BLC012-1 Fuji Electric holdings CO.,Ltd, BLC012-1 Datasheet - Page 12

• Manufacturer Part Number: BLC012-1
• Description: Low Voltage Fuses Blc, Cr And Cs Types Super Rapid Fuses
• Manufacturer: Fuji Electric holdings CO.,Ltd

Low Voltage Fuses
BLC, CR and CS types
Super Rapid Fuses
G
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.
If the current which flows in the
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 I
be n
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.
I
The total clearing I
important factor when considering the
protective coordination of the
semiconductor. This total clearing I
the value where the arcing I
to the melting I
necessary to satisfy the following
formula.
Fuse – total
clearing I
The total clearing I
upon the operational voltage and
interrupting current.
Therefore, for this reason if a 500 Volts
fuse is used in a 300 Volts circuit the
total clearing I
However, the reduction rate varies
according to the type of fuse
construction. This must be checked
and confirmed once more.
Example
I
All I
08/46
2
t
Rated current
Total clearing I
2
2
t values are ampere
·I
2
t and n
2
t
< =
2
2
t is reduced by 50–70%.
2
times the I
t. Therefore it is
2
t value of the fuse will
Semiconductor
I
2
2
2
t
2
t
t of fuse is a very
t of fuse depends
2
2
seconds.
t value of
2
t is added
2
t is
The I
thyristor elements are normally given in
their respective catalogs. If the A
data is not given in their catalog obtain
the value in the following manner. If
protection is needed for a 250V, 150A
(I
peak half sine wave current of 2700A, it
is important that the fuse has a total I
value lower than that of the diode.
Calculation
Maximum I
From the table ( Page 08/38 ), the fuse
with a total I
is the 260 Ampere fuse (CR 2L-260).
I
The rated interrupting current of the
fuse must exceed the maximum value
(Symmetrical RMS value) of the
estimated circuit fault current.
I
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.
I
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.
Available current is the current which
would flow if the fuse were not current-
limiting.
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.
How to find a let-thru current
– Example
Fuse: 200 Amps 500V
Available R.M.S symmetrical current:
o
) diode having a maximum allowable
100,000 Amps
Interrupting current
Peak arc voltage
Current limitation
2
t data for silicon diodes or
2
t diode = (
2
t nearest to 30,400A
= (
= 30,400A
1 Peak
2700
2
2
Fuji Electric FA components & Systems Co., Ltd./D & C Catalog
)
2
2
)
2
0.0167
Sec.
0.0167
2
2
S
Sec.
2
t
Let-thru peak current (Instantaneous):
Let-thru R.M.S. current
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).
105,000
Information subject to change without notice
11,600
M
11,600 Amps
11,600 ÷ 1.7 = 6,800 Amps
Available RMS symmetrical current(Ampere)
T
A
Peak let-thru current
(Actual short-circuit current)
Available
Let-thru
M:Melting time
A :Arcing time
T :Total clearing time
Peak current of first half cycle
of available peak current
(AC component)
Short circuit current that
would pass without current-
limiting fuse.
DC component
Time
100,000A
200A