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# Thread: TIG torch amp rating when using pulse

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1. Senior Member
Join Date
Aug 2009
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142

## TIG torch amp rating when using pulse

As an example, a WP9 family torch is rated at 125 amps DC (per the Weldcraft web site). If this torch is used on a pulse TIG power supply with the following settings:

Pulse frequency 100 Hz
Max amperage 200 amps
Peak time 50%
Background current 50 amps

the AVERAGE current through the torch would be 125 amps. Would it be correct to conclude that these parameters are within the capability of the torch?

TIA,

Ken

2. Well, I'm guessing the question is can you run 200 amps through a torch rated at 125 amps, Hmmmm, yes for a short while till the internal conduit has a melt down.

Not sure, try it and tell us how long your torch lasts.

3. Member
Join Date
Sep 2010
Location
Sorong, Indonesia
Posts
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@Taylor: I think its close. Like Cruizer said, try it and then let us know.

4. Lets say that 1 amp = 1 degree in heat

So if you were pulsing at 200/50 then your cord is running 200/50 degrees.

The problem with your theory is that you assume that the internal wiring of the tig torch cools down to 0 degrees between pulses. If your torch is running at 200 degrees for .01 second, then 50 degrees for .01 second, then back to 200... The laws of heat dissipation say that the torch would run around 180-185 degrees and would burn up because if the torch is rated for 125 amps / 125 degrees during the span of the duty cycle.

Now with that said, I ran a WP17 on my 200 DX with 50/50 helium at 200 full amps for about 5 hours. Yes the torch did end up dying at the end but I had to get the job done and the torch was put in the bill.

So yes its possible for a short time, yes you will damage the torch over the span of a half hour, yes you can just buy the head since that is what went on my line.

5. Senior Member
Join Date
Aug 2009
Posts
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Well I had some coffee this AM and thought through my question a little more. The issue is dissipation of heat by the torch. The heat generated by the flow of current through the torch may be represented as:

P=I^2 * R

Where P is the power or heat, I is the current and R is the resistance of the torch. If we assume that the resistance of the torch is constant and simply solve for the pulse max current which is equivalent to 125 amps constant current and using my original example:

I = max pulse current
I/4 = background current
1/2 = fraction of time at max or background current

125^2 = (I^2)/2 + ((I/4)^2)/2

multiplying both sides by 2 gives

31,250 = I^2 + (I/4)^2 = I^2 + .0625 * I^2

solving for I

I^2 = 31,250/1.0625

I = SQRT(31,250/1.0625) = SQRT(29,412) = 171.5 amps

How does that sound? I may send a question to Weldcraft and see what their answer is. And I need to ping Miller to see if they have duty cycle charts for the Dynasty 200 when used in pulse mode

Ken

6. Junior Member
Join Date
Feb 2011
Posts
20
Why don't you just upgrade the torch, and be done with it?

7. Originally Posted by taylorkh
Well I had some coffee this AM and thought through my question a little more. The issue is dissipation of heat by the torch. The heat generated by the flow of current through the torch may be represented as:

P=I^2 * R

Where P is the power or heat, I is the current and R is the resistance of the torch. If we assume that the resistance of the torch is constant and simply solve for the pulse max current which is equivalent to 125 amps constant current and using my original example:

I = max pulse current
I/4 = background current
1/2 = fraction of time at max or background current

125^2 = (I^2)/2 + ((I/4)^2)/2

multiplying both sides by 2 gives

31,250 = I^2 + (I/4)^2 = I^2 + .0625 * I^2

solving for I

I^2 = 31,250/1.0625

I = SQRT(31,250/1.0625) = SQRT(29,412) = 171.5 amps

How does that sound? I may send a question to Weldcraft and see what their answer is. And I need to ping Miller to see if they have duty cycle charts for the Dynasty 200 when used in pulse mode

Ken
Too much math for my pea brain...

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