heat sink

Hi: this is a question of 2 transistors mounted on the same heat sink...

Two 2N3055 transistors are mounted close together on a heasink consisting of a square piece of bright aluminum 1/8 thick. the transistors are electrically insulated from the heatsink by using MICA washers coated with silicone grease between each transistor and the heatsink. for a 2N3055. Qjc = 1.5 C/W and Q of a greased MICA washer = 0.5 C/W Tj Maximum of a 2N3055 is given as 200C. a) If each transistor is producing 20W of heat at its collector junction, calculate the maximum allowable value of Qsink if the ambient temp cam be as high as 45C b) if Qsink = 2.5 C/W, and one transistor is dissipating 30 Watts and the other is in cutoff, calculate the case temp of each transistor if T ambient = 50C. State any assumption made.

after filtering out the decorations... i figure the following are the useful part of the question:

2 transistor with Qjc = 1.5 C/W each is mounted on a Qsink = unknown heatsink between the heatsink and two transistors, there is 2 MICA Q = 0.5 C/W Tj(max) = 200C Tambient = 45C a) im looking for Qsink b) i have no clue :P

I did Tj(max) - Tambient = 200C - 45C = 155C (which is the heat allowed on my components) Total Power dissipation = 2 x 20W = 40 W

155C/40W = 3.875 C/W --> total thermal resistance I can have in my circuit then im stucked... lol... two transistor mounted on the same heatsink... so the total thermal resistance they have should be the sum of parrallel of the two?? the same for two MICAs...

and for (b)... i have totally no clue about it...

thanks for any help...

{ Kelvin@!!! }
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It may help you visualize this problem, by representing the thermal resistances as resistors, temperatures as voltages and power flows as currents. If you assume that the thermal resistance of the aluminum between the two transistors has an insignificant thermal resistance (one of those assumptions you have to state) then the problem can be analogized by the following schematic (viewed in fixed width font):

Qjc1 Qmica1 TJ1--/\\/\\/--/\\/\\/--+ Pj1-> Pj1-> |Tsink Qsink +-------/\\/\\/--Ta Qjc2 Qmica2| Pj1+Pj2->

Tj2--/\\/\\/--/\\/\\/--+ Pj1-> Pj1->

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John Popelish

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