sla4071 Sanken Electric Co., Ltd., sla4071 Datasheet

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sla4071

Manufacturer Part Number
sla4071
Description
Pnp Darlington With Built-in Flywheel Diode
Manufacturer
Sanken Electric Co., Ltd.
Datasheet

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Company
Part Number
Manufacturer
Quantity
Price
Part Number:
SLA4071
Manufacturer:
SK
Quantity:
20 000
SLA4071
Absolute maximum ratings
Equivalent circuit diagram
Symbol
V
V
V
V
I
T
V
I
FSM
V
P
CBO
CEO
EBO
I
CP
I
I
T
ISO
stg
Characteristic curves
C
B
CE
F
j-c
R
T
j
(sat)-I
0.5
–8
–7
–6
–5
–4
–3
–2
–1
–3
–2
–1
20
10
0
0
–0.3
5
1
I
0
1
C
C
1000 (Between fin and lead pin, AC)
-V
125 C
Temperature Characteristics (Typical)
75 C
CE
–0.5
j-a
–1
25 C
–5 (PW 0.5ms, D
Characteristics (Typical)
5
-PW Characteristics
–8 (PW 1ms, D
–8 (PW 10ms, single)
T
10
a
=–30 C
–1
–2
P
V
–40 to +150
5 (T
W
25 (T
I
CE
C
Ratings
(mS)
(A)
(V)
–100
–100
50 100
–0.5
120
a
150
–3
–6
–5
=25 C)
c
5
=25 )
(I
C
–4
/ I
u
–0.8mA
–0.6mA
–0.4mA
u
B
–5
R
=1000)
50%)
500 1000
25%)
1
: 3k typ R
–5
– 8
2
: 100 typ
PNP Darlington
With built-in flywheel diode
(T
V
Unit
C/W
a
W
V
V
V
A
A
A
A
A
V
rms
C
C
=25 C)
10000
5000
1000
V
500
100
50
–3
–2
–1
25
20
15
10
–0.03
CE
h
0
–0.3 –0.5 –1
5
0
–40
FE
Without Heatsink
(sat)-I
-I
Electrical characteristics
C
P
Diode for flyback voltage absorption
Characteristics (Typical)
Symbol
V
V
Symbol
–0.1
T
0
B
I
-T
CE
C
V
BE
I
=1A
I
h
CBO
EBO
V
V
Characteristics (Typical)
a
CEO
I
t
(sat)
FE
(sat)
R
rr
R
F
Characteristics
I
–5 –10
T
B
I
C
a
50
–0.5
(mA)
(A)
( C)
I
C
=–3A
–1
With Silicone Grease
Natural Cooling
Heatsink: Aluminum
in mm
–100
2000
min
min
120
100
I
C
–50 –100 –200
=–5A
(V
CE
Specification
Specification
typ
=–4V)
–5
150
–8
5000
–1.0
–1.6
100
typ
typ
15000
External dimensions
max
–1.5
–2.0
max
–10
–10
1.2
10
h
I
FE
C
-V
-I
10000
–0.05
–0.03
C
BE
5000
1000
–0.5
–0.1
–10
500
100
Temperature Characteristics (Typical)
–8
–6
–4
–2
–5
–1
50
–0.03
0
Temperature Characteristics (Typical)
–3
0
Single Pulse
Without Heatsink
T
a
Safe Operating Area (SOA)
=25 C
Unit
Unit
mA
ns
–5
V
V
V
V
V
A
A
–0.1
–10
–1
V
V
I
CE
BE
C
V
I
–0.5 –1
(A)
C
(V)
(V)
CE
A
=–3A, I
Conditions
Conditions
V
I
I
=–2V, I
F
V
V
C
–2
CB
I
= 100mA
R
• • •
=–10mA
R
EB
I
=10 A
F
=–100V
=120V
(V
(V
=1A
–50
=–6V
SLA (12-pin)
CE
CE
B
=–4V)
=–6mA
=–4V)
C
–5
=–3A
–100
(T
(T
–3
–8
a
a
=25 C)
=25 C)
25

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