# How is it possible for a welding machine to end up creating more power than what it started with, if that is physically impossible?

## Main Question or Discussion Point

How is it possible for a welding machine to end up creating more power than what it started with if that is physically impossible? For ex; 200 amp @ 28vdc welding machine. starts out with main voltage being 120v @ 20amps = 2400 watts then converts it to 200 amps at 28vdc which = 5,600 watts of power.

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200 amp @ 28vdc welding machine. starts out with main voltage being 120v @ 20amps = 2400 watts then converts it to 200 amps at 28vdc which = 5,600 watts of power.
It doesn't

davenn
Gold Member
2019 Award
For ex; 200 amp @ 28vdc welding machine. starts out with main voltage being 120v @ 20amps

For any high power device, it will use 240V and not 120V
Some one answered a similar Q in the last few days

@anorlunda @berkeman

200 A is the highest output current setting. Doesn't mean it occurs at 28 VDC

Guineafowl and Averagesupernova
berkeman
Mentor
How is it possible for a welding machine to end up creating more power than what it started with if that is physically impossible? For ex; 200 amp @ 28vdc welding machine. starts out with main voltage being 120v @ 20amps = 2400 watts then converts it to 200 amps at 28vdc which = 5,600 watts of power.
Welcome to the PF.

Please provide a link to the datasheet and user manual for this welder, or attach a picture of the nameplate. Something like this:

200 A is the highest output current setting. Doesn't mean it occurs at 28 VDC

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It doesn't
That's odd because the back of my welding machine says it does.

It doesn't
If it doesn't then that means every welding machine ever created is lying about how much amperage and voltage is being created, they would have to be lying by a huge margin to such a degree that it would warrant a massive lawsuit. If power in watts is volts X amps or Amps X volts then. Even if I start out with 220v @ 20amps = 4,400 watts of power, and I'm told I'm getting 200amps at 28v of output then I must be getting 5,600 watts. Or the welding machine along with every other machine is waay overated, or my math is wrong.

I believe there is a number that implies a 60% duty cycle (this must be PWM DC somehow?) So 3,360W out ? (did it in my head...always risky). Whew. Both the integrity of the welding folks and physics is preserved!

I believe there is a number that implies a 60% duty cycle (this must be PWM DC somehow?) So 3,360W out ? (did it in my head...always risky). Whew. Both the integrity of the welding folks and physics is preserved!
Duty cycle is in reference to the total amount of time you can weld continuously in a ten minute period before the machine has to cool down. It has nothing to do the amount of power it can produce just the total time peak power can be used before heat loss occurs thus loss of power.

I am unaware of the 10 minute stipulation in duty cycle definition. I'll bet the manufacturer also similarly uninformed. Where made?

Baluncore
2019 Award
I see the manufacturer's label includes 'Gooling: air cool'.
So I know it is “Chinese Junk Brand”, and that the numbers cannot be trusted.

berkeman and hutchphd
I am a very novice welder. Is there any possible utility to pulsating DC that would give higher instantaneous current at same average power?

I see the manufacturer's label includes 'Gooling: air cool'.
So I know it is “Chinese Junk Brand”, and that the numbers cannot be trusted.
The brand is irrelevant because all the American manufactures do the same. I'll post a link

I am unaware of the 10 minute stipulation in duty cycle definition. I'll bet the manufacturer also similarly uninformed. Where made?
China

hutchphd
28V is the open circuit voltage. 200A is the maximum (near short circuit) current. Once the arc is struck, the voltage drops.

To put it another way, 200A and 28V don’t occur together, so you can’t just multiply them.

Windadct, Merlin3189, NTL2009 and 7 others
Baluncore
2019 Award
The single phase, 50 Hz input sinewave will be rectified to high voltage DC in a capacitor.
The input is rated in kVA not kW because the rectifier input is not resistive.
220 V * 39 A = 8.580 kVA.
220 V * 30 A = 6.600 kW effective.

The continuous DC output is 155 A * 26.2 V = 4.061 kW.
The 60% duty output is 200 A * 28 V = 5.600 kW.

The output power is less than the input, so conservation of energy is OK. No problem.
I think it needs a higher current outlet. The OP assumed it to be standard.

russ_watters
Mentor
How is it possible for a welding machine to end up creating more power than what it started with if that is physically impossible? For ex; 200 amp @ 28vdc welding machine. starts out with main voltage being 120v @ 20amps = 2400 watts then converts it to 200 amps at 28vdc which = 5,600 watts of power.
Where do you see 20A? Isn't that the input parameters at the bottom of the nameplate photo? 78A at 110V?

zoki85, berkeman and hutchphd
I am a very novice welder. Is there any possible utility to pulsating DC that would give higher instantaneous current at same average power?
Nothing will occur in a welding machine that is not pre determined PWM or not. Inverter welding machines use pulse width modulation in order to provide a consistent arc but it won't cause the welding machine to exceed its max amperage or voltage rating.

I believe what your alluding to is voltage spiking and unnecessary chatter in the square or sign wave that is un accounted for. This is mitigated with the capacitor bank that comes with all inverter welding machines along with software programming. The capacitor bank in the circuit does two primary things, 1) provide consistent output and 2) removes unwanted noise, chatter or voltage spiking in the circuit.

hutchphd
The nameplate photo says 78A at 110V.

The single phase, 50 Hz input sinewave will be rectified to high voltage DC in a capacitor.
The input is rated in kVA not kW because the rectifier input is not resistive.
220 V * 39 A = 8.580 kVA.
220 V * 30 A = 6.600 kW effective.

The continuous DC output is 155 A * 26.2 V = 4.061 kW.
The 60% duty output is 200 A * 28 V = 5.600 kW.

The output power is less than the input, so conservation of energy is OK. No problem.
I think it needs a higher current outlet. The OP assumed it to be standard.
I got what your saying my next question would be:
Inverter welders use n channel mosfets in parallel attached to the same heatsink to increase total output of current and voltage then positive terminal is attached to the heatsink thus drawing out all the newly created power to weld with. Unless I’m wrong then to calculate total current of mosfets in parallel and voltage as well is as simple as adding them together, which means Theoretically I should be able to add an infinite amount of mosfets from the same power source to the same heatsink and keep increasing the power?

davenn and berkeman
I got what your saying my next question would be:
Inverter welders use n channel mosfets in parallel attached to the same heatsink to increase total output of current and voltage then positive terminal is attached to the heatsink thus drawing out all the newly created power to weld with. Unless I’m wrong then to calculate total current of mosfets in parallel and voltage as well is as simple as adding them together, which means Theoretically I should be able to add an infinite amount of mosfets from the same power source to the same heatsink and keep increasing the power?
MOSFETs don’t generate power.
The single phase, 50 Hz input sinewave will be rectified to high voltage DC in a capacitor.
The input is rated in kVA not kW because the rectifier input is not resistive.
220 V * 39 A = 8.580 kVA.
220 V * 30 A = 6.600 kW effective.

The continuous DC output is 155 A * 26.2 V = 4.061 kW.
The 60% duty output is 200 A * 28 V = 5.600 kW.

The output power is less than the input, so conservation of energy is OK. No problem.
I think it needs a higher current outlet. The OP assumed it to be standard.
This got me ferreting around a bit.
As it happens, I have a reasonably good quality 200A inverter stick welder. It came factory fitted with a standard UK 16A plug. At normal 240V, this will supply around 4kW (strictly kVA, I suppose). It welds perfectly happily at the 200A setting.

I can’t see the manufacturer supplying the machine with an under-rated plug. So those 30-39A figures might refer to brief surges, such as switch-on when the cap bank charges. The continuous current consumption at full power must be at or below 16A or the breaker would trip, and the plug would be overloaded, even taking into account some leeway from the breaker.

sysprog
Baluncore