Page 22 - Application Guide Semiconductor Fuse Link

Surge withstand of semiconductor devices

Example 2

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2
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A thyristor has an I t for fusing of 120,000 A s at 8.33 ms. What is the I t withstand for a surge of 1
ms duration ? Assume N = 3.
Solution

120000
= -- =3795 A
For the 8.33 ms surge the r.m.s. current I0
0,00833

3
3
If I0 t = K then K = (3795) x 0.00833 = 4.555 x 10 8

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At 1ms I0 will be (K/t) 0.333 = (4.555 x 10 / 0.001) 0.333 = 7694A

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so the I t withstand at 1ms will be 7694 x 0.001 = 59195 A s.

In general the above procedure can be summarized as

N-2
t - t 0,333
I t at time t = I0 t0 x ( - ) N = I0 t0 ( - ) for N = 3
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2
2
t t
0 0
IGBT and other power transistor circuits

When the current increases suddenly in a power transistor due to a short-circuit, the collector-
emitter voltage increases immediately to a high value which gives a rapid increase in internal power
dissipation and failure of the transistor. Electronic means of fault detection are often used to switch
off the device very quickly. High-speed fuses are recommended in these circumstances to limit the
destructive effects which occur if the electronic protection is not used or if it fails. These destructive
effects include case rupture of IGBTs and melting of the emitter connections, bonding wires and
other circuit conductors.

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Case-rupture I t values for IGBTs should be used as withstand values, but these are not always
available. In the absence of other data, short-circuit coordination with fuses can be done on the basis
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of the melting I t of the bonding wire. If, as is usual, copper wires are used, the melting I t can be
estimated using
2
I t = KMA 2

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-4
where KM is Meyer’s constant (approximately 100,000-110,000 A s-mm for copper) and A is the wire
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cross-sectional area in mm . However these values are very pessimistic. Experience shows that case
rupture values are signifi cantly higher. They can only be determined by test.

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