Pcb as heatsink

That's a lot of voltage drop for 2.5 amps. And a lot of power to dissipate in pcb copper.

Maybe you should look for a better diode, to reduce the power dissipation. 2.5 amps should be achievable with a schottky soldered to a couple of square inches of copper, or an internal plane.

Is there any air flow? That makes a big difference.

Is this going to be a multilayer board?

That's a dual diode. How is it going to be used?

John

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the Tjc is a bit worse that your diode but you should still be able to deduct something from the numbers in this:

117.pdf page 10 and forward

-Lasse

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Sounds like you're doing the same thing as me. I'm building a negative regulator with the LTC3757. The circuit is described better in the LT3704 datasheet. There is no way around the power loss in the diode. I found these app notes on PCB Heatsinks.

The diode gets hot. I'm using the ss8ph10 because of the high 175C junction temperature. But I'm thinking of looking for a part in a bigger package. I'd be interested in any PCB heatsink tricks you find out.

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The first step is: reduce the dissipation as much as possible. Vishay gives the the max. forward drop as 0.61v (not 0.76v) for a single leg of that dual diode, carrying 3A, t(j)=3D25=C2=BAC:

Paralleling the legs is better still. V(f) @ 1.5A per leg is roughly

0.5v. Plus, V(f) falls as the thing heats up, which helps too.

0.5v x 2.5A =3D 1.25W. That's better.

Single-sided? Better use a big copper pour, or a heat sink.

-- Cheers, James Arthur

It appears the OP mentioned its a single sided board - usual industry practice would be *at least* double sided and put a grid of vias under the device footprint to a copper area on the other side.

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Estimating from the OP's numbers, it looks like he's shooting for a

You can't parallel (these kind of) diodes. Better use some synchronous rectifier scheme if power really is an issue.

--
Failure does not prove something is impossible, failure simply
indicates you are not using the right tools...
nico@nctdevpuntnl (punt=.)
--------------------------------------------------------------

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Why not? Current sharing? From the data sheet, they've got about 100 milliohms effective series ballast resistance. That's 100mV/A. They might not share perfectly, but it doesn't look too bad. Besides, it's free, and anything is better than nothing.

-- Cheers, James Arthur

Negative temperature coefficient?

The problem is that these dual diodes are often rated by adding the current rating for each diode. The negative temperature coefficient is the killer. Dual diodes are very often two dies mounted in one package and there is no guarantee the dies where close on the wafer. So you'll need to consider the worst case scenario.

Dual power diodes are intended for half bridge rectifiers (or similar applications) not for parallelling. If you need more current, use a bigger diode.

--
Failure does not prove something is impossible, failure simply
indicates you are not using the right tools...
nico@nctdevpuntnl (punt=.)
--------------------------------------------------------------

Or an ideal diode controller if it's just a pass diode or switching slowly.

--
_____________________
Mr.CRC
crobcBOGUS@REMOVETHISsbcglobal.net
SuSE 10.3 Linux 2.6.22.17

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That's good practice in general. In this particular case it doesn't apply--each leg of the diode is fully rated to handle the OP's entire current, so there's no drawback to paralleling the diodes.

I paralleled schottkies in a switcher a *long* time ago, when MOSFETs were still boutique. A Collmer Semi unit, IIRC. I put some effort into the current-sharing question for that part, to make sure it was okay. YMMV.

I've seen dual schottkies rated as 6A, for example, with each leg rated at 3A. I took that as guaranteeing 6A parallel operation, 3A individually.

(Most recently I wound up using a MOSFET anyhow, for reverse supply protection, because of similar concerns as here--the schottky was marginally too hot, and I didn't have enough real estate to comfortably cool it.)

-- Cheers, James Arthur

A Rohm RB051L-40 is rated 0.45 volts max at 3 amps. Typ is more like

0.35. That would be less than half the power dissipation.

John

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Yeah, but they're both negative. As a practical matter, methinks the delta(Vf) will dominate, putting the delta(If) at something close to delta(Vf)/(r1+r2).

For [delta(Vf) =3D 100mV] and [r1=3Dr2=3D0.1 ohm], delta(If)=3D500mA.

That's off-the-cuff, assumes the diodes are reasonably isothermal, etc.

Looking at the datasheet, it looks like the thermal resistance between legs is as high as 10 C/W, and figure 1 indicates d(Vf)/dT is something like -1mV/C, so a) these are separate dice, and b) that potentially adds as much as 10mV to d(Vf).

It's free. Looks worth doing to me.

-- Cheers, James Arthur

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Good point. That's a 20v diode. A lower voltage diode helps (if he doesn't really need the original part's 60v breakdown).

James

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But that's the kink. That's not usually a good assumption with this sort of part. They are rarely monolithic.

a) usually are b) yes.

If one won't catch fire, two won't either. ;-)

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Not catching fire, that's gotta be worth somethin' :-)

John linked an awesome 20VDC (40V peak) diode with lower Vf:

See pg. 2 for the Vf dispersion graph--the spread is 8mV (incredibly tight!). Above I took delta(Vf) as 100mV, a mighty big number. Hopefully that's more than enough to cover many sins.

Even that 100mV assumption results in a 1.55/.95A split--well worth doing. It also parallels the thermal impedances, lowering Tj by another 2C/W or so, for free.

Another factor--the OP didn't spec whether that's 2.5A peak or average. If peak, avg. dissipation will be lower than we're calculating, possibly a lot lower (e.g. a boost switcher). Every little bit helps.

There are a few caveats--paralleling the diodes doubles capacitance. Leakage current doubles. If the standoff voltage is high and the diodes are hot, Ir x Vr may add more dissipation in reverse than is saved in forward conduction.

-- Cheers, James Arthur

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James it is a boost switcher.

Hi John yes it is single sided board, the diode is being used parallel up evently the board will be in something so no airflow, I think I have answered all the you ask.

Kind Regards