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> Mot Arc Welder, built
Hamlet
Posted: September 21, 2011 10:12 am
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Well if I cant get 5mm wire, then I'll use two 2.5mm parallel wires. Should be equivalent.


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AwesomeMatt
Posted: September 21, 2011 10:58 am
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Hmmm... check that math on that again.

Area = Pi *r^2
= pi * 25

= pi * 6.25

Double the second one, you're only half way there. 4 parallel 2.5mm.

Or more simply... put the two wires side by side. They now span 5mm. Are they a single 5mm cylinder, or is there a lot of material missing?
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CWB
Posted: September 21, 2011 11:30 am
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another way to look at this would be to look up the current carrying capacity for a given diameter of wire .
using two conductors would double the current carrying capacity ... look up this value on the chart to see which size conductor would carry this current .


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Hamlet
Posted: January 23, 2012 11:38 pm
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I'm going to rectify this machine, bought some 160A 1300V diodes for the purpose. Getting information from here:
http://www.tpub.com/neets/book7/27f.htm

As for the inductor, some other MOT is probably best, though I have some fat iron rod laying around (1 inch, 2.5cm).
Anyways, I wanted to simulate the scenario in LTSpice, in which I am noob. Here is what I do not understand:
user posted image
The green line is straight from the "+" and the blue from the load resistor.
It's a simple full wave bridge, but for some reason the voltage gets clipped at around 4 V.
Running it at transient setting, 25 V amplitude and 50 Hz.
D1 N001 N002 DCH251-160H-13 (This is my diode, some USSR stuff, the specs are as follows:
.model DCH251-160H-13 D(Vfwd=2.4 Is=.1u Rs=25m N=3 Cjo=10.5p M=.4 tt=5m Iave=160 Ipk=300 Vpk=1300 mfg=Estbel type=silicon
Is - saturation current
Rs - Ohmic resistance
N - Emission coefficient
Cjo - Zero-bias junction cap.
M - Grading coefficient
60% of this is my guesswork.
A 5 ohm resistor in series seems to solve the problem, giving a nice peak to peak waveform between the V-source and the series resistor, but between the bridge and resistor, the voltage gets clipped again...
user posted image
Blue - between V and R2
Red - between R2 and bridge
Green - at R1
Half wave works fine:
user posted image

Help needed smile.gif

Hamlet




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kellys_eye
Posted: January 24, 2012 12:54 am
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QUOTE (AwesomeMatt @ September 21, 2011 10:58 am)
Hmmm... check that math on that again.

Area = Pi *r^2
= pi * 25

= pi * 6.25

Double the second one, you're only half way there. 4 parallel 2.5mm.

Or more simply... put the two wires side by side. They now span 5mm. Are they a single 5mm cylinder, or is there a lot of material missing?

errrr, we're talking 'cross-sectional area' of the wire aren't we? (assuming the 2.5mm DOES refer to c.s.a.) - NOT width.

A wire with a csa of 5mm is exactly double the csa of 2.5mm therefore you CAN use two wires each of 2.5mm csa to equal a single one of 5mm csa.


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Sch3mat1c
Posted: January 24, 2012 03:04 am
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FYI about SPICE models:

Vfwd (usually VJ = built-in junction potential, I think?) is the voltage drop due *entirely to the theoretical junction itself*. This is typically 0.5 to 0.8V for silicon junction diodes. Especially for high voltage diodes (like yours), additional voltage drop appears from the resistivity of the silicon, which is modeled by RS. If you have a V-I plot for your diode, you'll see for small currents, the voltage is fractional (usually these are semilog plots, so it's a straight line in this region, meaning an exponential curve (because V = IS * exp(N * V/Vth)). At higher currents (usually around rated current), the curve steepens, and either bends over (semilog plots) or straightens out (linear or log-log plots), meaning it becomes resistive. The slope of the line in the resistive region is RS. The intercept of this line back to the I = 0 axis is VJ, the ideal diode drop (using the battery-and-resistor model commonly tought in school).

If you also have a plot of capacitance vs. applied voltage, you want CJO as close as you can to the value at 0V. For a 160A diode, this is probably going to be a few nF. M is the grading coefficient, which controls how fast CJO drops off; this can also be read from the C vs. V plot, but a typical value of 0.3-0.5 will be fine. These parameters don't really matter at line frequency anyway.

TT is the "transit time" of the diode, which isn't *actually* transit time, it's just a means of modeling it using nonlinear capacitance. Roughly speaking, TT is the reverse recovery time, which is usually a few microseconds for slow diodes (I think I once saw 4us for 1N4001; this characteristic depends more on doping than geometry, so it will be typical for most slow power diodes).

IS is the "saturation" current, which is pretty roughly just reverse leakage current. I would guess 100uA or 1mA would be more typical, but this won't really matter here anyway.

I'm not familiar with Iave or Ipk parameters, but Vpk is probably related to BV and IBV, the breakdown voltage parameters. These are pretty close to the avalanche / breakdown / sustain / maximum voltage given in the datasheet, along with an accompanying test current the parameter was measured at. I usually see IBV = 1mA and BV a little above rated maximum voltage (which will be typical data; the datasheet value is guaranteed minimum).

I don't remember what N (emission coefficient) goes into, but a number between 0.5 and 4 is typical for realistic models (I've seen >10 in some behavioral models). It basically scales junction voltage, so it can be used to manipulate forward voltage drop vs. reverse leakage current (IS).

Now to your sim:
You haven't labeled your nets, so I can't tell which one is which, but I'm guessing you're measuring one of the supply voltages, and the "DC" output. Each side of the supply goes to ~0V every half cycle because that side is being grounded by D2 or D4.

Beware: in cap-input rectifier circuits, the diodes are all off inbetween gulps of current. Where the supply voltage ends up, with respect to ground, doesn't really matter, and will end up somewhere in the middle, depending on which diode recovered first, and how much the leakage current varies between diodes. If the leakage is very small, the supply will be undefined (it's literally floating between "OFF" diodes), resulting in a singular matrix or timestep error. The solution is to set RSHUNT or add resistors or capacitors to one, or more, nodes around the source, back to something around ground, to "anchor" it to a known voltage.

By adding a series resistor, you're seeing a sine wave because most of the source's voltage is dropped across that resistor (and you're reading that voltage on the node between them). If you just want to see the source's sine wave, specify a math function, use the difference between the two ends of the source.

FYI: amazingly, there is a glaring, awful mistake on that NEETS page. Which surprises me, it seems to be a pretty informative resource. Figure 4-23 is missing the diode from ground to choke, which allows the choke's current to circulate while the source voltage goes below ground. Without this diode, the DC output will be approximately zero, with a source impedance roughly equal to the inductor's reactance. This is because the single diode stays on almost continuously, so the inductor never gets charged significantly.

Tim


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johansen
Posted: January 24, 2012 06:05 am
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For welder duty I find it reasonable to design for 2-3 joules of energy stored per kilogram of microwave transformer iron.

typically that means a 2-3mm physical gap depending on the size of the core.
efficiency follows kilograms provided the aspect ratio of the core (length and width) remains constant, so find the largest core you can.


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tekwiz
Posted: January 24, 2012 11:48 pm
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Forget the simulations...they require knowing too much data to get any type of accurate results. Rule of thumb is the best for this type of project, as well as looking at how welder manufacturers do it.


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Hamlet
Posted: January 25, 2012 02:25 am
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@Tim, thanks for the input, did not know about the Vfwd. I do not have any plots available for the diode. How do you set RSHUNT? LTSpice help did not give anything, I see it being mentioned on other forums.

@Tek, yes, I'm realizing this. Yesterday I did try to caluclate the specs for the inductor, but... well.. all that matters really is that the relative permeability of steel is 100 and electric steel (I presume MOT's E's are made of that) is 4000. Meaning, I'll go with a stripped MOT core for the inductor. Still good to have the knowledge for running a simulation, if for example, someone wants "documentation".

Rule of thumb for welding rectifiers: get the fattest capacitors (talking in the 100mF range, with really high surge rating) and fattest inductors (40H?), basically. The fewer turns on the inductor, the better, since it drops less voltage (as I understand, most of the high current birdges drop around 2V, adding to the loss). So you need high relative permeability for keeping up the 40H challenge.

Hamlet


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johansen
Posted: January 25, 2012 02:42 am
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lol. 40 millihenries would be far too much.
40H would store 200,000 joules at 100 amps, it would take a minute or 3 to reach full current...

I have used a 300 foot roll of 12 awg wire as an inductor and resistor for welding thin sheet metal, 5-10mH -- it would probably be about right for a TIG setup.


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Hamlet
Posted: January 25, 2012 09:22 am
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I might be off, that is what I read from:
http://www.tpub.com/neets/book7/27f.htm


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johansen
Posted: January 25, 2012 10:02 am
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I forgot to mention that 40H inductor would weigh about 40,000 kilograms smile.gif

You don't really need an inductor on the dc side of the welder.
While the inductor on the ac side of the bridge rectifier gives the welder an ac reactance and allows you to short circuit the output without blowing the breaker, the choke on the dc side does nothing to reduce the short circuit current aside from adding its own dc resistance.

If the inductor is sufficiently large enough to ensure that the arc current is only 30% ac current, then you're going to find that the arc doesn't follow the stick, and will move around at will, and will often wind up finding a shorter path an inch away from where you're welding. i'm not sure what the reason is but arc welding with an ideal constant current dc source is difficult.

A 40H inductor would be reasonable for a 100ma 20,000 volt power supply
the arc of a stick welder would be a resistance of say .5 ohms and as such you only need an ac resistance of 1 ohm to cut the ripple down to reasonable levels. thus, 2-3mH by the simple math of 1 ohm/ 2*pi*frequency.


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Hamlet
Posted: January 25, 2012 06:46 pm
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Well geesh, that really shows how much I know about inductors.
Anyways, I'm finding this miller schematic nice:
http://www.millerwelds.com/om/o316g_mil.pdf
(on page 16)
I found out what the diodes are in SR1. How about the caps, and Z1, SR1? Gonna surf.


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tekwiz
Posted: January 25, 2012 08:44 pm
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THat welder is an ordinary variable coupling buzzbox. As such, nothing is critical. You don't even need the inductor...many similar units don't have them. The inductor merely smooths the arc a bit; it will make virtually no difference in welding.
The caps are something like .1f, 1kv ceramic.
SN1 isn't a diode...it's something like an MOV. In fact, a MOV would work just fine. Minimum of 100V. Maximum 250V. Again, it's there to protect the diodes.
Use diodes of minimum 400V rating, with enough current capacity to handle your desired maximum current. More is OK.


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CWB
Posted: January 26, 2012 03:43 am
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those look like the old "transorb" legends .


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Hamlet
Posted: January 26, 2012 05:25 am
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Some text on protecting diodes with RC, if anyone ever is interested:
http://www.hagtech.com/pdf/snubber.pdf

QUOTE

Transorb was a modified Rho-class shuttle which had been upgraded discreetly by the Rebel Alliance to meet the Rebellion's demanding specifications for use in the Galactic Civil War.


Yea, definitely cool.gif

EDIT:
I did get my simulation running, together with a DC side series inductance - in the range 100-10 mH the smoothing effect is very noticeable. You get an rms value voltage out, say 23 V of the 40 V waveform input. The capacitor barely produces effect on reasonable (eg. 3000uF) levels, but that is on transient settings. .1F range starts to look like slightly underdamped oscillation, with a higher initial voltage knee, but with massive rise time of around 1-2s. Going to run some spikes on it. That 23V brings to a new point of starting a DC arc. I see that the Miller welder I attached has a small additional secondary for the DC operation, giving more voltage. Meaning I'll have to squeeze in a couple more turns onto the MOTs... or put a winding in phase with the MOT configuration at the output to give a jump-start to the arc.


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AwesomeMatt
Posted: January 26, 2012 06:06 am
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BTW... while surfing around looking at MOT projects on the net, I came across your youtube channel the other day. I started watching it and I thought "I swear I know this guy" when I saw the diagonal CDROM case for magnetic shielding, I knew it was you tongue.gif

I think it was the video where you dunked your leads into the water to measure the max load.
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Hamlet
Posted: January 26, 2012 06:23 am
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Oh noes, I'm explosed! tongue.gif


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tekwiz
Posted: January 26, 2012 10:14 pm
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The output may look smoother with an inductor, but you'll never notice the difference welding. However, a MOT core wound with #6 would make a decent filter inductor if you really want one.
Note that the filtering effect of capacitance will depend on the current being drawn.


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For to you, kings & armies are things mighty & enduring.
To him, mere toys of the moment, to be overturned at the flick of a finger.

Fortuna favet fortibus.
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Hamlet
Posted: January 26, 2012 10:30 pm
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Yeah, I guess I'm being a perfectionist, but it wont hurt to add the inductor, I've got a smaller MOT waiting to be used anyway (except for the losses). Around 60-100 A the cap still needs to be rather large, and the effect is not really there for the money that needs to be put into them. So I'll stick with rebuiling the cooling, adding a diode bridge and the inductor in the 10mH range. I'll thest the setups.


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