Energy Automation
7UT8 Transformer Differential
Protection:
- CT Requirements
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Content





Basics
Magnitude and angle errors of current transformers
Transients
Design
Summary of current transformers
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Basics
Types of current transformers
Resin-insulated
Indoor or outdoor
Generator-current transformer
For terminal box or main leads
Current transformers
insulator-type transformer
slip-over transformer
bushing-type current transformer
totally insulated
Cable-type current transformer
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Basics
Magnitude and angle errors
N
I‘p = Ip. 1
N2
Equivalent circuit:
Ip
N1
I‘p
N2
I0
L0
ideal
current
transformer

Rct
Is
Is
E
U2
Rba
Faulted part of current
transformer
Note:
The secondary current is leading
(positive angle error)
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Magnitude
error
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
Angle error
I0
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Basics
Defined errors at the P class
Definition:
5 P 20 30 VA
Rated burden
Rated accuracy limiting factor (ALFN old n)
core type P = protection
accuracy in % at ALFN  IN
Accuracy
class
at rated current
Magnitude
error
at primary overcurrent
Angle error
Angle error
Total error
%
minute
centigrade
%
5P
1
60
1,8
5
10P*
3


10
All values are valid for steady state conditions (symmetrical currents)
* fault angle is not specified in IEC 61869-2, therefore, the Class 10P for the differential protection is not recommended.
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Basics
Influence of the burden on the errors
Isecondary/In
burden < 15 VA
Under
burden
Improvement
of the saturation
degree
burden = 15 VA
10
Fi = 10%
Note:
The burden is mainly
determined by the instrument
leads.
burden > 15 VA
5
10P10, 15 VA
0
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30.06.2015
10
15
Over
burden
Iprimary/In
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Basics
Definition of the rated currents
Primary rated current
Orientation on the rated current of the protective object
Normally the CT current selection is a little bit higher
Grading according standard: Decimal multiple of
10 – 12,5 – 15 – 20 – 25 – 30 – 40 – 50 – 60 – 75 A
(CT manufacturer delivers also a different grading)
Example: IN,G = 7330 A  selected rated CT current: 8000 A
Secondary rated current
Dominant current are 1 A and 5 A (according standard 2 A is permissible)
5 A: Advantage: less number of winding turns; Disadvantage: low permissible
burden ( ZN = SN,CT/(IN,CT)2 = 15 VA/25 A2 = 0,6 Ohm)
1A: Advantage: Higher permissible burden ( ZN = SN,CT/(IN,CT)2 = 15 VA/1 A2 = 15
Ohm); Disadvantage: higher number of winding turns (larger internal burden)
Preferred secondary rated current: 1A
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Basics
Equivalent circuit and Magnetizing characteristic
Ip
I‘p
Is
RCt
E = secondary e.m.f. (rms)
(knee point voltage)
IP = primary current
Is = secondary current
I0
L0
TP
ideal
CT
E
US
RBC
Faulted part of current
transformer
I0 = Magnetizing current
RCt = sec. winding resistance
RBC= connected burden
Example: CT 1500/1A, 5P20, 50VA, Rct = 3Ω
US = sec. voltage (rms)
TP = primary time constant
(e.g. network time)
constant)
E
Note:
Magnetizing characteristic is nonlinear
E = Is·(Rct + Rba)
I0
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Basics
Special features of class P CTs
Specified is for the total error at accuracy limit current
(5P20  5% at twenty time of rated current and rated burden)
No requirements on remanence
Closed ring core
Large CT time constant (seconds to minutes)
Remanence
Hysterese
Initial
magnetizing
curve
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Problem:
 Current transformer is magnetized e.g. via
switching off of a transient current with a large
DC time constant. He remains in the resulting
remanence point.
 An new fault current can immediately lead to a
saturation. Reason: The process starts from the
remanence point (worst case: 80%)
 The actual value of remanence is unknown (e.g.
the fault current can also start in the opposite
direction). This leads to a reduction of
remanence)
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Basics
Transient currents
Primary- and
secondary
current
I‘p [A]
IS [A]
CT goes into
saturation
Magnetizing
current
Page 10
I0 [A]
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CT saturation
Critical at differential protection
Trip
45°
Block
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Current differential protection
Switching on of external loads can lead to problems 1/3
Generator differential protection trip on generator G1
In a industrial plant a transformer was switched on
69 kV
Industrial infeed
30 km
12 MVA
6,9 kV
G1
5,6 MVA
Page 12
G2
G
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G
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Current differential protection
Switching on of external loads can lead to problems 2/3
Instantaneous values from the protection
DC component in the phase current is clear visible
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Current differential protection
Switching on of external loads can lead to problems 3/3
0,15
0,18
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Different DC transient
transmission leads to a
small increased differential
current with a low
stabilizing current
Countermeasure:
Increasing of the
inclination of slope 1 from
0,15 to 0,25 (factory
setting)
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Current transformer dimensioning
Over dimensioning factor
To improve the transient performance of the CT the relay manufacturer specify a
transient dimensioning factor (Ktd).
Required actual accuracy limiting factor (ALF):
ALF  K td  K SSC
Required rated accuracy limiting factor (ALF’):
R BC  R Ct
ALF'  ALF
R BN  R Ct
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Transformer Differential CT calculation
Glossary of used abbreviations:
ALF'
Operational Accuracy Limit Factor
ALF'
ALF
Rct
Rb
Rba
Rs
Ipr
Isr
IInt.max
IExt.max
Ktd
KSSC
Ek
Vt
Vta
Page 16
Accuracy limit factor
Secondary winding resistance at 75°C
Ohmic rated burden
Rated resistive burden
(Rct + Rba)
Primary rated current of the transformer
Secondary rated current of the transformer
Maximum internal symmetric fault current
Maximum external symmetric fault current
Transient rated dimensioning factor
Factor of the symmetric rated short-circuit current = Ipsc/Ipr
Knee-point voltage
Rated terminal voltage with 20-fold rated current and rated burden Rb
Current terminal voltage with 20-fold rated current and current burden Rba
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Transformer Differential CT calculation
Recommendation:
As remanence leads to earlier saturation which, in general, is critical for the differential protection. For new
systems, Siemens recommends anti-remanence cores in the classes 5PR, PXR and TPY. In case of the high
direct-current components to be expected, Siemens recommends the class TPZ, so that an overfunction with
external faults is avoided.
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Current transformer dimensioning
CT verification
1. Internal Fault
31.5kA
110 kV, 50Hz
1/5
2. External Fault
l= 60m
A= 4mm2
7UT85
-T (T HV)
1250/1A
5P30
5 VA
Rct = 18Ω
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-T1
200MVA
110/30 kV
50 Hz
uk = 14%
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Current transformer dimensioning
CT verification
Transforme r rated current : I N ,Tr 
2/5
SN
200000kVA

 1050 A
3 U N
3 110kV
Maximum external symmetric fault current : I Ext. max  I N ,TR 
100%
100%
 1050 A 
 7498 A
uk
14%
31500 A
 25.2
1250 A
7498 A

 6.0
1250 A
Factor of the internal symmetric rated short - circuit current : K ssc(int) 
Factor of the external symmetric rated short - circuit current : K ssc( ext )
a.)IEC class 5P
31500 A
 12.6
1250 A
7498 A
2. External fault : ALF'  2 
 12 " but minimum 25"
1250 A
Result :
1. Internal fault : ALF'  0.5 
Operating accuracy limiting factor of  25, for example : 5P30, R b  Sr /I sr  connected burden
R ba for example : Sr  2.5 VA or 5 VA -  5P30 5VA
Ktd = 2 for external fault (given from the relay manufacturer)
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Current transformer dimensioning
CT verification
Rb 
3/5
Sn 5 VA

5Ω
2
I sn
1A 2
R ba  R Cable  R Relay 
2ρ l
 0.1 Ω
A
Ω mm 2
2  0.0175
 60 m
m

 0.1 Ω  0.625 Ω
4 mm 2
CT:
ALF  ALFN 
R Ct  R BN
18 Ω  5 Ω
 30 
 37.05
R Ct  R BC
18 Ω  0.625 Ω
Rb
Rba
ALF'
ALF
ρ
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ALF required = 25
ALF CT = 37.05
25 < 37.05
 CT Dimensioning is okay
Ohmic rated burden
Rated resistive burden
Operational Accuracy Limit Factor
Accuracy limit factor
Specific resistance of cupper
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Division Energy Management
Current transformer dimensioning
Avoiding of overfunction via remanence
Present practice:
Current applications live the risk of remanence (low probability assumed) and
trust the saturation detection in the relays
Additional over dimensioning of the P – core
K rem 
1
1
 worst case at class P :
5
1 - Kr
1 - 0,8
CT design is becomes larger and is more expensive
Current transformer with air gap and lower remanence
 PR
as class P , max. remanence factor 10%
 TPY
 TPZ
Page 23
(approx. factor 1,2 larger as P core)
IEC 61869-2:2012, max. remanence factor 10%
(Cl. 1 at IN, error is specified at the transient overcurrent)
IEC 61869-2:2012, no remanence
(angle error 180‘ time constant 60 ms; Rated burden must be
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connected on the CT)
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Appendix
Thank you for your
attention!
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Division Energy Management