This document provides an introduction and overview of current transformer performance analysis. It defines key terms related to current transformers like excitation curve, knee point, and accuracy class. It also outlines the steps to evaluate current transformer performance for phase faults, including selecting a CT ratio, relay tap, determining total burden, and analyzing performance using ANSI/IEEE standards and the excitation curve method. An example is provided to demonstrate calculating CT performance using the excitation curve for a fault current of 12500 amps.

1. INTRODUCTION
TO
CURRENT TRANSFORMER
PERFORMANCE ANALYSIS
Hands on workshop developed for field relay techs practical approach

2. Yellow Brick Road
• INTRODUCTION
• DEFINITIONS
• PERFORMANCE CALCULATIONS
• RATIO SELECTION CONSIDERATIONS
• VARIOUS TOPICS
• TEST

3. Z = V/I --- accurate value of I
DISTANCE ~ Z

4. INTRODUCTION
• IEEE Standard Requirements for Instrument
Transformers C57.13
• IEEE Guide for the Application of Current
Transformers Used for Protective Relaying
Purposes C37.110

5. INTRODUCTION
• Bushing, internal to Breakers and
Transformers
• Free standing, used with live tank breakers.
• Slipover, mounted externally on
breaker/transformers bushings.
• Window or Bar - single primary turn
• Wound Primary
• Optic

6. MAGNETO-OPTIC CT
• Light polarization passing through an
optically active material in the presence of a
magnetic field .
• Passive sensor at line voltage is connected to
substation equipment by fiber cable.
• Low energy output used for microprocessor
relays
• Eliminates heavy support for iron.

7. DEFINITIONS
• EXCITATION CURVE
• EXCITATION VOLTAGE
• EXCITATION CURRENT
• EXCITATION IMPEDANCE

8. DEFINITIONS
• EQUIVALENT CIRCUIT/DIAGRAM
• POLARITY
• BURDEN
• TERMINAL VOLTAGE
• CLASSIFICATIONS T AND C

9. DEFINITIONS
• KNEE POINT
• RELAYACCURACY CLASS
• MULTI-TAPS ACCURACY
• SATURATION ERROR - RATIO/ANGLE

10. EXCITATION CURVE

11. f
Ip
Ie Ze
Xp
Rp e Rs
Sec
g
h
c
d
Pri
Is
EQUIVALENT DIAGRAM
Ve = EXCITATION VOLTAGE Vef
Ie = CURRENT (read a few values)
Ze = IMPEDANCE
Vt = TERMINAL VOLTAGE Vgh
POLARITY - next

12. TYPICAL EXCITATION BBC
CURRENT vs VOLTAGE
V (volts) Ie(amps) Ze(ohms)
3.0 0.004 750
7.5 0.007 1071
15 0.011 1364
42 ------ -----
85 ------ -----
180 ------ ------
310 ------ 3100
400 0.25 1600
425 ------ ------
450 ------ ------
500 5.0 100.0
520 10.0 52.0

13. CURRENT vs VOLTAGE
V (volts) Ie(amps) Ze(ohms)
3.0 0.004 750
7.5 0.007 1071
15 0.011 1364
42 0.02 2100
85 0.03 2833
180 0.05 3600
310 0.1 3100
400 0.25 1600
425 0.5 850
450 1.00 450
500 5.0 100.0
520 10.0 52.0

14. N1
N2
I1 Ze
Ie
I2
Rsec
RB
LB
EXTERNAL
BURDEN {
Ie+I2
Zint
POLARITY
I1

15. DEFINITIONS
• EXCITATION CURVE
• EXCITATION VOLTAGE
• EXCITATION CURRENT
• EXCITATION IMPEDANCE
• EQUIVALENT CIRCUIT/DIAGRAM
• BURDEN - NEXT

16. BURDEN
• The impedances of loads are called BURDEN
• Individual devices or total connected load,
including sec impedance of instrument
transformer.
• For devices burden expressed in VA at
specified current or voltage, the burden
impedance Zb is:
• Zb = VA/IxI or VxV/VA

17. RB
LB
BURDEN
=
VA / I² {
EXTERNAL BURDEN
Burden: 0.27 VA @ 5A = …….. Ohms
2.51 VA @ 15A = …….. Ohms

18. I2
RB
CT winding resistance = 0.3 ohms
Lead length = 750 ft # 10 wire
Relay burden = 0.05 ohms
QUIZ

19. DEFINITIONS
• CLASSIFICATIONS T AND C

20. ANSI/IEEE STANDARD FOR
CLASSIFICATION T & C
• CLASS T: CTs that have
significant leakage flux within the
transformer core - class T; wound
CTs, with one or more primary-
winding turns mechanically
encircling the core. Performance
determined by test.

21. CLASS C
• CTs with very minimal leakage
flux in the core, such as the
through, bar, and bushing types.
Performance can be calculated.
KNEE POINT

22. DEFINITIONS
• KNEE POINT IEEE IEC - effective
saturation point
• Quiz- read a few knee point voltages and also
at 10 amps Ie.

23. ANSI/IEEE
KNEE POINT
Excitation
Volts
Knee
Point
Volts
45° LINE
QUIZ: READ THE KNEE POINT VOLTAGE

24. KNEE POINT OR EFFECTIVE
POINT OF SATURATION
• ANSI/IEEE: as the intersection of the curve
with a 45 tangent line
• IEC defines the knee point as the
intersection of straight lines extended from
non saturated and saturated parts of the
excitation curve.
• IEC knee is higher than ANSI - ANSI more
conservative.

25. IEC KNEE POINT
ANSI/IEE
KNEE POINT
EX: READ THE KNEE POINT VOLTAGE

26. DEFINITIONS
• EQUIVALENT CIRCUIT/DIAGRAM
• EXCITATION VOLTAGE, CURRENT,
IMPEDANCE
• TERMINAL VOLTAGE
• BURDEN
• CLASSIFICATIONS T AND C
• EXCITATION CURVE
• KNEE POINT IEEE IEC
• ACCURACY CLASS

27. CT ACCURACY CLASSIFICATION
The measure of a CT performance is its
ability to reproduce accurately the primary
current in secondary amperes both is wave
shape and in magnitude. There are two
parts:
• Performance on symmetrical ac component.
• Performance on offset dc component. Go over the
paper

28. ANSI/IEEE ACCURACY CLASS
• ANSI/IEEE CLASS DESIGNATION C200:
INDICATES THE CT WILL DELIVER A
SECONDARY TERMINAL VOLTAGE OF
200V
• TO A STANDARD BURDEN B - 2 (2.0 ) AT
20 TIMES THE RATED SECONDARY
CURRENT
• WITHOUT EXCEEDING 10% RATIO
CORRECTION ERROR. Pure sine wave
Standard defines max error, it does not specify the actual error.

29. ACCURACY CLASS C
STANDARD BURDEN
• ACCURACY CLASS: C100, C200, C400, & C800 AT POWER
FACTOR OF 0.5.
• STANDARD BURDEN B-1, B-2, B-4 AND B-8 THESE
CORRESPOND TO 1, 2, 4 AND 8.
• EXAMPLE STANDARD BURDEN FOR C100 IS 1 , FOR C200
IS 2 , FOR C400 IS 4  AND FOR C800 IS 8 .
• ACCURACY CLASS APPLIES TO FULL WINDING, AND ARE
REDUCED PROPORTIONALLY WITH LOWER TAPS.
• EFFECTIVE ACCURACY =
TAP USED*C-CLASS/MAX RATIO

30. AN EXERCISE
• 2000/5 MR C800 tap used*c-class/max ratio
TAPS KNEE POINT EFFECTIVE ACCURACY
2000/5 ……………….. ……………...
1500/5 ……………….. ……………...
1100/5 ……………….. ……………...
500/5 ……………….. ……………...
300/5 ……………….. ……………...

31. AN EXERCISE
• 2000/5 MR C800 tap used*c-class/max ratio
TAPS KNEE POINT EFFECTIVE ACCURACY
2000/5 590 800
1500/5 390 600
1100/5 120 440
500/5 132 200
300/5 78 120

32. AN EXERCISE
• 2000/5 MR C400 tap used*c-class/max ratio
TAPS KNEE POINT EFFECTIVE ACCURACY
2000/5 ……………….. ……………...
1500/5 ……………….. ……………...
1100/5 ……………….. ……………...
500/5 ……………….. ……………...
300/5 ……………….. ……………...

33. AN EXERCISE
• 2000/5 MR C400 tap used*c-class/max ratio
TAPS KNEE POINT EFFECTIVE ACCURACY
2000/5 220 400
1500/5 170 300
1100/5 125 220
500/5 55 100
300/5 32 60

34. CT SELECTION
ACCURACY CLASS
POINT OF SATURATION : KNEE
POINT
IT IS DESIRABLE TO STAY
BELOW OR VERY CLOSE TO
KNEE POINT FOR THE
AVAILABLE CURRENT.
Recap

35. ANSI/IEEE ACCURACY
CLASS C400
• STANDARD BURDEN FOR C400: (4.0 )
• SECONDARY CURRENT RATING 5 A
• 20 TIMES SEC CURRENT: 100 AMPS
• SEC. VOLTAGE DEVELOPED: 400V
• MAXIMUM RATIO ERROR: 10%
• IF BURDEN 2 , FOR 400V, IT CAN SUPPLY
MORE THAN 100 AMPS SAY 200 AMPS
WITHOUT EXCEECING 10% ERROR.

36. N1
N2
I1 Ze
Ie <10
Isec = 100
Rsec
RB
LB
EXTERNAL
BURDEN
Ie+Isec
Zint
ACCURACY ACLASS: C200 RATED SEC CURRENT = 5 A
EXTERNALBURDEN = STANDARD BURDEN = 2 .0 OHMS
Ve=200 V Isec = 100 A Ie <10 Amps.
I1

38. PERFORMANCE
CALCULATIONS

40. BUT
THE REST OF US
“SHOW US THE DATA”

41. PERFORMANCE CRITERIA
• THE MEASURE OF A CT
PERFORMANCE IS ITS ABILITY TO
REPRODUCE ACCURATELY THE
PRIMARY CURRRENT IN SECONDARY
AMPERES - BOTH IN WAVE SHAPE
AND MAGNITUDE …. CORRECT
RATIO AND ANGLE.

42. CT SELECTION AND PERFORMANCE
EVALUATION FOR PHASE FAULTS
600/5 MR Accuracy class C100 is selected
Load Current= 90 A
Max 3 phase Fault Current= 2500 A
Min. Fault Current=350 A
STEPS:
CT Ratio selection
Relay Tap Selection
Determine Total Burden (Load)
CT Performance using ANSI/IEEE Standard
CT Performance using Excitation Curve

43. PERFORMANCE CALCULATION
STEPS:
CT Ratio selection
Relay Tap Selection
Determine Total Burden (Load)
CT Performance using ANSI/IEEE Standard
CT Performance using Excitation Curve
STEPS:
CT Ratio selection
- within short time and continuous current – thermal limits
- max load just under 5A
Load Current= 90 A
CT ratio selection : 100/5

44. PERFORMANCE CALCULATION
STEP: Relay Tap Selection
O/C taps – min pickup , higher than the max. load
167%, 150% of specified thermal loading.
Load Current= 90 A for 100/5 CT ratio = 4.5 A sec.
Select tap higher than max load say = 5.0
How much higher – relay characteristics, experience and
judgment.
Fault current: min: 350/20 = 17.5
Multiple of PU = 17.5/5 = 3.5
Multiple of PU = 17.5/6 = 2.9

45. PERFORMANCE CALCULATION
STEP: Determine Total Burden (Load)
Relay: 2.64 VA @ 5 A and 580 VA @ 100 A
Lead: 0.4 Ohms
Total to CT terminals:
(2.64/5*5 = 0.106) + 0.4 = 0.506 ohms @ 5A
(580/100*100 = 0.058) + 0.4 = 0.458 ohms @ 100 A

46. PERFORMANCE CALCULATION
STEPS:
CT Ratio selection
Relay Tap Selection
Determine Total Burden (Load)
CT Performance using ANSI/IEEE
Standard
CT Performance using Excitation
Curve

47. PERFORMANCE CALCULATION
STEP: CT Performance using ANSI/IEEE Standard
Ip
Ie Ze
Xp
Rp e Rs
Sec
g
h
c
d
Pri
Is
Determine voltage @ max fault current CT must develop
across its terminals gh

48. PERFORMANCE CALCULATION
STEP: Performance – ANSI/IEEE Standard
Vgh = 2500/20 * 0.458 = 57.25
600/5 MR C100 CT used at tap 100/5 -- effective
accuracy class
(100/600) x 100 = ?
CT is capable of developing 16.6 volts.
Severe Saturation. Cannot be used.

49. PERFORMANCE CALCULATION
STEP: Performance – ANSI/IEEE Standard
For microprocessor based relay:
Burden will change from 0.458 to o.4
Vgh = 2500/20 * 0.4 = 50.0
600/5 MR C100 CT used at tap 100/5 -- effective
accuracy class
(100/600) x 100 = ?
CT is capable of developing 16.6 volts.
Severe Saturation. Cannot be used.

50. PERFORMANCE CALCULATION
STEP: Performance – ANSI/IEEE Standard
Alternative: use 400/5 CT tap:
Max Load = 90 A
Relay Tap = 90/80 = 1.125 Use: 1.5 relay tap.
Min Fault Multiples of PU=(350/80=4.38, 4.38/1.5= 2.9)
Relay burden at this tap = 1.56 ohms
Total burden at CT terminals = 1.56 + 0.4 = 1.96
Vgh = 2500/80 * 1.96 = 61.25
600/5 MR C100 CT used at tap 400/5-- effective accuracy
class is = (400/600) x 100 = ?
CT is capable of developing 66.6 volts. Within CT capability

51. PERFORMANCE CALCULATION
STEP: CT Performance using Excitation Curve
ANSI/IEEE ratings “ballpark”. Excitation curve method provides relatively exact
method. Examine the curve
Burden = CT secondary resistance + lead resistance +
relay burden
Burden = 0.211 + 0.4 + 1.56 = 2.171
For load current 1.5 A:
Vgh = 1.5 * 2.171 = 3.26 V Ie = 0.024
Ip = (1.5+0.024) * 80 = 123 A
well below the min If = 350 A (350/123=2.84 multiple of
pick up)

52. PERFORMANCE CALCULATION
STEP: CT Performance using Excitation Curve
For max fault current
Burden = CT secondary resistance + lead resistance + relay burden
Burden = 0.211 + 0.4 + 1.56 = 2.171
Fault current 2500/80 = 31.25 A:
Vgh = 31.25 * 2.171 = 67.84 V Ie = 0.16
Beyond the knee of curve, small amount 0.5% does not significantly
decreases the fault current to the relay.

53. I2
RB
CT winding resistance = 0.3 ohms
Lead length = 750 ft # 10 wire
Relay burden = 0.05 ohms as constant
Fault current = 12500A/18000A
CT CLASS = C400/C800
2000/5 MR current transformer
CT RATIO = 800/5
TEST
Determine CT performance using Excitation Curve
method:

54. AN EXAMPLE – C400
• CT RESISTANCE 0.3 OHMS
• LEAD RESISTANCE 1.5 OHMS
• IMPEDANCE OF VARIOUS DEVICES 0.05
OHMS
• FAULT CURRENT 12500 AMPS
• CT RATIO 800/5
• ACCURACY CLASS C400
• supply curves C400/800

55. CALCULATIONS for 12500 A – C400
• BURDEN = ( Z-LEAD + Z - CT SEC + D -
DEVICES)
• Ve = (1.5 + 0.3 + 0.05 ) 12500/160
• Ve = 144.5 VOLTS Plot on curve
• Plot on C400

56. CALCULATIONS for 18000 –C400
• BURDEN = ( Z-LEAD + Z - CT SEC + D -
DEVICES)
• Ve = (1.5 + 0.3 + 0.05 ) 18000/160
• Ve = 209 VOLTS Plot on curve
• Plot on C400

57. ANOTHER EXAMPLE C800
• CT RESISTANCE 0.3 OHMS
• LEAD RESISTANCE 1.5 OHMS
• IMPEDANCE OF VARIOUS DEVICES 0.05
OHMS
• FAULT CURRENT 12500 AMPS
• CT RATIO 800/5
• ACCURACY CLASS C800
• supply curves C400/800

58. CALCULATIONS for 12500 A – C800
• BURDEN = ( Z-LEAD + Z - CT SEC + D -
DEVICES)
• Ve = (1.5 + 0.3 + 0.05 ) 12500/160
• Ve = 144.5 VOLTS Plot on curve
• Plot on C800

59. CALCULATIONS for 18000 A –C800
• BURDEN = ( Z-LEAD + Z - CT SEC + D -
DEVICES)
• Ve = (1.5 + 0.3 + 0.05 ) 18000/160
• For 18,000 A (Ve =209 V) Plot on curve
• Plot on C800

60. FAULT CURRENT
MAGNITUDES
• 25 -33 KA 8
• 20 - 25 KA 10
• 12.5 -20 KA 46
• 20 - 25 KA 35
• 10 -12.5 KA 35
• <10 KA +150
REFER TO PAGE 6 OF PAPER

61. RED DELICIOUS
C400
ZONE1

62. Z = V/A
DISTANCE ~ Z

63. STANDARD DATA FROM
MANUFACTURER
• ACCURACY:
– RELAY CLASS C200
– METERING CLASS, USE 0.15%
– 0.3%, 0.6% & 1.2% AVAIALABLE BUT NOT
RECOMMENDED
– 0.15% MEANS +/- 0.15% error at 100%
rated current and 0.30% error at 10% of rated
current ( double the error)

64. STANDARD DATA FROM
MANUFACTURER
• CONTINUOUS (Long Term) rating
– Primary
– Secondary, 5 Amp ( 1Amp)
– Rating factor (RF) of 2.0 provides Twice
Primary and Secondary rating continuous at
30degrees

65. STANDARD DATA FROM
MANUFACTURER
• SHORT TIME TERMINAL RATINGS
Transmission Voltage Applications
– One Second Rating = 80% Imax Fault, based
on IxIxT=K where T=36 cycles & I=Max fault
current
Distribution Voltage Applications
One Second Rating = Maximum Fault Current
level

66. RATIO CONSIDERATIONS
• CURRENT SHOULD NOT EXCEED
CONNECTED WIRING AND RELAY
RATINGS AT MAXIMUM LOAD. NOTE
DELTA CONNECTD CT’s PRODUCE
CURRENTS IN CABLES AND RELAYS
THAT ARE 1.732 TIMES THE
SECONDARY CURRENTS

67. RATIO CONSIDERATIONS
• SELECT RATIO TO BE GREATER THAN
THE MAXIMUM DESIGN CURRENT
RATINGS OF THE ASSOCIATED
BREAKERS AND TRANSFORMERS.

68. RATIO CONSIDERATIONS
• RATIOS SHOULD NOT BE SO HIGH AS
TO REDUCE RELAY SENSITIVITY,
TAKING INTO ACCOUNT AVAILABLE
RANGES.

69. RATIO CONSIDERATIONS
• THE MAXIMUM SECONDARY
CURRENT SHOULD NOT EXCEED 20
TIMES RATED CURRENT. (100 A FOR
5A RATED SECONDARY)

70. RATIO CONSIDERATIONS
• HIGHEST CT RATIO PERMISSIBLE
SHOULD BE USED TO MINIMIZE
WIRING BURDEN AND TO OBTAIN
THE HIGHEST CT CAPABILITYAND
PERFORMANCE.

71. RATIO CONSIDERATIONS
• FULL WINGING OF MULTI-RATIO CT’s
SHOULD BE SELECTED WHENEVER
POSSIBLE TO AVOID LOWERING OF
THE EFFECTIVE ACCURACY CLASS.

72. TESTING
•Core Demagnetizing
– The core should be demagnetized as the final
test before the equipment is put in service.
Using the Saturation test circuit, apply enough
voltage to the secondary of the CT to saturate
the core and produce a cecondary currrent of 3-
5 amps. Slowly reduce the voltage to zero
before turning off the variac.

73. TESTING
•Saturation
– The saturation point is reached when there is a rise
in the test current but not the voltage.

74. TESTING
•Flashing
• This test checks the polarity of the CT
•Ratio
•Insulation test