2N5194_06 Datasheet, PDF (5/6 Pages) ON Semiconductor – Silicon PNP Power Transistors

2N5194, 2N5195
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
q_JC(max) = 3.12°C/W
0.2 0.3 0.5 1.0 2.0 3.0 5.0 10 20 30 50
t, TIME OR PULSE WIDTH (ms)
Figure 12. Thermal Response
100 200 300 500 1000
DESIGN NOTE: USE OF TRANSIENT THERMAL RESISTANCE DATA
t_P
P_P
P_P
t₁
1/f
DUTY CYCLE, D = t₁ f =
t1
tP
PEAK PULSE POWER = P_P
Figure 13.
A train of periodical power pulses can be represented by
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.
To find q_JC(t), multiply the value obtained from Figure 12
by the steady state value q_JC.
Example:
The 2N5193 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 junction temperature is therefore:
DT = r(t) x P_P x q_JC = 0.27 x 50 x 3.12 = 42.2_C
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