Page 27 - Application Guide Semiconductor Fuse Link

Converters with multiple parallel devices

Allowance for unbalance

Usually an adjustment should be made to allow for the fact that the parallel branches in the leg
do not share the current equally. If this is the case, the fuse rating must be based on the calculated
current through the fuse, as shown above, which is then increased by the expected imbalance. If
the total current is I and an unbalance of 10% is assumed, the fuse rating should be based on leg
current of 1.10 x (I / n). If there is a redundancy requirement in an unbalanced system then the
leg current of 1.10 x (I /( n-1)) should be used. (Note: In some very large power converters the
unbalance can reach 60%).

Protection against device explosion

For a fault such as that shown in Fig.19 the fuse which opens is in series with a device that has failed.
The fuse’s fi rst job is to clear the circuit before the faulty device explodes. In this case the total clearing
2
2
2
I t of the fuse must be less than the explosion I t of the device. The explosion I t of a device is usually
2
around 4 times the withstand I t for fusing (contact the device manufacturer for a precise value).
Device protection for an internal fault

In the event of an internal fault, the short-circuit current from the a.c. side fl ows through the fuse in
series with the faulty device. It also fl ows through the n devices in parallel in the healthy leg of the
bridge in series. The other job of the fuse in series with the faulty device is to protect the n devices
in the healthy leg in series. During a fault the sharing of the current between the parallel paths is
much better than the sharing of the current during normal operation. In practice the adjustment
for unequal sharing of the short-circuit current between the parallel paths is typically 5 to 10%.
Assuming a 5% unbalance ( a realistic value when there are 4 devices in parallel as shown in Fig.19)
the current in the most loaded branch of the healthy leg will be 1.05/n times the fault current.
2
The total clearing I t of the fuse must be less than (n/1.05)² times the device withstand I t (adjusted
2
for the fault duration). For large values of n this means that device protection is usually easy to
achieve. (Instead of using a typical 5 to 10 % unbalance, the SAF/PE software uses the same per
cent unbalance under short circuit conditions as under full load conditions, which yields a more
conservative solution). Since only one fuse will operate to clear the fault, it must be rated for the full
a.c. voltage (and sometimes the d.c. voltage), and no assistance from other fuses can be assumed
2
when calculating the fuse let-through I t.

Inter-leg fuse selectivity

To enable uninterrupted operation of the converter after the fault and to prevent the opening of
a large number of fuses, the fuse in series with the faulty device must clear the fault before any of
the fuses in the healthy leg of the bridge in series reach their melting point. Assuming, as in the
previous section, a 5% unbalance during the short circuit, the current in the most loaded branch
of the healthy leg will be 1.05/n times the fault current. This an occasional overload and the Cf3
coeffi cient on the current (see section 7) must be used, which becomes C 3 for I²t values. Then if P
2
f
2
is the melting (prearcing) I t of the fuse at a time equal to its total clearing time, and T is the total
2
clearing I t, it is necessary that:
1,05 2
T x ( - ) Cf3 x p
2
n
If n is large it is easy to meet this condition, while for n = 1 it is clearly impossible. Usually with n =
3 selectivity can be obtained, but if the converter is already operating with one device out due to a
previous fault, selectivity becomes very diffi cult and may not be possible.

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