T H E R M A L M A N A G E M E N T F O R T O T A L - A C E X T R E M E
Data Device Corporation
DS-BU-67301B-G
1/14
105
Where P
C
is the portion of the component power which flows through the top of the
case to the ambient,
θ
JC
is the thermal impedance of the component from the
junctions to the top of the case, and
θ
CA
is the thermal impedance of the system from
the top of the case to ambient. Similarly, P
B
is the portion of the component power
which flows through the bottom of the case into the PCB,
θ
JB
is the thermal
impedance of the component from the transistor junctions to the board, and
θ
BA
is the
thermal impedance of the system from the board under the component to ambient.
For the case of a BU-67301B0T0L-102 used as a monitor terminal, P=0.42W and
θ
JB
=46.9°C/W. This can be simplified by assuming that no heat flows out through the
top of the case and only measuring the rise above package temperature without
considering the whole path to ambient (ignoring
θ
BA
). The equation thus simplifies to:
∆
T = P*(
θ
JB
) = 0.42*(46.9+0) = 19.7°C
In order to maintain a junction temperature below 135°C, the case temperature must
be kept below 115°C. If operation with a package temperature of 125° is required,
then a heatsink may be added to the top of the component for extra heat transfer.
The transmit duty cycle of a remote terminal is typically under 25%, with a few
exceptions for high bandwidth devices like data loaders. Substituting the 25% duty
cycle power (P=0.58W) into the equation above leads to a temperature rise of
27.2°C, which lowers the maximum allowable case temperature to 105°C unless
provision is made to remove heat from the top of the component.
For a bus controller, higher duty cycles are common, so it will likely be necessary to
add some form of heat sink to remove heat from the top of the component. For a BU-
67301B0T0L-102 with a duty cycle of 50%, P=0.7W. If the cooling paths from the top
and bottom of the case to the ambient are assumed to be equally effective, ignoring
θ
CA
and
θ
BA
and setting the temperatures at both the top and bottom of the component
to be the same, then the equations become:
∆
T = P
C
*(
24.5
) = P
B
*(
46.9
)
And:
0.7 = P
C
+P
B
Which turns into:
P
C
= 0.7-P
B
Substituting for P
C
and
θ
Jx
in the equation for
∆
T, we get:
