357/

Gunner wrote:

Add it where, and how much do I start with? My power bill from
Pacific Greed and Extortion (PG&E) was a $100 higher this month from
what I suspect is the welders being used a fair amount..most of which
was simply idling.

Add capacitance right across the power lines, from L1 to L2; i.e. connect
each capacitor in parallel with the whole welder.

I'd scrounge some 10uF motor run caps and add one and turn on your welder
and measure the current draw (with a loop-type current meter) and then add
another cap and measure it again. If it goes down then just keep adding
capacitance. You might want to also add some 30 or 35 uF run caps just
so you can build up quicker.

If you need help scrounging motor run caps drop me an email. I have an
unimpeachable source.

Grant

Do you have one of the new "speed read" meters, or the old manual type?
When there is a sudden jump in the bill it's always a good idea to check
that the meter reading actually matches what the bill says.

Twice before they upgraded to the speed read meters I had to call in
corrected readings when the meter reader misread one of the upper digits
causing a 1,000 kwh jump in the bill. I probably didn't notice other
errors in lower digits and fortunately the next correct reading will put
things back on track anyway.

$100 jump would seem to be an awful lot of KWH, somewhere between 400
and 800 kwh depending on how bad your electric rates are. Even at 400
kwh that works out to on the order of 13 kwh / day more consumption
*every* day during the month.

Pete C.


Gunner wrote:

On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin
wrote:

Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Calculating isn't easy. Much easier to add some capacitance and check
current, add some more, keep testing until you find a minimum idle
current.

Add it where, and how much do I start with? My power bill from
Pacific Greed and Extortion (PG&E) was a $100 higher this month from
what I suspect is the welders being used a fair amount..most of which
was simply idling.

I believe that for optimal performance you would want SQRT(2*PI/LC) = 60 Hz.

You know everything in that equation except the C. That assumes you know
the effective inductance "looking into" the primary of the transformer.

I think that an LC network looks resistive at the resonant frequency, and
that the resonant frequency is SQRT(2*PI/LC). That's why if you add just
the right amount of capacitance for the inductance of your welder, then
the resonant frequency is the same as that of the power grid and you get
zero imaginary current; i.e. all of your current will be real current.

Real current and imaginary current are electrical engineering terms. Imaginary
current can burn up a wire just like real current, it just doesn't show up on
your power meter.

I used to know all this stuff ..


I never knew any of that stuff...shrug.

Grant

Gunnner

"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated." - Trefor Thomas

Gunner;
Have you checked to see if Miller sold Power factor Caps for your
machine? Lincoln supplied the ones in my Squarewave 275. It was about
$100 cdn at the time. Miller might give specs for theirs (if they sell them)

Pete

Gunner wrote:

Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


--
__
Pete Snell
Royal Military College
Kingston Ontario


The reasonable man adapts himself to the world;
the unreasonable one persists in trying to adapt the world to himself.
Therefore all progress depends on the unreasonable man.
- George Bernard Shaw

"Grant Erwin" wrote in message
...
Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?


snip


Real current and imaginary current are electrical engineering terms.
Imaginary
current can burn up a wire just like real current, it just doesn't show up
on
your power meter.


If so, then balancing isnt going to lower the operating costs anyways.....

--

SVL

"PrecisionMachinisT" wrote in message
...

"Grant Erwin" wrote in message
...
Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?


snip


Real current and imaginary current are electrical engineering terms.
Imaginary
current can burn up a wire just like real current, it just doesn't show
up
on
your power meter.


If so, then balancing isnt going to lower the operating costs anyways.....

This is true. Power delivered is power sold. IMO, it is highly unlikely
Gunner used $100 worth of juice in one month just heating up the wiring
during welder idle time. The extra charge is probably due to an error in
meter reading - or - Gunner just did a lot more welding in that month.

Bob Swinney

--

SVL

Grant Erwin wrote:

Gunner wrote:

Add it where, and how much do I start with? My power bill from
Pacific Greed and Extortion (PG&E) was a $100 higher this month from
what I suspect is the welders being used a fair amount..most of which
was simply idling.

Add capacitance right across the power lines, from L1 to L2; i.e. connect
each capacitor in parallel with the whole welder.

I'd scrounge some 10uF motor run caps and add one and turn on your welder
and measure the current draw (with a loop-type current meter) and then add
another cap and measure it again. If it goes down then just keep adding
capacitance. You might want to also add some 30 or 35 uF run caps just
so you can build up quicker.

If you need help scrounging motor run caps drop me an email. I have an
unimpeachable source.

Grant
You want to measure the current on the meter side of the caps; not on the
welder side.
--
Keith Bowers - Thomasville, NC

"Grant Erwin" wrote in message
...
Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Calculating isn't easy. Much easier to add some capacitance and check
current, add some more, keep testing until you find a minimum idle
current.

I believe that for optimal performance you would want SQRT(2*PI/LC) = 60
Hz.

More like: Resonant Frequency = [1 / (2*PI*SQRT(L*C) ] and solving for C =
SQRT[1 / (*60*60*4*PI*PI*L)]
But as stated you'd have to know the value of L . . . . .

Bob Swinney

You know everything in that equation except the C. That assumes you know
the effective inductance "looking into" the primary of the transformer.

I think that an LC network looks resistive at the resonant frequency, and
that the resonant frequency is SQRT(2*PI/LC). That's why if you add just
the right amount of capacitance for the inductance of your welder, then
the resonant frequency is the same as that of the power grid and you get
zero imaginary current; i.e. all of your current will be real current.

Real current and imaginary current are electrical engineering terms.
Imaginary
current can burn up a wire just like real current, it just doesn't show up
on
your power meter.

I used to know all this stuff ..

Grant

Power factor should not affect your electric bill. KWH meters ignore
reactive current.

On Tue, 14 Dec 2004 09:46:33 GMT, Gunner
wrote:

On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin
wrote:

Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?


Add it where, and how much do I start with? My power bill from
Pacific Greed and Extortion (PG&E) was a $100 higher this month from
what I suspect is the welders being used a fair amount..most of which
was simply idling.

"Robert Swinney" wrote in message
...

"PrecisionMachinisT" wrote in message
...

If so, then balancing isnt going to lower the operating costs
anyways.....

This is true. Power delivered is power sold.


Yeah, I know......I just that I wanted third party verification for Gunner's
sake.....

( Pretty sure he has me branded as being 'Liberal'--and so anything *I*
might say is *definately* suspect )

G

--

SVL

On Tue, 14 Dec 2004 14:26:49 GMT, "Pete C." wrote:

Do you have one of the new "speed read" meters, or the old manual type?
When there is a sudden jump in the bill it's always a good idea to check
that the meter reading actually matches what the bill says.

Twice before they upgraded to the speed read meters I had to call in
corrected readings when the meter reader misread one of the upper digits
causing a 1,000 kwh jump in the bill. I probably didn't notice other
errors in lower digits and fortunately the next correct reading will put
things back on track anyway.

$100 jump would seem to be an awful lot of KWH, somewhere between 400
and 800 kwh depending on how bad your electric rates are. Even at 400
kwh that works out to on the order of 13 kwh / day more consumption
*every* day during the month.

Pete C.

Gunner, you might want to check your rate structure. PG&E may have
something like a demand charge on its residential (I assume you're a
residential customer) rates that kicks in if you have a big surge in
demand over something like a 5-minute interval.

That sort of thing is not common in residential rates, but PG&E is
notorious for its creative rate plans.

--RC

Gunner wrote:

On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin
wrote:

Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Calculating isn't easy. Much easier to add some capacitance and check
current, add some more, keep testing until you find a minimum idle
current.

Add it where, and how much do I start with? My power bill from
Pacific Greed and Extortion (PG&E) was a $100 higher this month from
what I suspect is the welders being used a fair amount..most of which
was simply idling.

I believe that for optimal performance you would want SQRT(2*PI/LC) = 60 Hz.

You know everything in that equation except the C. That assumes you know
the effective inductance "looking into" the primary of the transformer.

I think that an LC network looks resistive at the resonant frequency, and
that the resonant frequency is SQRT(2*PI/LC). That's why if you add just
the right amount of capacitance for the inductance of your welder, then
the resonant frequency is the same as that of the power grid and you get
zero imaginary current; i.e. all of your current will be real current.

Real current and imaginary current are electrical engineering terms. Imaginary
current can burn up a wire just like real current, it just doesn't show up on
your power meter.

I used to know all this stuff ..


I never knew any of that stuff...shrug.

Grant

Gunnner

"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated." - Trefor Thomas

Projects expand to fill the clamps available -- plus 20 percent

On Tue, 14 Dec 2004 22:24:57 GMT, wrote:


$100 jump would seem to be an awful lot of KWH, somewhere between 400
and 800 kwh depending on how bad your electric rates are. Even at 400
kwh that works out to on the order of 13 kwh / day more consumption
*every* day during the month.

Pete C.

Gunner, you might want to check your rate structure. PG&E may have
something like a demand charge on its residential (I assume you're a
residential customer) rates that kicks in if you have a big surge in
demand over something like a 5-minute interval.

That sort of thing is not common in residential rates, but PG&E is
notorious for its creative rate plans.

OK, Ill check into it.

Since Gray(out) Davis locked California into the highest rates in the
US, this is a bit much. $200 a month power bills for regular
residential service is hard enough to swallow, but when it jumps to
$300..it really hurts.
We recalled his sorry ass, but the damage was already done. Im hoping
that burning about 50 rods worth of power doesnt cost me an extra $100
per month..

Gunner



"If I'm going to reach out to the the Democrats then I need a third
hand.There's no way I'm letting go of my wallet or my gun while they're
around."

"Democrat. In the dictionary it's right after demobilize and right
before demode` (out of fashion).
-Buddy Jordan 2001

On Tue, 14 Dec 2004 14:07:10 -0800, "PrecisionMachinisT"
wrote:


This is true. Power delivered is power sold.


Yeah, I know......I just that I wanted third party verification for Gunner's
sake.....

( Pretty sure he has me branded as being 'Liberal'--and so anything *I*
might say is *definately* suspect )

Only the political stuff you utter is suspect. I read your other
posts religiously G
Well..and some of the 2nd Am stuff is also suspect VBG

Gunner

G

--

SVL



"If I'm going to reach out to the the Democrats then I need a third
hand.There's no way I'm letting go of my wallet or my gun while they're
around."

"Democrat. In the dictionary it's right after demobilize and right
before demode` (out of fashion).
-Buddy Jordan 2001

If the power bill is painful, have you considered trying to trade for a
gas powered welder/generator? Again if power is that dear, it would
likely pay off long term.

TMT

"Gunner" wrote in message
...
On Tue, 14 Dec 2004 14:07:10 -0800, "PrecisionMachinisT"
wrote:

( Pretty sure he has me branded as being 'Liberal'--and so anything *I*
might say is *definately* suspect )

Only the political stuff you utter is suspect. I read your other
posts religiously G


"Political stuff" is always suspect, regardless of who it was uttered it.

--

SVL

" Conservative, n: A statesman who is enamored of existing evils, as
distinguished from the Liberal who wishes to replace them with others."
--Ambrose Bierce

Don Foreman wrote:
Power factor should not affect your electric bill. KWH meters ignore
reactive current.

True for now Don, talking residential that is. But probably not forever
here in the USA. New Zealand already charges residential customers for
KVARs, and I recall seeing some outfits already selling "whole house"
power factor correction boxes here, which IMO are a sham because they
don't automatically track the requirement and correct for it. Capacitors
of the right size at each load are the easiest solution and also reduce
the I^2*R losses in the house's wiring, giving a tiny additional savings
there.

BTW, there was some work done by some Navy guy a while back showing how
much power could be saved by adding a running capacitor to exixting dumb
old split phase induction motors. IIRC it increased the efficiency of
the average motor by about 10%.

Just my .02,

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"As long as there are final exams, there will be prayer in public
schools"

On Tue, 14 Dec 2004 06:12:36 GMT, Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300

Yes, I know how to calculate the required capacitor value, if you can give me
the idling inductance of the transformer primary. But you don't know that, and
neither to I.

I do know that Miller uses 250 uF of capacitance in their power factor correction
kit for the Synchrowave 250. That makes it tempting to say 1 uF per amp of output,
but that would be wrong. It depends on the particular transformer. Something in
the neighborhood of 250 uF for your machines should be about right though. You
don't have to hit it right on the nose to get a useful amount of PF correction.

Gary

Hey Gunner,

For what it's worth, I brought home some oil-filled caps from Apex
Surplus out in the Sun Valley area of LA. For my comments here, I'm
making San Fernando as east-west, and Lankershim as north-south.
Anyway, just in case you are NOT familiar with the place, it's on San
Fernando just east of Lankershim. Finding anything in there on your
own is a bit of a bitch though, but a real adventure. Ask at the
counter if you're short on time. The larger ones I got were at the
west end of the first or second row about waist high, and the smaller
ones were on the north wall due north of the large ones (but over two
rows). I haven't even unpacked mine yet though, so I don't know that
they are "good" yet. The guy said I could bring them back if they
didn't work, and he knew right away they were for a Rotary Phase
Convertor.

Take care.

Brian Lawson,
Bothwell, Ontario.
XXXXXXXXXXXXXXXXXXXXXX
On Tue, 14 Dec 2004 06:12:36 GMT, Gunner
wrote:

Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated." - Trefor Thomas

On Sun, 09 Jan 2005 23:24:40 -0500, Brian Lawson
wrote:

Hey Gunner,

For what it's worth, I brought home some oil-filled caps from Apex
Surplus out in the Sun Valley area of LA. For my comments here, I'm
making San Fernando as east-west, and Lankershim as north-south.
Anyway, just in case you are NOT familiar with the place, it's on San
Fernando just east of Lankershim. Finding anything in there on your
own is a bit of a bitch though, but a real adventure. Ask at the
counter if you're short on time. The larger ones I got were at the
west end of the first or second row about waist high, and the smaller
ones were on the north wall due north of the large ones (but over two
rows). I haven't even unpacked mine yet though, so I don't know that
they are "good" yet. The guy said I could bring them back if they
didn't work, and he knew right away they were for a Rotary Phase
Convertor.

Take care.

Brian Lawson,
Bothwell, Ontario.
XXXXXXXXXXXXXXXXXXXXXX

Thanks Brian.
Im well familiar with Apex. Someday I aspire to be 1% of them G

Sorry we didnt meet up this trip. I got your voice mail today.
I was up on the roof in the rain..sigh.

Next trip, try to make a little time to come visit the homestead. Im
sure we can find Stuff to fit in your suitcase G

Gunner

On Tue, 14 Dec 2004 06:12:36 GMT, Gunner
wrote:

Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated." - Trefor Thomas

"The French are a smallish, monkey-looking bunch and not dressed any better,
on average, than the citizens of Baltimore.
True, you can sit outside in Paris and drink little cups of coffee,
but why this is more stylish than
sitting inside and drinking large glasses of whiskey I don't know."
-- P.J O'Rourke (1989)
--------------------------------------------------------------------------------

Gunner wrote:



Thanks Brian.
Im well familiar with Apex. Someday I aspire to be 1% of them G

Sorry we didnt meet up this trip. I got your voice mail today.
I was up on the roof in the rain..sigh.

snip

Gunner


Speaking of roof - house floating, dab of tar here and there or 5000 sq ft. tarp ?
Hope it was simple!!!


Several years ago it was tar for me - I had a Coastal Redwood limb fall but end
first and drive a hole through my shop roof. I found it the next day checking the
house for leaks due to the massive wind storm and rain...

This time, Gunner gets the Pineapple express (sorry Hawaii! ) and I get a 24" oak
to fall across my driveway across the 220 feeder lines and wedge into some Redwoods.

Thank the storm god on that one, it was the tree next to the power pole and the
heavy weight - on the wires "threatened the pole" so the Power company PG&E called
a tree service - Davey - managed to cut the 40 or 50' tree and drop or swing
the chunks to the sloping driveway so they wouldn't roll. The driveway is between
30 degree and 45 degree slope depending on the side or location...

The shop is at the bottom of the slope (live on a hill side - 30' setback means
down hill house...) and a bedroom and office in danger if the bottom
end swings or kicks out and the top goes like a tossed log...

I got the tree down and then paid for a hauler to take just over 100 cu ft of cut and
crushed down Oak and Redwood limbs to the dump. The logs sit for now -
looking for a home.

I have a home for a few of them - anvil holders :-)

Martin (yet another near hit!)

--
Martin Eastburn, Barbara Eastburn
@ home at Lion's Lair with our computer
NRA LOH, NRA Life
NRA Second Amendment Task Force Charter Founder

Calculating the amount of capacitance needed to reduce the idle current
is actually fairly simple. I have posted how to do it at least twice.
Try seaching on google with "power factor correction "

What you do need to remember is that you really don't want to correct
to more than about 85% power factor. Correcting for that last 15% will
not change the idle current very much.

Okay, I will post this one more time. First measure the current drawn
with the welder idling and no power factor cap. Now add a power
factor cap right across the line in parallel with the welder.
Measure the current again. Now for the graphic solution. Use a
compass and swing an arc that is proportional to the first measurement
from point A. Shorten up the compass and swing a second arc ( also
from point A ) proportional to the current with the P.F. cap
installed. Now subtract the second current from the first current and
set the compass to this length. Now put the point of the compass on
the first arc and swing an arc so it intersects the second arc. Draw
lines from point A to where you had the compass on the first arc and
from point A to where the intersection is with the second arc. DRaw a
third line between the ends of the two lines. Now turn your drawing
around until that last line is vertical. Draw a line from point A
horizontal to underneath the vertical line. Extend the vertical line
to the horizontal line.

Okay? The horizontal line from point A to the vertical line is the
amount of current that is real. The vertical line is the imaginary
current. The length between the two arcs is how much imaginary current
flows through the P.F. cap. The line from point A anywhere on the
vertical line is the total current. Note that doubling the imaginary
current ( doubling the capacitance ) does not double the reduction is
the total current.

This is not easy to explain. So measure the current with no P.F. cap
and with a P.F. cap installed and post the results and I will do the
graphics and tell you what I get for results.

Dan




Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the
civilized,
merely the domesticated." - Trefor Thomas

Why do I bother.......... No one has made any comments on what I
posted.......... No even that it is Wrong, Wrong, wrong........

I did make a serious mistake. Where I said " subtract the second
current from the first current " I meant to connect just the power
factor cap across the power line and measure the current it draws.

But apparently no one read my post critically or else they would have
said " WTF ". I am hurt........8-(.

Dan



wrote:
Calculating the amount of capacitance needed to reduce the idle
current
is actually fairly simple. I have posted how to do it at least
twice.
Try seaching on google with "power factor correction
"

What you do need to remember is that you really don't want to correct
to more than about 85% power factor. Correcting for that last 15%
will
not change the idle current very much.

Okay, I will post this one more time. First measure the current
drawn
with the welder idling and no power factor cap. Now add a power
factor cap right across the line in parallel with the welder.
Measure the current again. Now for the graphic solution. Use a
compass and swing an arc that is proportional to the first
measurement
from point A. Shorten up the compass and swing a second arc ( also
from point A ) proportional to the current with the P.F. cap
installed. Now subtract the second current from the first current
and
set the compass to this length. Now put the point of the compass on
the first arc and swing an arc so it intersects the second arc. Draw
lines from point A to where you had the compass on the first arc and
from point A to where the intersection is with the second arc. DRaw
a
third line between the ends of the two lines. Now turn your drawing
around until that last line is vertical. Draw a line from point A
horizontal to underneath the vertical line. Extend the vertical line
to the horizontal line.

Okay? The horizontal line from point A to the vertical line is the
amount of current that is real. The vertical line is the imaginary
current. The length between the two arcs is how much imaginary
current
flows through the P.F. cap. The line from point A anywhere on the
vertical line is the total current. Note that doubling the imaginary
current ( doubling the capacitance ) does not double the reduction
is
the total current.

This is not easy to explain. So measure the current with no P.F. cap
and with a P.F. cap installed and post the results and I will do the
graphics and tell you what I get for results.

Dan




Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the
civilized,
merely the domesticated." - Trefor Thomas

On 15 Jan 2005 11:40:15 -0800, wrote:

Why do I bother.......... No one has made any comments on what I
posted.......... No even that it is Wrong, Wrong, wrong........

I did make a serious mistake. Where I said " subtract the second
current from the first current " I meant to connect just the power
factor cap across the power line and measure the current it draws.

But apparently no one read my post critically or else they would have
said " WTF ". I am hurt........8-(.

Dan


Dan, I read it, but its over my head so I simply nodded, and printed
it out to try in the near future.

Gunner



wrote:
Calculating the amount of capacitance needed to reduce the idle
current
is actually fairly simple. I have posted how to do it at least
twice.
Try seaching on google with "power factor correction
"

What you do need to remember is that you really don't want to correct
to more than about 85% power factor. Correcting for that last 15%
will
not change the idle current very much.

Okay, I will post this one more time. First measure the current
drawn
with the welder idling and no power factor cap. Now add a power
factor cap right across the line in parallel with the welder.
Measure the current again. Now for the graphic solution. Use a
compass and swing an arc that is proportional to the first
measurement
from point A. Shorten up the compass and swing a second arc ( also
from point A ) proportional to the current with the P.F. cap
installed. Now subtract the second current from the first current
and
set the compass to this length. Now put the point of the compass on
the first arc and swing an arc so it intersects the second arc. Draw
lines from point A to where you had the compass on the first arc and
from point A to where the intersection is with the second arc. DRaw
a
third line between the ends of the two lines. Now turn your drawing
around until that last line is vertical. Draw a line from point A
horizontal to underneath the vertical line. Extend the vertical line
to the horizontal line.

Okay? The horizontal line from point A to the vertical line is the
amount of current that is real. The vertical line is the imaginary
current. The length between the two arcs is how much imaginary
current
flows through the P.F. cap. The line from point A anywhere on the
vertical line is the total current. Note that doubling the imaginary
current ( doubling the capacitance ) does not double the reduction
is
the total current.

This is not easy to explain. So measure the current with no P.F. cap
and with a P.F. cap installed and post the results and I will do the
graphics and tell you what I get for results.

Dan




Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to
reduce idle current in a single phase 220 welder?

Gunner, Idealarc Tig250/250 and Miller Dialarc 300


"To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the
civilized,
merely the domesticated." - Trefor Thomas

"At the core of liberalism is the spoiled child -
miserable, as all spoiled children are, unsatisfied,
demanding, ill-disciplined, despotic and useless.
Liberalism is a philosphy of sniveling brats." -- P.J. O'Rourke

On 15 Jan 2005 11:40:15 -0800, wrote:

Why do I bother.......... No one has made any comments on what I
posted.......... No even that it is Wrong, Wrong, wrong........

I did make a serious mistake. Where I said " subtract the second
current from the first current " I meant to connect just the power
factor cap across the power line and measure the current it draws.

But apparently no one read my post critically or else they would have
said " WTF ". I am hurt........8-(.

Dan




Actually, this is starting to **** me off.
Are you saying that I can put 50 - 70 uF across L1 and L2 in a grey
box and it is the same as what Miller wants to sell me for $200 usd?!

Regards,
Jim

I have wondered about the high prices charged for adding PF correction,
and keep thinking there must be reason for the high price. That is
reasons beyond that Miller would be buying new caps and reselling them
at a profit, and then the distributor would be wanting to make a profit
too. There might be some reason beyond that they can and do sell some
for $200 usd. , but I sure don't know what it is.

Actually I would never recommend putting 50 - 70 uf in a grey box.
I am too cheap for that. I would see if there wasn't some space in
that big red box you already have. That is what I did with my Buzz box
years ago.

Dan


JK wrote:
On 15 Jan 2005 11:40:15 -0800, wrote:


Actually, this is starting to **** me off.
Are you saying that I can put 50 - 70 uF across L1 and L2 in a grey
box and it is the same as what Miller wants to sell me for $200
usd?!

Regards,
Jim

I did say to measure the current with no PF correction and with some PF
cap installed and I would help you with the rest. Let me now amend
that to those two measurements and also the current to just the PF cap.
I will help you understand it all if you get those three measurements.


Dan

Dan,

I routinely use a Miller 251 wire feed or a Miller 250 DX Tig machine.

Is there a ball park figure to use a for the power factor correction
capacitors that I missed somewhere. I built my own 7.5 to 17.5 hp
phase converter to run my shop equipment so I understand the dangers
associated with 220 v. I tuned it to optimize performance with
balancing capacitors.

regards,
Jim

http://www.lmphotonics.com/pwrfact.htm is the first site I found
by searching on Google for " Power Factor correction ". It give you
some rules of thumb for PF correction for motors and some idea of why
one does not try to get 100 % power factor. Take a look there and come
back and ask if you have questions.


Dan