Low Voltage Fuses
BLC, CR and CS types
Super Rapid Fuses
ꢁ
Rated current
The I2t data for silicon diodes or
Let-thru peak current (Instantaneous):
11,600 Amps
Let-thru R.M.S. current
The current values in fuses in the line
fuse system and the element fuse
system are different. Obtain the
correct current value from the table on
page 08/48 (Fig. 2).
When selecting the rated current of a
fuse choose a fuse having an
amperage rating greater than the
current which flows in the semi-
conductor if the load is continuous and
a fixed current.
thyristor elements are normally given in
their respective catalogs. If the A2S
data is not given in their catalog obtain
the value in the following manner. If
protection is needed for a 250V, 150A
(Io) diode having a maximum allowable
peak half sine wave current of 2700A, it
is important that the fuse has a total I2t
value lower than that of the diode.
11,600 ÷ 1.7 = 6,800 Amps
This example clearly shows that while
a 100kA (rms, sym) current is
available, the fuse limits the current let-
thru to 6,800 Amperes (rms, sym).
Peak let-thru current
(Actual short-circuit current)
Calculation
If the current which flows in the
Peak current of first half cycle
of available peak current
(AC component)
2
2
1 Peak
2
semiconductor is greater than the rated
current of the fuse connect the fuses in
parallel. However, in this case, if the
numbers of fuses arranged in parallel
are 'n', then the I2t value of the fuse will
be n2·I2t and n2 times the I2t value of
one fuse. This should be taken into
consideration when protective
coordination is taken into account.
In the case of the circuit where the load
rapidly varies the fuse element will
suffer from mechanical deterioration
and be damaged by thermal stress.
In loads of this type the deterioration
characteristics of the fuse must be
closely considered.
Moreover if the fuse current – time
characteristics of the fuse selected is
less than the overload characteristics
of the semiconductor element then
complete protection can be obtained.
However, if the semiconductor element
has a large capacity then protective
cooperation is very difficult to arrange.
The fuses are used to isolate the
shorted semiconductor element circuit
from sound operating circuits.
Maximum I t diode = (
) 0.0167
2700
= (
)2 0.0167
Short circuit current that
would pass without current-
limiting fuse.
2
= 30,400A2 Sec.
DC component
From the table (Page 08/38), the fuse
with a total I2t nearest to 30,400A2 Sec.
is the 260 Ampere fuse (CR 2L-260).
ꢀ Interrupting current
The rated interrupting current of the
fuse must exceed the maximum value
(Symmetrical RMS value) of the
estimated circuit fault current.
Time
M
A
T
M:Melting time
ꢀ Peak arc voltage
In the case of the current-limiting fuse
an arc voltage (overvoltage) is
generated at the time of interruption
due to its fusible element construction.
It is necessary to check that this peak
arc voltage does not exceed the
semiconductor's maximum (Non-
repetitive peak) reverse voltage value.
A :Arcing time
T :Total clearing time
Available
105,000
11,600
ꢀ Current limitation
Let-thru
200A
Select a fuse whose let-thru current
value does not exceed the allowable
1/2 cycle surge current of the
semiconductor. The allowable surge
current is the peak value of the current
which in case at 50Hz is allowed to
flow for 10ms. In the current-limiting
fuse the fault must be cleared in the
shortest possible time or in the first
1/2 cycle.
ꢀ Total clearing I2t
The total clearing I2t of fuse is a very
important factor when considering the
protective coordination of the
semiconductor. This total clearing I2t is
the value where the arcing I2t is added
to the melting I2t. Therefore it is
necessary to satisfy the following
formula.
100,000A
Available RMS symmetrical current(Ampere)
Available current is the current which
would flow if the fuse were not current-
limiting.
Fuse – total
Semiconductor
<
clearing I2t
I2t
=
The total clearing I2t of fuse depends
upon the operational voltage and
interrupting current.
This would cause damage to
equipment. Let-thru current is the
actual current allowed to flow by the
current limiting action of the fuse. A
number of let-thru current graphs are
given in this catalog and example is
given in the following paragraph. The
method of reading this graph is
provided for your reference.
Therefore, for this reason if a 500 Volts
fuse is used in a 300 Volts circuit the
total clearing I2t is reduced by 50–70ꢀ.
However, the reduction rate varies
according to the type of fuse
construction. This must be checked
and confirmed once more.
How to find a let-thru current
– Example
Fuse: 200 Amps 500V
Available R.M.S symmetrical current:
100,000 Amps
Example
I2t
All I2t values are ampere2 seconds.
Fuji Electric FA components & Systems Co., Ltd./D & C Catalog
Information subject to change without notice
08/46
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