Maximum percentage one breaker can take up of a subpanel?

Question

I am living in Ontario Canada so I am dealing with the Ontario electrical code. I am planning on installing a 240V welding receptacle in my garage. The garage subpanel is fed with a 60A 240V breaker from the main panel in our house. Both my welders pull 40A max at 240V, so to size the circuit properly I think I wire a 50A circuit so the maximum amperage is not more than 80% of rated capacity.

My question is, is there any issue with installing a 50A breaker on a circuit only fed with 60A from the main panel? 50A is more than 80% of 60A, but I'm not sure if this is actually an issue or not. I was hunting around in the Code this morning but I couldn't find anything helpful.

My guess is that there is no issue with it, but because I am not an electrician I really want to make sure I'm doing this right. And FYI, I am not likely to be running the welders at their maximum output often, or at all. If I do, I definitely won't be running other power tools at the same time so overloading the circuit shouldn't be an issue.

Either way I am excited to have more than 15A to play with because that is all I have at the moment. Thank you in advance for your help!

Edit thank you for the speedy responses folks. I will post pics of the nameplates tomorrow. Yes, welders are funky. I have a copy of the 2018 code and it only talks about transformer welders, motor-generator welders, and resistance welders.Both of mine are the inverter type. Maybe the 2021 code will mention them?

Also great point about the future owner potentially using it as a car charger, did not think of that. I am pretty sure all new builds in Ontario now need wiring capable of electric car charging, even if the current owners do not need it.

Here are the photos! The ESAB pulls 38A max and the cheap stick welder pulls 41.

Answer 1

Technically what you should do is a Load Calculation. Which if this is the only branch circuit is trivial: Expected Load = 40A = total = done! Where it starts to get complicated is if you do something like:

• 60A feed
• 50A welder for expected 40A load
• 15A lighting circuit
• 20A receptacle circuit (tools, etc.)

The lighting should be easy. With LEDs the real load might be 1A or less. But the receptacle circuit could be used up to 80% (16A), though more typically 12A (because most consumer tools, heaters, etc. are sized to fit on a 15A circuit) and real-world usage for you may be far less - e.g., a few A for portable tool charging. But the concern would be 40A + 1A + 12A = 53A > 80% of 60A (48A). Whether or not that would be a problem depends on how code treats all of this, which I honestly don't know.

One key thing to keep in mind before you jump to saying "but I know that I will never run anything more than a few Amps at the same time as the welder" is that the next owner might repurpose that 50A (40A continuous) circuit to use for EV charging. And running other tools, heater, etc. while your car is charging is a lot easier than running other tools while using a welder. Code does, to some degree, account for the variety of possible uses in the load calculation process.

Answer 2

I'd allocate 40A on a 50A breaker to the welder receptacle

For load calculation purposes, I would treat the welder receptacle as a 40A load. This covers both the current situation as neither of your welders has an I1eff (effective primary current) greater than 40A, and also permits the next occupant to use the receptacle for an EV charger instead. It also leaves you with 2 20A circuits of intermittent load, which is sufficient for LED lighting and a basic 120V tool circuit.