2N5194G ON Semiconductor, 2N5194G Datasheet - Page 5

TRANS PNP PWR GP 4A 60V TO225AA

2N5194G

Manufacturer Part Number
2N5194G
Description
TRANS PNP PWR GP 4A 60V TO225AA
Manufacturer
ON Semiconductor
Type
Powerr
Datasheets

Specifications of 2N5194G

Transistor Type
PNP
Current - Collector (ic) (max)
4A
Voltage - Collector Emitter Breakdown (max)
60V
Vce Saturation (max) @ Ib, Ic
1.4V @ 1A, 4A
Current - Collector Cutoff (max)
1mA
Dc Current Gain (hfe) (min) @ Ic, Vce
25 @ 1.5A, 2V
Power - Max
40W
Frequency - Transition
2MHz
Mounting Type
Through Hole
Package / Case
TO-225-3
Current, Collector
4 A
Current, Gain
10
Frequency
2 MHz
Package Type
TO-225AA
Polarity
PNP
Power Dissipation
40 W
Primary Type
Si
Resistance, Thermal, Junction To Case
3.12 °C/W
Voltage, Breakdown, Collector To Emitter
60 V
Voltage, Collector To Base
60 V
Voltage, Collector To Emitter
60 V
Voltage, Collector To Emitter, Saturation
1.4 V
Voltage, Emitter To Base
5 V
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Other names
2N5194GOS

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
2N5194G
Manufacturer:
ON
Quantity:
1 996
0.07
0.05
0.03
0.02
0.01
1.0
0.7
0.5
0.3
0.2
0.1
0.01
SINGLE PULSE
0.02
0.05
D = 0.5
0.02 0.03 0.05
0.2
0.1
t
1
P
DESIGN NOTE: USE OF TRANSIENT THERMAL RESISTANCE DATA
P
0.01
1/f
Figure 13.
t
P
0.1
DUTY CYCLE, D = t
PEAK PULSE POWER = P
0.2 0.3
P
P
0.5
Figure 12. Thermal Response
1
f =
2N5194, 2N5195
http://onsemi.com
1.0
t, TIME OR PULSE WIDTH (ms)
t P
t 1
P
2.0 3.0
5
the model shown in Figure 13. Using the model and the
device thermal response, the normalized effective transient
thermal resistance of Figure 12 was calculated for various
duty cycles.
by the steady state value q
The 2N5193 is dissipating 50 watts under the following
conditions: t
the reading of r(t
A train of periodical power pulses can be represented by
To find q
Example:
Using Figure 12, at a pulse width of 0.1 ms and D = 0.2,
The peak rise in junction temperature is therefore:
5.0
DT = r(t) x P
10
JC
1
(t), multiply the value obtained from Figure 12
= 0.1 ms, t
20
1
P
, D) is 0.27.
x q
30
JC
p
= 0.27 x 50 x 3.12 = 42.2_C
50
JC
= 0.5 ms. (D = 0.2).
.
q
JC(max)
100
= 3.12°C/W
200 300
500
1000

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