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參數(shù)資料
| 型號(hào)： | MURH8100E |
| 廠商： | MOTOROLA INC |
| 元件分類(lèi)： | 參考電壓二極管 |
| 英文描述： | ULTRAFAST RECTIFIER 8.0 AMPERES 1000 VOLTS |
| 中文描述： | 4 A, 1000 V, SILICON, RECTIFIER DIODE, TO-220AC |
| 文件頁(yè)數(shù)： | 4/6頁(yè) |
| 文件大?。?/td> | 140K |
| 代理商： | MURH8100E |
4
Rectifier Device Data
t0
t1
t2
t
VDD
ID
IL
BVDUT
MERCURY
SWITCH
Figure 9. Test Circuit
Figure 10. Current–Voltage Waveforms
+VDD
DUT
40 H COIL
VD
IL
S1
ID
The unclamped inductive switching circuit shown in
Figure 9 was used to demonstrate the controlled avalanche
capability of the new “E’’ series Ultrafast rectifiers. A mercury
switch was used instead of an electronic switch to simulate a
noisy environment when the switch was being opened.
When S1 is closed at t0 the current in the inductor IL ramps
up linearly; and energy is stored in the coil. At t1 the switch is
opened and the voltage across the diode under test begins to
rise rapidly, due to di/dt effects, when this induced voltage
reaches the breakdown voltage of the diode, it is clamped at
BVDUT and the diode begins to conduct the full load current
which now starts to decay linearly through the diode, and
goes to zero at t2.
By solving the loop equation at the point in time when S1 is
opened; and calculating the energy that is transferred to the
diode it can be shown that the total energy transferred is
equal to the energy stored in the inductor plus a finite amount
of energy from the VDD power supply while the diode is in
breakdown (from t1 to t2) minus any losses due to finite com-
ponent resistances. Assuming the component resistive ele-
ments are small Equation (1) approximates the total energy
transferred to the diode. It can be seen from this equation
that if the VDD voltage is low compared to the breakdown
voltage of the device, the amount of energy contributed by
the supply during breakdown is small and the total energy
can be assumed to be nearly equal to the energy stored in
the coil during the time when S1 was closed, Equation (2).
The oscilloscope picture in Figure 11, shows the test circuit
conducting a peak current of one ampere at a breakdown
voltage of 1300 volts, and using Equation (2) the energy ab-
sorbed is approximately 20 mjoules.
Although it is not recommended to design for this condi-
tion, the new “E’’ series provides added protection against
those unforeseen transient viruses that can produce unex-
plained random failures in unfriendly environments.
WAVAL
1
2LI2
LPK
BVDUT
BVDUT–VDD
WAVAL
1
2LI2
LPK
Figure 11. Current–Voltage Waveforms
CHANNEL 2:
IL
0.5 AMPS/DIV.
CHANNEL 1:
VDUT
500 VOLTS/DIV.
TIME BASE:
20 s/DIV.
EQUATION (1):
EQUATION (2):
CH1
CH2
REF
REF
CH1
CH2
ACQUISITIONS
SAVEREF SOURCE
1
217:33 HRS
STACK
A
20 s
953 V
VERT
500V
50mV
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