CT Saturation Tutorial
Presented by
Tony Giuliante
Physical Properties of Core
Length L
Area A
B-H Characteristic
BH
Flux to Volts per Turn
Flux to Volts per Turn
V
=
ω
•B
•A
N
Electric Field to Ampere Turns
Convert B-H Characteristic
B H V = ω • B • A N ΝΙ = H • LV/N vs. NI
V NCT Exciting Characteristic
V
SI
S2000:5
300:5
Simplified Bushing CT Circuit
N
I
PIS
Simplified Bushing CT Circuit
IM XMN
I
P IS REB RCTSimplified Bushing CT Circuit
ISFlux vs Voltage
d
Φ
dt
V
=
N
Φ
=
N
∫
V
•dt
1
Φ
=
N
∫
I
SR
TB •dt
1
Flux vs Voltage
Φ
=
N
∫
I
SR
TB •dt
1
Voltage Demand
I
S
R
TB
Voltage & Flux Waveforms
I
SR
TBFlux Design Limits
+
Φ
S-
Φ
SΦ
Secondary Current
No Saturation
Increased Voltage Demand
Five times I
S
R
TB
5*I
SR
TBFlux for Ideal CT
No Saturation
Current Output for Ideal CT
No Saturation
Amperes
Time (Seconds)
Primary Current Secondary Current
Flux Design Limits
+
Φ
S-
Φ
SFlux Design Limits
+
Φ
S-
Φ
SFlux Excursion
Current vs Flux
Φ
AC Saturation
• Large Fault Current
• Large Burden
• Low CT Kneepoint Voltage
AC Saturation
Relay Applications
• Large Fault Current
87UAT
Unit Auxiliary Transformers
GDC Offset
Offset Current vs Flux
Time (Seconds)
Primary Current Flux
Sec. Amperes or Flux Density Primary Current Secondary Current
Secondary Current
Secondary Current
Observations
• 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 50 100 Secondary Current Amps2 Magnetic Flux Density (B) ISEC
Time (Seconds) Amperes
Secondary Current Primary Current
Differential Current
Large Differential Current
DC Saturation Factors
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 Percentage in % of saturation of cts
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
Primary Current Secondary Current
50% Remanent Flux
Time (Seconds) Amperes Primary Current Secondary Current75% Remanent Flux
Amperes Primary Current Secondary CurrentRemanent Flux Results
Remanent flux Time-to-saturation
0 % 1+ cycles
50% 1/2 cycle
75% 1/3 cycle
IEEE Guide for the Application of
Current Transformers Used for
CT Classification
CT Accuracy Class
• ANSI defines accuracy rating classes by a
letter and number
C100, C800 or T100, etc.
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
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
2000:5
300:5 45o Tangent
A B
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
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
Example - Equivalent Circuit
IP IS= 100 A VS= 500 IE= 10 A VB= ? Z BExample - continued
– VB (voltage to the burden)
Examples
• Determine Accuracy Class
• Selecting CT Ratings
• Calculating Time to Saturation
V
X> I
S· Z
TB(1 + X/R)
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
Transient Response of
Current Transformers
Power Systems Relaying Committee
VK I F TBR ( I - K ) =R T -CT TS TCT - t TS - t
e
- e
+ 1 TCT TS ω}
{
V
SI
S 2000:5 300:5 Tangents IntersectV
K
VKSaturation Parameters
R = R + R +RTB CT LEADS DEVICES I = FAULT CURRENTV
SI
S 2000:5 300:5 45o TangentV
M
& I
M
VM IMCT Inductance
TCT = M RTB L VMDC Offsets
DC Offset Current
• Depends on where in the voltage wave the
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 HzFIA
Voltage Waveform
Power System
Z =
√ R
2+ X
2θ =
ARCTAN(
ω
L / R)
G
R
L
Power System
R
L
Fault at FIA =
θ
No Offset
G
R
L
-1 0 1 2 Cu rren tCurrent Waveform
No Offset FIA = θ
Fault at FIA =
θ ± 90
Max Offset
G
R
L
0 1 2Current Waveform
0 10 20 30 40 50 -3 -2 -1 0 1 2 Time - Milliseconds Cu rren t
Total Current
Equations
v(t) = Vmax * sin ( Wt + Close_Ang ) i (t) = i ss (t) + i trans (t)
Power System
Time Constants
L/R (MS) X/R Ang (deg) Power System
1 0.377 20.66 High Fault Resistance
2 0.754 37.02 5 1.885 62.05 Distribution Lines 10 3.770 75.14 Subtransmission Lines 30 11.310 84.95 EHV Lines 100 37.699 88.48 Transformers 200 75.398 89.24 Generators 400 150.796 89.62 1000 376.991 89.85 Large Generators 0 1 2
Subtransmission Line
L/R = 10 ms
0 10 20 30 40 50 -3 -2 -1 0 1 2 Time - Milliseconds Cu rren t
Generator
L/R = 200 ms
-1 0 1 2 Cu rren tEHV Line
L/R = 30 ms
0 10 20 30 40 50 -3 -2 -1 0 1 2 Time - Milliseconds Cu rren t