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# Thread: Lincoln Power Mig 140C Wiring

1. Originally Posted by Thatguy
33Vx140A=4620W=39A@120V.

A Q0 20A breaker can deliver 40A for 8 to 28 seconds before tripping.

Dunno'.

I don't weld, but I think they're depending on average current draw being within a 20A breaker's capacity most of the time. If you have some weld that takes 140A for more than 8 seconds you might be inconvenienced.

Can you post a schematic/parts list? I'm thinking the 20% duty cycle depends on the internal transformer's I-squared-T rating. The whole setup might depend on assumed I-squared-T ratings of wires, sockets, breakers, everything.

If anyone has a link to welding current vs. time I'd sure like to see it.
I will need an explanation to see how you are using the power factor for the secondary side of a transformer to calculate amperage on the primary AC side.

I am lost.

2. 20% would be 12 minutes in an hour. Most jobs have a fair amount of setup and cleanup in between when actually welding, so that may not be a major deal. An industrial unit that can run 100% of the time would likely be much more expensive.

3. Originally Posted by jwelectric
Install a 20 amp circuit with a 20 amp receptacle and plug in the welder and burn up some metal
Hey dude I think you might be on to something right here. Why else would the people that built that damn machine put a 15 amp rated plug on unless that is all it needs?

4. Gentlemen, thank you all very much for your assistance and insight. It was all very good information. I found the answer to requiring a 20 amp circuit breaker while also requiring 25 amps available to that circuit in order to obtain the full output of the welder. As others have suggested, it's in the circuit breaker design and test criteria. The following is an excerpt from a circuit breaker manufacturer that explains the current carrying capabilities of most circuit breaker:

The CSA test program as well as our inhouse test program confirms compliance with the regulatory requirements by conducting the following tests:

Initial Trip time confirmation

* The breaker is "over loaded" to a value of 135% of the current rating and the breaker must trip within one hour.
* The breaker is then "overloaded to a value of 200% of the current rating and the breaker must trip in typically in less than two minuets.

Temperature test

* The circuit breaker is overloaded to a current value of 6 times the rating ( but not less than 150 amp.) and switched 50 times.
* The circuit breaker is then placed in a 40 deg. C. ambient temperature and forced to carry 100% of the current rating until the circuit breaker reaches thermal equilibrium. Temperatures are recorded and must be below the maximum temp. limits of the requirements.

Considering the welder's duty cycle it will operate at the 140 amps advertised by Lincoln so my previous assumptions were wrong.

5. Originally Posted by jar546
I will need an explanation to see how you are using the power factor for the secondary side of a transformer to calculate amperage on the primary AC side.

I am lost.
No PF, just Pin = Pout and P=VI. I assumed resistive loads, and the xformer inductance and power loss to be negligible.
For this exercise it might be OK.

The V-I curve of an arc is supposed to exhibit (incrementally) negative resistance, so an exact analysis of this setup would be really weird.

6. Originally Posted by vstoyko
Gentlemen, thank you all very much for your assistance and insight. It was all very good information. I found the answer to requiring a 20 amp circuit breaker while also requiring 25 amps available to that circuit in order to obtain the full output of the welder. As others have suggested, it's in the circuit breaker design and test criteria. The following is an excerpt from a circuit breaker manufacturer that explains the current carrying capabilities of most circuit breaker:

The CSA test program as well as our inhouse test program confirms compliance with the regulatory requirements by conducting the following tests:

Initial Trip time confirmation

* The breaker is "over loaded" to a value of 135% of the current rating and the breaker must trip within one hour.
* The breaker is then "overloaded to a value of 200% of the current rating and the breaker must trip in typically in less than two minuets.

Temperature test

* The circuit breaker is overloaded to a current value of 6 times the rating ( but not less than 150 amp.) and switched 50 times.
* The circuit breaker is then placed in a 40 deg. C. ambient temperature and forced to carry 100% of the current rating until the circuit breaker reaches thermal equilibrium. Temperatures are recorded and must be below the maximum temp. limits of the requirements.

Considering the welder's duty cycle it will operate at the 140 amps advertised by Lincoln so my previous assumptions were wrong.
I an't sure but I think that someone wrote in here to just plug the thing and go to work didn't they? Well I guess they must have known a little something wouldn't you say?

7. Originally Posted by Thatguy
No PF, just Pin = Pout and P=VI. I assumed resistive loads, and the xformer inductance and power loss to be negligible.
For this exercise it might be OK.

The V-I curve of an arc is supposed to exhibit (incrementally) negative resistance, so an exact analysis of this setup would be really weird.
Yeah, and the flux capacitor may not help it either.

This is a 20% duty cycle nonmotor welder so a 20A circuit is more than sufficient.

8. Originally Posted by jar546
Yeah, and the flux capacitor may not help it either.
.
Plus I need some new dilithium crystals for the anti-matter core.

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