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I have a 12V power source that can deliver 25A (DC). I intend to use something around 15A .. 20A on a circuit that includes a diode. However, I can't find a diode that can handle that much current, and I've seen people recommending bridge rectifiers instead.

I happened to find a KBPC50005 bridge, which is rated 50Amp / 50V.

However, I'm not sure how to make the connections to get something equivalent to a 50Amp diode -- there are two ways, as far as I can see:

schematic

simulate this circuit – Schematic created using CircuitLab

This is what I suppose:

  • the second setup would result in a higher voltage drop (because I would be using two paths, each with two diodes in series). In the first setup (A-B), I would be using a single diode, so a lower voltage drop;
  • the second setup would be equivalent to diodes in parallel, so it would perhaps allow for an even higher current. The diodes are in a single package, so the usual problems with diodes in parallel would not happen;

Am I wrong?

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Think of the schematic of your bridge rectifier as 4 diodes

schematic

simulate this circuit – Schematic created using CircuitLab

You have multiple options:

  1. Use AC1 as Anode of D1 and Pos as Cathode of D1
  2. Use AC2 as Anode of D2 and Pos as Cathode of D2
  3. Use Neg as Anode of D3 or D4 and use AC1 or AC2 to select either D3 or D4
  4. Connect AC1 and AC2 together as the anodes of D1/D2 in parallel and Pos as the cathode D1/D2
  5. Connect AC1 and AC2 together as the cathodes of D3/D4 in parallel and Neg as the anode of D3/D4

In 4) and 5) above the diode pairs won't match perfectly, but you'd probably be able to achieve 1.5-1.6 of the rating of a single diode and achieve a lower Vf using the parallel diode pair. Within a bridge rectifier, the diodes will all be at roughly the same temperature, especially for the pairs with anode and terminal bonded together.

I left off the option to use Neg/Pos as a diode serial/parallel pair as this is not a good option.

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The first option is the better of the two as winny states. I have seen this applied in a Christchurch sweatshop in 1992. I thought that you will be slightly better off when it comes to voltage drop and hence losses if the AC terminals are tied together.

Remember that with burning bridges there are a lot of bulk resistance. Remember that from college a Si diode would drop 700 mV, but this was at say 1 mA and an ideal semiconductor junction. Real diodes can and do waste 1 volt or even more at your currents. The current sharing can not be expected to be perfect but there will be some current distribution so voltage drop and hence losses will be slightly lower. Reducing junction temperature is always good.

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  • \$\begingroup\$ A sweatshop, or a welder maker (that could also be a sweatshop), or welding shop maybe? \$\endgroup\$ – rackandboneman Apr 22 '17 at 11:59
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Diodes in parallel will not give you higher current cabability in general due to negative temperature coefficient. Since these are molded in the same package, they will be kept at resonably the same temperature but don't expect them to share current perfectly. Some silicon diodes which are bonded to the same copper substate exists (TO-247) but your bridge rectifier is plastic/epoxy so the thermal bond is not that good.

Your first setup is the way to go since you only get one voltage drop.

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    \$\begingroup\$ Where are silicone diodes produced? In Silicone Valley? \$\endgroup\$ – Peter Mortensen Apr 22 '17 at 11:39
  • \$\begingroup\$ @PeterMortensen I blaim autocorrect! :-) \$\endgroup\$ – winny Apr 22 '17 at 11:44
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    \$\begingroup\$ Getting a good matched pair is important in both cases.... \$\endgroup\$ – rackandboneman Apr 22 '17 at 12:00
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    \$\begingroup\$ @KalleMP Sure it might work, but unless the termal resistance between two diodes is explicitly stated in the datasheet, you can't be sure they will share current/put it in mass production. For a simple experiment, go for it. \$\endgroup\$ – winny Apr 22 '17 at 14:54
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    \$\begingroup\$ @Winny of course they will share current .....what you can't be sure of is that they will have absolutely equal current in paralleled diodes. Just look at the Vf characteristics and you will see that the differences will be only mV. If diodes didn't share current them you'd be unable to parallel devices such as LED's yet it's done all the time. The diodes in a bridge rectifier will most assuredly be the same devices, probably from the same batch, so will match nicely. It's good practice to use only a multiplier of 1.5-1.6 current rating when paralleling diodes, but quite safe at this level. \$\endgroup\$ – Jack Creasey Apr 22 '17 at 21:27

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