ct saturation tutorial
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Copyright ATG Consulting 2003 1
CT Saturation Tutorial
Presented byTony Giuliante
N TurnsIP
IS
Bushing CT
Copyright ATG Consulting 2003 2
Physical Properties of Core
Length L
Area A
B-H Characteristic
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B-H Characteristic
B
H
Flux to Volts per Turn
Φ = ∫s B • dA
Φ = B • A sin (ωt)
dΦ = ω • B • A cos (ωt)dt
V =N
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Flux to Volts per Turn
V = ω • B • AN
Electric Field to Ampere Turns
ΝΙ = ∫ H • dL
ΝΙ = H • L
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Convert B-H Characteristic
B
H
V = ω • B • AN
ΝΙ = H • L
V/N vs. NI
V
N
NI
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CT Exciting Characteristic
VS
IS
2000:5
300:5
Simplified Bushing CT Circuit
NIP
IS
REB
REB = RLEADS + RDEVICES
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Simplified Bushing CT Circuit
IM
XMNIP
IS
REB
RCT
Simplified Bushing CT Circuit
IM
XMV = IS RTB
NIP
IS
RTB
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Flux vs Voltage
dΦdtV = N
Φ = ∫ V • dtN1
Φ = ∫ IS RTB • dtN1
Flux vs Voltage
Φ = ∫ IS RTB • dtN1
Flux equals the AREA under the Voltage
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Voltage DemandIS RTB
Voltage & Flux Waveforms
Φ
IS RTB
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Flux Design Limits
+ ΦS
- ΦS
Φ
Secondary CurrentNo Saturation
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Increased Voltage DemandFive times IS RTB
5*IS RTB
Flux for Ideal CTNo Saturation
Φ
5*IS RTB
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Current Output for Ideal CTNo Saturation
Amperes
Time (Seconds)
Primary Current Secondary Current
Flux Design Limits
+ ΦS- ΦS
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Flux Design Limits
+ ΦS- ΦS
Flux Excursion
Φ
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Current vs Flux
Φ
AC Saturation
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AC Saturation
• Large Fault Current
• Large Burden
• Low CT Kneepoint Voltage
AC SaturationRelay Applications
• Large Fault Current– Unit Auxiliary Transformers
• Large Burden– High Impedance Bus Differentials
• Low CT Kneepoint Voltage– Compact Distribution Switchgear
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87UAT
Unit Auxiliary Transformers
G
DC OffsetTransformer
Fault current includes a dc component, or offset, that makes the current asymmetrical.
L/R = 100 ms X/R = 37.7
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Offset Current vs Flux
Time (Seconds)
Primary Current Flux
Sec. Amperes
or
Flux Density
Time (Seconds)
Amperes
Primary Current Secondary Current
Secondary Current
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Secondary CurrentObservations
• Secondary current is distorted due to the core flux saturation
• Secondary current distorts after a short time (time-to-saturation)
• Distortion slowly dissipates as primary dc offset decays
0 1 2 3 4 5 6 7-50
0
50
100Secondary Current
Am
ps
0 5 6 7-1
0
1
2Magnetic Flux Density (B)
Tesl
a
1 2 3 4Cycles
I SEC
I PRIM
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Time (Seconds)
Amperes
Secondary Current Primary Current
Differential Current
Large Differential Current
DC Saturation Factors
• Large DC Time Constant
• Large Burden
• Low CT Kneepoint Voltage
• High Remanent Flux
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Remanent Flux• Trapped magnetic flux in core if a previous
offset current is interrupted before reaching a symmetrical state
• High X/R ratios make remanent flux more likely due to the slow decay rates of offset current
Remanent Flux Survey
Remanent flux Percentagein % of saturation of cts
0 - 20 3921 - 40 1841 - 60 1661 - 80 27
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Remanent Flux Example
• CT data– 1200:5, C800, burden = 1.6 +j 0.7 ohm
• Fault current 24,000 amps with dc offset• X/R ratio = 19• Display ct secondary output current for
remanence of 0%, 50% and 75% of saturation
Time (Seconds)
Amperes
Primary Current Secondary Current
0% Remanent Flux
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50% Remanent Flux
Time (Seconds)
Amperes
Primary Current Secondary Current
75% Remanent Flux
Time (Seconds)
Amperes
Primary Current Secondary Current
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Remanent Flux ResultsRemanent flux Time-to-saturation
0 % 1+ cycles50% 1/2 cycle75% 1/3 cycle
IEEE Guide for the Application of Current Transformers Used for
Protective Relaying Purposes
C37.110-1996
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CT Classification
CT Accuracy Class
• ANSI defines accuracy rating classes by a letter and numberC100, C800 or T100, etc.
• Letter designates how the accuracy can be determined
• Number designates the minimum secondary terminal voltage under a standard burden
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Accuracy Class Letter
• “C” means by Calculation– non-gapped cores with negligible leakage flux,
such as bushing cts• “T” means by Test
– cts with leakage flux, such as cts with wound primaries
• Old classes “H” and “L”H T and L C
Accuracy Class Number
• Minimum secondary terminal voltage produced– at 20 times rated current– into a standard burden– without exceeding a 10% ratio correction factor
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What is a Standard Burden?
• IEEE Standard Requirements for Instrument Transformers C57.13-1993 the standard relaying burdens are 1, 2, 4 and 8 ohms at a lagging 0.5 p.f.
• 20 times rated secondary current of 5 A is 100 A, and 100 A times the standard burdens yield C ratings of 100, 200, 400 and 800 V
VS
IS
2000:5
300:5
45o Tangent
A B
CT Knee Point Voltages
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Knee Point Definitions
• Point A is the ANSI knee point voltage– point tangent to 45 degree slope line
• Point B is the IEC knee point– where a 10% increase in voltage causes a 50 %
increase in current• IEC knee point is higher than ANSI knee
point
CT Excitation Impedance
• Excitation curve represents the exciting impedance in terms of voltage and current
• The ANSI knee point (A) represents the point of maximum permeability of the iron core
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Examples
• Determine Accuracy Class
• Selecting CT Ratings
• Calculating Time to Saturation
Example - Find Accuracy Class
• Find the approximate ct accuracy class from the excitation curve– the C class is defined for a 10% ratio correction
factor at 20 times rated current• 10% of (20 X 5 A) is 10 A• for IE = 10 A, use the excitation curve to find VS as
about 500 V
– next find the ct terminal voltage by subtracting the internal voltage drop from VS
» (continued)
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Example - Equivalent Circuit
IP IS = 100 A
VS = 500IE = 10 A
VB = ? ZB
Example - continued– VB (voltage to the burden)
VB = VS - (IS X RS)VB = 500 - (100 X 0.61)VB = 439 V
– The approximate ct accuracy class is the next lowest ANSI class number (C400)
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Examples
• Determine Accuracy Class
• Selecting CT Ratings
• Calculating Time to Saturation
VX > IS · ZTB (1 + X/R)
VX = saturation voltage
IS = secondary current
ZTB = total ct secondary burden
Avoiding CT Saturation
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CTs for Generator Differentials• For generators, typically cts cannot be sized
to avoid saturation because of:– high fault current– high X/R ratio
• Common applications would:– select adequate ct primary rating– select highest practical C class– match manufacturer and types of cts
Examples
• Determine Accuracy Class
• Selecting CT Ratings
• Calculating Time to Saturation
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Transient Response of Current Transformers
Power Systems Relaying Committee
VKI F TBR
( I - K ) =RT -CT TS
TCT
- tTS
- t
e - e + 1TCT TSω }{
Time to Saturate Equation
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VS
IS
2000:5
300:5
TangentsIntersect
VK
VK
Saturation ParametersR = R + R +RTB CT LEADS DEVICES
I = FAULT CURRENT(SEC RMS AMPS)
F
K = .5 - .75 IRON CORE.1 AIR GAP
R
ω = 377
TCT =LMRTB
{
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VS
IS
2000:5
300:5
45o Tangent
VM & IM
VM
IM
CT Inductance
TCT =M
RTB
L
X =MVMI M
L =MXMω
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DC Offsets
DC Offset Current
• Depends on where in the voltage wave the fault occurs.
• Fault time is defined as:F I A => Fault Initiation Angle
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FIA
Voltage Waveform
0 45 90 135 180 225 270 345 360 Degrees
0 4.16 8.33 12.5 16.67 Time ms 60 Hz
FIA
Voltage Waveform
0 45 90 135 180 225 270 345 360 Degrees
0 5 10 15 20 Time ms 50 Hz
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Power System
Z = √ R2 + X2
θ = ARCTAN (ω L / R)G
R L
Power System
θ = Characteristic AngleG
R L
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Fault at FIA = θNo Offset
G
R L
0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
Current WaveformNo Offset FIA = θ
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Fault at FIA = θ ± 90Max Offset
G
R L
0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
Current WaveformMax Offset FIA = θ + 90
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0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
Total Current
Equations
v(t) = Vmax * sin ( Wt + Close_Ang )
i (t) = i ss (t) + i trans (t)
i ss (t) = [ Vmax / Z ] * sin ( Wt + Alpha )
Alpha = Close_Ang - ArcTan ( WL/R )
i trans (t) = [ e^ (-R/L) t ] * [ -Vmax / Z ] * sin ( Alpha )
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Power SystemTime Constants
L/R (MS) X/R Ang (deg) Power System1 0.377 20.66 High Fault Resistance2 0.754 37.025 1.885 62.05 Distribution Lines
10 3.770 75.14 Subtransmission Lines30 11.310 84.95 EHV Lines100 37.699 88.48 Transformers200 75.398 89.24 Generators400 150.796 89.62
1000 376.991 89.85 Large Generators
0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
Subtransmission LineL/R = 10 ms
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0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
GeneratorL/R = 200 ms
0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
EHV LineL/R = 30 ms
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0 10 20 30 40 50-3
-2
-1
0
1
2
Time - Milliseconds
Cur
rent
Distribution LineL/R = 5 ms
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