Welding Machine Increases Power: Physically Impossible?

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SUMMARY

The discussion centers on the apparent contradiction in power output from welding machines, specifically a 200 amp @ 28V DC welder that claims to produce 5,600 watts from an input of 2400 watts at 120V @ 20 amps. Participants clarify that the output current and voltage do not occur simultaneously, and the actual power output is less than the input due to energy conservation principles. The duty cycle, typically around 60%, limits continuous operation and is crucial for understanding the machine's performance. Misleading labels on machines and the importance of accurate specifications are emphasized throughout the conversation.

PREREQUISITES
  • Understanding of electrical power calculations (Watts = Volts x Amps)
  • Familiarity with welding machine specifications and duty cycles
  • Knowledge of inverter technology and pulse width modulation (PWM)
  • Basic principles of energy conservation in electrical systems
NEXT STEPS
  • Research "Welding Machine Duty Cycle" to understand operational limits
  • Explore "Pulse Width Modulation in Inverter Welders" for deeper insights
  • Learn about "Electrical Power Calculations" for accurate assessments
  • Investigate "Common Misconceptions in Welding Machine Specifications" to avoid pitfalls
USEFUL FOR

This discussion is beneficial for welding professionals, electrical engineers, and anyone involved in purchasing or operating welding machines, particularly those seeking to understand power output and machine specifications accurately.

nmatiiq
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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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nmatiiq said:
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
 
nmatiiq said:
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

please refresh my memory :wink:
 
200 A is the highest output current setting. Doesn't mean it occurs at 28 VDC
 
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nmatiiq said:
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:

https://lh3.googleusercontent.com/proxy/Wd_6YwLq3FoweHGgT8CmvMnuissrB_yX5N99dgdeflp1SoSB9rANJmNppsIT9Kjk8seUcIRSW3qci7jznt1-d1zNS0E1rk0boMuDe-b8wHw2xfwP9KxDjA-_4hAjXDOZ22I680JE

1583798865668.png
 
zoki85 said:
It doesn't
berkeman said:
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:

https://lh3.googleusercontent.com/proxy/Wd_6YwLq3FoweHGgT8CmvMnuissrB_yX5N99dgdeflp1SoSB9rANJmNppsIT9Kjk8seUcIRSW3qci7jznt1-d1zNS0E1rk0boMuDe-b8wHw2xfwP9KxDjA-_4hAjXDOZ22I680JE

View attachment 258411
 

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zoki85 said:
200 A is the highest output current setting. Doesn't mean it occurs at 28 VDC
 

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zoki85 said:
It doesn't
That's odd because the back of my welding machine says it does.
 
zoki85 said:
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.
 
  • #10
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!
 
  • #11
hutchphd said:
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.
 
  • #12
I am unaware of the 10 minute stipulation in duty cycle definition. I'll bet the manufacturer also similarly uninformed. Where made?
 
  • #13
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.
 
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  • #14
I am a very novice welder. Is there any possible utility to pulsating DC that would give higher instantaneous current at same average power?
 
  • #15
Baluncore said:
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
 
  • #16
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  • #17
hutchphd said:
I am unaware of the 10 minute stipulation in duty cycle definition. I'll bet the manufacturer also similarly uninformed. Where made?

China
 
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  • #18
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.
 
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  • #19
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.
 
  • #20
nmatiiq said:
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?
 
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  • #21
hutchphd said:
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.
 
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  • #22
russ_watters said:
The nameplate photo says 78A at 110V.
Duh...I didn't read that far...
 
  • #23
Baluncore said:
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?
Reply
 
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  • #24
nmatiiq said:
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?
Reply
MOSFETs don’t generate power.
Baluncore said:
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.
 
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  • #25
nmatiiq said:
... , 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?
Energy can be audited. The account must always balance.
Power is the rate of energy flow or energy conversion.

A MOSFET is a fast switch that has a limited switching current capability. That limits the power one MOSFET can convert. If you need more output current you will need more MOSFETs to do the switching, but the extra energy being converted must come from a sufficiently greater source of energy somewhere. And it will need proportionally more heatsink.
 
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  • #26
nmatiiq said:
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.
It doesn't. Read better nameplates, write better messages.
 
  • #27
I think OP might have bad/misleading labels on the machine. My welder for reference (only a small 90A mig by comparison):
1583870934599.png


Rated amps*rated volts 88A*18V = 1600W, it pulls 20A from 115V = 2300W, unfortunately no free energy here.

Note that the duty cycle in welder terms is a thermal limit. Due to the size of the transformer this has fairly slow time constant, and it usually is telling you how much you can weld in 10min:

1583871161890.png
 
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  • #28
nmatiiq said:
. 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?

Yes, you are wrong
You cannot produce more power out than power in, therefore you cannot just keep adding
mosfet's to create more power out of thin air
 
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  • #29
nmatiiq said:
That's odd because the back of my welding machine says it does.
davenn said:
Yes, you are wrong
You cannot produce more power out than power in, therefore you cannot just keep adding
mosfet's to create more power out of thin air
The point of having several MOSFETS is to protect the MOSFETS and not to produce more power. This Conservation of Energy thing really confuses people and that suits the manufacturers fine.

My cheap and cheerful MIG welder just has Hi/Low and 1/2 settings on it. It just about works but I'd love the chance to play on a proper set. ) Then I'd show 'em.
 
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  • #30
essenmein said:
I think OP might have bad/misleading labels on the machine. My welder for reference (only a small 90A mig by comparison):
View attachment 258495

Rated amps*rated volts 88A*18V = 1600W, it pulls 20A from 115V = 2300W, unfortunately no free energy here.

Note that the duty cycle in welder terms is a thermal limit. Due to the size of the transformer this has fairly slow time constant, and it usually is telling you how much you can weld in 10min:

View attachment 258496
Wow. Forgive me, but your welder must be very irritating to use. Most give the duty at full output, and also the output at 60% duty, which I assume is considered reasonably usable. In a typical project, the time taken between welds to set up, clamp, mark etc means a 60% DC won’t cramp your style.

Given that a 20% DC means 12s on, 48s off, over the whole project it would take the machine a full minute to do each 2” or so of weld bead.

When buying a welder, the advice is to get one that has at least 60% DC at the output you will tend to use, rather than to look at the maximum output. The quality can be estimated by how the DC falls off as output rises.
 

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