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INTRODUCTION
TO
CURRENT TRANSFORMER
PERFORMANCE ANALYSIS
Hands on workshop developed for field relay techs practical approach
Yellow Brick Road
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INTRODUCTION
DEFINITIONS
PERFORMANCE CALCULATIONS
RATIO SELECTION CONSIDERATIONS
VARIOUS TOPICS
TEST
Z = V/I --- accurate value of I
DISTANCE
~
Z
INTRODUCTION
• IEEE Standard Requirements for Instrument
Transformers C57.13
• IEEE Guide for the Application of Current
Transformers Used for Protective Relaying
Purposes C37.110
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
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.
DEFINITIONS
• EXCITATION CURVE
• EXCITATION VOLTAGE
• EXCITATION CURRENT
• EXCITATION IMPEDANCE
DEFINITIONS
• EQUIVALENT CIRCUIT/DIAGRAM
• POLARITY
• BURDEN
• TERMINAL VOLTAGE
• CLASSIFICATIONS T AND C
DEFINITIONS
• KNEE POINT
• RELAY ACCURACY CLASS
• MULTI-TAPS ACCURACY
• SATURATION ERROR - RATIO/ANGLE
EXCITATION CURVE
EQUIVALENT DIAGRAM
Ip
Rp
Xp
Rs
e
Is
g
c
Pri
Ie
Ze
Sec
h
d
f
Ve = EXCITATION VOLTAGE Vef
Ie = CURRENT (
)
Ze = IMPEDANCE
Vt = TERMINAL VOLTAGE Vgh
POLARITY read a few values
next
TYPICAL EXCITATION BBC
CURRENT vs VOLTAGE
V (volts)
3.0
7.5
15
42
85
180
310
400
425
450
500
520
Ie(amps)
0.004
0.007
0.011
--------------------0.25
----------5.0
10.0
Ze(ohms)
750
1071
1364
-------------3100
1600
----------100.0
52.0
CURRENT vs VOLTAGE
V (volts)
3.0
7.5
15
42
85
180
310
400
425
450
500
520
Ie(amps)
0.004
0.007
0.011
0.02
0.03
0.05
0.1
0.25
0.5
1.00
5.0
10.0
Ze(ohms)
750
1071
1364
2100
2833
3600
3100
1600
850
450
100.0
52.0
Rsec
Zint
I1
I2
Ie+I2
Ie
N1 I1
N2
Ze
{
RB
EXTERNAL
BURDEN
POLARITY
LB
DEFINITIONS
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EXCITATION CURVE
EXCITATION VOLTAGE
EXCITATION CURRENT
EXCITATION IMPEDANCE
EQUIVALENT CIRCUIT/DIAGRAM
BURDEN - NEXT
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
EXTERNAL BURDEN
Burden:
0.27 VA @ 5A = …….. Ohms
2.51 VA @ 15A = …….. Ohms
BURDEN
=
VA / I²
{
RB
LB
QUIZ
I2
RB
CT winding resistance = 0.3 ohms
Lead length = 750 ft # 10 wire
Relay burden = 0.05 ohms
DEFINITIONS
• CLASSIFICATIONS T AND C
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 primarywinding turns mechanically
encircling the core. Performance
determined by test.
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
DEFINITIONS
• KNEE POINT IEEE IEC - effective
saturation point
• Quiz- read a few knee point voltages and also
at 10 amps Ie.
Knee Point Volts
Excitation Volts
45° LINE
ANSI/IEEE
KNEE POINT
QUIZ: READ THE KNEE POINT VOLTAGE
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.
IEC KNEE POINT
ANSI/IEE
KNEE POINT
EX: READ THE KNEE POINT VOLTAGE
DEFINITIONS
• EQUIVALENT CIRCUIT/DIAGRAM
• EXCITATION VOLTAGE, CURRENT,
IMPEDANCE
• TERMINAL VOLTAGE
• BURDEN
• CLASSIFICATIONS T AND C
• EXCITATION CURVE
• KNEE POINT IEEE IEC
• ACCURACY CLASS
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
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.
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
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
………………..
……………...
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
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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
………………..
……………...
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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
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
ANSI/IEEE ACCURACY
CLASS C400
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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.
Rsec
Zint
Isec = 100
I1
Ie+Isec
Ie <10
N1 I1
N2
Ze
RB
EXTERNAL
BURDEN
LB
ACCURACY ACLASS: C200
RATED SEC CURRENT = 5 A
EXTERNALBURDEN = STANDARD BURDEN = 2 .0 OHMS
Ve=200 V
Isec = 100 A
Ie <10 Amps.
PERFORMANCE
CALCULATIONS
BUT
THE REST OF US
“SHOW US THE DATA”
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.
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
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
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
PERFORMANCE CALCULATION
STEP: Determine Total Burden (Load)
Relay: 2.64 VA @ 5 A
Lead: 0.4 Ohms
and
580 VA @ 100 A
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
PERFORMANCE CALCULATION
STEPS:
CT Ratio selection
Relay Tap Selection
Determine Total Burden (Load)
CT Performance using ANSI/IEEE
Standard
CT Performance using Excitation
Curve
PERFORMANCE CALCULATION
STEP: CT Performance using ANSI/IEEE Standard
Ip
Rp
Xp
Rs
e
Is
g
c
Pri
Ie
Ze
Sec
h
d
Determine voltage @ max fault current CT must develop
across its terminals gh
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.
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.
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
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)
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.
TEST
I2
RB
Determine CT performance using Excitation Curve
method:
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
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
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
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
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
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
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
FAULT CURRENT
MAGNITUDES
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25 -33 KA
20 - 25 KA
12.5 -20 KA
20 - 25 KA
10 -12.5 KA
<10 KA
8
10
46
35
35
+150
REFER TO PAGE 6 OF PAPER
RED DELICIOUS
C400
ZONE1
Z = V/A
DISTANCE
~Z
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)
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
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
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
RATIO CONSIDERATIONS
• SELECT RATIO TO BE GREATER THAN
THE MAXIMUM DESIGN CURRENT
RATINGS OF THE ASSOCIATED
BREAKERS AND TRANSFORMERS.
RATIO CONSIDERATIONS
• RATIOS SHOULD NOT BE SO HIGH AS
TO REDUCE RELAY SENSITIVITY,
TAKING INTO ACCOUNT AVAILABLE
RANGES.
RATIO CONSIDERATIONS
• THE MAXIMUM SECONDARY
CURRENT SHOULD NOT EXCEED 20
TIMES RATED CURRENT. (100 A FOR
5A RATED SECONDARY)
RATIO CONSIDERATIONS
• HIGHEST CT RATIO PERMISSIBLE
SHOULD BE USED TO MINIMIZE
WIRING BURDEN AND TO OBTAIN
THE HIGHEST CT CAPABILITY AND
PERFORMANCE.
RATIO CONSIDERATIONS
• FULL WINGING OF MULTI-RATIO CT’s
SHOULD BE SELECTED WHENEVER
POSSIBLE TO AVOID LOWERING OF
THE EFFECTIVE ACCURACY CLASS.
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 35 amps. Slowly reduce the voltage to zero
before turning off the variac.
TESTING
• Saturation
– The saturation point is reached when there is a rise
in the test current but not the voltage.
TESTING
• Flashing
• This test checks the polarity of the CT
• Ratio
• Insulation test