GE Industrial Solutions LC-LW015-Series - Manual (Page 17)

Lineage Power 17
Data Sheet
March 27, 2008 18 Vdc to 36 Vdc or 36 Vdc to 75 Vdc Inputs, 10 W and 15 W
LC/LW010- and LC/LW015-Series Power Modules:
Thermal Considerations
The power module operates in a variety of thermal
environments; however, sufficient cooling should be
provided to help ensure reliable operation of the unit.
Heat-dissipating components inside the unit are ther-
mally coupled to the case. Heat is removed by conduc-
tion, convection, and radiation to the surrounding
environment. Proper cooling can be verified by mea-
suring the case temperature. The case temperature
(T
C
) should be measured at the position indicated in
Figures 41 and 42.
8-1363(C).b
Note: Dimensions are in millimeters and (inches). Pin locations are
for reference only.
Figure 41. LW010 and LC010 Case Temperature
Measurement Location
8-1363(C).c
Note: Dimensions are in millimeters and (inches). Pin locations are
for reference only.
Figure 42. LW015 and LC015 Case Temperature
Measurement Location
Note that the views in Figures 41 and 42 are of the sur-
face of the modules. The temperatures at these loca-
tions should not exceed the maximum case
temperature indicated on the derating curve. The out-
put power of the module should not exceed the rated
power for the module as listed in the Ordering Informa-
tion table.
Heat Transfer Characteristics
Increasing airflow over the module enhances the heat
transfer via convection. Figures 43 through 45 show
the maximum power that can be dissipated by the mod-
ule without exceeding the maximum case temperature
versus local ambient temperature (T
A
) for natural con-
vection through 3.0 ms –1 (600 ft./min.).
Systems in which these power modules are used typi-
cally generate natural convection airflow rates of
0.25 ms
–1
(50 ft./min.) due to other heat dissipating
components in the system. Therefore, the natural con-
vection condition represents airflow rates of approxi-
mately 0.25 ms
–1 (50 ft./min.). Use of Figure 43 is
shown in the following example.
Example
What is the minimum airflow necessary for an LW010A
operating at 48 V, an output current of 2.0 A, and a
maximum ambient temperature of 91 °C?
Solution:
Given: V
I
= 48 V, I
O
= 2.0 A (I
O
,
max
), T
A
= 91 °C
Determine P
D
(Figure 58): P
D
= 2.5 W
Determine airflow (Figure 43): v = 2.0 ms
–1
(400 ft./min.)
8-1375(C).a
Figure 43. LW010/LC010 Forced Convection Power
Derating; Either Orientation
dc-dc POWER MODULE
15.2
(0.6)
+
-
+
-
OUT IN
LW010/LC010
10.2
(0.4)
dc-dc POWER MODULE
5.1
(0.2)
+
-
+
-
OUT IN LW015/LC015
5.1 (0.2)
UNITS PO
WER DISSIP
A
TION, P
D
(W)
50 60 70 80 90 100 110 40 45 55 65 75 85 95 105
0
3.5
1.5
1
0.5
2
2.5
3
MAX AMBIENT TEMPERATURE, T A (˚C)
MAXIMUM CASE TEMPERATURE
2.0 ms -1 (400 ft./min.)
3.0 ms -1 (600 ft./min.)
1.0 ms -1 (200 ft./min.)
NATURAL CONVECTION