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This document provides information about current transformers (CTs) including their function, construction, standards, ratings, errors, and types. CTs are used to reduce high power system currents to lower values that can be measured by instrumentation. They provide insulation between the primary and secondary circuits and allow the use of standard current ratings for secondary equipment. The performance of protective relays depends on the CT that drives it. The document discusses various CT constructions, standards, magnetization characteristics, saturation effects, and ratings parameters like rated burden, continuous and short time rated currents. It also defines current and phase errors that can occur in CTs.

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GRID Technical Institute This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.
GRID Technical Institute This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved. Current Transformers
s 3 Current Transformer Function  Reduce power system current to lower value for measurement.  Insulate secondary circuits from the primary.  Permit the use of standard current ratings for secondary equipment. REMEMBER : The relay performance DEPENDS on the C.T which drives it !
s 4 Instrument Transformer Standards IEC IEC 185:1987 CTs IEC 44-6:1992 CTs IEC 186:1987 VTs EUROPEAN BS 7625 VTs BS 7626 CTs BS 7628 CT+VT BRITISH BS 3938:1973 CTs BS 3941:1975 VTs AMERICAN ANSI C51.13.1978 CTs and VTs CANADIAN CSA CAN3-C13-M83 CTs and VTs AUSTRALIAN AS 1675-1986 CTs
s 5 Current Transformer Construction BAR PRIMARY WOUND PRIMARY Primary Secondary
s 6 Bar Primary Type Current Transformer(s) Primary Conductor Ring Type Current Transformer Primary Insulation Core Secondary Winding
s 7 Typical Protection Bar-Primary Current Transformer RELAY 1A ? 1000A ? 1000 turns sec. ? Insulation covered wire, giving inter-turn insulation & secondary to core insulation Generator, or system voltage source ‘Feeder’ or ‘Bus-bar’ forming 1 turn of primary circuit Insulation to stop flash-over from HV primary to core & secondary circuit Laminated ‘strip’ wound steel toroidal core
s 8 Polarity P2 P1 Is S2 S1 Ip Inst. Current directions :- P1  P2 S1  S2 Externally
s 9 Differential Protection (1) P2 P1 IFP1 IFS IFS1 S2 S1 P1 P2 S1 S2 R IFP2 IFS2 Operation for Internal Faults
s 10 Differential Protection (2) P2 P1 IFP IFS S2 S1 P1 P2 S1 S2 R IFS Stability for External Faults
s 11 Differential Protection (3) IFP IFS S2 S1 S1 S2 R Maloperation due to Incorrect Connections IFP IFS 2IFS
s 12 Basic Theory
s 13 Basic Theory (1) IP IS R 1 Primary Turn N Secondary Turns For an ideal transformer :- PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS  IP = N x IS
s 14 Basic Theory (2) IP IS R ES For IS to flow through R there must be some potential - ES = the E.M.F. ES = IS x R ES is produced by an alternating flux in the core. ES  dØ dt
s 15 Basic Theory (3) IP NP NS EK IS ZB ZCT VO/P = ISZB = EK - ISZCT
s 16 Basic Formulae Circuit Voltage Required : ES = IS (ZB + ZCT + ZL) Volts where :- IS = Secondary Current of C.T. (Amperes) ZB = Connected External Burden (Ohms) ZCT = C.T Winding Impedance (Ohms) ZL = Lead Loop Resistance (Ohms) Require EK > ES
s 17 Basic Formulae (1) Maximum Secondary Winding Voltage : EK = 4.44 x B A f N Volts …. 1 where :- EK = Secondary Induced Volts (knee-point voltage) B = Flux Density (Tesla) A = Core Cross-sectional Area (square metres) f = System Frequency (Hertz) N = Number of Turns
s 18 Simple Selection Example Example Calculation : C.T. Ratio = 2000 / 5A Max Flux Density = 1.6 T RS = 0.31 Ohms Core C.S.A = 20 cm2 IMAX Primary = 40 kA Find maximum secondary burden permissible if no saturation is to occur. Solution : N = 2000 / 5 = 400 Turns IS MAX = 40,000 / 400 = 100 Amps From equation 1 the knee point voltage is :- VK = (4.44 x 1.6 x 20 x 50 x 400) = 284 Volts 104 Therefore Maximum Burden = 284 / 100 = 2.84 Ohms Hence Maximum CONNECTED burden : 2.84 - 0.31 = 2.53 Ohms
s 19 Knee Point Material : Cross Flux Density (B) Tesla (Wb/m2) 0 0.02 0.04 0.06 0.08 0.1 0.12 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 Magnetizing Force (H) Ampere-Turns / mm (Magnetic Field Strength : H, A/m) Mag Curves
s 20 ES (Secondary E.M.F.) (Magnetising Current) Ie ES = 4.44 N f A B = Kv B where :- Kv = 4.44 N f A (B in Wb/m2 ; A in m2) Ie = H . L = Ki . H N where :- Ki = L/N L = mean magnetic path in metres H = amp. Turns / metre Ie = Amps Mag Curves Cont...
s 21 Multiply By K V to obtain Volts Multiply By Ki to obtain Amps B H KV x B = Volts (E) Ki x H = Amps Units : KV = E = 4.44 N f A (m2 x turns) B Ki = Ie = L (m / turns) H N Note : L = Mean Magnetic Path Mean Magnetic Path = 2r r = ro - ri + ri 2 ro ri r Mag Curves Cont...
s 22 Magnetisation Characteristic (1) C.T Ratio 100/5 A connected to CDG11. Relay Burden = 2 VA (at 10% rated I) Problem :- Calculate the necessary values of Kv and Ki to provide the necessary output at ten times setting (Assume maximum flux density before saturation = 1.6 Tesla)
s 23 Magnetisation Characteristic (2) (i) Bar Type Primary CDG current setting = 10% = 0.5 A Volts required to operate relay = 2 / 0.5 = 4 V Therefore voltage required at 10 times setting = 10 x 4 = 40 V Hence for a flux density of 1.6T the C.T. should have an output of at least 40 V. With bar primary, number of turns = 20. Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 20 x 50 A = 40 / 0.71 = 56.3 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 61.2 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 10.2cm KV = AN / 45 = (56.3 x 20) / 45 = 25 Ki = L / N = 24TT / 20 = 3.77 cm / turn
s 24 Magnetisation Characteristic (3) (ii) Wound Type Primary Assume primary has 5 turns - Therefore secondary has 100 turns Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 100 x 50 A = 40 / 3.55 = 11.26 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 12.24 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 2.04cm KV = AN / 45 = (11.26 x 100) / 45 = 25 Ki = L / N = 24TT / 100 = 0.754 cm/turn
s 25 Low Reactance Design With evenly distributed winding the leakage reactance is very low and usually ignored. Thus ZCT ~ RCT
s 26 Knee-Point Voltage Definition Iek +10% Vk + 50 % Iek Vk Exciting Current (Ie) Exciting Voltage (V S )
s 27 C.T. Equivalent Circuit Zb N P1 ZCT S1 Vt Ze Is Ip Ip/N Ie Es Ip = Primary rating of C.T. Ie = Secondary excitation current N = C.T. ratio Is = Secondary current Zb = Burden of relays in ohms Es = Secondary excitation voltage (r+jx) Vt = Secondary terminal voltage ZCT = C.T. secondary winding across the C.T. terminals impedance in ohms (r+jx) Ze = Secondary excitation impedance in ohms (r+jx)
s 28 Phasor Diagram Ic Ep Ip/N Ie Is Ie Im Es  Ep = Primary voltage Im = Magnetising current Es = Secondary voltage Ie = Excitation current  = Flux Ip = Primary current Ic = Iron losses (hysteresis & Is = Secondary current eddy currents)
s 29 Saturation
s 30 Steady State Saturation (1) 100A 100/1 1A E 1 ohm E = 1V 100A 100/1 1A E 10 ohm E = 10V 100A 100/1 1A E 100 ohm E = 100V 100A 100/1 1A E 1000 ohm E = ?
s 31 Steady State Saturation (2) Time +V -V 0V A  Assume :- Zero residual flux Switch on at point A
s 32 Steady State Saturation (3) Time +V -V 0V C  Assume :- Zero residual flux Switch on at point C
s 33 Steady State Saturation (4) Time +V -V 0V B  Assume :- Zero residual flux Switch on at point B
s 34 Steady State Saturation (5) Mag Core Saturation +V -V 0V Mag Core Saturation Time
s 35 Steady State Saturation (6)  Time +V -V 0V +V -V 0V Mag Core Saturation Mag Core Saturation Output lost due to steady state saturation Actual Output Voltage Prospective Output Voltage
s 36 Transient Saturation L1 R1 Z1 i1 v = VM sin (wt + ) v = VM sin (wt + ) TRANSIENT STATE STEADY e . ) - ( sin - ) - (wt sin e . ) - ( sin Z V ) - (wt sin Z V i 1 L / t 1 R - 1 1 1 L / t 1 R - 1 M 1 M 1                    ˆ ˆ 1 M 1 1 1 1 - 1 2 1 1 Z V R wL tan L w R Z - : where       ˆ 2 2
s 37 Transient Saturation : Resistive Burden FLUX CURRENT Primary Current Secondary Current Actual Flux Mag Current Required Flux ØSAT 0 0 10 20 30 40 50 60 70 80 M
s 38 CT Types
s 39 Current Transformer Function Two basic groups of C.T.  Measurement C.T.s  Limits well defined  Protection C.T.s  Operation over wide range of currents Note : They have DIFFERENT characteristics
s 40 Measuring C.T.s B Protection C.T. Measuring C.T. H Measuring C.T.s  Require good accuracy up to approx 120% rated current.  Require low saturation level to protect instruments, thus use nickel iron alloy core with low exciting current and knee point at low flux density. Protection C.T.s  Accuracy not as important as above.  Require accuracy up to many times rated current, thus use grain orientated silicon steel with high saturation flux density.
s 41 CT Errors
s 42 Current Transformer Errors ZS Zb Ze Is Ip Ie Es Ip/N Ie Iq Ie Es  Ep Ic’ Ip/N Ie Is Is
s 43 Current Transformer Error Transformer Error vs Primary Current Error Rated Current Primary Current Phase Error Current Error Errors can be reduced by :- 1. Using better quality magnetic material 2. Shortening the mean magnetic path 3. Reducing the flux density in the core
s 44 Current Transformer Errors Current Error Definition. Error in magnitude of the secondary current, expressed as a percentage, given by :- Current error % = 100 (kn IS - Ip) Ip kn = Rated Transmission Ratio Ip = Actual Primary Current Is = Actual Secondary Current Current Error is :- +ve : When secondary current is HIGHER than the rated nominal value. -ve : When secondary current is LOWER than the rated nominal value.
s 45 Current Transformer Errors Phase Error Definition. The displacement in phase between the primary and secondary current vectors, the direction of the vectors being chosen so the angle is zero for a perfect transformer. Phase Error is :- +ve : When secondary current vector LEADS the primary current vector. -ve : When secondary current vector LAGS the primary current vector.
s 46 Current Transformer Ratings
s 47 Current Transformer Ratings (1) Rated Burden  Value of burden upon which accuracy claims are based  Usually expressed in VA  Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA Continuous Rated Current  Usually rated primary current Short Time Rated Current  Usually specified for 0.5, 1, 2 or 3 secs  No harmful effects  Usually specified with the secondary shorted Rated Secondary Current  Commonly 1, 2 or 5 Amps
s 48 Current Transformer Ratings (2) Rated Dynamic Current Ratio of :- IPEAK : IRATED (IPEAK = Maximum current C.T. can withstand without suffering any damage). Accuracy Limit Factor - A.L.F. (or Saturation Factor) Ratio of :- IPRIMARY : IRATED up to which the C.T. rated accuracy is maintained. e.g. 200 / 1A C.T. with an A.L.F. = 5 will maintain its accuracy for IPRIMARY < 5 x 200 = 1000 Amps
s 49 Choice of Ratio Clearly, the primary rating IP  normal current in the circuit if thermal (continuous) rating is not to be exceeded. Secondary rating is usually 1 or 5 Amps (0.5 and 2 Amp are also used). If secondary wiring route length is greater than 30 metres - 1 Amp secondaries are preferable. A practical maximum ratio is 3000 / 1. If larger primary ratings are required (e.g. for large generators), can use 20 Amp secondary together with interposing C.T. e.g. 5000 / 20 - 20 / 1
s 50 Current Transformer Designation Class “P” Specified in terms of :- i) Rated burden ii) Class (5P or 10P) iii) Accuracy limit factor (A.L.F.) Example :- 15 VA 10P 20 To convert VA and A.L.F. into useful volts Vuseful  VA x ALF IN
s 51 BS 3938 Classes :- 5P, 10P. ‘X’ Designation (Classes 5P, 10P) (Rated VA) (Class) (ALF) Multiple of rated current (IN) up to which declared accuracy will be maintained with rated burden connected. 5P or 10P. Value of burden in VA on which accuracy claims are based. (Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA) ZB = rated burden in ohms = Rated VA IN 2
s 52 Protection Current Transformers Table 3 - Limits of Error for Accuracy Class 5P and 10P Phase Displacement at rated primary current Accuracy Class Current Error at rated primary current (%) Minutes Centiradians Composite Error (%) at rated accuracy limit primary current 5P 10P 1 3 60 1.8 5 10
s 53 Interposing CT
s 54 Interposing CT NP NS ZB ZCT LINE CT Burden presented to line CT = ZCT + ZB x NP 2 NS 2
s 55 ‘Seen’ by main ct :- 0.1 + (1)2 (2 x 0.5 + 0.4 + 1) = 0.196 (5) Burden on main ct :- I2 R = 25 x 0.196 = 4.9VA Burden on a main ct of required ratio :- Total connected burden = 2 x 0.5 + 1 = 2 Connected VA = I2 R = 2  The I/P ct consumption was about 3VA. NEG. 1A 0.1 0.1  0.4  1VA @ 1A  1.0 R 0.5  5A 500/5 0.5  R 1.0  500/1
s 56 Current Transformer Designation
s 57 Current Transformer Designation Class “X” Specified in terms of :- i) Rated Primary Current ii) Turns Ratio (max. error = 0.25%) iii) Knee Point Voltage iv) Mag Current (at specified voltage) v) Secondary Resistance (at 75°C)
s 58 Choice of Current Transformer  Instantaneous Overcurrent Relays  Class P Specification  A.L.F. = 5 usually sufficient  For high settings (5 - 15 times C.T rating) A.L.F. = relay setting  IDMT Overcurrent Relays  Generally Class 10P  Class 5P where grading is critical Note : A.L.F. X V.A < 150  Differential Protection  Class X Specification  Protection relies on balanced C.T output
s 59 Selection Example
s 60 Burden on Current Transformers RCT RCT RCT IF IF RL RL RL RL Rr Rr Rr Rr RCT RCT RCT IF IF RL RL RL RL Rr Rr Rr Rr 1. Overcurrent : RCT + RL + Rr 2. Earth : RCT + 2RL + 2Rr
s 61 Overcurrent Relay VK Check Assume values : If max = 7226 A RCT = 0.26  C.T = 1000 / 5 A Rr = 0.02  7.5 VA 10P 20 RL = 0.15  Check to see if VK is large enough : Required voltage = VS = IF (RCT + Rr + RL) = 7226 x 5 (0.26 + 0.02 + 0.15) = 36.13 x 0.43 = 15.54 Volts 1000 Current transformer VK approximates to :- VK ~ VA x ALF + RCT x IN x ALF In = 7.5 x 20 + 0.26 x 5 x 20 = 56 Volts 5 VK > VS therefore C.T VK is adequate
s 62 Earth Fault Relay VK Check Assume values : As per overcurrent. Note For earth fault applications require to be able to pass 10 x relay setting. Check to see if VK is large enough : VK = 56 Volts Total load connected = 2RL + RCT + 2Rr = 2 x 0.15 + 0.26 + 2 x 0.02  Maximum secondary current = 56 = 93.33A 0.6 Typical earth fault setting = 30% IN = 1.5A Therefore C.T can provide > 60 x setting C.T VK is adequate
s 63 Voltage Transformers
s 64 Voltage Transformers  Provides isolation from high voltages  Must operate in the linear region to prevent accuracy problems - Do not over specify VT  Must be capable of driving the burden, specified by relay manufacturer  Protection class VT will suffice
s 65 Typical Working Points on a B-H Curve Flux Density ‘B’ Magnetising Force AT/m Cross C.T.’s & Power Transformers Saturation V.T.’s Protection C.T. (at full load) ‘H’ 1000 2000 3000 1.6 1.0 0.5 Tesla
s 66 Types of Voltage Transformers Two main basic types are available:  Electromechanical VT`s  Similar to a power transformer  May not be economical above 132kV  Capacitor VT`s (CVT)  Used at high voltages  Main difference is that CVT has a capacitor divider on the front end.
s 67 Electromagnetic Voltage Transformer ZB (burden) NP / NS = Kn VP RP IP LP LM IM IC Re VS EP = ES Ie RS IS LS
s 68 Basic Circuit of a Capacitor V.T. C1 C2 VP VC2 Vi L T VS ZB
s 69 Ferro-resonance  The exciting impedance of auxiliary transformer T and the capacitance of the potential divider form a resonant circuit.  May oscillate at a sub normal frequency  Resonant frequency close to one-third value of system frequency  Manifests itself as a rise in output voltage, r.m.s. value being 25 to 50 per cent above normal value  Use resistive burden to dampen the effect
s 70 VT Earthing  Primary Earthing  Earth at neutral point  Required for phase-ground measurement at relay  Secondary Earthing  Required for safety  Earth at neutral point  When no neutral available - earth yellow phase (VERY COMMON)  No relevance for protection operation
s 71 VT Construction  5 Limb  Used when zero sequence measurement is required (primary must also be earthed)  Three Single Phase  Used when zero sequence measurement is required (primary must also be earthed)  3 Limb  Used where no zero sequence measurement is required  V Connected (Open Delta)  No yellow phase  Cost effective  Two phase-phase voltages  No ground fault measurement
s 72 VT Connections a b c a b c A B C N da dn a b c n V Connected Broken Delta
s 73 VT Construction - Residual  Used to detect earthfault  Useful where current operated protection cannot be used  Connect all secondary windings in series  Sometimes referred to as `Broken Delta`  Residual Voltage is 3 times zero sequence voltage  VT must be 5 Limb or 3 single phase units  Primary winding must be earthed
s 74 Voltage Factors Vf  Vf is the upper limit of operating voltage.  Important for correct relay operation.  Earthfaults cause displacement of system neutral, particularly in the case of unearthed or impedance earthed systems.
s 75 Protection of VT’s  H.R.C. Fuses on primary side  Fuses may not have sufficient interrupting capability  Use MCB
GRID Technical Institute This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.

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Current-Transformer-ppt.ppt

  • 1. GRID Technical Institute This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.
  • 2. GRID Technical Institute This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved. Current Transformers
  • 3. s 3 Current Transformer Function  Reduce power system current to lower value for measurement.  Insulate secondary circuits from the primary.  Permit the use of standard current ratings for secondary equipment. REMEMBER : The relay performance DEPENDS on the C.T which drives it !
  • 4. s 4 Instrument Transformer Standards IEC IEC 185:1987 CTs IEC 44-6:1992 CTs IEC 186:1987 VTs EUROPEAN BS 7625 VTs BS 7626 CTs BS 7628 CT+VT BRITISH BS 3938:1973 CTs BS 3941:1975 VTs AMERICAN ANSI C51.13.1978 CTs and VTs CANADIAN CSA CAN3-C13-M83 CTs and VTs AUSTRALIAN AS 1675-1986 CTs
  • 5. s 5 Current Transformer Construction BAR PRIMARY WOUND PRIMARY Primary Secondary
  • 6. s 6 Bar Primary Type Current Transformer(s) Primary Conductor Ring Type Current Transformer Primary Insulation Core Secondary Winding
  • 7. s 7 Typical Protection Bar-Primary Current Transformer RELAY 1A ? 1000A ? 1000 turns sec. ? Insulation covered wire, giving inter-turn insulation & secondary to core insulation Generator, or system voltage source ‘Feeder’ or ‘Bus-bar’ forming 1 turn of primary circuit Insulation to stop flash-over from HV primary to core & secondary circuit Laminated ‘strip’ wound steel toroidal core
  • 9. s 9 Differential Protection (1) P2 P1 IFP1 IFS IFS1 S2 S1 P1 P2 S1 S2 R IFP2 IFS2 Operation for Internal Faults
  • 10. s 10 Differential Protection (2) P2 P1 IFP IFS S2 S1 P1 P2 S1 S2 R IFS Stability for External Faults
  • 11. s 11 Differential Protection (3) IFP IFS S2 S1 S1 S2 R Maloperation due to Incorrect Connections IFP IFS 2IFS
  • 13. s 13 Basic Theory (1) IP IS R 1 Primary Turn N Secondary Turns For an ideal transformer :- PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS  IP = N x IS
  • 14. s 14 Basic Theory (2) IP IS R ES For IS to flow through R there must be some potential - ES = the E.M.F. ES = IS x R ES is produced by an alternating flux in the core. ES  dØ dt
  • 16. s 16 Basic Formulae Circuit Voltage Required : ES = IS (ZB + ZCT + ZL) Volts where :- IS = Secondary Current of C.T. (Amperes) ZB = Connected External Burden (Ohms) ZCT = C.T Winding Impedance (Ohms) ZL = Lead Loop Resistance (Ohms) Require EK > ES
  • 17. s 17 Basic Formulae (1) Maximum Secondary Winding Voltage : EK = 4.44 x B A f N Volts …. 1 where :- EK = Secondary Induced Volts (knee-point voltage) B = Flux Density (Tesla) A = Core Cross-sectional Area (square metres) f = System Frequency (Hertz) N = Number of Turns
  • 18. s 18 Simple Selection Example Example Calculation : C.T. Ratio = 2000 / 5A Max Flux Density = 1.6 T RS = 0.31 Ohms Core C.S.A = 20 cm2 IMAX Primary = 40 kA Find maximum secondary burden permissible if no saturation is to occur. Solution : N = 2000 / 5 = 400 Turns IS MAX = 40,000 / 400 = 100 Amps From equation 1 the knee point voltage is :- VK = (4.44 x 1.6 x 20 x 50 x 400) = 284 Volts 104 Therefore Maximum Burden = 284 / 100 = 2.84 Ohms Hence Maximum CONNECTED burden : 2.84 - 0.31 = 2.53 Ohms
  • 19. s 19 Knee Point Material : Cross Flux Density (B) Tesla (Wb/m2) 0 0.02 0.04 0.06 0.08 0.1 0.12 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 Magnetizing Force (H) Ampere-Turns / mm (Magnetic Field Strength : H, A/m) Mag Curves
  • 20. s 20 ES (Secondary E.M.F.) (Magnetising Current) Ie ES = 4.44 N f A B = Kv B where :- Kv = 4.44 N f A (B in Wb/m2 ; A in m2) Ie = H . L = Ki . H N where :- Ki = L/N L = mean magnetic path in metres H = amp. Turns / metre Ie = Amps Mag Curves Cont...
  • 21. s 21 Multiply By K V to obtain Volts Multiply By Ki to obtain Amps B H KV x B = Volts (E) Ki x H = Amps Units : KV = E = 4.44 N f A (m2 x turns) B Ki = Ie = L (m / turns) H N Note : L = Mean Magnetic Path Mean Magnetic Path = 2r r = ro - ri + ri 2 ro ri r Mag Curves Cont...
  • 22. s 22 Magnetisation Characteristic (1) C.T Ratio 100/5 A connected to CDG11. Relay Burden = 2 VA (at 10% rated I) Problem :- Calculate the necessary values of Kv and Ki to provide the necessary output at ten times setting (Assume maximum flux density before saturation = 1.6 Tesla)
  • 23. s 23 Magnetisation Characteristic (2) (i) Bar Type Primary CDG current setting = 10% = 0.5 A Volts required to operate relay = 2 / 0.5 = 4 V Therefore voltage required at 10 times setting = 10 x 4 = 40 V Hence for a flux density of 1.6T the C.T. should have an output of at least 40 V. With bar primary, number of turns = 20. Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 20 x 50 A = 40 / 0.71 = 56.3 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 61.2 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 10.2cm KV = AN / 45 = (56.3 x 20) / 45 = 25 Ki = L / N = 24TT / 20 = 3.77 cm / turn
  • 24. s 24 Magnetisation Characteristic (3) (ii) Wound Type Primary Assume primary has 5 turns - Therefore secondary has 100 turns Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 100 x 50 A = 40 / 3.55 = 11.26 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 12.24 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 2.04cm KV = AN / 45 = (11.26 x 100) / 45 = 25 Ki = L / N = 24TT / 100 = 0.754 cm/turn
  • 25. s 25 Low Reactance Design With evenly distributed winding the leakage reactance is very low and usually ignored. Thus ZCT ~ RCT
  • 26. s 26 Knee-Point Voltage Definition Iek +10% Vk + 50 % Iek Vk Exciting Current (Ie) Exciting Voltage (V S )
  • 27. s 27 C.T. Equivalent Circuit Zb N P1 ZCT S1 Vt Ze Is Ip Ip/N Ie Es Ip = Primary rating of C.T. Ie = Secondary excitation current N = C.T. ratio Is = Secondary current Zb = Burden of relays in ohms Es = Secondary excitation voltage (r+jx) Vt = Secondary terminal voltage ZCT = C.T. secondary winding across the C.T. terminals impedance in ohms (r+jx) Ze = Secondary excitation impedance in ohms (r+jx)
  • 28. s 28 Phasor Diagram Ic Ep Ip/N Ie Is Ie Im Es  Ep = Primary voltage Im = Magnetising current Es = Secondary voltage Ie = Excitation current  = Flux Ip = Primary current Ic = Iron losses (hysteresis & Is = Secondary current eddy currents)
  • 30. s 30 Steady State Saturation (1) 100A 100/1 1A E 1 ohm E = 1V 100A 100/1 1A E 10 ohm E = 10V 100A 100/1 1A E 100 ohm E = 100V 100A 100/1 1A E 1000 ohm E = ?
  • 31. s 31 Steady State Saturation (2) Time +V -V 0V A  Assume :- Zero residual flux Switch on at point A
  • 32. s 32 Steady State Saturation (3) Time +V -V 0V C  Assume :- Zero residual flux Switch on at point C
  • 33. s 33 Steady State Saturation (4) Time +V -V 0V B  Assume :- Zero residual flux Switch on at point B
  • 34. s 34 Steady State Saturation (5) Mag Core Saturation +V -V 0V Mag Core Saturation Time
  • 35. s 35 Steady State Saturation (6)  Time +V -V 0V +V -V 0V Mag Core Saturation Mag Core Saturation Output lost due to steady state saturation Actual Output Voltage Prospective Output Voltage
  • 36. s 36 Transient Saturation L1 R1 Z1 i1 v = VM sin (wt + ) v = VM sin (wt + ) TRANSIENT STATE STEADY e . ) - ( sin - ) - (wt sin e . ) - ( sin Z V ) - (wt sin Z V i 1 L / t 1 R - 1 1 1 L / t 1 R - 1 M 1 M 1                    ˆ ˆ 1 M 1 1 1 1 - 1 2 1 1 Z V R wL tan L w R Z - : where       ˆ 2 2
  • 37. s 37 Transient Saturation : Resistive Burden FLUX CURRENT Primary Current Secondary Current Actual Flux Mag Current Required Flux ØSAT 0 0 10 20 30 40 50 60 70 80 M
  • 39. s 39 Current Transformer Function Two basic groups of C.T.  Measurement C.T.s  Limits well defined  Protection C.T.s  Operation over wide range of currents Note : They have DIFFERENT characteristics
  • 40. s 40 Measuring C.T.s B Protection C.T. Measuring C.T. H Measuring C.T.s  Require good accuracy up to approx 120% rated current.  Require low saturation level to protect instruments, thus use nickel iron alloy core with low exciting current and knee point at low flux density. Protection C.T.s  Accuracy not as important as above.  Require accuracy up to many times rated current, thus use grain orientated silicon steel with high saturation flux density.
  • 43. s 43 Current Transformer Error Transformer Error vs Primary Current Error Rated Current Primary Current Phase Error Current Error Errors can be reduced by :- 1. Using better quality magnetic material 2. Shortening the mean magnetic path 3. Reducing the flux density in the core
  • 44. s 44 Current Transformer Errors Current Error Definition. Error in magnitude of the secondary current, expressed as a percentage, given by :- Current error % = 100 (kn IS - Ip) Ip kn = Rated Transmission Ratio Ip = Actual Primary Current Is = Actual Secondary Current Current Error is :- +ve : When secondary current is HIGHER than the rated nominal value. -ve : When secondary current is LOWER than the rated nominal value.
  • 45. s 45 Current Transformer Errors Phase Error Definition. The displacement in phase between the primary and secondary current vectors, the direction of the vectors being chosen so the angle is zero for a perfect transformer. Phase Error is :- +ve : When secondary current vector LEADS the primary current vector. -ve : When secondary current vector LAGS the primary current vector.
  • 47. s 47 Current Transformer Ratings (1) Rated Burden  Value of burden upon which accuracy claims are based  Usually expressed in VA  Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA Continuous Rated Current  Usually rated primary current Short Time Rated Current  Usually specified for 0.5, 1, 2 or 3 secs  No harmful effects  Usually specified with the secondary shorted Rated Secondary Current  Commonly 1, 2 or 5 Amps
  • 48. s 48 Current Transformer Ratings (2) Rated Dynamic Current Ratio of :- IPEAK : IRATED (IPEAK = Maximum current C.T. can withstand without suffering any damage). Accuracy Limit Factor - A.L.F. (or Saturation Factor) Ratio of :- IPRIMARY : IRATED up to which the C.T. rated accuracy is maintained. e.g. 200 / 1A C.T. with an A.L.F. = 5 will maintain its accuracy for IPRIMARY < 5 x 200 = 1000 Amps
  • 49. s 49 Choice of Ratio Clearly, the primary rating IP  normal current in the circuit if thermal (continuous) rating is not to be exceeded. Secondary rating is usually 1 or 5 Amps (0.5 and 2 Amp are also used). If secondary wiring route length is greater than 30 metres - 1 Amp secondaries are preferable. A practical maximum ratio is 3000 / 1. If larger primary ratings are required (e.g. for large generators), can use 20 Amp secondary together with interposing C.T. e.g. 5000 / 20 - 20 / 1
  • 50. s 50 Current Transformer Designation Class “P” Specified in terms of :- i) Rated burden ii) Class (5P or 10P) iii) Accuracy limit factor (A.L.F.) Example :- 15 VA 10P 20 To convert VA and A.L.F. into useful volts Vuseful  VA x ALF IN
  • 51. s 51 BS 3938 Classes :- 5P, 10P. ‘X’ Designation (Classes 5P, 10P) (Rated VA) (Class) (ALF) Multiple of rated current (IN) up to which declared accuracy will be maintained with rated burden connected. 5P or 10P. Value of burden in VA on which accuracy claims are based. (Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA) ZB = rated burden in ohms = Rated VA IN 2
  • 52. s 52 Protection Current Transformers Table 3 - Limits of Error for Accuracy Class 5P and 10P Phase Displacement at rated primary current Accuracy Class Current Error at rated primary current (%) Minutes Centiradians Composite Error (%) at rated accuracy limit primary current 5P 10P 1 3 60 1.8 5 10
  • 54. s 54 Interposing CT NP NS ZB ZCT LINE CT Burden presented to line CT = ZCT + ZB x NP 2 NS 2
  • 55. s 55 ‘Seen’ by main ct :- 0.1 + (1)2 (2 x 0.5 + 0.4 + 1) = 0.196 (5) Burden on main ct :- I2 R = 25 x 0.196 = 4.9VA Burden on a main ct of required ratio :- Total connected burden = 2 x 0.5 + 1 = 2 Connected VA = I2 R = 2  The I/P ct consumption was about 3VA. NEG. 1A 0.1 0.1  0.4  1VA @ 1A  1.0 R 0.5  5A 500/5 0.5  R 1.0  500/1
  • 57. s 57 Current Transformer Designation Class “X” Specified in terms of :- i) Rated Primary Current ii) Turns Ratio (max. error = 0.25%) iii) Knee Point Voltage iv) Mag Current (at specified voltage) v) Secondary Resistance (at 75°C)
  • 58. s 58 Choice of Current Transformer  Instantaneous Overcurrent Relays  Class P Specification  A.L.F. = 5 usually sufficient  For high settings (5 - 15 times C.T rating) A.L.F. = relay setting  IDMT Overcurrent Relays  Generally Class 10P  Class 5P where grading is critical Note : A.L.F. X V.A < 150  Differential Protection  Class X Specification  Protection relies on balanced C.T output
  • 60. s 60 Burden on Current Transformers RCT RCT RCT IF IF RL RL RL RL Rr Rr Rr Rr RCT RCT RCT IF IF RL RL RL RL Rr Rr Rr Rr 1. Overcurrent : RCT + RL + Rr 2. Earth : RCT + 2RL + 2Rr
  • 61. s 61 Overcurrent Relay VK Check Assume values : If max = 7226 A RCT = 0.26  C.T = 1000 / 5 A Rr = 0.02  7.5 VA 10P 20 RL = 0.15  Check to see if VK is large enough : Required voltage = VS = IF (RCT + Rr + RL) = 7226 x 5 (0.26 + 0.02 + 0.15) = 36.13 x 0.43 = 15.54 Volts 1000 Current transformer VK approximates to :- VK ~ VA x ALF + RCT x IN x ALF In = 7.5 x 20 + 0.26 x 5 x 20 = 56 Volts 5 VK > VS therefore C.T VK is adequate
  • 62. s 62 Earth Fault Relay VK Check Assume values : As per overcurrent. Note For earth fault applications require to be able to pass 10 x relay setting. Check to see if VK is large enough : VK = 56 Volts Total load connected = 2RL + RCT + 2Rr = 2 x 0.15 + 0.26 + 2 x 0.02  Maximum secondary current = 56 = 93.33A 0.6 Typical earth fault setting = 30% IN = 1.5A Therefore C.T can provide > 60 x setting C.T VK is adequate
  • 64. s 64 Voltage Transformers  Provides isolation from high voltages  Must operate in the linear region to prevent accuracy problems - Do not over specify VT  Must be capable of driving the burden, specified by relay manufacturer  Protection class VT will suffice
  • 65. s 65 Typical Working Points on a B-H Curve Flux Density ‘B’ Magnetising Force AT/m Cross C.T.’s & Power Transformers Saturation V.T.’s Protection C.T. (at full load) ‘H’ 1000 2000 3000 1.6 1.0 0.5 Tesla
  • 66. s 66 Types of Voltage Transformers Two main basic types are available:  Electromechanical VT`s  Similar to a power transformer  May not be economical above 132kV  Capacitor VT`s (CVT)  Used at high voltages  Main difference is that CVT has a capacitor divider on the front end.
  • 67. s 67 Electromagnetic Voltage Transformer ZB (burden) NP / NS = Kn VP RP IP LP LM IM IC Re VS EP = ES Ie RS IS LS
  • 68. s 68 Basic Circuit of a Capacitor V.T. C1 C2 VP VC2 Vi L T VS ZB
  • 69. s 69 Ferro-resonance  The exciting impedance of auxiliary transformer T and the capacitance of the potential divider form a resonant circuit.  May oscillate at a sub normal frequency  Resonant frequency close to one-third value of system frequency  Manifests itself as a rise in output voltage, r.m.s. value being 25 to 50 per cent above normal value  Use resistive burden to dampen the effect
  • 70. s 70 VT Earthing  Primary Earthing  Earth at neutral point  Required for phase-ground measurement at relay  Secondary Earthing  Required for safety  Earth at neutral point  When no neutral available - earth yellow phase (VERY COMMON)  No relevance for protection operation
  • 71. s 71 VT Construction  5 Limb  Used when zero sequence measurement is required (primary must also be earthed)  Three Single Phase  Used when zero sequence measurement is required (primary must also be earthed)  3 Limb  Used where no zero sequence measurement is required  V Connected (Open Delta)  No yellow phase  Cost effective  Two phase-phase voltages  No ground fault measurement
  • 72. s 72 VT Connections a b c a b c A B C N da dn a b c n V Connected Broken Delta
  • 73. s 73 VT Construction - Residual  Used to detect earthfault  Useful where current operated protection cannot be used  Connect all secondary windings in series  Sometimes referred to as `Broken Delta`  Residual Voltage is 3 times zero sequence voltage  VT must be 5 Limb or 3 single phase units  Primary winding must be earthed
  • 74. s 74 Voltage Factors Vf  Vf is the upper limit of operating voltage.  Important for correct relay operation.  Earthfaults cause displacement of system neutral, particularly in the case of unearthed or impedance earthed systems.
  • 75. s 75 Protection of VT’s  H.R.C. Fuses on primary side  Fuses may not have sufficient interrupting capability  Use MCB
  • 76. GRID Technical Institute This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.