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- 1. Current Transformers CT Basics – Construction – Theory of Operation – Polarity – Equivalent Model – Open Circuit Voltage – Accuracy CT Transient Performance CT Construction Bar-Type – A fixed insulated straight conductor that is a single primary turn passing through a core assembly with a permanently fixed secondary winding. Bushing Type – A secondary winding insulated from and permanently assembled on an annular core with no primary winding or insulation for a primary winding.
- 2. CT Construction Window Type – A secondary winding insulated from and permanently assembled on the core with no primary winding but with complete insulation for a primary winding. Wound Type – A primary and secondary winding insulated from each other consisting of one or more turns encircling the core. Constructed as multi-ratio CTs by the use of taps on the secondary winding. Theory of Operation Op Ip - Primary Current Ip Np - Primary Winding Turns Is - Secondary Current Np Es - emf Induced φp - Magnetic Flux Due to Primary Ip φs - Magnetic Flux Due to Secondary Is φe - Magnetic Flux Due to Exciting Ie Oe Zb - Burden Impedance Is Zb Vs Ns Os
- 3. Theory of Operation When a time varying current Ip flows, a magnetomotive force Fp (mmf) is developed by: Fp = Ip * Np. The primary mmf creates a magnetic flux Φp in the core given by: Φp = Fp/Rm where Rm is the core reactance. The direction of Φp is determined by the right hand rule. Φp Links the secondary winding, inducing an electromotive force Es (emf), resulting in a secondary current Is flowing through burden Zb. Theory of Operation The magnetomotive force Fs developed is due to Is and given by: Fs = Is * Ns. Fs creates an opposing flux Φs to Φp by Lenz’s law resulting in a magnetic flux Φe in the core by: Φe = Φp - Φs. Φe is sufficient to maintain Es at a given Ip and load impedance. Since the magnetic flux is proportional to the mmf we get: – Fe = Fp - Fs or – Ie * Np = Ip * Np - Is * Ns dividing by Ns – Ie * Np/Ns = Ip * Np/Ns - Is – Is = Ip * Np/Ns if Ie is small
- 4. Polarity The CT primary and secondary terminal is physically marked with a polarity. The marking indicates the instantaneous direction of the secondary current in relation to the primary current. When current flows in at the marked primary, current is flowing out of the marked secondary: PRI. SEC. Hint: Direction of the secondary current can determined as if the two polarity terminals formed a continuous circuit Equivalent Model The transformation of current induces errors. Some energy form the primary winding is used to: – Establish magnetic flux in the core. – Change the direction of the magnetic flux in the core named hysteresis losses. – Generate heat due to eddy currents. – Establish leakage flux. To account for losses a fictitious component is introduced, the exciting current Ie.
- 5. Equivalent Model Ip/n Is 1:n Ie/n Zpn**2 Zs Ic/n Im/n Zb Rcn**2 Xmn**2 The primary current Ip is stepped down in magnitude by 1:n through a no-loss transformer. – Zpn**2 - primary winding impedance – Zs - secondary winding impedance – Rcn**2 - hysteresis and eddy current losses referred to the secondary – Xmn**2 - magnetic reactance accounting for losses to establish flux referred to the secondary Equivalent Model Ip/n Is 1:n Ie/n Rs Ic/n Im/n Es Us Zb Rcn**2 Xmn**2 If the secondary winding is uniformly distributed on the core, Zs is resistive = Rs. The voltage drop across the primary winding is negligible to the source voltage to which it is connected and does not effect current flow, Zp/n**2 = 0. The secondary current is reduced by the shunting current of the exciting branch. The greater Ie the less accurate Is represents Ip.
- 6. I Equivalent model Ie' Ip' Vector relationship from equivalent model – Is is reference. Is – Voltage across burden is: Us = Is * Zb. – emf induced is: Es = Us + Is * Rs = Is * (Zb + Rs) – The angle α between Us and Is defines the burden power factor – Treating the exciting reactance as linear, Im lags Es by 90 – The magnetic flux Φe from Im lags Es by 90 – The Vectorial sum Im and Ic define Ie – Ip is the Vectorial sum of Is and Ie Oe – Is is less then Ip by a delta I and with a phase Es IsXb angle error of β. Us IsRb Ie' Ic' Im' Open Circuit Voltage Open secondary causes Φs to go to zero. Ip drives the core to saturation each half cycle. The action of Ip changing from maximum to zero back to maximum causes Φp to change from saturation in one direction to its saturated value in the opposite direction. The rapid rise of Φp induces high voltage spikes in the secondary winding. A formula for peak voltage derived from CT tests is: Vpeak = 35 × Zb × Ip / n . Tests have shown values ranging fron 500 to 11,000 volts.
- 7. Accuracy Definition: Ability to reproduce the primary current in secondary amperes in both wave-shape and magnitude. ANSI/IEEE C57.13 designates two rating classes C or T describing capability. – C: the ratio can be calculated, leakage flux is negligible due to uniform distribution of secondary winding. – T: the ratio must be determined by test, leakage flux is appreciable due to undistributed windings. Designations are followed by a terminal voltage rating that the CT can deliver to a standard burden at 20 times rated secondary current without exceeding 10% ratio correction. – Voltage classes are 100, 200, 400, 800 Accuracy The burdens are in ohms and at a .5 pf. – Standard burdens are B-1, B-2, B-4, B-8 Example: – C800: 800 V/ 5 A * 20 = 8 – If current is lower the ohmic burden can be higher in proportion. – Accuracy applies to the full winding. If a lower tap of a multiratio CT is used the voltage capability must be reduced proportionally.
- 8. CT Transient Performance CT Saturation CT Burden CT Parallel Operation Residual Magnetism CT Saturation Type C class CTs performance can be calculated from the excitation characteristics. – The excitation curve specifies the relationship of the exciting current Ie to exciting voltage Vs. – The point of the curve where the tangent is at a 45 to the abscissa is called the knee. Above the knee is the saturated region where the change in Ie no longer results in an appreciable change to Es. Below the knee is the unsaturated region where Ie is negligible.
- 9. CT Saturation Type T class CTs performance must be determined from test curves of primary to secondary current at standard burdens. Factors influencing the threshold of saturation under steady state are Zb and Is. – Transformer operation is given by: Es = 4.44 fNAB max substitute Φe = AB max Φe = Es / 4.44 fN where Es = Is( Zb + Rs) – Decrease Zb or Is (through the turns ratio) will limit Es and thus Φe. CT Saturation Steady State Analysis – Saturation by the AC component is avoided by selecting the proper turns ratio, decreasing the burden, or choosing the proper CT accuracy class. – Criteria: The product of Is and Zb does not exceed the saturation or knee point voltage of the CT. – Procedure: Determine the secondary current(Ip/n) from the primary fault current at the desired turns ratio. Determine the total secondary burden, Zt = Zb + Zlead + Zct. Calculate the required secondary voltage, Vs = (Ip/n) * Zt. Determine the secondary excitation current Ie required for the value of Vs from the excitation characteristic curve. Determine the approximate burden current Is by subtracting Ie from Ip/n. Check the effective ratio with the desired ratio to see if the performance is within the intended accuracy.
- 10. CT Saturation Steady State Example: CT ratio 500:5 C100, If = 12000 A – Ip/n = 12000/100 = 120 – Zct = 80*.005 = .4 – Zlead is 200 feet of full circuit run of #10 Awg, Zl = e**(.232(10)-2.32)*200/1000 = .2 – Relay burden of .15 – Zt = .4 + .2 +.15 = .75 – Vs = 120 * .75 = 90 – Vs of 90 is approximately Ie = 18 – Is = 120 - 18 = 102 – Effective ratio is 12000/102 = 117.6 or 588.23:5 CT Saturation Steady state analysis – An alternate procedure to check CT performance is to use Ks the saturation factor and a criteria of Ks = Vk/Vs. Where Vk is the effective knee point voltage of the CT from the excitation characteristic curve and Vs is the voltage across the CT secondary. Determine Vk from the excitation characteristic curve and CT ratio. Calculate Vs from (Ip/n)*Zt. Determine Ks and check criteria.
- 11. CT Saturation DC Saturation – When a fault occurs the current usually contains a DC component. The total flux required to produce the offset drives the CT into saturation. – Flux reaches a saturation during the positive cycle and the exciting reactance decreases shunting the primary current thus distorting the secondary. – During the negative cycle of the primary current, the core becomes unsaturated. – As the DC component decays the negative cycle of the primary current and flux become greater and the core eventually runs out of saturation during a complete cycle returning to steady state. – The DC component time constant is the X/R ratio of the primary circuit. CT Saturation DC Saturation Analysis – If saturation is to be avoided the secondary voltage requirement Vs must be (1 +X/R) times the voltage required for the AC component. – Example: CT 2000:5 C800 tap 1500:5, If = 19349 A with a source impedance of 81 deg. Ip/n = 19349/300 = 64.5 Zct = 300 * .0025 = .75, Zlead = .2, Zb = .15 Zt = .75 + .2 +.15 = 1.1 X/R = tan(81) = 6.31 Vs = (6.311 +1) * 64.5 * 1.1 = 518.9 Vs of 518.9 is approximately Ie = 20 Is = 64.5 - 20 = 44.5 Effective ratio is 19349/44.5 = 434 or 2175:5 – Due to 1500:5 tap the voltage must be reduced by the proportional amount giving the CT a 600 volt rating.
- 12. CT Burden Burden is defined as the total impedance of the secondary circuit. This includes: – CT winding resistance – Leads resistance from CT to relay and back – Impedance of the connected relays CT winding resistance – Higher ratio CTs (3000:5) resistance is .0025 Ω/turn – Lower ratio CTs (300:5) resistance is .005 Ohms/turn Resistance of leads is based on the AWG gage given by: .232G−2.32 Ω / 1000' = e residual relays. Note: Most microprocessor relays calculate I residual. CT Burden Relay Impedance – Microprocessor based relay almost negligible – Electromechanical given in VA @ rated I Burden is also influenced by CT connection and type of fault. – Example: Y-connected CTs 3Ph fault is balanced, current through the phase relay is the only burden. SLG fault is unbalanced, current must go through the phase and
- 13. Parallel Operation Parallel interconnection increases the burden seen by each individual relay. – The increase is dependent upon the type of connection, number of transformers, and distribution of current between transformers. Parallel interconnection can be used to supply a high burden when low ratios are required. – Primaries are connected in series and secondaries in parallel. Residual Magnetism (Remanence) Caused by current interruption and the magnetizing force becoming zero while the flux density in the core is operating at a high level. Occurs when: – Direct current is passed through a winding – Application of a high overcurrent interrupted at a peak magnitude Effect of residual magnetism is the accuracy of the secondary when the CT is next energized. How effects: – The flux changes start at the remnant value near saturation distorting the waveform. – The primary current is required for the excitation reducing the secondary output and increasing the CT error.