MPSA13RLRAG ON Semiconductor, MPSA13RLRAG Datasheet - Page 5

TRANS NPN DARL SS 30V TO92

MPSA13RLRAG

Manufacturer Part Number
MPSA13RLRAG
Description
TRANS NPN DARL SS 30V TO92
Manufacturer
ON Semiconductor
Datasheet

Specifications of MPSA13RLRAG

Transistor Type
NPN - Darlington
Current - Collector (ic) (max)
500mA
Voltage - Collector Emitter Breakdown (max)
30V
Vce Saturation (max) @ Ib, Ic
1.5V @ 100µA, 100mA
Dc Current Gain (hfe) (min) @ Ic, Vce
10000 @ 100mA, 5V
Power - Max
625mW
Frequency - Transition
125MHz
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
30 V
Emitter- Base Voltage Vebo
10 V
Collector- Base Voltage Vcbo
30 V
Maximum Dc Collector Current
0.5 A
Maximum Collector Cut-off Current
0.1 uA
Power Dissipation
625 mW
Maximum Operating Temperature
+ 150 C
Continuous Collector Current
0.5 A
Dc Collector/base Gain Hfe Min
5000
Minimum Operating Temperature
- 55 C
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Current - Collector Cutoff (max)
-
Lead Free Status / Rohs Status
Lead free / RoHS Compliant
Other names
MPSA13RLRAGOS
MPSA13RLRAGOS
MPSA13RLRAGOSTR

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
MPSA13RLRAG
Manufacturer:
ON Semiconductor
Quantity:
25 973
Part Number:
MPSA13RLRAG
Manufacturer:
ON Semiconductor
Quantity:
295
0.07
0.05
0.03
0.02
0.01
1.0
0.7
0.5
0.3
0.2
0.1
0.1
0.2
D = 0.5
0.2
0.1
0.05
0.5
1.0
SINGLE PULSE
Design Note: Use of Transient Thermal Resistance Data
1.0 k
SINGLE PULSE
700
500
300
200
100
70
50
30
20
10
2.0
0.4
Figure 13. Active Region Safe Operating Area
0.6
T
A
= 25°C
V
5.0
CE
1.0
, COLLECTOR−EMITTER VOLTAGE (VOLTS)
Figure 12. Thermal Response
MPSA13, MPSA14
DUTY CYCLE + t 1 f +
PEAK PULSE POWER = P
http://onsemi.com
CURRENT LIMIT
THERMAL LIMIT
SECOND BREAKDOWN LIMIT
10
t
1
2.0
FIGURE A
P
P
1/f
t
20
P
t, TIME (ms)
5
4.0
6.0
50
T
t 1
t P
C
P
= 25°C
10
100
P
1.0 s
P
1.0 ms
Z
Z
100 ms
qJC(t)
qJA(t)
20
200
= r(t) • R
= r(t) • R
40
500
qJC
qJA
T
T
J(pk)
J(pk)
1.0 k
− T
− T
A
C
= P
= P
2.0 k
(pk)
(pk)
Z
Z
qJA(t)
qJC(t)
5.0 k 10 k

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