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參數(shù)資料
| 型號(hào)： | MM74C908 |
| 廠商： | National Semiconductor Corporation |
| 英文描述： | Dual CMOS 30V Relay Driver |
| 中文描述： | 雙30V的繼電器驅(qū)動(dòng)器的CMOS |
| 文件頁(yè)數(shù)： | 4/5頁(yè) |
| 文件大?。?/td> | 55K |
| 代理商： | MM74C908 |
www.fairchildsemi.com
4
M
Power Considerations
Calculating Output “ON” Resistance (R
L
>
18
)
The output “ON” resistance, R
ON
, is a function of the junc-
tion temperature, T
J
, and is given by:
R
ON
=
9 (T
J
25) (0.008)
+
9:
and T
J
is given by:
T
J
=
T
A
+
P
DAV
θ
JA
,:
(2)
where T
A
=
ambient temperature,
θ
JA
=
thermal resistance,
and P
DAV
is the average power dissipated within the
device. P
DAV
consists of normal CMOS power terms (due
to leakage currents, internal capacitance, switching, etc.)
which are insignificant when compared to the power dissi-
pated in the outputs. Thus, the output power term defines
the allowable limits of operation and includes both outputs,
A and B. P
D
is given by:
P
D
=
I
OA2
R
ON
+
I
OB2
R
ON
,
where I
O
is the output current, given by:
(1)
(3)
(4)
V
L
is the load voltage.
The average power dissipation, P
DAV
, is a function of the
duty cycle:
P
DAV
=
I
OA2
R
ON
(Duty Cycle
A
)
+
I
OB2
R
ON
(Duty Cycle
B
)
where the duty cycle is the % time in the current source
state. Substituting equations (1) and (5) into (2) yields:
T
J
=
T
A
+
θ
JA
[9 (T
J
25) (0.008)
+
9]:
[I
OA2
(Duty Cycle
A
)
+
I
OB2
(Duty Cycle
B
)]
simplifying:
(5)
(6a)
Equations (1), (4), and (6b) can be used in an iterative
method to determine the output current, output resistance
and junction temperature.
For example, let V
CC
=
15V, R
LA
=
100
, R
LB
=
100
,
V
L
=
0V, T
A
=
25
°
C,
θ
JA
=
110
°
C/W, Duty Cycle
A
=
50%,
Duty Cycle
B
=
75%.
Assuming R
ON
=
11
, then:
and
and R
ON
=
9 (T
J
25) (0.008)
+
9
=
9(52.6
25) (0.008)
+
9
=
11
Applications
(See AN-177 for applications)
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