Answer / tariq

Because semiconductors has negative resistance temperature
co-efficient. When current flows in it, its temperature
increases due to I^2*R losses but resistance decrease
accordingly. The decrease in resistance allows more current
to flow in it which produces more heat. It will melt
ultimately.

Answer / j.kesavan

IT IS A FAST ACTING FUSE ,KA RATING IS VERY LOW, IF ANY
PROBLEM IN THE FIELD DRIVE WILL NOT AFFECT.THATS WHY WE AR
USING SEMICONDOCTOR FUSE IN VSD DRIVES

Answer / jane lal

IT IS A FAST ACTING FUSE ,KA RATING IS VERY LOW, IF ANY
PROBLEM IN THE FIELD DRIVE WILL NOT AFFECT.THATS WHY WE AR
USING SEMICONDOCTOR FUSE IN VSD DRIVES

Answer / harish nair

Basically a drive means cluster of thyristors, these thyristors may fail when subject to 10mS(one half cycle in a 50 Hz system)pulse of only 10 times the RMS value of normal current.A normal fuse requires 15 to 30 times of its nominal ampere rating to melt with in 10mS, which is not fast enough to provide the protection, whereas a semiconductor fuse requires only 5 to 6 times it ampere rating to defuse thus ensuring fast action & protection to thyristor. It has a exponentional decay graph the greater the current the lesser operation time.

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why transformer rated in KVA not in KW ? we know that W= v*i

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