ABB RELION 620 Series Technical Manual page 453

1MAC504801-IB E
620 series ANSI
Technical Manual
of the phase displacement is 60 minutes. The limit of the composite error at the rated
accuracy limit primary current is 5 percent.
The approximate value of the accuracy limit factor F
burden can be calculated on the basis of the rated accuracy limit factor F
rated burden, the rated burden S
current transformer can be calculated.
S + S
in n
F = F ×
a n
S + S
in a
GUID-25DCBB76-7EB4-4CDE-9234-AEEA0DECEAC2 V1 EN
Example 1
In the example, the rated burden S
5A, the internal resistance R
corresponding to the rated burden is 20 (5P20). The internal burden of the current
transformer is S = (5A)2 x 0.07 Ω = 1.75 VA. The input impedance of the relay at a rated
in
current of 5A is < 20 mΩ. If the measurement conductors have a resistance of 0.113 Ω, the
actual burden of the current transformer is S
the accuracy limit factor F
The CT burden can grow considerably at the rated current of 5A. At the rated current of
1A, the actual burden of the current transformer decreases, while the repeatability
simultaneously improves.
At faults occurring in the protected area, the fault currents can be very high compared to
the rated currents of the current transformers. Due to the instantaneous stage of the
differential function block, it is enough that the current transformers are capable of
repeating, during the first cycle, the current required for instantaneous tripping.
The current transformers should be able to reproduce the asymmetric fault current without
saturating within the next 10 ms after the occurrence of the fault, to secure that the
operating times of the protection relay comply with the times stated in the section
"Technical Data".
The accuracy limit factors corresponding to the actual burden of the phase current
transformer to be used in differential protection fulfill the requirement of the equation.
, the internal burden S
n
of the CTs 5P20 is 10 VA, the secondary rated current
n
= 0.07 Ω and the accuracy limit factor F
in
=(5A)2 x (0.113 + 0.020) Ω = 3.33 VA. Thus
a
corresponding to the actual burden is 46.
a
Section 4
Protection functions
corresponding to the actual CT
a
(ALF) at the
n
and the actual burden S
in
(ALF)
n
of the
a
(Equation 49)
447