Looking for diodes for montgomery ward welder

[patrick_g]

Thanks everyone for your input.
I called centrury and they gave me a part number on the diodes 214 004 666
Has anybody have a ref. book to see what size diodes this is?
I have four diodes in my welder. The sales rep. told me all four have the same part number.
If i upsize, how big can i upsize? and
Do i need to upsize all four diodes?
Again thanks everyone for your input.
All info. will be thankful.
samshine

You would need to upsize all of the diodes to have the mod make sense.

There is really no theoretical limit but practicality implies not exceeding maybe twice the forward current capability. Ditto the PIV rating if you want to raise the PIV as a sort of insurance consideration for line surges and such. Beyond that, to my way of thinking, you would be throwing money away. Even upgrading the forward current 30-50% could extend the life of the diodes considerably.

Good air flow and an upgraded heatsink might do as much or more for you as upgrading the diodes. If you can keep the stock parts cooler you will extend their life expectancy considerably. Of course the Cadillac solution is better diodes AND heat sinks with a fan if you don't have one in the unit. Upgrading the fan alone would be an improvement.

I searched the DigiKey and Allied sites for diode part number 214 004 666 and got nothing. I suspect you were given a Century "in-house" part number which may not be easy to cross reference to industry standard part numbers. If you can't cross the Century numbers to industry standard numbers you may find upgrading the heat sink and fan is the easiest way to improve the situation.

Patrick

 

[Samshine]

I want to thank everyone for helping me.
As well thank you Patrick for giving me understanding of diodes as well as others. Thank you all.
samshine

 

[patrick_g]

I want to thank everyone for helping me.
As well thank you Patrick for giving me understanding of diodes as well as others. Thank you all.
samshine

It is my pleasure to help you. Consider it a federal tax rebate since much of my time at university was paid for by the Federal Government starting with the Viet Nam GI bill.

Patrick

 

[Transit]

Do not be overly concerned with the diodes voltage ratings, the secondary side of the transformer is below 100 volts AC, a 200 volt diode should do nicely for a full wave bridge, two diodes in series would equal 400 volts. You want the current rating to be equal or grater than the original. Have you looked into what may have caused the diodes to fail? Replacing expensive components to only fail again is pointless. Look for shorts, cooling fan working, heat sinks large enough. Was the duty cycle exceeded?
Patrick has many good points. Heat sinks need to be thick in order to have the heat dissipate outward through the metal. If you must drill the heat sinks to fit the studs, be sure to de-burr the holes for a good foot print and you may want to use heat sink grease under the studs, you do not need much. And if you replace or add a cooling fan choose a ducted fan. Some cheep fans have big paddle wheels that do not draw air through the case.

Heat sink - Wikipedia, the free encyclopedia

 

[patrick_g]

Do not be overly concerned with the diodes voltage ratings, the secondary side of the transformer is below 100 volts AC[/url]

HMMM I think I would rethink that for at least a couple good reasons. If the transformer were in say a DC power supply running a simple load then OK but... Striking the arc, sticking the rod, and breaking it free and so forth could make some pretty big inductive kicks. I should expect to see voltages well above the no load output winding voltage. Exceeding the PIV rating of the diodes to destroy them does not take much current just mostly voltage so even low energy transients can "punch" a hole in the junction and lead to catastrophic failure. Diodes in general are pretty unforgiving in that regard.

Also note that the PIV has to do with any instantaneous voltage not the RMS of the transformer's no load output. Even with no inductive transients a 100 VAC (RMS) voltage would have voltage peaks over 140 volts. (Peak voltage of an RMS sine wave value is 1.414 times the RMS value.

Pat

 

[Transit]

It has been my experience that exceeding the current or heat ratings have always caused diodes to fail. In a full wave bridge two blocking diodes are always in series adding the PIV limit. One cannot always guarantee that the impressed voltage will be equally divided, however the total PIV should be adequate. A small 0.1 uF cap across each diode should suppress any transient spikes. The other alternative solutions are expensive.

 

[Steve Bosshard]

Hi Transit,

Once upon a time after the days of selenium stacks, we would take a string of silicon diodes, and shunt each diode with a 470 K resistor and .001 uf capacitor to equally divide the reverse voltage. This was in another life.

On the welder the main concern is current when the operator strikes an arc. My YG212 mig machine (totally different animal) has a filter bank, plus a pretty healthy choke. The choke should reduce instantaneous high current spikes at the cost of possibly some pretty big voltage spikes (that the capacitor bank should buffer).

Sure wish we could find something other than a house number for the diodes. Something like a 1NXXX or the like that would cross.

Also like several have said before, it would be great to know what made the first set fail. Sure hate to let the smoke out of $120 worth of diodes.

All the best, steve

 

[Samshine]

I am going to have to look inside my welder tomorrow and find out were the heat sinks are.
After finding what a heat sink is. I know i don't have any next to the diodes. The four diodes are bolted on a i/4 inch alum. plate. Two on one plate horz. and two on the other plate vertical. The plate is bent into a channel shape. One overlaying the other with a 3/8 inch thick plastic or fiber material separating th two plates. The fan is working good. Like i said earlier this welder used to weld 8 to 10 to 12 hours straight in 100 degrees F. when we build a 57 foot steel hull. But them days are over. Most this welder will be used a one time is a hour. And thats pushing it.
I got a question when ohm the positive and negative diodes with a volt meter.
Is it true you will get a reading on one diode and not the other diode?
And if this is true which one will get a reading, or which one will not get a reading?
Thanks for all your help.
Samshine

 

[irvingj]

Just think of a diode as a check valve, a one-way valve for electricity.

It will pass DC current (from your ohmeter's battery) one direction but not the other. Connecting the ohmeter's leads across a single, isolated, diode, you should get a resistance one way and an open circuit (infinite resistance) the other way; if you can get a resistance reading both ways, the diode's shorted internally- fried.

Please note that, even if an ohmeter --very low voltage/current-- indicates a diode's OK, that doesn't necessarily mean it's really good; under a load it could still break down. But if you get readings in both directions, you know it's bad.

There- that's the extent of my electronic expertise!!

PS-- I don't suppose your welder's manufacturuer would give you the specs on your diodes? (Nah, guess that would be asking too much!)

 

[patrick_g]

I am going to have to look inside my welder tomorrow and find out were the heat sinks are.
After finding what a heat sink is. I know i don't have any next to the diodes. The four diodes are bolted on a i/4 inch alum. plate. Two on one plate horz. and two on the other plate vertical. The plate is bent into a channel shape. One overlaying the other with a 3/8 inch thick plastic or fiber material separating th two plates. The fan is working good. Like i said earlier this welder used to weld 8 to 10 to 12 hours straight in 100 degrees F. when we build a 57 foot steel hull. But them days are over. Most this welder will be used a one time is a hour. And thats pushing it.
I got a question when ohm the positive and negative diodes with a volt meter.
Is it true you will get a reading on one diode and not the other diode?
And if this is true which one will get a reading, or which one will not get a reading?
Thanks for all your help.
Samshine

You have found the heatsinks. They are the aluminum thingies to which the diodes are mounted. Larger aluminum thingies, at least as thick as the ones there but with more surface area (more flanges, fins, whatnot) will dissipate the heat generated by the diodes better and keep the diodes cooler. Make sure the air flow gets a good shot at the heatsinks (aluminum diode mount thingies.)

Also, you are getting a lot of technical advice that you may not be able to easily digest. Those mentioning small capacitors AND resistors across the diodes for transient suppression and reverse voltage equalization are giving you good advice. These components may be already in the circuit AND if so they should be checked before you fire the thing up. If they are not there (MFGs do try to skimp sometimes) adding them is inexpensive insurance and dead simple.

Oh, and one other thing (as Columbo used to say) these sorts of technical discussions tend to bring out lots of esoterica and general technobabble among those of us with some background who take turns one upping each other with arcane and likely marginally important FACTS. Not to worry. You really don't need to know/understand a lot of it. Please do not begrudge us our sport. We intend no harm. At any time you can request a straight answer as to what you actually need to do and any number of us can temporarily withdraw from tilting at the list long enough to give practical advice.

Pat