[hpsdr] LTC 2208 A/D Thermal Characteristics

[Ray Anderson]

Scott-

As rob mentioned .24deg/watt sounds awfully small. It was probably the
theta(jc) number (junction to case). You need the theta(ja) number
(junction to ambient).

Junction Temp - Ambient Temp = Theta(Ja) * Power dissipated

Therefore:
Junction Temp = Theta(Ja)* Power Dissipated + Ambient Temp

A quick review of the LTC 2208 data sheet shows a number of 20deg/watt.

So 1.5 watts * 20deg/watt = 30 degrees C

If the ambient is 25degrees C then the die temp would be 30+25 = 55
degrees

iF your ambient could be as high as 50 and the die would be around 80
degrees.


Note that the thermal resistances given in vendor datasheets are close
but not exact. They are determined with JEDEC thermal test fixture which
probably resemble no real world application. The numbers are useful for
comparing different packages etc. but for absolute numbers you should
always count on applying a large amount of conservatism.(If your
calculation says the die will be 100 degrees it may actually be
120.....).


The theta(jA) NUMBERS  ASSUME PROPER HEATSINKING AS DEFINED BY THE
VENDOR

In any case, even with pessimistic assumptions it looks likeMercury is
probably ok.
  

-Ray  WB6TPU



Raymond Anderson
Senior Signal Integrity Staff Engineer
Programmable Platforms Development
Silicon Technology Group
Signal Integrity and Package Development Department
Xilinx Inc.
2100 Logic Drive
San Jose, California  95124
(408) 626-6277



> -----Original Message-----
> From: hpsdr-bounces at lists.hpsdr.org
[mailto:hpsdr-bounces at lists.hpsdr.org]
> On Behalf Of Rob Frohne
> Sent: Wednesday, January 14, 2009 3:20 AM
> To: Scott Cowling
> Cc: HPSDR Reflector
> Subject: Re: [hpsdr] LTC 2208 A/D Thermal Characteristics
> 
> ***** High Performance Software Defined Radio Discussion List *****


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