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Hi I am making a 130 amp DC welder with IGBT transistors.
i have some 56 amp diodes for the bridge rectifier.
my question is do the 4 diodes in the bridge share the current which would be ok for the 130 amps . if not can i parellel them to increase the current
thanks Peter
NO! It's a little complicated to determine the exact current you can get out of a your diodes in a bridge, but 130 amps would be excessive.
If you have specs on the diodes you will see that ideally they are rated for average current, RMS current, and peak surge current for some limited time period (like a half-sine wave pulse). Search some app notes from the diode manufacturers on how to calculate the RMS and average currents.
Your diodes do not share the current in a full bridge, although only two diodes are conducting for each half sine wave. You would have to compute the average and RMS current to get an accurate idea of what your diodes could carry.
Beware, however, that the current ratings are for a specific junction temperature or heat sink and air temperature and ventilation condition that may be optomistic.
You can parallel identical diodes, but to have the diodes share the current equally (to get double the current of a single diode) they have to be matched for forward voltage at high current AND must be kept at the same temperature - not a simple task in real life. Manufacturers parallel diodes to lower costs of production of low end machines. However, they have the capability of accurately matching the diodes they install in parallel. In the absence of such matching capability, you would be asking for trouble to parallel diodes from a small sample.
many thanks for the reply
the diodes are 56 amps RMS and I will sourse higher current ones for my project. i bought 6 x 200amp 200 volt transistors on a big heat sink for $30 from the scrap metal yard , which is the basis for a DC tig welder, so no problems spending a little extra money.
regards peter
Yes, that's true the manufacturers can insure that all of the diodes were cut from exactly the same peice of silicone. You can't do this so you'll have to buy a bag of 100 or more to pick out a close enough match for four diodes.
Don't worry, this doesn't apply it you're building a bridge rectifer, just if you want to parallel some diodes. The good thing about a bridge is because only two diodes conduct simultaniously you can use doule the RMS current.
Yes, that's true the manufacturers can insure that all of the diodes were cut from exactly the same peice of silicone. You can't do this so you'll have to buy a bag of 100 or more to pick out a close enough match for four diodes.
As with anything, you can't guarantee exact matching - so you shouldn't try. Simply add low value resistors in series with each diode - this shares the current equally and helps prevent premature diode failure.
Obviously with the currents involved here, these resistors need to be VERY low values.
Yes, that's another approach. However, I'm not sure it would be worth the hassle and cost just to allow you to use diodes on hand that are not really best suited to the job.
You will have losses in the series resistors. About the lowest value resistor that is readily available "off the shelf" is 0.005 ohms (in the Mouser catalog, although I presume a search will locate lower values from some other source). 50 amps through 0.005 ohms gives a 0.25 volt drop and will dissipate 12.5 watts. Not a major problem, but a value that cannot be ignored. A suitable resistor is the Caddock MP2060-0.005-5, 18 WATT, 5% (Mouser Stock No. 684-MP2060-0.005-5) at $9.33 each. The 0.25 volt drop would be more than adequate to swamp out any variations in forward voltage drop of the diodes.
I think your money and effort are better invested in obtaining diodes rated for the application. (Now I fully expect Nigel to come up with a source for 50 cent resistors with a value of .001 ohms. Such a resistor would give a 0.05 volt drop for current equalization and dissipate 2.5 watts at 50 amps. That would change the cost/convenience equation.)
I think your money and effort are better invested in obtaining diodes rated for the application. (Now I fully expect Nigel to come up with a source for 50 cent resistors with a value of .001 ohms. Such a resistor would give a 0.05 volt drop for current equalization and dissipate 2.5 watts at 50 amps. That would change the cost/convenience equation.)
No, I would suggest making resistors, rather than buying them! - short lengths of resistance wire, or longer lengths of normal copper wire would be all that's requied.
I'd recommend a value that would give double the range of volt drops in the diodes. Suppose you need 1A and one diode is 0.65W, and the other is 0.7V then use a restor that will give a 100mV drop which 0.1 ohms. Am I right?
I'd recommend a value that would give double the range of volt drops in the diodes. Suppose you need 1A and one diode is 0.65W, and the other is 0.7V then use a restor that will give a 100mV drop which 0.1 ohms. Am I right?
U can buy or scavange SCR's from a tram motor controler. Those are quite OK matched and have high curents. U can make cheap resistors like Nigel said and is feasable. Use very good ventilation and heatsink.
Hi Guys
A visit to the scrap yard I came across 4 x 500 amp diodes on a large heat sink which i bought for $5 , also bought 6x 200 amp IGBT's on very large heat sink at the scrap alli cost for $30
regards peter
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