Current Transformers Power System Protection for BTech and MTech

Current Transformers Power System Protection for BTech and MTech

1. Power System Protection:
Current and Voltage
Transformers

2. Current and Voltage Transformers
Transducers/ Sensors
These essential components serve two critical functions in power systems:
1.Scaling down power system high currents or voltages to levels compatible with relays and
2.Providing galvanic isolation between high-voltage power networks and protection equipment.
Let’s explore certain things to ensure proper selection and application in protection systems.
•CT operation,
•Equivalent circuits,
•Burden calculations, and
•Classification standards

3. CT/ VT Fundamentals
CTs are magnetically coupled, multi-
winding transformers
i.e. CTs use magnetic coupling to
transform high primary currents to
manageable secondary levels.
•Current transformer (CT) secondary windings are
rated for 5 A or 1 A
•Voltage transformer (VT) secondary windings are
rated at 110 or 120 V for phase-to-phase voltage
connections, or, equivalently, at 63.5 or 69.3 V for
phase-to-neutral connection
•Transducers/sensors must be designed to tolerate
higher values for abnormal system conditions.
•Thus, CTs are designed to withstand fault currents
(which may be as high as 50 times the load current)
for a few seconds,
•While VTs are required to withstand overvoltages of
the order of 20% above the normal value for long
duration
CT Construction
•Primary: System conductor passing
through core
•Secondary: Multiple turns wound
around magnetic core
•Multiple cores possible in one unit for
different applications (protection,
metering, etc.)

4. Types of Current Transformers
1 2
3
Standard Current Transformer
Primary conductor passes
through a magnetic core with
secondary windings. The number
of secondary turns determines
the transformation ratio. Common
in high voltage applications.
Flux Summing CT
All three phase conductors pass
through a toroidal core with
proper insulation. Sums the
currents (I
a + I
b + I
c = 3I
0), providing
residual current measurement.
CT ratio is independent of load
current.
Multi-Core CT
Contains several cores, each with
its own secondary winding for
different applications. Example:
Core 1 & 2 for distance/ differential
protection, Core 3 for metering,
Cores 4 & 5 for bus-bar differential
protection.
The selection of CT type depends on
the specific application requirements in
the power system protection scheme.
A high-voltage current transformer may
contain several cores, each with a
secondary winding, for different
purposes (such as metering circuits,
control, or protection)
Application-In one substation in
Indian Grid
i)Core-1 Distance/differential
ii)Core-2 Distance/ differential
iii)Core-3 Metering
iv)Core-4 Bus-bar differential
v)Core-5 Bus-bar differential

5. CT Ratios and Selection Criteria
•Current Transformation Ratio Expressed as I₁:I₂ (e.g., 600:5)
•Turn ratio = 1:N (e.g., 1:120 for 600:5 CT)
•Standard CT ratios available per IEEE standards.
•As per IEEE C57.13-2016, Standard multi-ratio current transformer taps for 600:5 are 50:5, 100:5,
150:5, 200:5, 250:5, 300.5, 400:5, 450:5, 500:5, 600:5
•Other available CTs are 1200:5 2000:5 3000:5 4000:5 5000:5
•Selection should be based on system requirements and expected fault levels.
CT Selection Criteria
•Primary rating must meet maximum load current requirements (continuous rating)
•CT ratio should be large enough so secondary current doesn't exceed 20 times rated current during
maximum fault conditions

6. CT Burden Calculation
What is CT Burden?
The load connected to the secondary winding, expressed in VA:
CT Burden (VA) = I²₂ × Zb
Where Zb is the equivalent impedance of connecting leads and
elements connected to the secondary winding
Burden Comparison: 5A vs 1A CT
For example a CT lead runs 200 m, a typical distance for
outdoor EHV substation,
could have a loop resistance of approximately 3 ohm
For 5A CT with 3Ω lead: 5² × 3 = 75 VA (lead burden)
For 1A CT with 3Ω lead: 1² × 3 = 3 VA (lead burden)
Adding relay burden (10 VA for electromechanical, <1 VA for
numerical):
5A CT total: 85 VA vs 1A CT total: 13 VA
With 5A CT: Such a burden would require
the CT to be very large and expensive,
particularly if a high accuracy limit factor
were also applicable.
With 1A CT: It requires thinner conductors
due to lower burden, making them more
economical for long lead runs.
However, they require more secondary
turns than 5A CTs, which can be a
consideration for very high primary
currents.
Above 2000A, a CT of higher secondary
rating can be used, to limit the number of
secondary turns. In such a situation
secondary ratings of 5A may be preferred.
In extreme cases a 20A secondary CT may
be used, followed by a 20/1 interposing CT.

7. Polarity Marking
•Polarity marking of transformer windings is for describing the relative
directions in which the two windings are wound on the transformer core.
•The terminals identified by solid marks indicate the starting ends of the two
windings, if these are considered to be the starting points, then both windings
will go around the core in the same sense (clockwise or anti-clockwise).
•In a transformer, if one of the winding currents is considered to be flowing into
the marked terminal, the current in the other winding should be considered to
be leaving its marked terminal. The two currents will then be almost in phase
with each other.
•Similarly, the voltages of the two windings, when measured from the unmarked
terminal to the marked terminal, will be almost in phase with each other.
•Or label the primary winding terminals H
1, and H₂, and the secondary winding
terminals X
1, and X
2. H
1, and X
1, may be assumed to have the polarity mark on
them.
Proper polarity
connections are
essential for
differential
protection and
other schemes
requiring phase
comparison.

8. Requirement
•The current transformer output signal (=secondary current=input to relays) should be accurate
reproduction of the corresponding primary current.
•Current transformers used for relaying are designed to have small errors during faulted conditions,
while their performance during normal steady-state operation, when the relay is not required to
operate, may not be as accurate

9. CT Equivalent Circuit and Steady State Performance
Ideal vs. Actual Performance
In an ideal CT, the secondary current (I₂) would be perfectly proportional to the primary current (I₁).
However, the magnetizing current (I
m) introduces error.
The load impedance Z
b
= the impedance of all the relays
and meters connected to
secondary winding including the
leads
= the burden on the CT.
Equivalent Circuit
•Primary current (I₁) acts as a
current source
•Secondary leakage
impedance (Z
x2)
•Magnetizing impedance (Z
m)
•Burden impedance (Z
b)
CTs equivalent circuit (a) and its simplification (b)
Using the turns ratio (1:n) of the ideal transformer, we have
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10. Steady State Performance of CT
The voltage E
m across the magnetizing impedance Z
m
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I₁ = I₂ + I
m
the per unit current transformer error defined by
Error (ε) = (I₁ - I₂)/I₁ = I
m/I₁
The error ε is small for smaller Z
b and vice versa. Theoretically,
when Z
b = 0 and Z
x2 small, error ε tends to 0 and I
1 = I
2
•Ratio Correction Factor (R) = 1/(1-ε)
Note & and R are complex numbers, sometimes their
magnitudes are used
The equivalent circuit shows how the
magnetizing current (I
m) diverts some of
the primary current, causing the
secondary current to be less than the
ideal value. This creates an error that
increases with burden and decreases
with higher magnetizing impedance.

11. CT Error Analysis
1
Example: CT with Resistive Burden
CT with 600:5 ratio, Zx2 = (0.01+j0.15)Ω, Zm
= (5+j17)Ω, Zb = 1Ω
Calculations yield:
Error (ε) = 0.056∠-62.24°
Ratio Correction Factor (R) = 1.025∠-2.91°
2
Example: CT with Inductive Burden
Same CT with Zb = j1Ω
Calculations yield:
Error (ε) = 0.061∠14.93°
Ratio Correction Factor (R) = 1.062∠0.957°
3
Key Findings
Inductive burden produces higher error than
equivalent resistive burden
Error increases as burden increases
Error decreases as magnetizing impedance
increases
CT saturation causes severe errors (as high
as 47.5% in extreme cases)
These examples demonstrate how burden type and magnitude affect CT accuracy. For protection
applications, it's critical to ensure the burden doesn't cause excessive errors that could
compromise protection functions.

12. CT Classification Standards
10%
Maximum Error
Maximum allowable error at 20× rated current
20×
Current Rating
Maximum secondary current for which accuracy is specified
400V
Voltage Rating
Secondary voltage at which accuracy is specified
CT selection should ensure that maximum fault current doesn't exceed
20× rated current and burden voltage doesn't exceed the accuracy class voltage.
CT classification- protection class and
metering/measurement class
The IEEE class designation of a CT consists of
two integer parameters, separated by the letter
'C' or 'T': for example, 10C400 or 10T300.
The first integers (10 here) describe the upper
limit on the error made by the CT when the
voltage at its secondary terminals is equal to the
second integer, while the current in the
transformer is 20 times its rated value.
The letter 'C' in the class designation implies that
the transformer design is such that the CT
performance can be calculated.

13. CT Classification Standards
The letter 'T' signifies some uncertainties in the transformer design, and the performance of the CT
must be determined by testing the CT.
Accuracy is specified by a percent ratio error e that is not to be exceeded when the CT is subjected to
a stated burden and limit current.
Standard Burdens
IEC standard burdens: 1, 2, 4, 8Ω (all with 60° impedance angle)
However, the CT is classified by the voltage across the burden impedance at the limit current of 100 A
(5A CT).
Corresponding voltages standard classifications : 100V, 200V, 400V, 800V

14. CT Application Guidelines
Calculate Maximum Burden
Example: For a 500:5 CT with class 10C400
and primary current of 6000A:
Maximum burden impedance = 4.44Ω
Exceeding this value will cause errors greater
than 10%
Evaluate Saturation Effects
High burden impedances can cause severe
saturation
Example: A CT with (8+j3)Ω burden showed
47.5% error
Saturation significantly distorts the
secondary current waveform
Follow Best Practices
•Keep burden as low as possible
(especially for digital relays)
•Prefer resistive over inductive burden of
same magnitude
•Ensure maximum fault current doesn't
exceed 20× rated current
•Verify burden voltage doesn't exceed
accuracy class voltage
Proper CT selection and application are critical for reliable protection system operation.
Understanding the CT's limitations and ensuring it operates within its accuracy class will prevent
protection failures during fault conditions.