Tyco Electronics Corp.

3

Technical Note

July 1996

250 W—300 W Board-Mounted Power Modules

Thermal Management for FC- and FW-Series

Convection Without Heat Sinks

Increasing airflow over the module enhances heat

transfer via convection. Figures 4 and 5 show the maxi-

mum power that can be dissipated by the module with-

out exceeding the maximum case temperature versus

through 800 ft./min. A natural convection condition is

produced when air is moved only through the buoyancy

effects produced by a temperature gradient between

the module and surrounding air. In the test setup used,

natural convection airflow was measured at 10 ft./min.

to 20 ft./min., whereas systems in which these power

modules may be used typically generate natural con-

vection airflow rates of 60 ft./min. due to other heat dis-

sipating components in the system. The 100 ft./min. to

800 ft./min. curves are for airflow added externally to

the test setup, usually through the use of fans. Note

that there is a thermal performance improvement when

the long axis of the module is perpendicular to the air-

flow direction (transverse orientation).

8-1314

Figure 4. Convection Power Derating with No

Heat Sink; Airflow Along Length (Lon-

gitudinal)

8-1315

Figure 5. Convection Power Derating with No

Heat Sink; Airflow Along Width (Trans-

verse)

Figures 4 and 5 can be used to determine the appropri-

ate airflow for a given set of operating conditions as

shown in the following examples.

Example 1: Airflow Required to Maintain Tc

What is the minimum airflow necessary for a FW300A1

in the transverse orientation, operating at 54 V input, an

output current of 50 A, and a maximum ambient tem-

perature of 35 °C?

Solution:

Determine Airflow (Figure 5): v = 800 ft./min.

Example 2: Maximum Power Output

What is the maximum power output for a FW300A1 in

the longitudinal orientation, operating at 54 V input, in

an environment that provides 600 ft./min. with a maxi-

mum ambient temperature of 40 °C?

Solution:

Determine Io (Figure 2): Io = 40 A

Calculate Po = (Vo) * (Io) = 5 x 40 = 200 W

Although the above two examples use 100 °C as the

operating case temperature, for extremely high reliabil-

ity applications, one may design to a lower case tem-

perature as shown later in Example 4.

0

10

203040

100

0

40

60

70

°C)

30

20

10

90

80

70

60

50

800 ft./min.

700 ft./min.

600 ft./min.

500 ft./min.

400 ft./min.

300 ft./min.

200 ft./min.

100 ft./min.

50

20 ft./min. (NAT. CONV.)

0

10

203040

100

0

40

60

70

°C)

30

20

10

90

80

70

60

50

800 ft./min.

700 ft./min.

600 ft./min.

500 ft./min.

400 ft./min.

300 ft./min.

200 ft./min.

100 ft./min.

50

20 ft./min. (NAT. CONV.)