TIP30BG ON Semiconductor, TIP30BG Datasheet - Page 4

• Manufacturer Part Number: TIP30BG
• Description: Power 1a 100v Pnp Transistor Complementary Silicon Plastic Power Transistors
• Manufacturer: ON Semiconductor

APPROX
V
+11 V
APPROX
EB(off)
3.0
0.1
0.1
10
500
300
100
7.0
5.0
+11 V
70
50
30
10
1.0
V
0.03
V
in
in
CURVES APPLY BELOW
RATED V
0
Figure 5. Active Region Safe Operating Area
TURN-OFF PULSE
TURN-ON PULSE
Figure 3. Switching Time Equivalent Circuit
T
J
0.05
= 150°C
T
V
J
CEO
CE
= 150°C
0.07
25°C
SECOND BREAKDOWN LIMITED
THERMALLY LIMITED @ T
BONDING WIRE LIMITED
, COLLECTOR-EMITTER VOLTAGE, (VOLTS)
t
t
1
2
- 55°C
Figure 1. DC Current Gain
0.1
I
4.0
C
, COLLECTOR CURRENT (AMP)
t
3
100 < t
DUTY CYCLE ≈ 2.0%
APPROX - 9.0 V
t
V
t
1
3
CC
V
≤ 7.0 ns
< 15 ns
in
2
10
< 500 ms
0.3
TIP29, A, B, C (NPN), TIP30, A, B, C (PNP)
R
DESIRED CURRENT LEVELS.
C
B
= 25°C
0.5
C
and R
20
TIP29, 30
TIP29A, 30A
TIP29B, 30B
TIP29C, 30C
R
jd
B
<< C
0.7
C
R
VARIED TO OBTAIN
C
eb
- 4.0 V
1.0
dc
40
V
1 ms
CE
= 2.0 V
http://onsemi.com
5 ms
100
SCOPE
3.0
4
a transistor: average junction temperature and second
breakdown. Safe operating area curves indicate I
operation; i.e., the transistor must not be subjected to greater
dissipation than the curves indicate.
variable depending on conditions. Second breakdown pulse
limits are valid for duty cycles to 10% provided T
v 150°C. At high case temperatures, thermal limitations
will reduce the power that can be handled to values less than
the limitations imposed by second breakdown.
0.07
0.05
0.03
0.07
0.05
0.03
0.02
3.0
2.0
1.0
0.7
0.5
0.3
0.2
0.1
2.0
1.0
0.7
0.5
0.3
0.1
There are two limitations on the power handling ability of
The data of Figure 5 is based on T
0.03
0.03
t
f
0.05
0.05
@ V
t
r
@ V
CC
0.07
0.07
t
f
= 10 V
CC
@ V
t
r
0.1
0.1
I
@ V
I
= 10 V
C
Figure 2. Turn-Off Time
C
Figure 4. Turn-On Time
, COLLECTOR CURRENT (AMP)
, COLLECTOR CURRENT (AMP)
CC
CC
= 30 V
= 30 V
0.2
t
0.3
0.3
d
t
s
@ V
′
EB(off)
0.5
0.5
J(pk)
0.7
= 2.0 V
0.7
1.0
1.0
= 150°C; T
I
I
t
T
B1
C
s
J
′ = t
/I
I
T
= 25°C
C
B
= I
J
/I
= 10
= 25°C
s
B
B2
- 1/8 t
= 10
C
2.0
- V
J(pk)
f
C
3.0
3.0
CE
is