The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: "red rover"
Subject: Re: Selecting a heatsink?
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Tue, 14 Jan 2003 21:21:53 -0500
NNTP-Posting-Date: Tue, 14 Jan 2003 21:19:47 EST
Organization: Bell Sympatico
> Say you will settle for a 50C rise over ambient, and taking your 1.7W
> 50/1.7 - (2.2 + 0.5) = 29.4 - 2.7 = 26.7C/W, so any heatsink better
> (lower) than that will do.
> I'd be tempted to double that 1.7W, heat is generated in switching, as
> somebody said. A 12C/W heatsink is pretty small & cheap.
Heatsink calculation is actually fairly easy if
you are familiar with electrical circuits and their
equivalent thermal models. The power, Q, is like
a current source, the Temperature is like voltage
and Thermal Resistance is like Electrical resistance.
Junction to ambient is a Red Herring here. That number
assumes there no heatsink and tells you Rjc + Rca = Rja.
If you didn't put a heatsink on the device it would have
a junction temp rise of 1.7x 62 = 105. Now assume you
are sailing the yacht in a balmy tropical 35C.
Also, assume this board is tucked into an enclosure or some such and
that the ambient temperature inside the box is 10 or 20 C above ambient
(20 to be conservative). Then without a heatsink the ambient
of the device is 35+20= 55C. And the junction is 105 over ambient =
105+55 = 160 = too damn hot for reliable operation.
So aim for a junction temperature of say 120C to 125C or below. You know
you need a heatsink. The Wakefield catalog doesn't say it clearly
but it is a 65C rise above ambient when passing 6W = 65/6 =
10.8C/W thermal resistance for the heatsink-to-ambient (some
Wakefield charts say this, others say this is the rise of the tab which
only saves you the .5C/W).
So you have Rjc+Rcs+Rsa = Rja = 2.2+.5+10.8 = 13.5 c/w.
Your circuit isn't switching and is turned on hard so I assume you've
go the power number correctly calculated at 1.7W (not turned on hard,
or in switching mode requires more calculation).
So Tj = Ta + 1.7x 13.5 = 55 + 22.9C = 77.9 C (very good).
And the case will be Q x Tsa + Tamb = 1.7x10.8+55 = 73C
Sometimes you want this number because the problem is not
the junction temperature but people getting burned by the heatsink.
So, how small a heatsink could you use? Well if you go
with the 120 Tj then working backwards would give 35.8C/W max.
Now if Wakefield doesn't give this number directly it means
you need a number like 72C @ 2W or smaller.
Is there a reason you are not using a mechanical switch or relay?
That could save you power if this is simply an on/off device.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup