Catalog Add On: BETA Low-Voltage Circuit Protection ...

Occasional surge load from preloading with unknown surge outcome
Specifying the required rated current In of the fuse link is carried out
in two steps:
1. Specifying the rated current In based on the previous load
current Iprev :
1
In > Ivor × ----------------------------------------------------ku
× kq × kλ × kl × WL
Permissible load current I n' of the selected fuse link:
I n' = k u x k q x k λ x k l x WL x I n
2. Checking the permissible overload duration of the surge
current I surge
Melting time t vs (time/current characteristic curves) x
residual value factor RW ≥ surge wave duration t surge
To do this, you require the previous load ratio
V =
Ieff ------
In ′
as well as the characteristic curve "permissible overload and
melting time for previous load" (page 5/58, curve a or b) and the
"time/current characteristic curve" for the selected fuse link.
If a determined overload duration is less than the required overload
duration t surge , then you need to select a fuse link with a greater
rated current I n (taking into account the rated voltage U n and the
permissible breaking I 2 t value) and repeat the check.
La
Load current
'
Condition:
t interval ≥ 3 x t surge
t interval ≥ 5 min
0
J
surge
surge
J interval
Sample selections
For a converter assembly in circuit (B6) A (B6) C, whose rated direct
current is I dn = 850 A, fuse links that can be installed as arm fuses
should be selected. The choice of fuse is shown for different operating
modes of the converter assembly.
Data for converter assembly
Supply voltage
U N = 3 50 Hz 400 V AC
Recovery voltage
U W = 360 V = U N x 0.9 (for shoot-throughs)
Thyristor T 508N (from eupec),
l 2 t value
∫ I 2 dt = 320 x 10 3 A 2 s (10 ms, cold)
Fuse links, natural air cooling
Ambient temperature ϑ u = +35 °C
Conductor cross-section for copper fuse link: 160 mm 2
Conversion factor
Direct current I d/fuse load current I La: I La = I d x 0.58
For the following examples, it is assumed, in the case of loads that
exceed the rated direct current of the converter assembly, that the
converter assembly is rated for this load.
prev
J
I2_12652
© Siemens AG 2007
Continuous, no-break load
=
BETA Protecting
SITOR Semiconductor Fuses
" ' ! )
Siemens ET B1 AO · 2008, Characteristic curves
Configuration
Direct current Id = Idn = 850 A
ILa = Id x 0.58 = 493 A
Selected:
3NE3 335 SITOR fuse link
(560 A/1000 V), WL = 1
breaking I 2 t value
I 2 tA = 360 x 10 3 x 0.53 = 191 x 10 3 A 2 s
Test cross-section to page 5/54: 400 mm 2
The following correction factors are to be applied:
ku = 1.02 (ϑu = +35 °C)
kq = 0.91 (conductor cross-section, double-ended, 40 % of test
cross-section)
kλ = 1.0 (conduction angle λ = 120°)
kl = 1.0 (no forced-air cooling)
Required rated current In of the SITOR fuse link:
1
In ≥ ILa × ------------------------------------------------- = 493 A
ku × kq × kλ × kl × WL
1
493 A × --------------------------------------------------------------------- = 531 A
102 , × 0, 91 × 10 , × 10 , × 1, 0
Unknown varying load, but with known maximum current
=
= N " ! # )
Max. direct current I dmax = 750 A
Max. fuse current I max = I dmax x 0.58 = 435 A
Selected:
3NE3 334-0B SITOR fuse link
(560 A/1000 V), WL = 1
Breaking I 2 t value
I 2 t A = 260 x 10 3 x 0.53 = 138 x 10 3 A 2 s
Test cross-section to page 5/54: 400 mm 2
The following correction factors are to be applied:
k u = 1.02 (ϑ u = +35 °C)
k q = 0.91 (conductor cross-section, double-ended, 40 % of test
cross-section)
k λ = 1.0 (conduction angle λ = 120°)
k l = 1.0 (no forced-air cooling)
Required rated current I n of the SITOR fuse link:
1
In ≥ Imax × ------------------------------------------------- = 493 A
ku × kq × kλ × kl × WL
1
435 A × --------------------------------------------------------------------- = 469 A
102 , × 0, 91 × 10 , × 10 , × 1, 0
J
J
1 $ # !
1 $ # "
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