paralleing bridge rectifiers

How will you ensure equal sharing of the current?

I don't have to ensure equal sharing... I just need to ensure that the max rating isn't exceeded. I would imagine that between two identical diodes that they would share approximately the same current.

yes. it won't work. which ever bridge starts conducting first will be the one doing most of the work. it's not what you want to do. diodes have a region of no current flow. When it starts flowing, the resistance between the joints decreases as current increases. so if one should start before the other, then it will be the one doing the work. For power systems, you need to have a single bridge large enough for the job, or join them together via power resistors which most likely is not what you want due to power loss.

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I wouldn't assume that. If you run each via a one foot length of wire you'll probably be closer to sharing as it will act as a resistor.

This is not true. There is always current flowing. (diodes are not ideal switches)

The point is not that one will start before the other but if they will approximately equalize. I don't need to push each diode close to its max but, say, if I had 100 diodes each with a max of 1A and I paralleled them then I would expect I could run a few amps through them with no problem. Now if one blew then maybe there would be a cascade effect and obviously the more than blew the more that will blow.

e.g., it doesn't concern me if one diode uses 80% of its max and the other

50% as long as its well below 100%. Ofcourse if one diode went bad then you would be running 100% of the current through just one diode. I could put a fuse on each diode to prevent this(that is, let me know when one goes bad) or even maybe add some current limiting device to each diode?

that is, maybe a foldback current limiter on each branch that is maybe 80% of its max?

ultimately the best solution is to have the proper components but I don't. I have many 4A bridge rectifiers but no 12A ones. Now I can use the 4A but rather get about 8A out of it. Maybe paralleling 3 or 4 of them would be safe? putting fuses on each one for 4A would be even safer.

Ofcourse if this runaway current draw that your talking actually happens then it won't work. That is, if only one diode will conduct at the start while all others will not then it will obviously draw all the current through it. But if there is some sort of distribution that is approximately average then I can deal with it and it will work for my circumstances.

Thanks, Jon

Maybe theres some way to use negative feedback between two bridges so that the current between them will equalize? Something like a comparator that adds resistance to one branch if it has more current than the other and vice versa.

Of course it would be more work and money than just buying the proper bridge but would be interesting to see.

Thanks, Jon

For low powers they reckon on average you'll get a 30:70 balance on two diodes running in parallel. This implies something like 3 bridges needed to double the current. (you only see a 60mV change in a diodes ON voltage for it to be passing 10x the original current) But ... As as current goes up the diodes start generating noticable heat and the loadsharing starts getting much worse. Resulting in the initial 'best diode' in the batch hogging more and more of the current the hotter it becomes and the hotter it becomes then the more current it hogs and on so on. That best diode is losing 2mV forward voltage drop for every degreeC increase in it's temperature which means the other poorer diodes with their higher mV on-voltages don't get a look in as they are running cool with low currents and higher forward voltages and can only remain that way. Low diode currents gives some semblance of order. Beyond a critical current and a clear winner emerges by positive thermal feedback even though all the diodes may be physically separate. (a chaotic function?)

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As replied, you can't rely on better than about 30% to 70% sharing, if connected directly together, and mounted on a common heat sink. If you can add a low resistance (one that will drop about a tenth of a volt at average rated current, which will drop more at instantaneous peak current) you can force the sharing to be much closer. You can put the resistor in either AC leg, but you need to put it in the same leg of both bridges. Now you have the temperatures of two more components to worry about. But there may be an up side, besides the sharing. They will tend to slightly lower the peak current in total, helping the transformer and capacitors reduce their RMS current for a given DC average output current. The other down side is the slightly lower output voltage (that should be less than a volt).

only thermal runaway and/or current hogging.

If you arrange some series resistance to balance the current (winding resistance should be enough) all should be fine.

Bye. Jasen

Buy the bigger rectifier. Under load, any amount of difference in their forward voltage will cause positive feedback. As the diode heats up, its forward voltage decreases, increasing its share of the current, and you get thermal runaway. One hogs all of the current, melts, and then the other gets stuck with all of the current, and melts.

Or, you could use current-balancing resistors, which will waste power, but could protect the diodes; but bottom line, it'll be cheaper in the long run to just spring for the bigger bridge.

Hope This Helps! Rich

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I think the best way would be to put Peltier junctions and
thermocouples on each of the diodes and then as the diodes started
warming up to cool them all off so they were all at the same
temperature as the one with the lowest temperature which was being
cooled the least. 

Either that or to run them in a bath of circulating deionized
refrigerated water.

Greetings John, I built a power supply for a CNC machine retrofit using the original 3 phase xmfr and 5 full wave rectifiers. I used 4 1/2 of the

5 because the xmfr was actually 9 windings wound on 3 cores. Anyway, the machine is only using the original contactors, spindle motor, servos and the xmfr that the machine came with new . All the other electric and electronic parts are new. I did not use any resistance with the rectifiers to force the diodes to share the current more equally because I didn't know I should. The machine runs fine and the conversion has hundreds of hours on it. I think the rectifiers are rated at 20 amps. The servos draw 15 amps max before the control shots down. This has only happened once when I jammed one of the axes. The xmfr is wired to the rectifiers with 10 gauge wires a little over 3 feet long each. Some questions: How does the resistance help spread the load across all the diodes? Why should the resistance be located between the xmfr and the diodes? Why haven't the diodes in my setup been destroyed? Is this because of the long 10 gauge wires? Thank You, Eric

Without drawing pictures, I am having a little difficulty imaging that schematic. Is the secondary array essentially

3 Ys or 3 Deltas that all feed one DC supply? If this is the case, and you use separate rectifiers for each secondary, then the impedance of those secondaries acts as the current balancing resistance (and inductance), to help force all the rectifiers to share the load current.

So the 20 amp rated rectifier is usually running well below its rating. That's good.

It just adds additional voltage drop to he rectification that is proportional to the instantaneous current, unlike the very nonlinear and negative tempco drop of the diode junctions. It helps to make the overall drop more proportional to current and with a less negative tempco. It doesn't force perfect sharing, but only makes the sharing less unequal.

I was afraid you would ask me that. Ideally, you would add a bit of linear (and optimally, with a positive tempco) resistance in series with each diode junction. I was picturing the parallel rectifiers being fed from a single secondary, and powering a single storage capacitor. So it may work as well in series with any of the 4 legs of the bridge, since each leg carries either all the AC current, or all the DC current.

I was not taking the time to visualize all the permutations and their relative performance, but just told you about the first one I thought about. I stand by the requirement that if there is one AC source, and one DC load, the sharing resistors all need to be in the same led of the bridge.

Perhaps it is because, most of the time, the average current is pretty low, and the diodes have a fair thermal mass to absorb the big thumps. Most rectifier diodes have impressive surge current ratings.

And, yes, the wires are low value resistors with positive temperature coefficient.

Greetings John, I don't know what the configuration of the secondarys is. When connected to the original three servo amps the 9 taps from the xmfr were connected such that each winding had one wire going to each servo amp. I think. Anyway, what I did was connect those 9 taps to the rectifiers so each winding does have it's own rectifier. And the outputs from the rectifiers are all tied together. This is because the new servo amp configuration has all three amps on one board and they share the DC supply.

Thanks for the explanations John. I will keep them in mind next time I need to build a similar power supply. And I'm sure there will be a next time. Cheers, Eric

If the three secondaries all produce the same voltage you should be safe.

Bye. Jasen

If you're going to that sort of length, isn't there already some commercial technology for 100A+ rectifiers (from International Rectifier or the Hexfet people?) that uses a funny geometry so as to gain the right sort of temperature coefficient inherently. You can just parallel these things up as much as you like, they sort it out amongst themselves.

As to paralleling a pair of rectifiers to share current on a breadboard lash-up, then this is an old hack and works fine -- just use a matching pair (same part, same manu.) and don't expect that the total rating will be twice that of a single.

As to why you'd do it for production, then I don't know. High power rectifiers are cheaper than low power ($/A)

...

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I guess I should have punctuated it like this:

I think the best way would be to put Peltier junctions and
thermocouples on each of the diodes and then as the diodes started
warming up to cool them all off so they were all at the same
temperature as the one with the lowest temperature which was being
cooled the least. :^)

Uh, you're joking, right?

Mark

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OMIGOD! You mean it wouldn\'t work???

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