Highish Current/Lowish Voltage Rectification

The DS specifies a 22 x 22 x 5 cm heatsink. Is that a single block of aluminum or something with fins in one direction ? Dissipating 100 W at 10 .. 30 C heatsink/ambient temperature difference is a bit challenging without fans :-)

I think welding transformers often limit the current by deliberately allowing a high leakage inductance in the transformer, pretty much equivalent to an inductor in series with the secondary.

If your arc impedance increases suddenly whilst a large current is flowing in the windings, I would expect a very high voltage to be induced in the leakage inductance. You might want to add something to your circuit to protect whichever diodes are reverse biased from breakdown due to the voltage spike. Perhaps a very large varistor across the rectifier (though this may eventually wear out), or a big capacitor, or some active circuit.

Chris

Close enough.

Dan

It's called 'second breakdown' and usually discussed ion the context of transistors (really, the transistor emitter is where it happens).

That's an unfortunate example. For a welder application, the tempco should be examined only over the range of allowable repetitive peak currents. That implies that one is never operating in the zero-to-positive tempco region above 100A.

The fact that the fig. 1 semilog plot has a curve, though, does indicate that this is not an ideal diode, but includes significant series resistance. The series resistor, not the ideal diode, dominates at high forward bias.

They are not really in parallel, there is one in series with each bridge rectifier to improve load sharing.

?-)

One can parallel up diodes by using a small resistor in series with each one,to balance currents and prevent "hogging".

I've actually allowed for a fair bit of derating. For example, I'll be using 5 of these hunks which arrived yesterday:

And when I stated I would like to retain 200Amps of output capability, in practice I expect to use no more than half that amount for TIG and only one quarter for, or 50A if you prefer, for the plasma cutting configuration.

I'm more concerned about generating the correct amount of plasma-starting voltage, which I don't expect for one minute to be able to get from 80V without some form of step-up. I have yet to nail down the exact specifics in this one remaining area. Maybe a transformer from a microwave oven would do the trick....

I've got some 20" of 14AWG nichrome resistance wire for that purpose. :-)

The curvature is way past the stated operating current. Also, did it "become ohmic" or is the resistance that was already there now becoming more significant in the VI curve.

How hard is it just to get a diode at the rated current? I see no logic to use a part that doesn't meet the spec. You get current crowding, some hot spot occurs, and then it fries. This doesn't seem like a plan to me.

I suppose if you like to play in the FA lab, then stressing a part is a good plan.

Sorry, that should be 14 *SWG* resistance wire.

Bit more research indicates the plasma starting requirment to be between 5kv-10kv at 2Mhz. Have to have a think about how best to obtain that from the 230v 50Hz mains standard here.

google hf start

-Lasse

You don't need 2Mhz, that is just the convenient frequency for the coupling transformer in use. There are a few ways for generating the arc to the tip, the above is the most practical way.

Jamie

You might be disappointed; those are rated at low temperature, and with 200A current, it won't stay low long. At 2.0C per watt, 200A (even split 5 ways) means 225C operating temperature (assuming an INFINITE HEAT SINK).

SCRs with 110A and 400V ratings are about $50 each, and four of 'em might be your best bet. They won't need any ballast resistors, just connect 'em as a bridge and tickle the gates appropriately. These

have nice beefy integral studs, they heatsink very nicely.

As has been pointed out, the 2MHz has nothing to do with the arc itself.

Starting an arc involves getting electrical conduction through a gas. Initi ally you need enough of a voltage drop across the gap so that any passing c harge carrier - usually an electron produced by a cosmic ray or a little po tassium in the vicinity

gets accelerated enough to ionise an air molecule when it hits it, giving y ou avalanche breakdown.

In air at standard temperature and pressure, you need 3.4MV/metre, or 3.4kV over a 1mm gap. As the gap shrinks to 7.5 micron the electric field you ne ed increases, and the minimum voltage to start an arc in air is 327V at 7.5 micron. Narrower gaps need a higher voltage - they are enough narrower tha n the mean free path of an electron in air at STP that most of the electron s make it across the gap without hitting an air molecule to generate new ch arge carriers.

The initial spark discharge is initially sustained as a glow discharge, rat her than an arc - the electrons that carry most of the current are created at the cathode by positive ion bombardment, which needs a couple of hundred volts to work.

The surface of the cathode gets very hot, very rapidly - within a few micro seconds - and - if the current density is high enough to sustain an arc dis charge the surface gets close enough to melting to distort into a spiked st ructure under the electric field, and you get thermally augmented field emi ssion from the tips of the spikes, which only needs about a 20V drop to sus tain the discharge.

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Bill Sloman, Sydney

Wow! Useful advice from Bill Sloman for the first time in 16 years! You even sound as if you know what you're talking about, too. Try to get the formatting sorted out, though!

itially you need enough of a voltage drop across the gap so that any passin g charge carrier - usually an electron produced by a cosmic ray or a little potassium in the vicinity

g you avalanche breakdown.

4kV over a 1mm gap. As the gap shrinks to 7.5 micron the electric field you need increases, and the minimum voltage to start an arc in air is 327V at 7.5 micron. Narrower gaps need a higher voltage - they are enough narrower than the mean free path of an electron in air at STP that most of the elect rons make it across the gap without hitting an air molecule to generate new charge carriers.

rather than an arc - the electrons that carry most of the current are creat ed at the cathode by positive ion bombardment, which needs a couple of hund red volts to work.

croseconds - and - if the current density is high enough to sustain an arc discharge the surface gets close enough to melting to distort into a spiked structure under the electric field, and you get thermally augmented field emission from the tips of the spikes, which only needs about a 20V drop to sustain the discharge.

The formatting looks fine here. It isn't the first time that I've posted th is information here - I had to dig it out of text-books when I was trying to s tart and drive an arc lamp back in 1972. The contexts that call for this in formation do vary. And I have posted other useful stuff. John Larkin usuall y found it tedious because it didn't address the problem that was obsessing him at the time, or endorsed an approach that he'd already rejected.

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Bill Sloman, Sydney