paralleling diodes

Discussion in 'General Electronics Chat' started by mwmorgan, May 5, 2013.

  1. mwmorgan

    Thread Starter New Member

    May 5, 2013
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    Many say you can parallell diodes for higher rms current. Is this really the case since no two diodes are doped perfctly and have the exact same forward drop.Even if made on the same day on the same machinery and in the same package.Wouldnt the one withslightly less drop handle more of the load than the one beside it,thus creating an imbalance?If it blows the other blows right after.Also with fast schottky, wont it introduce noise?
     
  2. #12

    Expert

    Nov 30, 2010
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    1) Yes.
    2) I don't know.

    Power diodes have a curved graph of voltage drop per amp. It is almost certain that no two power diodes will match exactly, but the one that conducts more readily will have an increased voltage drop because of the higher current and the weaker diode will be right behind it. A few percent of imbalance doesn't make this principle useless.
     
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  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I wonder if the negative temperature coefficient of the forward voltage will work against the success of the parallel configuration.
     
  4. #12

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    If the application is all that critical, you can add a few tenths of an ohm in series with each diode.
     
  5. GopherT

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    Nov 23, 2012
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    Diodes are available in many different current carrying ranges. 50 amp bridge rectifiers are widely available and Digikey carries discrete diodes that an handle up to 1200 amps.

    There is no need to parallel diodes in a rectifier. LEDs, on the other hand, is another story and it is commonly done with the resister technique mentioned by #12 above (usually with slightly larger resistors than he suggests).
     
  6. #12

    Expert

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    Gopher is right. It's hard to imagine a need to parallel power diodes when you can buy them the size of hockey pucks.

    As for the noise, what kind of circuit is this that might get Shottky diode noise?
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Yes it will. The one with the slightly lower ON voltage will conduct more current, creating more heat and increasing the junction temperature, which further reduces its ON voltage. Thus the imbalance will tend to significantly increase, especially if the two diodes are not in close thermal coupling on the same heat-sink.
     
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  8. THE_RB

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    Feb 11, 2008
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    I've seen it done in commercial devices.

    Normally the very gentle V/I knee of silicon diodes outweighs the thermal temp coefficient.

    One technique to minimise the thermal issue is to make the diode bodies touching and wrap their legs together, so they will be at a very similar temperature as the internal die is closely bonded to the metal legs.
     
  9. WBahn

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    Mar 31, 2012
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    I think the practical bottom line is that if you are wanting to parallel two (or more) diodes together because your don't have a single diode that can handle the intended current, it can be done but should not be done blindly. If done blindly you encourage a thermal runaway event. If you are careful, either by adding balast resistors to provide negative feedback (and it doesn't take much of a resistor - as long as you can get a tenth of a volt or so at the maximum current you are probably good) or thermally coupling all of the diodes, then you can get away with it. But it is not an approach to be taken without some amount of thought and planning.
     
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  10. WBahn

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    The one that conducts more readily will have the same voltage drop as the other (they are in parallel, remember) but will be conducting more current at that same voltage drop and thus heating up more than it would if it were carrying just its share of the current. The other one will have less current and thus be heating up less. As the diode heats up, it will pull even more current at that same voltage drop and so it's current will go up, the current in the other diode will go down, and the overall voltage will do down. But the increased current more than offsets the lower voltage and so it heats up even more and the process continues. This is called thermal runaway and, in the end, only one diode will be conducting nearly all of the current. It will burn out and now the other diodes repeat this process and the next strongest diode burns out and so on down the line. The results can be pretty spectacular!
     
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  11. mwmorgan

    Thread Starter New Member

    May 5, 2013
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    That was my first post and after reading all the replys I now know how this goes,wealth of information.Yes, the reason I wanted to do it was because it was after4GMT(midnite Here) and I didnt have large enough diode and sometimes I am the kind of guy who cant stop till im falling asleep over my hot iron, adding varry small damping resistance prior to each diode makes the most sense. Thanx to all.
     
  12. bug13

    Well-Known Member

    Feb 13, 2012
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    I might be too late, but if it might be useful for someone tomorrow:

    Quote
    "Equal distribution of current in each diode can be achieved either by using the doublewould reactors that induce a reactor voltage so as to keep m.m.f balance or by adding additional resistors so that the IR drop predominates over diode characteristic differences."

    - Power Electronics and its Applications by Alok Jain, P431
     
  13. WBahn

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    Mar 31, 2012
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    I have no idea what a "doublewould reactor" is. Could it be "double wound"?

    The use of resistors has been discussion by two or three of us.
     
  14. bug13

    Well-Known Member

    Feb 13, 2012
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    Arrr, must be my typo, the main thing I want to introduce here is using of an inductor in series with a power diode (I am thinking rectify application only, but I would be very wrong).

    I don't really understand how it might work, but my power electronics tutor gave me this solution when I asked the same question as OP.
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    If you have an AC current, then even though it is being rectified it is still an AC current (the term "alternating" does not have to involve actually reversing direction, just changing with time). So you will have an L(di/dt) voltage across the inductors and the leg with the highest current will have the highest inductive voltage, which will reduce the voltage across the diode in that leg, relative to the others, resulting in a reduction in current, relative to the others.
     
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