High current pcb [Design]

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From: Uwe Hercksen on 16 Jul 2010 10:54

George Jefferson schrieb:

> I want to replace a CT full wave rectifier for a 200A+ battery charger
> with a active rectifying system.
>
> http://www.nxp.com/documents/data_sheet/PSMN2R5-30YL.pdf
>
> Are some 100A 80W logic level fets which means I can parallel 3 or 4 of
> them and increase the efficiency of the charger(there are other reasons
> for doing it too).

Hello,

these fets tolerate 80 W only at an mounting base temperature of 25 �C,
see figure 2 in the data sheet.
When the pcb should do the job of the heat sink too, there must be
enough copper area to get the heat out of the fets and the pcb.

Bye

From: John Larkin on 16 Jul 2010 11:17

On Fri, 16 Jul 2010 16:54:27 +0200, Uwe Hercksen
<hercksen(a)mew.uni-erlangen.de> wrote:

>
>
>George Jefferson schrieb:
>
>> I want to replace a CT full wave rectifier for a 200A+ battery charger
>> with a active rectifying system.
>>
>> http://www.nxp.com/documents/data_sheet/PSMN2R5-30YL.pdf
>>
>> Are some 100A 80W logic level fets which means I can parallel 3 or 4 of
>> them and increase the efficiency of the charger(there are other reasons
>> for doing it too).
>
>Hello,
>
>these fets tolerate 80 W only at an mounting base temperature of 25 �C,
>see figure 2 in the data sheet.
>When the pcb should do the job of the heat sink too, there must be
>enough copper area to get the heat out of the fets and the pcb.
>
>Bye

A small part like this doesn't conduct heat into pcb pours very well.
If you stick a part to a relatively thin thermally conductive sheet,
theta goes up as the part footprint area goes down [1]. That's a
really bad deal when you're working with copper foil and dpak-type
parts. Multiple internal pour patches and lots of thermal vias can
help increase the part's effective footprint.

A TO-220 on an infinite sheet of 0.062" thick aluminum has a theta of
roughly 2K/W. That fact can be used to do some very rough scaling,
factor-of-three type stuff maybe. Thermal design like this is very
messy. Lateral heat spreading, namely hot-spot effects, are important
even on extruded heat sinks; they become dominant on thin copper foil.
Now add in current crowding and the positive TCR of copper and
convection issues, eventually you have a serious math problem.

ftp://jjlarkin.lmi.net/Infinite_Sheet.jpg

A PCB is an expensive way to make a heat sink, and a bad way to
conduct/terminate a lot of current.

John

[1] anybody know the exact relation?

From: Tim Williams on 16 Jul 2010 11:44

"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:86t046tr9cu6p5n0f0c658sh1517m3p8so(a)4ax.com...
> A small part like this doesn't conduct heat into pcb pours very well.
> If you stick a part to a relatively thin thermally conductive sheet,
> theta goes up as the part footprint area goes down [1].
<snip>
> [1] anybody know the exact relation?

Easy to approximate. Assume a circular footprint (cf. spherical chicken). Assume heat dissipation at the center is zero (fair for an infinnitessimal segment, blatantly false for an infinite number of them). If heat diffuses through copper out to infinity, temperature drops inversely with distance (because cross sectional area increases linearly). It looks like a point charge in space, and the temperature is defined by Gauss' law.

Of course, heat diffuses through two or three means, with strange temperature-dependent coefficients besides. So it's not at all true that, the device itself, and the little bit of copper surrounding it that doesn't have a quite circular temperature profile, isn't dissipating any power. In fact, it could be dissipating a considerable amount of power. If the heat source were an infinnitessimal point, it would have infinite temperature, and therefore radiate infinite power density (power can still be finite, since the area is infinnitessimal).

However, it is true that heat diffuses out, one way or another, so maybe the power dissipation is just a little higher in the center, and spreads out in a slightly-steeper-than-inverse relationship, eventually going to zero at infinity all the same. The trouble is deriving the exponent and coefficient of that power law.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

From: Spehro Pefhany on 16 Jul 2010 12:02

From: John Larkin on 16 Jul 2010 12:03

On Fri, 16 Jul 2010 10:44:40 -0500, "Tim Williams"
<tmoranwms(a)charter.net> wrote:

>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:86t046tr9cu6p5n0f0c658sh1517m3p8so(a)4ax.com...
>> A small part like this doesn't conduct heat into pcb pours very well.
>> If you stick a part to a relatively thin thermally conductive sheet,
>> theta goes up as the part footprint area goes down [1].
><snip>
>> [1] anybody know the exact relation?
>
>Easy to approximate. Assume a circular footprint (cf. spherical chicken). Assume heat dissipation at the center is zero (fair for an infinnitessimal segment, blatantly false for an infinite number of them). If heat diffuses through copper out to infinity, temperature drops inversely with distance (because cross sectional area increases linearly). It looks like a point charge in space, and the temperature is defined by Gauss' law.
>
>Of course, heat diffuses through two or three means, with strange temperature-dependent coefficients besides. So it's not at all true that, the device itself, and the little bit of copper surrounding it that doesn't have a quite circular temperature profile, isn't dissipating any power. In fact, it could be dissipating a considerable amount of power. If the heat source were an infinnitessimal point, it would have infinite temperature, and therefore radiate infinite power density (power can still be finite, since the area is infinnitessimal).
>
>However, it is true that heat diffuses out, one way or another, so maybe the power dissipation is just a little higher in the center, and spreads out in a slightly-steeper-than-inverse relationship, eventually going to zero at infinity all the same. The trouble is deriving the exponent and coefficient of that power law.
>
>Tim

Nice rant, but still no answer.

Given a perfectly thermally conductive puck attached to an infinite
sheet of thin [1] finite-thermal-conductivity material, and assuming
conduction cooling only, what is the relationship of puck theta to
puck diameter?

This is relevant to situations where you have a choice of, say, SOT89
versus DPAK versus D2PAK and you're heatsinking to copper foil.

John

[1] thin relative to puck diameter

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