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RC4195
PRODUCT SPECIFICATION
7
Let’s look at an application where a user is trying to deter-
mine whether the RC4195 in a high temperature environ-
ment will need a heatsink.
Given:
T
J
at thermal shutdown = 150
°
C
T
A
= 125
°
C
q
J-A
= 41.6
°
C/W, K (TO-66) pkg.
V
IN
= 40V
V
OUT
= 30V
I
Q
= 1 mA + 75
m
A/V
OUT
x 30V
= 3.25 mA**
Solve for I
O
,
= 60 mA – 13 mA ~ 47 mA
If this supply current does not provide at least a 10% margin
under worst case load conditions, heatsinking should be
employed. If reliability is of prime importance, the multiple
regulator approach should be considered.
In Equation 1,
q
J-A
can be broken into the following compo-
nents:
q
J-A
=
q
J-C
+
q
C-S
+
q
S-A
where
q
J-C
= junction-to-case thermal resistance
q
C-S
= case-to-heatsink thermal resistance
q
S-A
= heatsink-to-ambient thermal resistance
In the above example, let’s say that the user’s load current is
200 mA and he wants to calculate the combined
q
C-S
and
q
S-A
he needs:
Given: I
O
= 200 mA,
= 11.75
°
C/W
Given
q
J-C
= 7.15
°
C/W for the 4194 in the K package,
q
C-S
+
q
S-A
= 11.75
°
C/W – 7.15
°
C/W
= 4.6
°
C/W
When using heatsink compound with a metal-to-metal
interface, a typical
q
C-S
= 0.5
°
C/W for the K package.
The remaining
q
S-A
of approximately 4
°
C/W is a large
enough thermal resistance to be easily provided by a number
of heatsinks currently available. Table 1 is a brief selection
guide to heatsink manufacturers.
q
J
A
–
T
------P
T
A
–
D
=
P
D
T
------------------
T
A
–
q
J
A
–
=
V
IN
V
OUT
–
(
)
I
O
V
IN
I
Q
′
+
′
=
I
O
T
IN
(
T
A
–
–
q
J
A
–
V
OUT
)
----------------V
V
IN
I
Q
′
V
OUT
–
(
)
-V
–
=
I
O
°
C
°
C/W
125
°
C
10V
′
–
41.6
40
3.25
10
′
10
3
–
′
–
=
q
J
A
–
T
T
I
O
′
V
IN
V
OUT
–
(
)
V
IN
I
Q
′
+
----------------------------------–
=
°
C
125
°
C
40
′
–
+
10V
200mA
′
3.25
10
3
–
′
----------------------50
=