
1MRS757208 EN
Technical Note
ABB Oy, Distribution Solutions
Issued: June 2010
Revision: C / 26 May 2020
High-Impedance Differential Protection
Principle, calculations and CT requirements
Contents:
1
Scope ..................................................................................................................... 2
2
Principle of the high-impedance differential protection .................................. 3
2.1 Requirements for the protection relay .......................................................... 6
3
Calculations and CT requirements .................................................................... 8
3.1 Calculation of the stabilizing voltage ........................................................... 8
3.2 CT knee-point requirement........................................................................... 9
3.3 Minimum possible relay operating current ................................................. 10
3.4 Calculation of the stabilizing resistor ......................................................... 10
3.5 Actual sensitivity of protection................................................................... 11
3.6 Need for Voltage Dependent resistor ......................................................... 12
4
Example calculations ......................................................................................... 13
4.1 Phase segregated high-impedance differential protection for generator .... 13
4.1.1 Stabilization voltage and CT knee-point requirement .................... 13
4.1.2 Stabilizing resistor ........................................................................... 14
4.1.3 Actual sensitivity of protection ....................................................... 15
4.1.4 Voltage Dependent Resistor ............................................................ 15
4.2 High-impedance based restricted earth-fault protection for transformer ... 15
4.2.1 Stabilization voltage and CT knee-point requirement .................... 16
4.2.2 Stabilizing resistor and actual sensitivity ........................................ 17
4.2.3 Voltage dependent resistor .............................................................. 17
5
List of symbols ................................................................................................... 19
6
List of equations ................................................................................................. 21
Document history, Disclaimer and Copyrights, Trademarks, Contact information .... 22
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High-Impedance Differential Protection
Principle, calculations and CT requirements
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1MRS757208 EN
Scope
This document describes the high-impedance differential protection principles, calculations
and CT requirements. This scheme is used both in phase segregated high-impedance
differential and high-impedance based restricted earth-fault protection. Typical areas of
applications are generator, motor, reactor, bus-bar and transformer protection.
First the operation principle of a high-impedance scheme is studied with explanatory figures
and diagrams. The need of the stabilizing and voltage dependent resistors is explained.
Further, the requirements of a protection relay to be used in high-impedance scheme are
listed.
Next, the rules of calculating the stabilizing voltage, relay setting, stabilizing resistor etc. are
presented.
Finally, the calculating rules are applied in protection application examples. Both the phase
segregated high-impedance differential and high-impedance base restricted earth-fault
protection is handled.
The rules given in this document are applicable for any high-impedance differential
protection relay/function, where setting is given in current.
KEYWORDS: high-impedance protection, restricted earth-fault protection, busbar
protection, differential protection
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High-Impedance Differential Protection
Principle, calculations and CT requirements
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1MRS757208 EN
Principle of the high-impedance differential protection
The high-impedance principle has been used through many years for differential protection
due its capability to manage through-faults also in case of current transformer (CT)
saturation.
Figure 2.-1 shows an example connection of a phase segregated high-impedance differential
protection scheme. The phase currents are measured from the both sides of the protected
object. The current transformers (CT) secondary in each phase are connected in parallel
along with a relay measuring branch. Thus the relay measures only the difference of the
currents. In an ideal situation there will be differential current to operate the relay only if
there is a fault between the CTs, i.e. inside of the protected zone.
In case there is a fault outside of the zone, a high current can go through the protected object
(known as through fault current). This can cause partial saturation in the CTs. The relay
operation is avoided by means of a stabilizing resistor (Rs) in the relay measuring branch.
The Rs increases the impedance of relay, hence the name high-impedance differential
scheme.
P1
P2
P2
P1
s1
s2
s2
s1
A
B
C
Protected object
IL1
IL2
IL3
Voltage
dependent
resistor
Ru
Stabilizing
resistor
Rs
Id1
Id2
Id3
Three-phase differential protection
Fig.2.-1. Phase segregated differential protection based on high-impedance principle.
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Principle, calculations and CT requirements
1MRS757208 EN
Figure 2.-2 shows a single phase representation of the circuit. CT secondary winding
resistances (Rin) and connection wire resistances (Rm/2) are also shown.
Fig.2.-2. Single phase representation.
We can further simplify the circuit to its equivalent as done in Figure 2.-3. For simplicity
reasons the voltage dependent resistor (Ru) is not shown. Wiring resistances are presented as
total wiring resistances (Rm1 and Rm2) and. The lower part of the figure shows the voltage
balance while there is no fault in the system, and no CT saturation.
Fig.2.-3. Equivalent circuit when there is no fault or CT saturation
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Principle, calculations and CT requirements
1MRS757208 EN
As the figure 2.-3 shows, when there is no fault the CT secondary currents, and thus its emf
voltages (E1 and E2) are opposite and the relay measuring branch has no voltage/current.
In case of in-zone fault, the secondary currents have the same direction. The relay measures
the sum of the currents as a differential and trips the circuit breaker. If the fault current goes
through one CT only, its secondary emf magnetizes the other side CT, i.e. E1  E2 (Fig. 2.4).
Fig.2.-4. Equivalent circuit in case of in-zone fault.
Figure 2.-5 shows CT saturation at through fault (i.e. out-of-zone) situation. The
magnetization impedance of a saturated CT is almost zero. Thus the saturated CT winding
can be presented as a short circuit, see Fig. 2.-5. When one CT is saturated the current of the
non-saturated CT has now two paths to follow: through the relay measuring branch (Rs +
relay) and through the saturated CT (Rm2+Rin2).
The relay should not operate during the saturation. This is achieved by increasing relay
impedance by means of the stabilizing resistor (Rs) which forces the majority of the
differential current to flow through the saturated CT. As a result relay operation is avoided,
i.e. the relay operation is stabilized against CT saturation at through fault current. The
voltage Us, known as stabilizing voltage, shown in the figure is the basis of all calculations.
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High-Impedance Differential Protection
Principle, calculations and CT requirements
1MRS757208 EN
Fig.2.-5. Equivalent circuit in case of CT saturation at through fault.
Note: The CT saturation can and most likely will happen also in case of in-zone fault. This
is not a problem because although the operation remains stable (non-operative) during the
saturated parts of the CT secondary current waveform, the non-saturated part of the current
waveform will cause the protection to operate.
I
Saturated part
Non-saturated part
Fig. 2.-6 Secondary waveform of a saturated CT.
Because of the stabilizing resistance and CT saturation the secondary circuit voltage can
easily exceed the isolation voltage of the CTs, connection wires and the relay. In order to
limit this voltage a voltage dependent resistor (VDR, Ru) is used as shown in Figures 2.-1
and 2.-2 earlier.
2.1
Requirements for the protection relay
It is commonly asked if a normal overcurrent or earth-fault relay could be used for highimpedance scheme. There are three requirements that any protection relay or function must
fulfil if it is used in high-impedance scheme
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High-Impedance Differential Protection
Principle, calculations and CT requirements
1MRS757208 EN
1. The operation must be based on the peak-to-peak measurement or have special peak
detection algorithm for ensuring fast and reliable operation with disturbed secondary
current of saturated CT.
2. Speed of operation. Typically very short relay operation time is required.
3. Sensitivity in current setting, i.e. relay minimum setting must be sufficient.
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High-Impedance Differential Protection
Principle, calculations and CT requirements
3
1MRS757208 EN
Calculations and CT requirements
The calculating and CT selection is usually an iterative process:
-
-
3.1
First a suitable looking CT is chosen.
Then, the stabilization voltage is calculated and compared to the CT knee-point
voltage requirement. Should the knee-point voltage of the CT be too small, another
CT must be chosen, and the process is started from the beginning.
Next, the relay setting and stabilizing resistor is calculated. Further, the actual
sensitivity of the protection, which will be affected by CT magnetization etc, is
calculated. In case the actual sensitivity is not satisfactory, a CT with bigger core
(lower magnetizing current) must be chosen and the process starts again from the
beginning.
Finally, the need of the voltage dependent resistor (VDR) is checked.
Calculation of the stabilizing voltage
The sensitivity and reliability of the protection is much dependent on the characteristics of
the current transformers. The CTs must have identical transformation ratio. It is further
recommended that all current transformers has equal burden, characteristics and are of same
type, preferably from the same manufacturing batch (i.e. identical construction should be
used if possible).
If the CT characteristics and burden values are not equal, calculation for each branch in the
scheme should be done separately and the worst case result is then used. Consider the case as
shown if figure 3.-1, CT winding resistance and burden of the branches are not equal, and
hence the maximum burden equal to 3.2Ω should be used for calculating stabilized voltage.
P1
P2
P2
P1
s1
s2
s2
s1
Rin1 = 1.2Ω
Rin2 = 0.8Ω
A
B
C
IL1
Rm1 = 2Ω
Rm2 =1.9Ω
Ru
Rin1 + Rm1 = 3.2Ω
Rin2 + Rm2 = 2.7Ω
Rs
Id1
Three-phase differential protection
Fig. 3.-1. Example of different CT burden value on each branch.
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Principle, calculations and CT requirements
1MRS757208 EN
The stabilizing voltage (i.e. the voltage appearing across the measuring branch during out-ofzone fault) is calculated assuming that one of the parallel connected CT is fully saturated.
The stabilizing voltage can be calculated by the formula
I k max
(1)
( Rin + Rm )
n
where, Ikmax = the highest through-fault current in primary Amps. I.e. the highest shortcircuit or earth-fault current when the fault is outside of the protected zone
n
= the turns ratio of the CT
Rin = the secondary winding resistance of the CT in ohms
Rm = the total resistance of the secondary circuit wires in ohms
US =
In principle, the highest through-fault current should be known. However, when the
necessary data is not available, the following approximates can be used:
- Small power transformers: Ikmax = 16 x In (corresponds zk = 6% and infinite grid)
- Large power transformers: Ikmax = 12 x In (corresponds zk = 8% and infinite grid)
- Generators & Motors:
Ikmax = 6 x In
Where In = rated current and zk = short-circuit impedance of the protected object.
3.2
CT knee-point requirement
The current transformers must be able to force enough current to operate the relay through
the differential circuit during a fault condition inside the zone of protection. To ensure this,
the knee-point voltage Ukn should be at least 2 times higher than the stabilizing voltage US.
U kn  2  U S
(2)
The factor two in the eq.2 is used for ensuring reliable and fast operation of
the protection. In some cases, it is possible to achieve stability with knee-point
voltage slightly lower than stated in the formula. The conditions in the
network, however, have to be known well enough to ensure the stability.
•
•
•
If Ukn >= 2 x US, fast relay operation is secured.
If Ukn >= 1.5 x US and < 2 x US relay operation can be prolonged and
should be studied case by case.
If Ukn <1.5 x US the relay operation is jeopardized. Another CT should
be chosen.
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High-Impedance Differential Protection
Principle, calculations and CT requirements
1MRS757208 EN
If other than class X CTs are used, an estimate for Ukn can be calculated from the rated
accuracy limit factor

Sn 
U kn = 0.8  Fn  I 2 n   Rin + 2 
I 2n 

where, Fn
= the rated accuracy limit factor of the CT
I2n = the rated secondary current of the CT
Rin = the secondary winding resistance of the CT
Sn
= the rated burden (VA rating) of the CT
(3)
Example: A CT 500/5, 10P20, 15VA, Rin 0.2ohm is used. I.e. the Fn=20, I2n=5A and
Sn=15VA. The knee-point voltage is

15VA 
 = 64V
U kn = 0.8  20  5 A   0.2 +
(5 A)2 

3.3
Minimum possible relay operating current
Should the protection be required to be as sensitive as possible, the following points must be
checked
-
What is the minimum possible setting in the relay?
What is the CT accuracy in through-fault case? The relay should not be set below the
CT accuracy.
What is the CT magnetization current (Im) at the stabilizing voltage? The relay setting
should be higher than the sum of the magn. currents of all parallel connected CTs.
3.4
Calculation of the stabilizing resistor
As the impedance of the protection relay alone is low, a stabilizing resistor is needed. The
value of the stabilizing resistor is calculated using the formula
US
I relay
where, Irelay = relay set operating current in secondary Amps
RS =
(4)
The continuous power of the stabilizing resistor is
P=
U kn2
Rs
(5)
The stabilizing resistor should be capable to dissipate high energy within very short time.
Therefore, a wire wound type resistor should be used.
Often the manufacturer of the resistor allows 10 times rated power for 5 sec. Then the
required power of the resistor can be calculated as
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High-Impedance Differential Protection
Principle, calculations and CT requirements
P=
1MRS757208 EN
U kn2
Rs  10
(6)
Because of possible CT inaccuracy, which might cause some current through
the stabilizing resistor in normal load situation, the rated power should be
25W minimum.
3.5
Actual sensitivity of protection
The relay set operating current does not directly define the selectivity of the protection, i.e.
the operation point in primary Amps. This is because of the CTs magnetizing currents and
the leakage current of the voltage dependent resistor. The value of the primary current (Iprim)
at which the protection operates at certain setting can be calculated using the formula:
I prim = n  ( I relay + I u + m  I m )
(7)
where, n
= turn ratio of the CT
Irelay = relay set operation current in secondary Amps
Iu
= leakage current through the Voltage Dependent Resistor (VDR) at
stabilizing voltage Us
m
= number of CTs connected in parallel in the secondary circuit
Im
= magnetizing current of the CT at stabilizing voltage Us
Note: Typically, in CT manufacturers catalogues, the magnetizing (=excitation) current of
the CT is given at knee-point voltage.
-
-
CT magnetizing current at the stabilizing voltage can be read from the CT
magnetizing curve, if available.
The magnetizing current can also be measured from a real CT: when primary is open,
inject AC voltage equal to stabilizing voltage to the CT secondary and measure
current. Note that current will increase rapidly if injected voltage exceeds CT kneepoint voltage.
Assuming linear magnetizing curve up to the knee-point, the Im can also be estimated
Us
 Ie
U kn
where, Ie
= excitation current of the CT at Ukn
Im 
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High-Impedance Differential Protection
Principle, calculations and CT requirements
3.6
1MRS757208 EN
Need for Voltage Dependent resistor
First, voltage Umax, ignoring the CT saturation during the fault is calculated as follows:
I k max in
I
 (Rin + Rm + Rs )  k max in  Rs
n
n
(9)
where, Ikmaxin = maximum fault current when fault is inside the zone, in primary Amps
n
= turn ratio of the CT
Rin = CT internal secondary winding resistance
Rm = resistance of the lead wires (longest loop/highest value)
Rs
= resistance of the stabilizing resistor
U max =
Relay impedance was ignored because it is negligible in comparison the other resistances.
Next the peak voltage (û), which includes the CT saturation, is estimated by using formula
given by P.Mathews 1955

u = 2  2  U kn  (U max − U kn )
(10)
The VDR is recommended when the peak voltage û ≥ 2kV, which is the generally used
insulation level of the secondary circuit.
Sometimes, if the Rs would have little smaller value, the VDR could perhaps be avoided. But
the value of Rs depends on the relay operation current and stabilizing voltage, thus we must
either use a higher setting in the relay, or try to lower the stabilizing voltage (which depends
on Rin and Rm).
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High-Impedance Differential Protection
Principle, calculations and CT requirements
1MRS757208 EN
4
Example calculations
4.1
Phase segregated high-impedance differential protection for generator
Figure 4.1.-1 shows schematically a phase segregated high-impedance differential
protection.
Sn = 8 MVA
Un = 6 kV
P1
P2
P2
P1
s1
s2
s2
s1
Rin
Rin
A
B
C
IL1
Rm
Rm
Ru
Rs
Id1
Three-phase differential protection
Fig. 4.1.-1. Differential protection for generator (only one phase is presented in detail).
The protected generator has the following values and CT data (CTs are equal)
-
Generator
8 MVA, 6 kV, In = 770A
-
Max. Through fault current (out of zone fault)
Max. fault current at in zone fault
Ikmax = 6 x In = 4620A
Ikmaxin = 12 x In = 9240A
-
CT ratio
CT secondary winding resistance
CT knee-point voltage
CT excitation current (at knee-point voltage)
n = 1000/1 A
Rin = 15.3 ohm
Ukn = 323 V
Ie = 0.035 A @ Ukn
-
Required sensitivity
5% of 770 A = 38.5 A
4.1.1 Stabilization voltage and CT knee-point requirement
The longest distance between the CT and relay is 100 m, giving a connection wire loop of
200 m. The area of the cross-section is 2.5 mm2 (resistance at 75 °C as 0.00865 Ω/m).
Rm = 0.00865  / m  200 m  1.73 
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Principle, calculations and CT requirements
1MRS757208 EN
First the stabilizing voltage is calculated based on eq.1
Us =
4620 A
 (15.3 + 1.73)  78 .7 V
1000 / 1
The requirement for the CT knee-point voltage (eq.2) is fulfilled because Ukn > 2xUs.
4.1.2 Stabilizing resistor
The value of the stabilizing resistor is calculated using eq.4. The required sensitivity of
protection is 38.5A in primary, 0.0385A in secondary.
RS =
US
78 .7V
=
= 2044 
I relay 0.0385 A
Note that this is the minimum value for Rs with relay setting 0.0385A. If suitable resistance
cannot be found, somewhat higher value can be used. From the opposite point of view, Rs =
2044Ω means that 0.0385A is the lowest possible setting in the relay, i.e. setting value can
be later increased but not decreased.
For the purpose of checking, we calculate what would be the minimum
possible setting from the CT magnetizing current point of view. Using eq.7,
the magnetizing current at Us is
𝐼𝑚 =
𝑈𝑠
78.7𝑉
× 𝐼𝑒 =
× 35𝑚𝐴 ≈ 8.5𝑚𝐴
𝑈𝑘𝑛
323𝑉
To obtain adequate protection stability, the setting current Irelay should be at
minimum of the sum of the magnetising currents of all connected CTs. As we
have two CT in parallel we get
Irs = 2  8.5mA  17 mA
Then the resistance of the stabilizing resistor is calculated based on eq.4
RS =
78 .7 V
 4629 
0.017 A
By using slightly higher relay setting like 20 mA, the Rs becomes 3900 Ω.
Using 2044 ohm, the power of the stabilizing resistor for continuous usage is (eq.5)
P=
(323V ) 2
 51W
2044 
Or, if the manufacturer of the resistor allows 10 times rated power for 5 sec. and using eq.6
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High-Impedance Differential Protection
Principle, calculations and CT requirements
P=
1MRS757208 EN
U kn2
= 5.1W
Rs  10
Note: Because of possible CT inaccuracy, which might cause some current through the
stabilizing resistor in normal load situation, the rated power should be 25 W minimum.
4.1.3 Actual sensitivity of protection
The actual sensitivity of the protection depends on the relay setting, CT magnetizing currents
and the leakage resistance of the VDR. The CT magnetizing current (Ie) at Us was calculated
earlier 8.5 mA. Using relay setting 0.0385A, neglecting the effect of VDR (current Iu) and
using eq.7
I prim = n  ( I relay + I u + m  I m ) = 1000  (0.0385 A + 2  0.0085 A + 0 A) ) = 55 .5 A in primary
This is higher than the required sensitivity. If requested, this can be compensated by using a
lower relay setting (keeping in mind the minimum possible setting). But then the value of the
stabilizing resistance must be re-calculated.
4.1.4 Voltage Dependent Resistor
Using eq.9 and 10
U max =
12  770 A
 (2044  + 15.3 + 1.73 ) 19 kV
1000
uˆ = 2 2  323V  (19000V − 323V )  7 kV
In this example, the VDR (one for each phase) is a must because the voltage during the fault
is much higher than 2 kV.
If the VDR supplier gives the leakage current through the VDR at the stabilizing voltage the
sensitivity of the protection can be re-calculated taking into account the leakage current.
4.2
High-impedance based restricted earth-fault protection for transformer
Figure 4.2.-1 shows an example of high-impedance based restricted earth-fault protection of
a power transformer winding. The secondary side of the power transformer system is solidly
earthed.
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High-Impedance Differential Protection
Principle, calculations and CT requirements
1MRS757208 EN
Fig. 4.2.-1. Restricted earth-fault protection for transformer.
The protected transformer has the following values and CT data (all CT are equal)
-
Transformer
20 MVA, 11 kV, In = 1050A
solidly earthed network
-
Max. fault current (in or out of zone)
Ikmax = 12 x In = 12600A
-
CT ratio
CT secondary winding resistance
CT knee-point voltage
CT excitation current (at knee-point voltage)
n = 1200/5A
Rin = 0.47 ohm
Ukn = 81 V
Ie = 0.055 A @ Ukn
-
Required sensitivity
5% of 1050A = 52.5 A
Note: The maximum fault current in this application is always the highest possible earthfault or short-circuit current, whichever is higher. This applies even though the transformer
neutral would have been earthed through a neural earthing resistor (NER).
4.2.1 Stabilization voltage and CT knee-point requirement
The longest distance between CT and relay is 40 m and the whole loop of connection wires
being thus 80 m and the area of the cross-section is 6 mm2 (resistance at 75 °C is
0.0036Ω/m).
Rm = 0.0036  / m  80 m  0.288 
The stabilising voltage is calculated based on eq.1
Us =
12600 A
 (0.47  + 0.288 )  40 V
1200 / 5
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High-Impedance Differential Protection
Principle, calculations and CT requirements
1MRS757208 EN
In this case the requirement for the current transformer knee-point voltage (eq.2) is fulfilled
because Ukn > 2Us.
4.2.2 Stabilizing resistor and actual sensitivity
Required sensitivity is 52.5 A in primary, 0.219 A in secondary.
We can already take account the effect of the CT magnetizing currents on the actual
sensitivity of the protection, by subtracting the CT magnetizing currents. In this scheme,
there are four CT secondary in parallel. The magnetizing current for one CT at Us is (eq.8)
𝐼𝑚 =
40𝑉
× 55𝑚𝐴 ≈ 27𝑚𝐴
81𝑉
Thus, taking into account the four CT magnetizing currents, the relay setting should be
Irelay = 219 mA – 4 x 27mA = 111 mA
The minimum possible setting must also be checked!
To obtain adequate protection stability, the setting current Irelay should be at
minimum of the sum of the magnetising currents of all connected CTs. In this
example 4 x 27 mA = 108 mA
The resistance of the stabilising resistor is calculated based on eq.4
RS =
40 V
 360 
0.111 A
The continuous power of the stabilising resistor is (eq.5)
P=
(81V ) 2
 18 W
360 
Note: Because of possible CT inaccuracy, which might cause some current through the
stabilizing resistor in normal load situation, the rated power should be 25 W minimum.
4.2.3 Voltage dependent resistor
Using eq.9 and 10
U max =
12600 A
 360  = 18 .9 kV
1200 / 5
uˆ = 2 2  81V  (18900V − 81V )  3.5 kV
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Principle, calculations and CT requirements
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VDR is needed because the peak voltage is higher than 2 kV.
Because the Us is small, the leakage current through the VDR most likely is negligible and
does not affect the actual sensitivity.
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1MRS757208 EN
List of symbols
CT
Current Transformer
EF
Earth-fault
Fn
rated accuracy limit factor of the CT
I2n
rated secondary current of the CT
Id1, Id2, Id3
Differential currents
Ie
CT excitation (=magnetizing) current at Ukn
Ikmax
highest through-fault current when the fault is outside the protected zone
Ikmaxin
highest fault current when the fault is inside the protected zone
IL1, IL2, IL3
CT secondary currents
Im
magnetizing current of the CT at CT knee-point voltage (Us)
In
relay rated secondary current
Iprim
primary current at which protection operates
Irelay
relay set operating current in secondary Amps
Iu
leakage current of the VDR at Us
M
number of CTs connected in parallel in the secondary circuit
n
CT ratio
P
stabilizing resistor power rating
Rin
CT internal secondary winding resistance
Rm
total secondary wiring lead resistance (known as loop resistance)
Rs
stabilizing resistor
Ru
voltage dependent resistor (VDR)
Sn
rated burden (VA rating) of the CT
û
peak voltage in the secondary
Ukn
CT knee-point voltage
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Principle, calculations and CT requirements
1MRS757208 EN
Umax
theoretical maximum voltage in CT secondary (ignores CT saturation)
Us
stabilizing voltage
VDR
Voltage Dependent Resistor (Ru), also known as varistor
© Copyright 2010-2020 ABB Oy, Distribution Solutions, Vaasa, FINLAND
20 (22)
High-Impedance Differential Protection
Principle, calculations and CT requirements
6
1MRS757208 EN
List of equations
Number Description
Equation
I k max
( Rin + Rm )
n
1
Stabilizing voltage
US =
2
CT knee-point requirement
U kn  2  U S
3
P class CT knee-point voltage

Sn 
U kn = 0.8  Fn  I 2 n   Rin + 2 
I 2n 

4
Stabilizing resistor
RS =
5
Continuous power of the
stabilizing resistor
P=
6
Short time power of the
stabilizing resistor
U kn2
P=
Rs  10
7
Actual sensitivity of protection
I prim = n  ( I relay + I u + m  I m )
8
CT magnetizing current at CT
knee-point (Us)
Im 
9
CT maximum secondary voltage
U max =
10
CT peak voltage during saturation
u = 2  2  U kn  (U max − U kn )
US
I relay
U kn2
Rs
Us
 Ie
U kn
I k max in
I
 (Rin + Rm + Rs )  k max in  Rs
n
n

© Copyright 2010-2020 ABB Oy, Distribution Solutions, Vaasa, FINLAND
21 (22)
Document revision history
Document revision/date
A / 09 June 2010
B / 24 October 2019
C / 26 May 2020
History
First revision
- Misspelling in p.10 example corrected, Fn=20, not 5
Company name changed, last line in Scope added
Chapter 3.5 some text added regarding CT magn.current
Chapter 4.2.2 1st equation, typo corrected
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