 ### Network # heat sink

K

#### Kelvin

Jan 1, 1970
0
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

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 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...

J

#### John Popelish

Jan 1, 1970
0
Kelvin said:
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

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 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...

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->