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Электронный компонент: BDP956E6327

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BDP952...BDP956
1
Jul-06-2001
PNP Silicon AF Power Transistors
For AF driver and output stages
High current gain
Low collector-emitter saturation voltage
Complementary types: BDP951...BDP955 (NPN)
VPS05163
1
2
3
4
Type
Marking
Pin Configuration
Package
BDP952
BDP954
BDP956
BDP 952
BDP 954
BDP 956
1 = B
1 = B
1 = B
2 = C
2 = C
2 = C
3 = E
3 = E
3 = E
4 = C
4 = C
4 = C
SOT223
SOT223
SOT223
Maximum Ratings
Parameter
Symbol
BDP952 BDP954 BDP956 Unit
Collector-emitter voltage
V
CEO
80
100
120
V
Collector-base voltage
V
CBO
100
120
140
Emitter-base voltage
V
EBO
5
5
5
DC collector current
I
C
3
A
Peak collector current
I
CM
5
Base current
I
B
200
mA
Peak base current
I
BM
500
Total power dissipation
, T
S
= 99 C
P
tot
3
W
Junction temperature
T
j
150
C
Storage temperature
T
stg
-65 ... 150
Thermal Resistance
Junction - soldering point
1)
R
thJS
17
K/W
1For calculation of R
thJA
please refer to Application Note Thermal Resistance
BDP952...BDP956
2
Jul-06-2001
Electrical Characteristics at T
A
= 25C, unless otherwise specified.
Parameter
Symbol
Values
Unit
min.
typ.
max.
DC Characteristics
Collector-emitter breakdown voltage
I
C
= 10 mA, I
B
= 0

BDP952
BDP954
BDP956
V
(BR)CEO
80
100
120
-
-
-
-
-
-
V
Collector-base breakdown voltage
I
C
= 100 A, I
B
= 0

BDP952
BDP954
BDP956
V
(BR)CBO
100
120
140
-
-
-
-
-
-
Emitter-base breakdown voltage
I
E
= 10 A, I
C
= 0
V
(BR)EBO
5
-
-
Collector cutoff current
V
CB
= 100 V, I
E
= 0
I
CBO
-
-
100
nA
Collector cutoff current
V
CB
= 100 V, I
E
= 0 , T
A
= 150 C
I
CBO
-
-
20
A
Emitter cutoff current
V
EB
= 4 V, I
C
= 0
I
EBO
-
-
100
nA
DC current gain 1)
I
C
= 10 mA, V
CE
= 5 V
I
C
= 500 mA, V
CE
= 1 V
I
C
= 2 A, V
CE
= 2 V
h
FE
25
40
15
-
-
-
-
475
-
-
Collector-emitter saturation voltage1)
I
C
= 2 A, I
B
= 0.2 A
V
CEsat
-
-
0.8
V
Base-emitter saturation voltage 1)
I
C
= 2 A, I
B
= 0.2 A
V
BEsat
-
-
1.5
AC Characteristics
Transition frequency
I
C
= 50 mA, V
CE
= 10 V, f = 100 MHz
f
T
-
100
-
MHz
Collector-base capacitance
V
CB
= 10 V, f = 1 MHz
C
cb
-
40
-
pF
1) Pulse test: t
=
300
s, D = 2%
BDP952...BDP956
3
Jul-06-2001
Total power dissipation
P
tot
= f (T
S
)
0
20
40
60
80
100
120 C
150
T
S
0
0.4
0.8
1.2
1.6
2
2.4
W
3.2

P
tot
Permissible Pulse Load
R
thJS
= f (t
p
)
10
-6
10
-5
10
-4
10
-3
10
-2
10
0
s
t
p
-2
10
-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
P
totmax
/ P
totDC
= f (t
p
)
10
-6
10
-5
10
-4
10
-3
10
-2
10
0
s
t
p
0
10
1
10
2
10
3
10
-

P
totmax
/ P
totDC
D = 0
0.005
0.01
0.02
0.05
0.1
0.2
0.5
DC current gain
h
FE
= f (I
C
)
V
CE
= 2V
10
0
10
1
10
2
10
3
10
4
mA
I
C
1
10
2
10
3
10
-

h
FE
-50C
25C
100C
BDP952...BDP956
4
Jul-06-2001
Collector cutoff current
I
CBO
= f (T
A
)
V
CB
= 45V
0
20
40
60
80
100
120 C
150
T
A
-1
10
0
10
1
10
2
10
3
10
4
10
5
10
nA

I
CBO
max
typ
Collector-emitter saturation voltage
I
C
= f (V
CEsat
), h
FE
= 10
0.0
0.1
0.2
0.3
0.4
0.5
0.6
V
0.8
V
CEsat
0
10
1
10
2
10
3
10
4
10
mA

I
C
100C
25C
-50C
Collector current
I
C
= f (V
BE
)
V
CE
= 2V
0.0
0.2
0.4
0.6
0.8
1.0
V
1.3
V
BE
0
10
1
10
2
10
3
10
4
10
mA

I
C
-50C
25C
100C
Base-emitter saturation voltage
I
C
= f (V
BEsat
), h
FE
= 10
0.0
0.2
0.4
0.6
0.8
1.0
V
1.3
V
BEsat
0
10
1
10
2
10
3
10
4
10
mA

I
C
-50C
25C
100C