Spot welding

[Nick Hulme]

The last thing you want for spot welding is more volts, heat generated in a conductor (anything you are welding) is relative to current squared times resistance.
This is why long distance power transmission is at higher voltages, lower losses for a given conductor.
What you need is the lowest possible voltage source, this increases the current & thus heat, for any given power input,
Regards,
Nick

 

[Kludge]

The last thing you want for spot welding is more volts, heat generated in a conductor (anything you are welding) is relative to current squared times resistance.

Darn - power equation. Hadn't even considered that. Thank, Nick.

BEst regards,

Kludge

 

[rawblacksmith]

i'm so behind in learning electronics and higher math.

really thankful for the internet ..the best course for mastery comes from finding out how much ther is to know!

eh i'm glad i got someone with a math degree in the house!

 

[batt-man]

The last thing you want for spot welding is more volts, heat generated in a conductor (anything you are welding) is relative to current squared times resistance.
This is why long distance power transmission is at higher voltages, lower losses for a given conductor.
What you need is the lowest possible voltage source, this increases the current & thus heat, for any given power input,
Regards,
Nick

I'm not sure i understand your staement above. Current is equal to volts divided by resistance; in other words for any given weld (which will have resistance, the value being dependent on the materials being welded plus the resistance between the materials where they touch) the ONLY way to get more current, and therefore heat, is to increase voltage.

 

[gunboatbay]

If you're not locked into building your own, you might want to take a look at this:
http://www.harborfreight.com/cpi/ctaf/displayitem.taf?Itemnumber=45690

Every once in a whaile they have it on sale for $139.95 US

 

[Mcgyver]

I'm not sure i understand your staement above. Current is equal to volts divided by resistance; in other words for any given weld (which will have resistance, the value being dependent on the materials being welded plus the resistance between the materials where they touch) the ONLY way to get more current, and therefore heat, is to increase voltage.

sort of, it definitely obeys ohms law but what you're missing is that there is always a practical constraint on power and transformer behaviour means you wont necessarily drop all the open circuit potential across the arms when closed.

lets say resistance is .001 ohms.

at 1.5 volts (a realistic spot welder potential) it'll flow 1500 amps.....say about 7.5 amps on 220V circuit, more than doable.

if you increase the voltage, yes amps increase but look at what happens to power...

at 11.5 volts, it'll flow 15,000 amps, yeah that'll spot weld and then some, but where's that going to come from? you'll instantly blow the breaker.

its amps that spot weld, so you've got to step down voltage so that you can get amps high enough (for mine I'm shooting for 3000) without hitting the wall on how much power you're trying to put through the panel/wires/etc. Transformers are rated in volt/amps so in the context of spot welders, when someone says you're better of with lower voltage it kind of means because then you are increase the amps out of the transformer (I know its not 1:1). For example if I put an extra turn on I'd be doubling the voltage and halving the amps before the transformer saturated whereas what i want is to increase the amps.

Things get more complex in design as you are not necessarily dropping all the potential across the arms, there are transformer losses etc....your open and closed circuit voltage reads on say the arms can be quite different.

the electrical engineers can elaborate but I hope that gives you some hints on it

 

[Nick Hulme]

I'm not sure i understand your staement above. Current is equal to volts divided by resistance; in other words for any given weld (which will have resistance, the value being dependent on the materials being welded plus the resistance between the materials where they touch) the ONLY way to get more current, and therefore heat, is to increase voltage.

Nope,
you're making an assumption that your power supply is capable of delivering infinite current limited only by the externa load and you are completely wrong.

While V = I*R and V/I =R and V/R =I are all true, they are in this context irrelevant, what I was pointing out was that heat generated in a conductor (including the workpeice) is relative ( or proportional ) to Current * Resistance.

1. Heat generated in a conductor is relative to Resistance Times Current, I'm not making this up, it's a rather inconvenient Physics fact, and as I stated is why long distance power transmission is done at higher voltages.

2. Double your voltage & you get half the curent for the same power so given that the above statement is true (and it is) if you double your voltage you need four times the power to do the same job when resistance welding - think about it.

Your statement would be correct when the voltage delivered is insufficient to drive a resistance welding current through the work, however the problem is almost always insufficient current delivery not insufficient voltage,

Regards,
Nick

 

[Kludge]

Hold on a second. Since what I was talking about was a cap discharge spot welder, the power source has a while to charge the caps between shots plus isn't providing anything during a weld. With appropriate limiting resistors to prevent a massive surge at the beginning of the charge cycle, we can set the maximum current at the wall socket to some reasonable level.

The capacitors and wiring will provide some resistance during the welding process, however slight it may be, but it will add to the total load about the same amount whether it's low voltage or high voltage. THis is independent of the power source used to charge the caps.

One other thing that will affect the "zap" is the capacitance. Less capacitance, less stored charge at a given voltage. Crank up the voltage and a few factors change and I think one of which is the internal resistance of the capacitor. I need to review here a bit but I still think there would be a net increase in current with increased voltage though I'm not sure how much caacitance I'd need at 450 volts to match a 40v charge on one farad, taking into account circuit resistance.

Just some thoughts from a med-fuzzed brain.

Best regards,

Kludge

 

[Nick Hulme]

Kludge,
yes Capacator discharge is a different matter to transformer, you have a shot at very high currents, for very small light jobs I think it's do able.
For anything structural I suspect caps with the spec required have a good chance of outstripping the cost of an off the shelf spot welder,
Regards,
Nick

 

[Kludge]

Well, work at the sizes I do, cap discharge can even be overkill sometimes though it's nice for assembling battery packs. I really have no idea where the cross over would be as far as cost compared to more conventional ways. Since I can put together a fairly fat condensor (what we called them when we had saber tooth tigers for pets ) pack rated at 450 volts vs 48 volts using more current technology without spending a whole horrid amount of cash, it may be interesting to try a few experiments to see what happens.

*sigh* ... my experiment list is growing. Is anyone keeping track of where I am on what?

Best regards,

Kludge

 

[cfellows]

Well, work at the sizes I do, cap discharge can even be overkill sometimes though it's nice for assembling battery packs. I really have no idea where the cross over would be as far as cost compared to more conventional ways. Since I can put together a fairly fat condensor (what we called them when we had saber tooth tigers for pets ) pack rated at 450 volts vs 48 volts using more current technology without spending a whole horrid amount of cash, it may be interesting to try a few experiments to see what happens.
Kludge

Kludge,

What do you make your capacitors from?

Chuck

 

[Bluechip]

Hi Folks

This seems to be going on for ages. I would have thought you had 'em all built by now. I think Kludge mentioned 'Computer Grade' capacitors before. Seen this link ??, someone with worse hand drawn schematics than me, it might have some pointers. No interest in blasting caps. off the face of the planet myself, so you're on you're own.

You need 'Low ESR' caps. of some description, and VERY expensive !! At least in UK. Sterling approaching parity with cat-crap

http://www.philpem.me.uk/elec/welder/

PS the energy in a cap. is 'half x C x V squared'. In joules. IF you don't know.


http://www.philpem.me.uk/elec/welder/images/schematic.png

Edit Sorry, wrong link. :













Hope you survive ...

Dave

 

[Kludge]

What do you make your capacitors from?

Oops, bad wording. I apologize.

I don't make the caps, I happily join banks of them together to create what I need. It's fun and entertaining and keeps the ward attendants on their toes. ;D

Best regards,

Kludge

 

[Kludge]

This seems to be going on for ages.

Well, yeah. Good topics don't die. They just ... ummm ... actually, I have no clue what they do.

You need 'Low ESR' caps. of some description, and VERY expensive !! At least in UK. Sterling approaching parity with cat-crap

They're not so very cheap here either. Well, nothing's cheap when "here" = "Hawaii" but these caps are also uncheap when "here" = "US."

http://www.philpem.me.uk/elec/welder/

"Computer grade" caps just shrug off extremely low resistance discharge paths but I'm not convinced the "auto grade" caps can say the same thing. THis is one area for testing along with newer tech electrolytics for the HV versions. They're small, cheap and available.

PS the energy in a cap. is 'half x C x V squared'. In joules. IF you don't know.

Ah, yes. THis looks familiar. C in farads, unless I want to play with fun decimal adjusting. Thanks!

BEst regards,

Kludge