參數(shù)資料
型號: AV60A-048L-033D025-8
元件分類: 電源模塊
英文描述: 2-OUTPUT 75 W DC-DC REG PWR SUPPLY MODULE
封裝: 61 X 57.90 MM, 12.7 MM HEIGHT, HALF BRICK PACKAGE-10
文件頁數(shù): 15/25頁
文件大?。?/td> 273K
代理商: AV60A-048L-033D025-8
A
A
AV
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V 6
6
6 0
0
0 A
A
A D
D
D U
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U A
A
A L
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L O
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O U
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U T
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T P
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P U
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U T
T
T H
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H A
A
A L
L
L F
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F --B
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B R
R
R IIIIC
C
C K
K
K P
P
P O
O
O W
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W E
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E R
R
R C
C
C O
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O N
N
N V
V
V E
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E R
R
RT
T
T E
E
E R
R
R S
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3 6
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6 V
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V D
D
D C
C
C T
T
TO
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O 7
7
7 5
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5 V
V
V D
D
D C
C
C IIIIN
N
N P
P
P U
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U T
T
T,,,, 7
7
7 5
5
5 W
W
WA
A
AT
T
T T
T
T O
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O U
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U T
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T P
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P U
U
U T
T
-22-
USA
Europe
Asia
TEL:
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velocity have been determined experimentally
and shown in figure 41. The highest values on
each curve represents the point of natural con-
vection.
Figure 41 is used for determining thermal per-
formance under various conditions of airflow
and heat sink configurations.
Example 4. How to determine the allowable
minimum airflow to heat sink combinations
necessary for a module under a desired Tc
and a certain condition?
Although the maximum case temperature for
the AV60A dual output series converters is
100° C, you can improve module reliability by
limiting Tc,max to a lower value. How to
decide? For example, what is the allowable
minimum airflow for AV60A-048L-050D033(N)
heat sink combinations at desired Tc of 80 °C?
The working condition is as following:
Vin = 48V
IO1 = 1.5 A
IO2 = 13.5 A
TA = 40 °C.
Determine PD ( Fig.30 )
PD = 13.5 W
Then solve RCA
::
RCA =
TC, max / PD
RCA =
(TC TA)
/ PD
RCA =
(80 40)
/ 13.5 = 3 °C/W
determine air velocity from figure 41:
If no heat sink:
v = 2.7 m/s (540 ft./min.)
If 1/4 in. heat sink:
v = 1.9 m/s (380 ft./min.)
If 1/2 in. heat sink:
v = 1.2 m/s (24 ft./min.)
If 1 in. heat sink:
v = 0.4 m/s (80 ft./min.)
Example 5. How to determine case tempera-
ture ( Tc ) for the various heat sink configu-
rations at certain air velocity?
What is the allowable Tc for AV60A-048L-
033D025(N) heat sink configurations at desired
air velocity of 2.0 m/s, and it is operating at a 48
V line voltage, a total output current of 15A, a
40 °C maximum ambient temperature?
Determine PD ( Fig. 32. ) with condition:
Vi = 48V
IO1 = 1.5 A, IO2 = 13.5 A
TA = 40 ° C
v = 2.0 m/s (400 ft./min.)
Get: PD = 11.5 W
Determine TC: TC = (RCA x PD) + TA
Determine the corresponding thermal resis-
tances ( RCA ) from Figure 41:
No heat sink: RCA = 3.8 ° C/W
TC = (3.8 x 11.5) + 40 = 83.7 ° C
1/4 in. heat sink: RCA = 2.8 ° C/W
TC = (2.8 x 11.5) + 40 = 72.2 ° C
1/2 in. heat sink: RCA = 2.0 ° C/W
TC = (2 x 11.5) + 40 = 63 ° C
1 in. heat sink: RCA = 1.2 ° C/W
TC = (1.2 x 11.5) + 40 = 53.8 °C
In this configuration, the module does not need
the heat sink and the power module does not
exceed the maximum case temperature of
100° C.
0
0.5(100)
1.0(200)
1.5(300)
2.0(400)
2.5(500)
3.0(600)
0
1
5
6
7
8
Air Velocity m/s (ft./min.)
4
3
2
Case-Ambient
Thermal
Resistance
R
CA
(°C/W)
1 in. heat sink
1/2 in. heat sink
1/4 in. heat sink
NO heat sink
Fig.41 Case-to-Ambient Thermal Resistance
Curves; Either Orientation
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