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From: John Larkin on 16 Jul 2010 12:08
On Fri, 16 Jul 2010 12:02:15 -0400, Spehro Pefhany
<speffSNIP(a)interlogDOTyou.knowwhat> wrote:

>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
>
>Since it's a full wave bridge, average current through each MOSFET is
>going to be 1/6 or 1/8 of the total, so only 33A or 25A, which looks
>within the realm of possibility. Power dissipation from the MOSFETs
>alone might be ~50W though, more if it's a nasty load.

And I'd guess that a battery charger will be a "nasty load." At 200
amps average, unless there's a big inductor available, 600 amp peaks
wouldn't shock (punalert) me.

John


From: Ecnerwal on 16 Jul 2010 12:40

In the "one-or-only-a-few-off for lab work" area, we have been known to
bend up some heavy copper wire and solder it to a trace to increase
current carrying capacity. But the problems of heat dissipation from the
transistors mentioned by others probably rule the pile of problems here,
and make choosing a serious power package rather than an unsuitable
surface mount package the better option. Sometimes smaller is not better.

--
Cats, coffee, chocolate...vices to live by
From: George Herold on 16 Jul 2010 14:51
On Jul 16, 12:03 pm, John Larkin
<jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
> On Fri, 16 Jul 2010 10:44:40 -0500, "Tim Williams"
>
> <tmoran...(a)charter.net> wrote:
> >"John Larkin" <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote in messagenews: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?

Hmm, I think I understand the question. I might try breaking up the
sheet into pie shaped pieces. (cylindrical coordinates) The pointed
end of the pie piece is cut off becasue it's touching the puck. Now
you've got to do the integral for a pie piece with different size
bites taken out of the pointy end. But if you insist on an infinte
sheet then the answer is infinity no matter what the puck size. So
you need a bit better boundary condition for the size of the sheet.

George H.
>
> 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

From: Tim Williams on 16 Jul 2010 14:57
"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:0301461dk2vlvg2o8f4tj5pviiiiho57fd(a)4ax.com...
> > It looks like a point charge in space, and the temperature is defined by
> > Gauss' law.
^ ^ ^ ^ ^
>
> Nice rant, but still no answer.

No? Gauss' law is easy. Set up the integral.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
From: Jan Panteltje on 16 Jul 2010 15:13
On a sunny day (Fri, 16 Jul 2010 13:57:42 -0500) it happened "Tim Williams"
<tmoranwms(a)charter.net> wrote in <FY10o.9935$KT3.8735(a)newsfe13.iad>:

>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in =
>message news:0301461dk2vlvg2o8f4tj5pviiiiho57fd(a)4ax.com...
>> > It looks like a point charge in space, and the temperature is =
>defined by
>> > Gauss' law.
> ^ ^ ^ ^ ^
>>
>> Nice rant, but still no answer.
>
>No? Gauss' law is easy. Set up the integral.
>
>Tim
>
>--
>Deep Friar: a very philosophical monk.
>Website: http://webpages.charter.net/dawill/tmoranwms

If you want the theoretical side, with calculations, and a lot of integrals,
google for ahttv131.pdf, the MIT heat transfer textbook.
Only 762 pages...
But it gives you that feeling that you know it all :-)

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