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Электронный компонент: BAT165E6433

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BAT165
Jul-06-2001
1
Silicon Schottky Diode
Medium current Schottky rectifier diode
For low-loss, fast-recovery, meter protection,
bias isolation and clamping applications
Miniature plastic package for surface
mounting (SMD)
VPS05176
1
2
ESD
: Electrostatic discharge sensitive device, observe handling precaution!
Type
Marking
Pin Configuration
Package
BAT165
White/C
1 = C
2 = A
SOD323
Maximum Ratings
Parameter
Symbol
Value
Unit
Diode reverse voltage
V
R
40
V
Forward current
I
F
750
mA
Average forward current (50/60Hz, sinus)
I
FAV
500
Surge forward current (t
10ms)
I
FSM
2.5
A
Total power dissipation
, T
S
= 66 C
P
tot
600
mW
Junction temperature
T
j
150
C
Storage temperature
T
stg
-65 ... 150
Maximum Ratings
Junction - soldering point
1)
R
thJS
140
K/W
1For calculation of R
thJA
please refer to Application Note Thermal Resistance
BAT165
Jul-06-2001
2
Electrical Characteristics at T
A
= 25C, unless otherwise specified
Parameter
Symbol
Values
Unit
min.
typ.
max.
Characteristics
Reverse current
V
R
= 30 V
I
R
-
-
50
A
Reverse current
V
R
= 30 V, T
A
= 65 C
I
R
-
-
900
Forward voltage
I
F
= 10 mA
I
F
= 100 mA
I
F
= 250 mA
I
F
= 750 mA
V
F
-
-
-
-
0.305
0.38
0.44
0.58
0.4
-
0.7
-
V
AC characteristics
Diode capacitance
V
R
= 10 V, f = 1 MHz
C
T
-
8.4
12
pF
Reverse current I
R
= f (V
R
)
T
A
= Parameter
0
4
8
12
16
20
24
V
30
V
R
-11
10
-10
10
-9
10
-8
10
-7
10
-6
10
-5
10
-4
10
A
I
R
-40C
-20C
0C
25C
45C
65C
Forward current I
F
= f (V
F
)
T
A
= parameter
0.00 0.10 0.20 0.30 0.40 0.50 0.60
V
0.80
V
F
-5
10
-4
10
-3
10
-2
10
-1
10
0
10
A
I
F
TA= -40 C
= -20
= 0
= 25
= 45
= 65
= 85
BAT165
Jul-06-2001
3
Forward current I
F
= f (T
S
)
0
20
40
60
80
100
120 C
150
T
S
0
100
200
300
400
500
600
mA
800

I
F
Diode capacitance C
T
= f (V
R
)
f = 1MHz
0
2
4
6
8
10
12
V
15
V
R
6
10
14
18
22
26
30
34
pF
40
C
T
Permissible Pulse Load R
thJS
= f(t
p
)
10
-6
10
-5
10
-4
10
-3
10
-2
10
0
s
t
p
-1
10
0
10
1
10
2
10
3
10
K/W

R
thJS
0.5
0.2
0.1
0.05
0.02
0.01
0.005
D = 0
Permissible Pulse Load
I
Fmax
/ I
FDC
= f(t
p
)
10
-6
10
-5
10
-4
10
-3
10
-2
10
0
s
t
p
0
10
1
10
2
10
-

I
Fmax
/
I
FDC
D = 0
0.005
0.01
0.02
0.05
0.1
0.2
0.5