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AN320 数据表(PDF) 7 Page - STMicroelectronics

部件名 AN320
功能描述  In the field of parallel protection, the devices used have two main functions in transient operation
PDF  13 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

AN320 数据表(HTML) 7 Page - STMicroelectronics

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AN320
Trisil characteristics
Doc ID 5649 Rev 3
7/13
the power distribution network. This type of protection is required by certain standards used
in telecommunications.
Figure 9.
Long duration overload test
1.5
Response time
The response time of the component is the time it requires to limit the voltage. From this
point of view the Trisil has exactly the same behavior as a Transil. The time is that required
to switch from the standby operating point to the avalanche voltage. This is almost
instantaneous.
This time should not be confused with that required to pass from the breakover point (VBO)
to the conducting characteristic. This time is longer but does not influence the limiting
capability of the device.
1.6
Operation within the avalanche area
This section concerns the segment VBR - VBO (see Figure 3) of the Trisil characteristic
between the blocked state and the conducting state at low VON.
This portion of the characteristic is identical to that of an avalanche diode. Thus within this
area, DC, AC or pulse-type operations are permitted. The currents are limited depending on
the possibilities of junction-ambient air heat dissipation. The maximum current is defined by
the following:
Equation 2:
Tj = TA + RTH VBO IMAX ≤ TjMAX = 150 °C
The condition when the Trisil is not triggered is defined as follows:
Equation 3:
IMAX < IBO
The main differences from Equation 1 are:
Maximum junction temperature which is now that given by the catalogue, i.e. 150 °C
Voltage which is that of the avalanche mechanism
Continuous thermal resistance replacing the transient thermal impedance
In AC operation, although Equation 2 still holds good, the voltage-current diagram as a
function of time shown in Figure 10 is clearer.
t(s)
I(A)
10
–10
F = 50 Hz
10 pair of pulses
1
12
1
12
180



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