I’m more than a bit confused about the behaviour of my spot welder transformer…

the secondary winding has only a couple of turns of very thick gauge wire with an open circuit voltage of ~2v. Current i measured (via a long convoluted way) to be circa 370a.

if i reduce the turns to a single turn, the open circuit voltage drops to ~1v, but the current i measure is less at about 200a. Surely the current should increase??!

similarly when i increase the number of turns and get about 5v, i can no longer measure current (too high for my meter), but the cables all start heating up massively, smoke comes out and the whole thing vibrates like mad, which leads me to conclude that im getting even more amps, maybe 600-800a??

so im v confused whats going on. Im sure if i increase the turns more with volts increasing current will start dropping but i wasnt expecting this sort of non linear behaviour. Any thought would be great as this is really bugging me now!

 

That ridiculously thick Secondary-Winding is not very efficient in it's design and implementation.

It simply becomes marginally more "efficient" with more turns of Wire.

The Mass, and the particular design, and the materials chosen, to manufacture the Transformer-Core,
will determine the maximum amount of PEAK-POWER that can be transferred through the Transformer.

The ability of the Windings to dissipate HEAT will determine the long-term POWER
that can be transferred through the Transformer.
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Ahhh i see, cant tell you how long this has been bugging me, thanks!

i did wonder about efficiency, but ruled it out since the results i saw (doubling amps with doubling voltage) would mean that efficieny improved four times (i think)..

i also wondered about resistance changing with current but the effect of that would be opposite to what was observed. (Its also what you refer to i think re heat management).

finally, i think i know what the shunts are for now!! I removed them and thats why the single turn maybe gives such a poor efficiency. I will close the gaps with shunts to see if they help..

will report back, thanks again!

 

A MOT has a very poorly designed Transformer-Core and a Primary-Winding that
does not have enough Primary-Turns to limit the "Magnetizing-Current" to a reasonably low level.

It is not very efficient, but the price is right.
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if i reduce the turns to a single turn, the open circuit voltage drops to ~1v, but the current i measure is less at about 200a. Surely the current should increase??!

Half the voltage into the same load and you expect the current to go up? That would assume the current was limited by the transformer turns ratio only, there is also resistance limiting the current, which, in this case, dominated.

It sounds like the original number of turns was chosen exactly correctly.

 

if i reduce the turns to a single turn, the open circuit voltage drops to ~1v, but the current i measure is less at about 200a. Surely the current should increase??!

Why would you think that??!.

The current is determined by the transformer open-circuit voltage divided by the resistance of the circuit load (Ohm's Law).
So, since the load resistance doesn't change, it's no surprise that the current changes as you have seen, with changes in the output turns and voltage.

 

my reasoning was that if we assume 100% efficiency (i know a wild assumption)
power in~700W (@240v, 3a), power out 700W @2v implies lower current (350a) than @1v (700a)

this doesnt hold since efficiency changing significantly at different windings/voltages? and also resistance increasing with higher currents?

otherwise surely, i just keep increase turns/volts and current increases too? if i assume resistance stays the same, i will get 9000a at 50v? (50/0.005) (see below)

also, @2v, i get more than 100% efficiency?! 720W Pin, 740W Pout (see below)

[confused face]

	volt	amp (measured)		Power (calculated)
primary	240​	3​		720​
	volt (measured)	amp (measured)	resistance (calculated)
secondary	1​	200​	0.0050	200
	2​	370​	0.0054	740
	50​	9,250	0.0054	462,500
		calc amps	assume same resitance

 

my reasoning was that if we assume 100% efficiency (i know a wild assumption)
power in~700W (@240v, 3a), power out 700W @2v implies lower current (350a) than @1v (700a)

You "wild assumption" does not apply to what you were testing.

 

Typical industrial spot welders have one single turn on the secondary.
If using a MOT version, ensure the magnetic shunt is punched out/removed,
Also effectiveness of the weld current is also somewhat affected by the dia of the weld points.

 

Typical industrial spot welders have one single turn on the secondary.
If using a MOT version, ensure the magnetic shunt is punched out/removed,
Also effectiveness of the weld current is also somewhat affected by the dia of the weld points.

shunts are removed, i thought this led to the low efficiencies and i was intending to put them back in??!

with a single turn i get 1v/200a, not enough to spot weld paper!

i got reasonable output from ~5 turns, (5v/~900a?).

im looking to upgrade the system by putting two microwave transformers in series (secondary), and using thicker gauge wire.

i wanted to understand the dynamics (V, I, R, eff etc) better before doing so...

[i have also had a headache with measuring these currents since my clamp meter only goes up to 100a. i will post seperatley about that!]

 

Also insure that secondary conductor cross section area matches the current.

 

A Spot-Welder's performance can easily be seriously reduced, or even stopped all together,
by the Resistance of the Welding-Probes, or even surface-rust on the parts being welded.

This is why the Welding-Probes are generally set up so that several-hundred pounds
of force can be applied to insure that a reasonably low Resistance
contact can be established through the parts being welded together.

200-Amps is serious business,
IF You can actually get the Secondary-Circuit-Resistance low enough to actually cause
200-Amps to flow in the completed-Circuit.

It seems that You are making too many assumptions with all of your Math-Calculations.

If You can get the Resistance low enough,
your biggest problem will be not melting the insulation off of
the Secondary-Winding in a matter of seconds.

The actual amount of Current flowing in the Secondary-Circuit,
caused by the unknown, "actual", Secondary-Circuit-Resistance,
will always be somewhat less than what You may predict it to be with a Math-Formula,
because the overall Secondary-Circuit-Resistance is extremely difficult to measure,
and can easily change because of a variety of factors.
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