MPSA06RLRAG ON Semiconductor, MPSA06RLRAG Datasheet - Page 5

• Manufacturer Part Number: MPSA06RLRAG
• Description: TRANS NPN GP SS 80V TO92
• Manufacturer: ON Semiconductor

Specifications of MPSA06RLRAG

• Transistor Type: NPN
• Current - Collector (ic) (max): 500mA
• Voltage - Collector Emitter Breakdown (max): 80V
• Vce Saturation (max) @ Ib, Ic: 250mV @ 10mA, 100mA
• Current - Collector Cutoff (max): 100nA
• Dc Current Gain (hfe) (min) @ Ic, Vce: 100 @ 100mA, 1V
• Power - Max: 625mW
• Frequency - Transition: 100MHz
• Mounting Type: Through Hole
• Package / Case: TO-92-3 (Standard Body), TO-226
• Configuration: Single
• Transistor Polarity: NPN
• Mounting Style: Through Hole
• Collector- Emitter Voltage Vceo Max: 80 V
• Emitter- Base Voltage Vebo: 4 V
• Continuous Collector Current: 0.5 A
• Maximum Dc Collector Current: 0.5 A
• Power Dissipation: 625 W
• Maximum Operating Frequency: 100 MHz
• Maximum Operating Temperature: + 150 C
• Dc Collector/base Gain Hfe Min: 100 at 10 mA at 1 V
• Minimum Operating Temperature: - 55 C
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: MPSA06RLRAGOS; MPSA06RLRAGOS; MPSA06RLRAGOSTR

−0.8
−1.2
−1.6
−2.0
−2.4
−2.8
1.0
0.8
0.6
0.4
0.2
0.07
0.05
0.03
0.02
0.01
0
1.0
0.7
0.5
0.3
0.2
0.1
0.5
0.05
1.0
10 mA
Figure 14. MPSA05/06 Collector Saturation
I
0.05
C
D = 0.5
SINGLE PULSE
0.1
50 mA
1.0
0.2
0.1
=
I
C
Figure 16. MPSA05/06 Base−Emitter
=
R
2.0
qVB
2.0
0.2
for V
Temperature Coefficient
I
C
, COLLECTOR CURRENT (mA)
5.0
BE
0.5
I
Figure 18. MPSA05, MPSA06, MPSA55 and MPSA56 Thermal Response
B
5.0
100 mA
, BASE CURRENT (mA)
NPN − MPSA05, MPSA06*; PNP − MPSA55, MPSA56*
I
SINGLE PULSE
C
0.02
=
1.0
10
Region
NPN
10
0.01
2.0
20
250 mA
I
20
C
=
5.0
50
100
10
500 mA
50
I
C
T
=
200
J
20
= 25°C
http://onsemi.com
100
t, TIME (ms)
500
50
200
5
−1.0
−0.8
−0.6
−0.4
−0.2
−0.8
−1.2
−1.6
−2.0
−2.4
−2.8
P
0
−0.5
−0.05
(pk)
500
DUTY CYCLE, D = t
Figure 15. MPSA55/56 Collector Saturation
−10 mA
−1.0 −2.0
−0.1
t
I
1
C
1.0 k
−50 mA
Figure 17. MPSA55/56 Base−Emitter
=
t
2
I
R
C
−0.2
qVB
=
for V
2.0 k
Temperature Coefficient
I
C
−5.0
, COLLECTOR CURRENT (mA)
−0.5
BE
1
I
/t
B
2
, BASE CURRENT (mA)
−100 mA
−1.0
−10
5.0 k
I
Region
C
PNP
=
−20
−2.0
10 k
−250 mA
Z
T
Z
T
D CURVES APPLY FOR
POWER PULSE TRAIN
SHOWN READ TIME AT t
(SEE AN469)
qJC
J(pk)
qJA
J(pk)
I
C
−5.0
−50
=
(t) = r(t) • R
(t) = r(t) • R
− T
− T
20 k
A
C
−100
−10
= P
= P
−500 mA
I
(pk)
(pk)
C
qJC
qJA
=
T
−200
−20
Z
Z
J
50 k 100 k
qJA
qJC
= 25°C
(t)
(t)
−500
1
−50