49666442-ct-sizing

Current Transformer Sizing

(August 19th, 2010)By: Siva Singupuram

James Trinh

Agenda

• Current Transformer (CT) introduction• Mathematical Modeling• Saturated Waveforms• CT saturation (Hysteresis Curve)• Remnant Flux, High Fault Currents, DC offset• Saturation Voltage, • CT ratings, Selection and Applications• ElectroMagnetic Transients Program (EMTP)/

PSCAD

Introduction CT types

Donut

Bus Bar

Hall Effect CT Magneto-Optical

Typical Current Transformers

Introduction Application• Current Transformers (CT’s) are instrument transformers that are used to

supply a reduced value of current to metering, protective relays, and other instruments. CT’s provide galvanic isolation from the high voltage primary, permit grounding of the secondary for safety, and step-down the magnitude of the measured current to a value that can be safely handled by the instruments (high fault currents).

• To achieve the above goals CT contains1) Iron Core2) Secondary windings3) Primary conductor4) External insulation

CT Symbols• When current flows in the CT from the H1 lead polarity (±) lead through,

the burden (load), and return to the secondary X2 non-polarity lead. The next half cycle the current will reverse.

ANSI IEC

Current Transformer (CT) Principle

Electrical Schematic 3D-Representation

CT Connections

ANSI Standard Letter Designation• B-The CT is a metering type• C-The CT has low leakage flux and the excitation characteristic can be

used directly to determine performance. (Accuracy can be calculated before manufacturing)

• K-Same as the C rating, but the knee point voltage must be at least 70% of the secondary terminal voltage rating.

• T-The CT can have significant leakage flux (Accuracy must be determined by testing at the factory)

• H-The CT accuracy is applicable within the entire range of secondary currents from 5 to 20 time the nominal CT rating. (Typically wound CTs)

• L-Same as C class except there were two accuracy classes 2.5% and 10%. The ratio accuracy can be up to four times greater than the listed value, depending on connected burden and fault current. (Typically window, busing, or bar-type CTs)

• L and H ANSI rating are applicable to old CT that where manufactured before 1954

ANSI Standard Terminal Voltage Class C CTs

C Class ZSTD (Ω) VSTD (V)

C100 1.0 100

C200 2.0 200

C400 4.0 400

C800 8.0 800

STDNSTD ZIV 20

For IN = 5A secondary

Metering Accuracy Classifications• Metering Accuracy CT’s are used where a high degree of accuracy is

required from low-load values up to full-load of a system. These are utilized by utility companies for revenue metering.

Available in Maximum Ratio Error Classes of: ±0.3%, ±0.6%, ±1.2%, ±2.4%

For Burdens (Loads) of: 0.1Ω, 0.2Ω, 0.5Ω, 0.9Ω, 1.8Ω

Volt-amperes (va) equivalent: 2.5va, 5.0va, 12.5va, 22.5va, 45va

Typical Number

0.3 B 0.2

Burden Ohms (Burden)Max Ratio Error ±%

Relaying Accuracy Classifications• Relaying Accuracy CT’s are used for supplying current to protective relays.

In this application, the relays do not normally operate in the normal load range, but they must perform with a reasonable degree of accuracy at very high overload and fault-current levels which may reach 20x the full-load amplitude.

Class C (C for calculated) is low leakage reactance type – typical of donut units – Formerly Class L (L for Low Leakage)

Typical Number

10 C 800

Low Leakage UnitMax secondary voltage developed at 20x rated current without exceeding the +10% ratio error

10% Max Ratio Error at 20x Rated Current

Class T (T for tested) is high leakage reactance type – typical of bar-type units – Formerly Class H (H for High Leakage)

Available secondary voltages: 100V, 200V, 400V, 800V

Burden support (ZSTD): 1.0Ω, 2.0Ω, 4.0Ω, 8.0Ω

CT Knee point

ANSI Method

CT Knee point

ANSI MethodIEC Method

Shorting of CT when not used

• CT’s can be very dangerous devices! Stepping down the current means that the voltage is automatically stepped up the same ratio. The secondary's of the CT’s are, therefore; always kept shorted when not in use and/or when no low impedance ammeter is connected! On the open-circuited secondary of a CT, just a few volts on the primary voltage would become a lethal thousands of volts on the secondary.

CT Equivalent Circuit

LOADsssss ZIjXRIV )(

es

ps I

NI

I

Hysteresis/Residual FluxReferring to figure on the left, if H is gradually increased from zero to H1, the magnetization follows path oa. Decreasing H back to zero will leave the material magnetized at a B-value corresponding to point b. Distance ob on the B axis is called the “Residual Flux Density”. To bring B back to zero requires a negative value of H corresponding to point c on the cure. Distance oc on the H-axis is called the Coercive Force. Decreasing H to H2=-H1 results in the B-value of point d. Finally increasing H back to the original H1 now causes the magnetization to traverse path defa. The latter path is nearly equivalent to path abcd, but flipped about the B and H axes. Path abcdefa is called a hysteresis loop which occurs when an AC voltage is applied to a coil on a ferromagnetic core. As the voltage and current cycle between their max and min values, the core magnetization repeatedly traverses a hysteresis loop.

Remanent Flux (Information Only)

Remanent Flux % Percentage of cts

0-20 39

21-40 18

41-60 16

61-80 27

Based on IEEE survey of 141 cts on a 230kV system

DC offset

Saturation Occurs

CT Errors impact

Errors to in currents magnitude and angle will have a significant effect on protection relays particularly (directional (67), distance (21), and differentials (87))

t tLR

SBF

tdtetRZIvdtBANN

00))(cos(2

RLe

RLtRZI

tLR

SBF

sin12

RXe

RXtRZI t

XR

SBF

sin)(2

RXe

RXtRZIBAN

tXR

SBF

sin)(2

RLe

RLtRZI t

LR

SBF

sin)(2

CT terminal Voltage (Vgh)

maxmax SVBAN

RXRZIBAN SBF 1)(2max

:Recall

RXRZIV SBFS 1)(2max

RXZIV BFgh 12max_

RXZI

VBF

gh 12max_


RXZI

VBF

gh 12max_

STDNSTDgh ZIV

V20

2max_

STDBF VRXZI

1

STDNSTD ZIV 20:Recall

STDNBF ZIRXZI 201

201

RX

ZZ

II

STD

B

N

F


201

RX

ZZ

II

STD

B

N

F 201

bf ZIR

X

CT primary rating if IF is given in primary Amps orNI

Standard Burden = 1, 2, 4, 8 Ω (at 60°)STDZ

Max Fault in per unit of CT secondary nominal CurrentfI

Burden in per unit of standard burdenbZ

CT secondary rating if IF is given in secondary Amps

Max Fault CurrentFI Burden from the CT’s point of view (CTwdg + Cable Leads + Relays)BZ

To Avoid CT saturation

Limitation of CT Selection Criterion

• We face high X/R ratio and high fault currents near generation sites. This becomes impractical to size the CT to avoid saturation during a asymmetrical fault. The formula derived earlier should not be used for this study and we must size the CT with reasonable sensitivity for line end faults

• For this situation we should analyze in further detail via EMTP simulation.

Example #1

C8002000/5 (full ratio)If = 3.07 puZb = 0.5 puX/R = 12

19.955 ≤ 20

201

bf ZIR

X

(1+12)(3.07)(0.5) ≤ 20 ?

CT OK!

Example #1


201

bf ZIR

X

19.955 ≤ 20

Example #2


49.985 > 20

201

bf ZIR

X

(1+12)(3.07)(0.5) ≤ 20 ? (1+12)(7.69)(0.5) ≤ 20 ?

CT OK! CT Saturates

RX

K

TTs

s

1

1ln1 RXT

1

primary system time constant1T the saturation factor Vx/Vs, where Vx is the excitation voltage at 10A and, Vs is the saturation voltage IF x ZB

sK

Resistance of the primary system to the point of the faultR

Natural log functionln time to saturatesT

Time to Saturate

Reactance of the primary system to the point of the faultX2πf, where f is the system frequency

Waveforms

Multi-Ratio CTs

1200/5, C800 tapped at 600/5C400 effective rating4.0 Ω effective standard burden

4008001200600

atingEffectiveR

481200600__ BurdenStdEffective

CT winding Resistance

Typical turn ratio resistance = 2.5mΩ/turn

0.15.25

20005/2000 mFor example a 2000:5 ratio CT;

Lead Resistance CalculationFormula to determine Copper lead resistanceLead resistance, Ω/1000’ = e0.232G-2.32

where, G = AWG number

For a 10AWG cable at (305m = 1000’)R = e0.232(10) -2.32 = e0 = 1Ω/1000’

R = 1Ω/1000’x500’ = 0.5Ω10AWG cable at (152m = 500’) will have a resistance of

Similarly for 8AWG cable at (305m = 1000’)R = e0.232(8) -2.32 = 0.629Ω/1000’

CT Application

ConnectionType of Fault

3 ph or ph-to-ph Ph-to-ground

Wye (connect at CT) Z = RS + RL + ZR Z = RS + 2RL + ZR

Wye (connected at switchhouse) Z = RS + 2RL + ZR Z = RS + 2RL + ZR

Delta (connected at switchhouse) Z = RS + 2RL +3 ZR Z = RS + 2RL +3 ZR

Delta (connected at CT) Z = RS + 3RL +3 ZR Z = RS + 2RL +2 ZR

Z is the effective impedance seen by the CTRS is the CT secondary winding resistance and CT lead resistance; also

includes any relay impedance that is inside the delta connection (Ω)RL is the circuit one-way lead resistance (Ω)ZR is the relay impedance in the CT secondary current path (Ω)

IEEE typical X/R Ratios

Parameters for X/R & SCC Example Given ParametersSCC = 7316A @ V=240kVX/R = 12Transformer size = 50MVA; 240/138kVZxmfr = 8.0%

Solve for X/R at 138kV.Solve for SCC at 138kV.

Example for X/R

%644.110009.204,041,3/000,50% xkVAkVAZ source

Step #1: Calculate the equivalent source impedance

SCCkVKVA LLitshortcircu 3

100/% xKVAKVAZ itshortcircurtransformesource

kVAAkVKVA itshortcircu 09.204,041,373162403

Example for X/RStep #2: Source R and X

644.1% sourceZ12/ RX

24.8512tan 1

%638.1%)644.1(*)24.85( SinX

%138.0%)644.1(*)18.85( CosR

RX1tan ZSinX *

ZCosR *

Example for X/RStep #3: Transformer R and X

0.8% xmfrZ15/ RX

19.8615tan 1

%982.7%)0.8(*)19.86( SinX

%531.0%)0.8(*)19.86( CosR

(From IEEE typical value)

Example for X/RStep #4: Total Resistance and Reactance at 138kV

%669.0%531.0%138.0 xmfrsourcetotal RRR

%62.9%982.7%638.1 xmfrsourcetotal XXX

Step #5: X/R at138kV

4.14%669.0%622.9

total

total

RX

RX

Example for SCC

%644.110009.204,041,3/000,50% xkVAkVAZ source

Step #1: Calculate the equivalent source impedance

SCCkVKVA LLitshortcircu 3

100/% xKVAKVAZ itshortcircurtransformesource

kVAAkVKVA itshortcircu 09.204,041,373162403

Example for SCCStep #2: Calculate the 138kV side full load current

LL

xmfrLV kV

kVAFLA

3

AkVkVAFLALV 18.209

)138(3000,50

Example for SCCStep #3: Calculate the short circuit current on the 138kV side

)%(%100

sourcexmfr

LVLV ZZ

FLASCC

AASCCLV 06.169,2%)644.1%0.8(

10018.209

Global Example

C200400/5 (full ratio)IF = 7,316A (HV side)X/R = 15 (HV side faults)One-way RL length = 152m (500’)Determine what CT Class and Ratio is validon the 138kV

240kV 138kV

Global Example

4.14%669.0%622.9

total

total

RX

RX

Case #1) From previous slide for X/R and SCC on the 138kV side.

Rone-way = 1Ω/1000’x500’ = 0.5Ω

For a C200: ZSTD = 2.0Ω

ZB = RS + 2RL + ZR

ZB = 0.2Ω + 2(0.5Ω) + 0.1Ω=1.3

Note: 0.1Ω = digital & 1.0 Ω for electromechanical

puZZZSTD

Bb 7.0

0.23.1

Global Example

AASCCLV 06.169,2%)644.1%0.8(

10018.209

?201

bf ZIR

X

From previous slide for X/R and SCC on the 138kV side.

puAA

IIIN

Ff 5.5

400169,2

607.05.54.141 CT Saturates

Global Example

RX

K

TTs

s

1

1ln1

sRXT 038.0

3774.14

1

33.63.35

230

VV

VVK

s

Xs

VVx 230

VAZIV BFS 3.35)3.1(5/400

2169

Taken from excitation Curve at IE = 10A (full ratio)

)1(19

15.8

1ln038.0 cyclemsRXTs

Global ExampleCase #2 Choosing C400 (2000/5 tapped at 600/5).

puZZZSTD

Bb 18.1

2.141.1

2.142000600__ BurdenStdEffective

puAA

IIIN

Ff 62.3

600169,2

?201

bf ZIR

X

6618.162.34.141 CT Saturates

ZB = 1.4Ω due to 600/5 ratio CT winding resistance.

Global Example

RX

K

TTs

s

1

1ln1

sRXT 038.0

3774.14

1

8.55.25

150

VV

VVK

s

Xs

VVx 150

VAZIV BFS 5.25)41.1(5/600

2169

Taken from exication Curves and at IE = 10A (600/5)

cyclemsRXTs 111

18.5

1ln038.0

Global ExampleCase #3 Choosing C400 (2000/5 full ratio).

puZZZSTD

Bb 525.0

0.41.2

puAA

IIIN

Ff 08.1

2000169,2

?201

bf ZIR

X

73.8525.008.14.141 CT OK

For a C400: ZSTD = 4.0Ω

ZB = 2.1Ω due to 2000/5 ratio CT winding resistance.

Global ExampleCase #4 Choosing C400 (2000/5 full ratio with 80% remenant flux).

?4%)801(*201

bf ZIR

X

73.8525.008.14.141 CT Saturates

Case #4 Choosing C400 (2000/5 full ratio with 50% remenant flux).

?10%)501(*201

bf ZIR

X

73.8525.008.14.141 CT OK

IEEE PSRC CALCULATOR

Vs

a m p s rm s

slope= 1/S

lo g - lo g p lo t ,e q u a l

d e c a d es p a c in g

vo l tsrm s

I e

mf gr 'sd a taV e

1 0

Saturation Curve

iee VIS

V loglog1log

The Saturation Curve above is represented by the following equation

Where Vi is the value of Ve for Ie=1, that is for (log Ie=0)

Measuring the ‘S’

0

1

2

3

4

5

6

7

8

9

1

0

slope = 0.7/9.3 = 1/S

S = 9.3/0.7 = 13

Ruler, inches or centimeters

Use a ruler to determine the slope.Do NOT read currents or voltages for this purpose.(We assume equal decade spacing of the axes)

HOW TO MEASURE THE SLOPE

IEEE PSRC CALCULATOR EXAMPLECase #1) From previous slide for X/R and SCC on the 138kV side.

EMTP/PSCAD SIMULATION• The CTs are subject to saturation during DC transient

of fault current , hence there is growing interest in simulations like EMTP/PSCAD.

• The IEEE-PSRC CT calculator is only accurate at high fault currents and burden , hence to get better idea of CT performance at low fault currents/ burden we need to model in PSCAD/EMTP.

• The primary aim of PSCAD simulation is to obtain digitized records, which can be recognized as secondary analog signal using D/A conversion and amplification, for the purpose of relay testing.

EMTP/PSCAD SIMULATION

PSCAD -CT Modeling

QUESTIONS?

49666442-ct-sizing

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