BCR10 [INFINEON]

NPN/PNP Silicon Digital Tansistor Array (Switching circuit, inverter, interface circuit, drive circuit); NPN / PNP硅数字Tansistor阵列(开关电路,逆变器,接口电路,驱动电路)
BCR10
型号: BCR10
厂家: Infineon    Infineon
描述:

NPN/PNP Silicon Digital Tansistor Array (Switching circuit, inverter, interface circuit, drive circuit)
NPN / PNP硅数字Tansistor阵列(开关电路,逆变器,接口电路,驱动电路)

开关 驱动
文件: 总5页 (文件大小:67K)
中文:  中文翻译
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BCR 10PN  
NPN/PNP Silicon Digital Tansistor Array  
• Switching circuit, inverter, interface circuit,  
drive circuit  
• Two (galvanic) internal isolated NPN/PNP  
Transistor in one package  
• Built in bias resistor (R =10k , R =10k )  
1
2
Tape loading orientation  
Type  
Marking Ordering Code Pin Configuration  
Package  
BCR 10PN W1s  
Q62702-C2411 1=E1 2= B1 3=C2 4=E2 5=B2 6=C1 SOT-363  
Maximum Ratings  
Parameter  
Symbol  
Values  
Unit  
Collector-emitter voltage  
Collector-base voltage  
Emitter-base voltage  
Input on Voltage  
V
CEO  
V
CBO  
V
EBO  
V
i(on)  
50  
V
50  
10  
20  
DC collector current  
I
100  
mA  
mW  
°C  
C
Total power dissipation, T = 115°C  
P
250  
S
tot  
j
Junction temperature  
Storage temperature  
T
T
150  
- 65 ... + 150  
stg  
Thermal Resistance  
1)  
Junction ambient  
R
R
275  
K/W  
thJA  
Junction - soldering point  
140  
thJS  
1) Package mounted on pcb 40mm x 40mm x 1.5mm / 0.5cm2 Cu  
Semiconductor Group  
1
Nov-26-1996  
BCR 10PN  
Electrical Characteristics at T =25°C, unless otherwise specified  
A
Parameter  
Symbol  
Values  
typ.  
Unit  
min.  
max.  
DC Characteristics  
Collector-emitter breakdown voltage  
V
V
V
(BR)CEO  
(BR)CBO  
CBO  
I = 100 µA, I = 0  
50  
50  
-
-
-
-
-
-
-
-
-
C
B
Collector-base breakdown voltage  
I = 10 µA, I = 0  
-
C
B
Collector cutoff current  
= 40 V, I = 0  
I
I
nA  
mA  
-
V
CB  
100  
0.75  
-
E
Emitter cutoff current  
= 10 V, I = 0  
EBO  
V
EB  
-
C
DC current gain  
I = 5 mA, V = 5 V  
h
FE  
30  
-
C
CE  
Collector-emitter saturation voltage 1)  
I = 10 mA, I = 0.5 mA  
V
V
V
V
CEsat  
i(off)  
0.3  
1.5  
C
B
Input off voltage  
I = 100 µA, V = 5 V  
0.8  
C
CE  
Input on Voltage  
I = 2 mA, V = 0.3 V  
i(on)  
1
-
2.5  
13  
C
CE  
Input resistor  
Resistor ratio  
R
7
10  
1
k
-
1
R /R  
0.9  
1.1  
1
2
AC Characteristics for NPN Type  
Transition frequency  
f
MHz  
pF  
T
I = 10 mA, V = 5 V, f = 100 MHz  
-
-
130  
3
-
-
C
CE  
Collector-base capacitance  
= 10 V, f = 1 MHz  
C
cb  
V
CB  
1) Pulse test: t < 300µs; D < 2%  
Semiconductor Group  
2
Nov-26-1996  
BCR 10PN  
NPN TYPE  
DC Current Gain h = f (I )  
Collector-Emitter Saturation Voltage  
V = f(I ), h = 20  
CEsat  
FE  
C
V
CE  
= 5V (common emitter configuration)  
C
FE  
10 2  
10 3  
-
hFE  
IC  
V
10 2  
10 1  
10 0  
10 1  
10 0  
10 -1  
10 0  
10 1  
mA  
IC  
0.0  
0.2  
0.4  
0.6  
V
1.0  
VCEsat  
Input on Voltage V  
= f(I )  
Input off voltage V = f(I )  
i(off) C  
i(on)  
C
V
CE  
= 0.3V (common emitter configuration)  
V
CE  
= 5V (common emitter configuration)  
10 2  
mA  
10 1  
mA  
IC  
IC  
10 1  
10 0  
10 -1  
10 0  
10 -1  
10 -2  
10 -1  
10 0  
10 1  
V
0.0  
0.5  
1.0  
1.5  
2.0  
V
3.0  
Vi(on)  
Vi(off)  
Semiconductor Group  
3
Nov-26-1996  
BCR 10PN  
PNP TYPE  
DC Current Gain h = f (I )  
Collector-Emitter Saturation Voltage  
V = f(I ), h = 20  
CEsat  
FE  
C
V
CE  
= 5V (common emitter configuration)  
C
FE  
10 2  
10 3  
-
hFE  
IC  
mA  
10 2  
10 1  
10 0  
10 1  
10 0  
10 -1  
10 0  
10 1  
mA  
IC  
0.0  
0.2  
0.4  
0.6  
V
1.0  
VCEsat  
Input on Voltage V  
= f(I )  
Input off voltage V = f(I )  
i(off) C  
i(on)  
C
V
CE  
= 0.3V (common emitter configuration)  
V
CE  
= 5V (common emitter configuration)  
10 2  
mA  
10 1  
mA  
IC  
IC  
10 1  
10 0  
10 -1  
10 0  
10 -1  
10 -2  
10 -1  
10 0  
10 1  
V
0.0  
0.5  
1.0  
1.5  
V
2.5  
Vi(on)  
Vi(off)  
Semiconductor Group  
4
Nov-26-1996  
BCR 10PN  
Total power dissipation P = f (T *;T )  
tot  
A
S
* Package mounted on epoxy  
300  
mW  
Ptot  
TS  
TA  
200  
150  
100  
50  
0
0
20  
40  
60  
80  
100 120 °C 150  
TA,TS  
Permissible Pulse Load R  
= f(t )  
Permissible Pulse Load P  
/ P = f(t )  
totDC p  
thJS  
p
totmax  
10 3  
K/W  
10 3  
-
RthJS  
10 2  
Ptotmax/PtotDC  
D = 0  
0.005  
0.01  
0.02  
0.05  
0.1  
10 2  
10 1  
10 0  
10 1  
0.2  
0.5  
0.5  
0.2  
0.1  
0.05  
0.02  
0.01  
0.005  
D = 0  
10 0  
10 -1  
10 -6  
10 -5  
10 -4  
10 -3  
10 -2  
10 -1 s 10 0  
tp  
10 -6  
10 -5  
10 -4  
10 -3  
10 -2  
10 -1 s 10 0  
tp  
Semiconductor Group  
5
Nov-26-1996  

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