Pcb as heatsink

65=C2=B0C/W

of

)

Looks similar to what I am trying to achieve very interesting article.

Thanks

Reply to
Paul Taylor
Loading thread data ...

1.65=C2=B0C/W

ad of

R4)

ay

y

Thanks James I will be paralleling up the legs as you described.

Reply to
Paul Taylor

sage

C/W

of

o

the

James I am interested in how you got the 1.25W, I am guessing it is because of paralleling up the legs? But the formula behind it.

Thanks

Reply to
Paul Taylor

sage

C/W

of

o

the

I was lead to believe and national also say you can parallel these diodes up because they are matched as they are in the same package, single ones is definitely a no - no for paralleling up as you get different vf for each one so one could drop more and hence get hotter than the other, eventually failing.

Reply to
Paul Taylor

e:

21.65=C2=B0C/W
e

tead of

(FR4)

e

..

shay

hly

The temp drop (delta theta per inch) along the copper as you move furher away from the device leg is highest where the leg meets the copper. So leaving out the solder mask to bulk up around the leg can help pcb thermal conduction significantly. Maskless lines radiating out also help. Its also possible to solder in copper wires through hole to increase both conduction and dissipation area. If these arent enough, another option is to sit a small piece of sheet ali on the top off the pcb and bolt the diode to it and teh pcb.

NT

Reply to
NT

ote:

=3D 21.65=C2=B0C/W

ize

nstead of

s (FR4)

ive

e.

m...

nd

Vishay

ughly

another trick is to position the diode next to an electrolytic and simply cable tie the 2 together. The non-flat contact isnt great but gives some useful thermal diss. I forget the figures but I think I dissipated a watt or so this way one time. The cap didnt get hot enough to be an issue with reliability.

NT

Reply to
NT

essage

5 C/W
e

d of

4)

to

r the

k,

I guesstimated that paralleled legs would share current, with the heavier leg pulling 1.5A or less. That was intuition. I justified that in a later post.

I eyeballed V(f) from the datasheet (Fig. 1) @ 1.5A per leg is roughly

0.5v. 0.5v x 2.5A =3D 1.25W.

Don't forget the caveats. You said elsewhere it's a boost converter. What's the input and output voltage?

-- Cheers, James Arthur

Reply to
dagmargoodboat

Find a suitable tab mount (or stud mount) device?

--
Paul Hovnanian     mailto:Paul@Hovnanian.com
------------------------------------------------------------------
Happily doing the work of 3 Men ... Moe, Larry & Curly
Reply to
Paul Hovnanian P.E.

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.