Rectifier Schottkies in parallel

Hello,

many Schottky rectifier diodes contain two diodes within a single package, usually with the cathodes connected. Does it make sense to connect the anodes too in order to get a diode which can withstand higher current? Is a 2x30A diode more or less equivalent to a single

60A diode? I am thinking of VBT6045CBP if it helps.

The application: a high-current 50Hz rectifier.

Best regards, Piotr

Reply to
Piotr Wyderski
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Is there any way of knowing if the two diodes are on the same piece of silicon? (That would help for thermal/ current balancing.)

George H.

Reply to
George Herold

On a sunny day (Mon, 27 Oct 2014 11:54:43 -0700 (PDT)) it happened George Herold wrote in :

x-ray

Reply to
Jan Panteltje

I don't know if this is true of Schottky diodes (I think it is). With ordinary junction diodes you can't parallel them because the temperature coefficient of the forward voltage is negative -- that means that whatever diode happens to be getting more current will get hotter, lower its forward drop, and get even MORE current, until sooner or later it pops.

You can do it, but you have to ballast the diodes with series resistance, and work hard to thermal match the diodes, and generally do a whole bunch more messing around than you'd do just getting a bigger diode.

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Tim Wescott 
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Reply to
Tim Wescott

I don't see why not. Disclaimer: offset in Vf, TJ, etc. manifests as a somewhat constant offset in current. So, if you derate the total by the worst case difference (maybe, say, 10A), you still get a beefy sum (50A) without worrying about breaking things.

Advice of this nature applies most importantly to separate diodes, BJTs or IGBTs in parallel. Probably, co-pack devices are close enough not to mind, let alone monolithic pairs (if they are such).

Here's an example from my junk bin:

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SR3540PT

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Cracked open, two dies. Placed pretty far apart, good to know for heatsink design. Possibly done to minimize lead length?

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Breaking the lead off, it's apparent the bond wire isn't so much a wire as a strip (of thin copper, apparently). This looks to be soldered, welded or brazed to the lead itself, and soldered to the die (or some other sort of die-attach). The die to base plate bond doesn't look like much, but may be solder as well. (The brittleness of the bond-to-lead joint probably suggests die-attach rather than metallurgical methods?)

Tim

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Reply to
Tim Williams

You're basically correct, but the problem is mitigated in dual Schottky diodes by way of them being roughly isothermal--the hotter diode can't get that much hotter than the cooler diode, which limits the hogging. Depending on the operating point the diodes are somewhat resistive, which may help too.

ISTR Motorola spec sheets for dual Schottkies rated some for 2x current if the diodes were paralleled, while other spec sheets had no such guarantee.

YMMV.

Cheers, James Arthur

Reply to
dagmargoodboat

It's done, a lot. And, it fails, a lot (in switching power supplies, this is a key check for shorts and failures). The old recommendation on high current paralleling of diodes is to put a little series inductance in each; this makes a current hog diode have a little less terminal voltage, and fights thermal imbalance. For this application, you do NOT want a lossy ferrite.

Reply to
whit3rd

The OP's part is here:

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30A @ Vf=0.54v (typ) is 18 milliohms, or 9 milliohms for two perfectly matched sections in parallel.

Your inductor in series is a neat trick for SMPS.

At 50Hz, I'd be tempted to make a MOSFET synchronous rectifier if ultimate low loss is important, or just parallel the Schottkies and add some small ballast resistors if he really wants 60A and loss doesn't matter.

For 30A I'd parallel the two bare diodes and be glad for whatever heat-spreading it gave me for free.

Cheers, James Arthur

Reply to
dagmargoodboat

That's qualitative. Just because there's some positive feedback doesn't mean that the system runs away.

As the current goes up, the ohmic loss inside the diode growns bigger relative to the ideal PN diode drop. The PN drop has a negative tempco but the ohmic drop tempco is positive. At some current, the TC is zero, and at higher currents it's positive. The crossover current is sometimes at a useful point, sometimes not, but the ohmic positive-TC effect mitigates the current-hogging effect.

Some big potted diode assemblies are multiple chips in parallel.

A single large diode will not be isothermal, so arguments about runaway can apply to regions of a single diode.

I'd go for 50 amps for the case at hand.

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John Larkin         Highland Technology, Inc 
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Reply to
John Larkin

** Scottkys have a similar negative tempco, about -1.6mV/C.

-- that means that whatever

** You are merely theorising aren't you ?

If you ever tried it you will find it works rather well.

** Nope.

Just soldering the leads into the same PCB area or twisting them together provides all the thermal coupling needed.

.... Phil

Reply to
Phil Allison

I'm suggesting some homework for the OP. I know that the whole parallel diode thing works sometimes, and sometimes it doesn't, and that making sure that it will in any given circuit is a Good Thing to do.

I spent a summer once in my college years as a service tech, ripping out paralleled sets of 2N3055 transistors and replacing them with some heftier transistor. The comments by the engineer, in reference to the previous engineer (who had ECO'd the second 2N3055 in) were both priceless and educational.

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Tim Wescott 
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Reply to
Tim Wescott

** So you have never tried it.
** The above drivel alludes to the alleged opinion of some alleged expert the OP alleges is source of wisdom. While he himself has none.

FYI A paralleled a pair of 2N3055 (in TO3 pack) provides more *usable* dissipation capacity that any alternative, single TO3 BJT available.

... Phil

Reply to
Phil Allison

Of course it would be good to check it in vivo, but I am just considering the alternatives, so it would not be possible to try a given circuit without actually building it. I want 40A and all the available diodes in sensible packages are actually

2x30A, not 60A, so the question is how would a paralleled 2x30A diode work. Is it 60A, 30A or something in between? According to Tim and John it is ~50A, which should be fine.

Best regards, Piotr

Reply to
Piotr Wyderski

The rectifier *was* synchronous and worked like a charm. For about 10 minutes. :-/ I don't know yet what *really* happened, but a rapid change in the load (~15A) was enough to expose the internals of the upper 180A key... :-)

It could have been a dI/dt surge which exceeded the 40V MOSFET rating or EMI that caused the steering circuit to hang up. Whatever the reason, the transformer didn't even notice. :-)

What I want is a rock-solid replacement in case of the next failure.

Best regards, Piotr

Reply to
Piotr Wyderski

Well, there are these things called data sheets, and this thing called mathematics...

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Reply to
Tim Wescott

I've seen it done in brownwares quite a bit. I wouldn't count on double the current rating, because there may be some imbalance. Using ballast resitor s kind defeats the purpose of using Schottkies in the first place doesn't i t ?

Alos consider the price. That is part of engineering. If you can get your 6

0 amp unit cheaper than two units, all else being equal, you know what to d o.

I suspect the reason they do it in brownwares is cost. They are on a whole different scale of operations and probably manufacture more than one produc t of course. So they can use these things when they need two diodes or one, doesn't matter. Buy then by the metric ton. Get stock in the diode's manuf acturer if you're big enough, but on here I doubt anyone is into that kind of mass production. Correct me if I am wrong - any LG, Funai or Samsung eng ineers here ? (if so I got a few bones to pick with you lol)

Reply to
jurb6006

If you are feeding them from a transformer, you may be able to use two smaller secondary windings, one per diode, instead of one large one. That way, the unavoidable winding resistances and stray inductances of the transformer will tend to balance the currents, with no need for any additional components or losses.

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Reply to
Adrian Tuddenham

Don't forget in standard silicon there is a 25mV/Amp (I think) bulk resistance (remember V/I=R) so as the current goes up, the resistance does also, acting kind of like a small ballast resistor. As most of these kinds of questions, the answer is VERY dependant on the application.

Reply to
WangoTango

When I saw that it was out of phase. Alot of times you can see just by looking at the foil on the board they are simply in parallel.

Probably had less effective series inductance at least.

Reply to
jurb6006

[snip]

Well, rock-solid would be a 45L(R) :-)

Klaus

Reply to
Klaus Bahner

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