Monday, August 20, 2007

DIY Arc Welder - Adding DC part 2

The rectifier bridge and the choke are installed and correctly wired. It occured to me that I may need to explain how I constructed the rectifier bridge, since it looks complicated.

It's really not - how it works can be found here. The basic construction consists of four 100-amp diodes with independant heatsinks wired in the configuration explained in the article. I tapped the heatsinks and bolted the rectifiers on. A liberal application of arctic silver heatsink paste was used to ensure good heat transfer. Then the heatsinks were attached to an insulating plastic base which is attached to a wood plate I varnished for longevity. All this effort is to ensure that the heatsinks do not short circuit on each other or anything else, and remain firmly in place when vibration, heat, etc are present.

You may be wondering, isnt this welder capable of more than 100 amps? Won't this exceed the diode current ratings? Ah! the secret, look back at the wikipedia entry - only two diodes are actually on at any given time, so the average current works out to be 100 amps for each diode even if the welder is providing 200 amps! The diodes also have a peak repetive surge current rating of 200 amps, no there should be no problem. I actually would have bought beefier diodes but these were a great ebay auction at only $20 for all 4 - couldn't pass it up.

I have measured the output voltage and it is now indeed DC.

I also received a voltmeter and ammeter to monitor the power output of the welder when welding. This will also help me gauge how well the voltage-sensing MIG wire controller is working when I build it. I bought a beefy 200 amp shunt to provide the 50mV to the ammeter to measure the current.

The shunt is pictured above (its the large thing in the lower right corner), and is mounted with fiberglass strips to keep it isolated from the wooden chassis. It is inline with the welder's grounding cable.

Shunts are curious- they are just big resistors that work by providing a very precise and low resistance (basically a short) but even shorts have a measurable resistance. All resistance causes a voltage drop, and this particular shunt is designed to provide a 50mV (yes, milla-volt) drop when 200amps passes through. 50mV just happens to be the full deflection range for an analog meter element. So if we hook these two tiny screws up to a meter, it will deflect proportionatley with the current. If the meter has a 200amp scale printed on it, this will tell us how many amps are passing. Cool, eh?

The main thing to remember here is that we are working with ohm's law. Since we know two values, we can calculate the third.

Fully deflected,
V=IR, so R=V/I= 50mV/200A=.00025 ohms....or 250 nano-ohms.
Not much eh?

In fact, you can see the calibration notch on this shunt in the lower part of the black strip in the middle (it is a thick copper strip painted black). Some technican built this thing, and then hacksawed it a bit to get it juuuuuust right. So why so low a resistance? Well mainly, we want to minimize how much power is dissipated in the shunt. Power = Volts * Amps, and a larger resistor will result in a larger voltage drop.

1 comment:

benthomasson said...

how does the shunt work?