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MOT spot welder / discussion
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All of my electronics is rusty, so my observations are as a curious spectator but not with any authority
By your own math, you're pushing 4kW through a system only rated for 1.8kW (2 900W transformers). Since your pulsing and it's not consistent duty cycle it's probably okay. But I would worry about it when you're talking about doubling your power draw.
From a theoretical position, doubling your voltage should double your power, so that put you at 8 kW instantaneous draw). What's your 220 volt input circuit rated for?
Have you thought about charging capacitors and then drawing from those capacitors for the instantaneous load?
Assuming you have the power capacity for your individual welding loads, I don't see why double the voltage wouldn't be better, as long as there's not delicate electronic circuitry that could be damaged from an inrush voltage you should be golden
Hmm, maybe I should have clarified. You would be correct if I was talking about doubling the overall power. I meant simply adding another turn to my secondary coils so as to raise the voltage and lower the amps a bit or to connect the 2 transformers in series.
Caps seem like a decent idea actually. The house is rated for something north of 40 amps so I should be fine for a ms load.
I did some basic googling, and there's a bunch of different spot welding materials amperage and voltage tables. Ideally you'd want to use the minimum amount of amperage, and voltage for the material you're welding, it's thickness, and the attenuation size of the rest of the material.
Those look up tables, at least some of them, have demonstration welds at the different power configurations which are really interesting
Could you send it over? Seems good to check it out. I'll be on the lookout for a more powerful transformer in the meantime. I also went out and bought a more reputable SSR still rated at 40 amps.
*current draw. That would be four times the power (I^2^R)
P⁰ =i*V⁰
V' = 2*V⁰
P' = i * V'
P' = i * 2V⁰
P' =2 ( i * V⁰)
P' = 2 P⁰