2N5190G Datasheet, PDF (5/6 Pages) ON Semiconductor – Silicon NPN Power Transistors

2N5190G, 2N5191G, 2N5192G
10
5.0
T_J = 150°C
2.0
5.0 ms
100 ms
1.0 ms
dc
1.0
SECONDARY BREAKDOWN LIMIT
0.5
THERMAL LIMIT AT T_C = 25°C
BONDING WIRE LIMIT
CURVES APPLY BELOW RATED V_CEO
0.2
2N5191
0.1
1.0 2.0
2N5192
5.0
10
20
50
100
V_CE, COLLECTOR-EMITTER VOLTAGE (VOLTS)
Figure 11. Rating and Thermal Data
Active−Region Safe Operating Area
There are two limitations on the power handling ability of
a transistor; average junction temperature and second
breakdown. Safe operating area curves indicate I_C − V_CE
limits of the transistor that must be observed for reliable
operation; i.e., the transistor must not be subjected to greater
dissipation than the curves indicate.
The data of Figure 11 is based on T_J(pk) = 150_C; T_C is
variable depending on conditions. Second breakdown pulse
limits are valid for duty cycles to 10% provided T_J(pk)
≤ 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.
1.0
0.7 _{D = 0.5}
0.5
0.3
0.2
0.2
0.1
0.1
0.05
0.07
0.05 ^0.02
0.03
0.01
SINGLE PULSE
0.02
0.01
0.01 0.02 0.03 0.05 0.1 0.2 0.3 0.5 1.0 2.0 3.0 5.0 10 20
t, TIME OR PULSE WIDTH (ms)
Figure 12. Thermal Response
q_JC(max) = 3.12°C/W — 2N5190-92
50 100 200
500 1000
DESIGN NOTE: USE OF TRANSIENT THERMAL RESISTANCE DATA
t_P
P_P
P_P
t₁
1/f
Figure A
DUTY CYCLE, D = t₁ f -
t1
tP
PEAK PULSE POWER = P_P
A train of periodical power pulses can be represented by
the model shown in Figure A. Using the model and the
device thermal response, the normalized effective transient
thermal resistance of Figure 12 was calculated for various
duty cycles.
To find q_JC(t), multiply the value obtained from Figure 12
by the steady state value q_JC.
Example:
The 2N5190 is dissipating 50 watts under the following
conditions: t₁ = 0.1 ms, t_p = 0.5 ms. (D = 0.2).
Using Figure 12, at a pulse width of 0.1 ms and D = 0.2,
the reading of r(t₁, D) is 0.27.
The peak rise in function temperature is therefore:
DT = r(t) × P_P × q_JC = 0.27 × 50 × 3.12 = 42.2_C
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