Capacitor Powered Spot Welder

guido

Well-known Member
Hello,

I Had these capacitors around, so I paralleled and charged them. The charger stopped at 17 volts or so. They powered an 1157 bulb for a few seconds, the voltage then dropped to zero. I'm wondering if the charging time make a difference on the capacity?

Guido.
a187433.jpg
 
I was taught the charging time is a function of capacitance and resistance, one time constant is T = RC. Resistance times capacitance. It takes 5 time constants to charge to about 99% of applied voltage.
 
I once worked with a man that was bad as a coon about picking things up to fiddle with them. I laid a charged ignition condenser on my desk for him to examine when he droped by.
 
The reason they stopped charging at 17 volts is because that's the peak voltage of the full-wave rectified output of your battery charger. The only way to further charge the caps is to raise the charge voltage.
 
Well sir that is the way I was taught also. So you have a battery charger with some internal impedance (resistance plus any reactive/regulatory components) and a peak output voltage capability (essentially at zero current....when the C is fully charged.....99% as you stated) as Mark B stated.

The circuit in question is unlike an RC time constant in an electronic circuit unless you are talking about power supply capacitors for that circuit. So it takes multiple pulses of power (rectified power line AC) to fill the "joule capacity" of the capacitor but is still can only go as high as the power supply (battery charger) can put out. Only variable here is the "RC time constant" as to how long that will take. Obviously the higher the charger capacity in output amperage, the faster it will occur as will happen as the size of the C is reduced.

We're on the same sheet.

Mark
 
Once the caps were charged to the peak voltage available from your charger as shown, the rest is up to the size of the capacitors in Farads (unit of capacitance derived by Faraday a brilliant scientist many many years ago) and the size of the "load"....thing draining amperes out of the caps...in your case the light bulb. How long you let the caps charge (after you reach your 17v with this battery charger) is immaterial as they have all the capacity they're going to get at that applied voltage.

Google says a 1157 bulb consumes 27 watts when at normal rated brightness. Your capacitors, with the voltage on your meter will contain so many "watt-seconds" of energy (power consumption waiting to happen). As the energy is consumed, the voltage will fall and as a result so will the current and your light will get dimmer and dimmer till it is no longer visible and finally deplete the caps of all their stored energy.

The equation for that is Wc (energy stored in the capacitors) is 1/2 the product of capacitor(s) value (in Farads) x the applied voltage (in volts) squared. So if you assume a 35 uf (uf is the notation for micro farads which is 0.000001 Farad) cap, x 2 caps you have 70 exponent-6 x 17 volts squared all divided by 2 for a watt-second rating of .01 Joule (Joule...unit of watt seconds of energy) .

If it were a straight line consumption, which it's not as the bulb has resistance and the consumption follows the RC time constant logarithm we discussed in your other post, you would exhaust the contents of the capacitors in less than a second. As it is, with consumption being as it is it will take a few seconds as you realized.

The kicker here is voltage. Since it is a second order equation, it, like previous discussions of 2 squared is 4 and 3 squared is 9, a change in voltage has a tremendous effect on how much energy your caps can hold, like in the example below, a 10x change in voltage gets you a 100x change in stored energy and a 100x improvement in how long your bulb will burn...course you can't use a 12v rated bulb for that.

For example since you had a Joule capacity of .01 at your 17v, if you upped the voltage by an order of 10x (170v) your stored energy would be (70 exp-6 x 170 squared)/2 or 1 Joule......an increase in capacity of 100x. When you get up into the thousands of volts, you can store a whole lot of energy in a very small (high voltage rated) capacitor which saves space and is used in things like airborne electronic systems.

So, now that we got through all that, what about a capacitor spot welder? As I said I know nothing but I'll dig around on the www and educate myself. What is the basis for your question so that I will know what to look for? What do you want to do?

Mark
 
I have used capacitive discharge stud welders at work. For that application they work pretty well. But it was very finicky about grounds. Also you had to make POSITIVE that the leads were straight and not coiled up. The added inductance made it not weld correctly. It was a great way to add #10 studs to sheet metal.

Cliff(VA)
 
A fellow built a capacitor welder in school to weld a hair diameter hot wire into some air flow thing he was building. Trial and error getting the charger right.
 
One thing to remember is that electrolytic filter capacitors do not like sudden heavy discharges, they frequently short if discharged too quickly. You need special capacitors designed for the high sudden current if you expect it to work repeatedly.
 
Hey Guido. Check out:" http://zeva.com.au/Projects/SpotWelder/". I see what you want to do and this looks like a pretty good step by step process.

Pay particular attention to the values of the components.....where capacitors are normally measured in microfarads (0.000001 x farads), these capacitors, 6 of them, are 670 FARADS which are a million times larger in capacitance than normal capacitor ratings. Only saving grace is the DCWV is only 17v. Otherwise they would be the size of your shop. I was wondering how you were going to do much welding with a couple of capacitors and this is how. It takes a lot of Joules and for that it takes a lot of capacitance at a low voltage as I mentioned earlier.

You use the same calculation as shown there, energy for one pulse measured in Joules (watt-seconds......volt x amp x number of seconds or parts of a second of the pulse) = 1/2 the capacitance value in Farads x the voltage to which the capacitors are charged squared.

He has the formula in his paper, but uses the symbol E for Voltage. E is actually field intensity and in my line of work related to electromagnetic waves radiated in space. Volts is what we used as volts is the difference of potential between 2 points where 1 volt will drive a current of 1 ampere through a resistance of 1 ohm, or a current of 1 ampere through a resistor of 1 ohm will develop 1 volt across it.....aka Ohm's Law.

Good luck,
Mark
 
Something to keep in mind is personal safety. These things hold a huge amount of energy and if you get it across you it can kill you! It would be well past the milliamp levels that makes your heart wobble. It would burn you, externally and internally, just like it makes the weld. A 12v battery could do the same thing if properly connected to you but we seldom if ever have the right things exposed across the battery terminals for that to be a problem.

Mild case in point. Went fishing one day. Before launching saw a guy with a boat battery problem. Went to his car and took the car battery to the boat. Was standing waist deep in the salt water and was trying to put the positive terminal wire on the battery with the negative side of the battery connected to his outboard which was in the water as he was. It didn't occur to him that his body resistance was lowered adequately to pass enough current through his body, even with 12v to give him a good shock. Lucky he was able to jerk his fingers off when it bit him.

Be careful.

Mark
 
Hello Texasmark1,

I will read all the replies, when I get a chance. Things got crazy busy in 24 hours around here. Know that I started small when it comes to capacitors. Check out the picture. They are NEC 0.10 farad 5.5volts caps. The bigger one is rated at 5 volts 0.22 farad. Thanks for your info. A 1157 bulb will draw 1.75 amps at 12volts for one element, and about 1/2 amp on the other I think. have not checked recently, but I think it is close? I have both elements soldered together.
Got to run.........


Guido.
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