TIP115G ON Semiconductor, TIP115G Datasheet - Page 4

• Manufacturer Part Number: TIP115G
• Description: TRANS DARL PNP 2A 60V TO220AB
• Manufacturer: ON Semiconductor
• Type: Medium Power, Switchr

Specifications of TIP115G

• Transistor Type: PNP - Darlington
• Current - Collector (ic) (max): 2A
• Voltage - Collector Emitter Breakdown (max): 60V
• Vce Saturation (max) @ Ib, Ic: 2.5V @ 8mA, 2A
• Current - Collector Cutoff (max): 2mA
• Dc Current Gain (hfe) (min) @ Ic, Vce: 1000 @ 1A, 4V
• Power - Max: 2W
• Mounting Type: Through Hole
• Package / Case: TO-220-3 (Straight Leads)
• Current, Gain: 500
• Current, Input: 50 mA
• Current, Output: 2 A
• Package Type: TO-220AB
• Polarity: PNP
• Power Dissipation: 50 W
• Primary Type: Si
• Resistance, Thermal, Junction To Ambient: 62.5
• Voltage, Collector To Emitter, Saturation: 2.5 V
• Voltage, Input: 5 V
• Voltage, Output: 60 V
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Frequency - Transition: -
• Lead Free Status / Rohs Status: RoHS Compliant part Electrostatic Device
• Other names: TIP115GOS

approx
+ 8.0 V
approx
-12 V
V
R
D
V
t
DUTY CYCLE = 1.0%
r
2
1N5825 USED ABOVE I
MSD6100 USED BELOW I
1
, t
B
1
, MUST BE FAST RECOVERY TYPE, eg:
0
f
& R
0.07
0.05
0.03
0.02
0.01
≤ 10 ns
1.0
0.7
0.5
0.3
0.2
0.1
0.01
C
VARIED TO OBTAIN DESIRED CURRENT LEVELS
Figure 2. Switching Times Test Circuit
0.02
0.01
D = 0.5
0.05
0.02
0.2
0.1
25 ms
SINGLE PULSE
TIP110, TIP111, TIP112 (NPN); TIP115, TIP116, TIP117 (PNP)
B
≈ 100 mA
B
0.05
≈ 100 mA
51
for t
and V
to obtain desired test currents.
For NPN test circuit, reverse diode,
polarities and input pulses.
d
0.1
and t
2
R
= 0, R
D
B
1
+ 4.0 V
r
, D
B
1
0.2
is disconnected
and R
≈ 8.0 k
C
are varied
TUT
0.5
≈ 60
R
Figure 4. Thermal Response
C
- 30 V
V
http://onsemi.com
CC
1.0
SCOPE
2.0
t, TIME (ms)
4
4.0
2.0
1.0
0.8
0.6
0.4
0.2
0.04
Z
R
D CURVES APPLY FOR POWER
PULSE TRAIN SHOWN
READ TIME AT t
T
5.0
qJC(t)
J(pk)
qJC
= 2.5°C/W MAX
- T
0.06
= r(t) R
C
10
= P
qJC
(pk)
0.1
1
PNP
NPN
Figure 3. Switching Times
Z
I
20
C
qJC(t)
, COLLECTOR CURRENT (AMP)
t
0.2
s
50
P
0.4 0.6
V
I
(pk)
C
CC
/I
B
DUTY CYCLE, D = t
100
= 30 V
= 250
t
1
t
2
200
1.0
I
T
B1
J
= 25°C
t
= I
d
t
f
@ V
B2
2.0
1
BE(off)
/t
500
2
t
r
= 0
1.0 k
4.0