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參數(shù)資料
| 型號(hào)： | 2N5745 |
| 廠商： | ON SEMICONDUCTOR |
| 元件分類： | 功率晶體管 |
| 英文描述： | PNP Silicon High-Power Transistor(80V（集電極-發(fā)射極）硅PNP大功率晶體管) |
| 中文描述： | 20 A, 80 V, PNP, Si, POWER TRANSISTOR, TO-204AA |
| 封裝： | METAL, TO-3, 2 PIN |
| 文件頁(yè)數(shù)： | 6/8頁(yè) |
| 文件大?。?/td> | 104K |
| 代理商： | 2N5745 |
2N4398 2N4399 2N5745
http://onsemi.com
6
RATINGS AND THERMAL DATA
100
1.0
Figure 8. Active Region Safe Operating Area
V
CE
, COLLECTOR-EMITTER VOLTAGE (VOLTS)
50
20
10
5.0
2.0
1.0
0.5
0.1
2.0
3.0
5.0
10
20
30
100
50
0.2
I
100
μ
s1.0 ms
2N5745
2N4398, 2N4399
dc
5.0ms
Secondary Breakdown Limited
Bonding Wire Limited
Thermal Limitations
Pulse Duty Cycle
≤
10%
T
J
= 200
°
C
2N4398
2N4399
2N5745
T
C
= 25
°
C
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 8 is based on T
J(pk)
= 200 C; T
C
is
variable depending on conditions. Second breakdown pulse
limits are valid for duty cycles to 10% provided T
J(pk)
200 C. T
J(pk)
may be calculated from the data in
Figure 9. At high case temperatures, thermal limitations will
reduce the power that can be handled to values less than the
limitations imposed by second breakdown.
Figure 9. Thermal Response
t, TIME OR PULSE WIDTH (ms)
1.0
0.7
0.5
0.010.01
0.3
0.2
0.1
0.07
0.05
0.03
0.02
0.02
r
T
0.05
0.1
0.2
0.5
1.0
2.0
5.0
10
20
50
100
200
1000
500
STEADY STATE VALUES
θ
JC
(
∞
) = 0.875
°
C/W
θ
JC
(t) = r(t)
θ
JC
(
∞
)
D = 0.5
0.2
0.05
0.01
0 (SINGLE PULSE)
0.1
DESIGN NOTE: USE OF TRANSIENT THERMAL RESISTANCE DATA
t
P
P
P
P
P
t
1
1/f
DUTY CYCLE, D = t
1
f -
t1
tP
PEAK PULSE POWER = P
P
A train of periodical power pulses can be represented by
the model as shown in Figure A. Using the model and the
device thermal response, the normalized effective transient
thermal resistance of Figure 9 was calculated for various
duty cycles.
To find
θ
JC
(t), multiply the value obtained from Figure 9
by the steady state value
θ
JC
(
Example:
The 2N4398 is dissipating 100 watts under the following
conditions: t
1
= 1.0 ms, t
P
= 5.0 ms. (D = 0.2)
Using Figure 9, at a pulse width of 1.0 ms and D = 0.2, the
reading of r (t) is 0.28.
The peak rise in junction temperature is therefore
T = r(t) x P
P
x
θ
JC
(
) = 0.28 x 100 x 0.875 = 24.5 C
).
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