Current Transformer Design By Colonel Wm. T. McLyman Current transformers are used to measure or monitor the current in the lead of an ac power circuit. The current transformer, shown in Figure 1, is very useful in high power circuits where the currents are large, i.e., higher than the rating of the so-called, self-contained current meters. The current transformers, used by aerospace engineers, are for applications, such as over-current detection, under-current detection, peak current, average current, and regenerative base drive. Is I in R1 Vo T1 Figure 1. Typical, Current Transformer Monitor. A simplified, equivalent circuit of a current transformer is shown in Figure 2, and represents the important elements of a current transformer, where the ratio of primary to secondary turns is: Ns , [turns ratio] Np T1 I in Rs Io n= Ip Lc Im Np Ns Re Ro Figure 2. Simplified, Equivalent Circuit for a Current Transformer. The input current, Iin, is divided into several components in a current transformer. By considering input in the primary winding in terms of these various components, a better understanding of a current transformer behavior may be achieved. Only the ampere-turn (Iin Np) drives the magnetic flux around the core. The magnetizing ampere-turn (Im Np) provides the core loss. The secondary, ampere-turns, (Is Ns), balance the remainder of the primary ampere-turns. The magnetizing current, Im , in Figure 2, determines the maximum accuracy that can be achieved with a current transformer. 1 The magnetizing current Im may be defined as: Im = H ( MPL ) 0.4π N H ( MPL ) , [amps] H = [oersteds] Im = , [amps] N H = [amp-turn] MPL = [Magnetic Path Length] [cm] The magnetizing current, Im , in Figure 2, determines the maximum accuracy that can be achieved with a current transformer. The magnetizing current may be defined as the portion of the primary current, which satisfies the hysteresis and eddy current losses of the core. If the value of Lc and Re , in Figure 2, is too small, because the permeability of the core material is low and the core loss is too high, then, a different magnetic material must be used or a change must be made in the turns ratio. The magnetizing current, Im , is split off from the input current, Iin, in order to excite the core. What remains is the primary current, Ip, as shown in Figure 3. Rs Iin Im Io Ro Figure 3. Input Current, Output Current Relationship. Only part of the current, (n Ip), will flow in the output load, Ro. Figure 3 shows how the magnetizing current relates to the output. If accuracy is required, as it usually is for instrumentation, the loss and magnetizing currents must be kept to a minimum. The B-H loops, shown in Figure 4, represent two extremely different magnetic materials that are used in the design of current transformers. The two magnetic materials, shown in Figure 4, are Silicon Iron, a relatively low permeability, and Permalloy 80, a high permeability, Ni-Fe material. If accuracy is required, the losses and magnetizing currents must be kept small. 2 Silicon Permalloy 80 Figure 4. Silicon and Permalloy 80, BH Loop Comparison. The magnetizing impedance, Re , determines accuracy, because it shunts part of input current, Iin, away from the primary current, Ip, and thus, produces an error, as shown in Figure 5. Core material with the lowest value of H achieves the highest accuracy. Im Iin Io Figure 5. Input Current, Vector Phase Relationship. The current transformer function is different than that of a voltage transformer. A current transformer operates with a set primary current and will try to output a constant current to the load, independent of the load. The current transformer will operate into either a short circuit or a resistive load until the voltage induced is enough to saturate the core or cause a voltage breakdown. For this reason a current transformer should never operate into an open circuit, as a voltage transformer should never operate into a short circuit. The primary current of a current transformer is not dependent of the secondary load current. The current is really injected into the primary by an external load current, Iin. If the load current, Io, on the current transformer is removed from the secondary winding, while the external load current, Iin, is still applied, the flux in the core will rise to a high level, because there is not an opposing current in the secondary winding to prevent this. A very high voltage will appear across the secondary. A current transformer, like any other transformer, must satisfy the amp -turn equation: 3 Ip Is = Ns Np The secondary load resistor, Ro, the secondary winding resistance, Rs, and the secondary current, Is, determine the induced voltage of a current transformer. Vs = I s ( Ro + Rs ) , [volts] This voltage determines the number of turns. The value of, Bm , should be chosen at the highest level where the material gives maximum permeability. The core cross-sectional area required to generate a voltage in the secondary circuit is: I s ( Ro + Rs ) 10 K f N sf B m Ac = Note #1 ( ), 4 [cm 2 ] For this article, the magnetic core data and wire data has been taken from the author's book, Magnetic Core Selection for Transformers and Inductors, Second Edition. The author will refer to chapters and page numbers in the book for a quick reference. Some of the design information has been described in past articles, in Wound Magnetics, formerly Wound Magnetics Journal . For a typical design example, assume an output filter circuit, as shown in Figure 6, with the following specifications: T1 I in Is CR 1-4 Ro Vo C1 Figure 6. Current Transformer with DC Output. Current Transformer Design specification 1. 2. 3. 4. 5. 6. 7. 8. Frequency ................................................................................................................ f = 10 kHz Input Current ........................................................................................................... Iin = 0 to 5 amps Output Voltage ....................................................................................................... Vo = 5 volts Output Resistance .................................................................................................. Ro = 500 ohms Operating flux density ........................................................................................... Bm = .1 tesla Primary Turns .......................................................................................................... Np = 1 turns Core Loss ................................................................................................................ Pfe > 3 % Magnetic Material .................................................................................................. Supermalloy 4 Step No. 1 Calculate the secondary current, Is. Is = Is = Vo Ro , [amps] 5 , [amps] 500 I s = 0.010, [amps] Step No. 2 Calculate the secondary turns, Ns. Ns = Ns = Ip N p Is , [turns] [turns] (5 )(1) , ( 0.01) N s = 500, [turns] Step No. 3 Calculate the core iron cross section, A c , allowing for a 1.0, Vd diode drop. Kf = 0.4, (square wave) Ac = Ac = (Vo + 2Vd ) (104 ) K f N sf B m , [cm 2 ] 4 ( 4.0)( 500)(10000)( 0.1) (5.0 + 2 (1.0) ) (10 ) , [cm 2 ] Ac = 0.035, [cm 2 ] Step No. 4 Select from the Cased Toroidal Core Data on page 206, a 2 mil toroidal core with a value of A c closest to the one calculated in Step No. 3. Core number ................................................................................................................ 52057-2 Manufacturer .............................................................................................................. Magnetics Magnetic path length ................................................................................................ MPL = 5.5, cm Core weight ................................................................................................................. W tfe = 1.8, grams Copper weight ............................................................................................................. W tcu = 13.6, grams Mean length turn ........................................................................................................ MLT = 2.8, cm Iron area ....................................................................................................................... A c = 0.043, cm2 Window Area .............................................................................................................. W a = 1.37, cm2 Area Product ............................................................................................................... A p = 0.059, cm4 Core geometry ............................................................................................................. Kg = 0.000362, cm5 Surface area ................................................................................................................. A t = 24.2, cm2 5 Step No. 5 Calculate the effective window area, W a(eff0. A typical value for S3 is 0.75. See Note #1. 2 Waeff ( ) = Wa S 3 , [cm ] 2 Waeff ( ) = (1.37 )( 0.75 ) , [cm ] Waeff ( ) = 1.03, [cm ] 2 Secondary Winding Area Wa(sec ) = Waeff ( ) 2 = 0.052, [cm 2 ] Step No. 6 Calculate the secondary wire area, A w with insulation. A typical fill factor value for S2 is 0.6. See Note #1 at the beginning of this design example. Wa(sec ) S2 Ns Aw = Aw = , [cm 2 ] [cm 2 ] 2 ( 0.52)( 0.6) , ( 500 ) Aw = 0.000624, [cm ] Step No. 7 Select a wire area, A w, with insulation, from Table 11, closest to the value calculated in Step No. 6. See Note #1 at the beginning of this design example. AWG, No. 30 = 0.000678, [cm 2 ] µΩ = 3402 cm The rule is that when the calculated wire size does not fall within 10% to those listed in the Table, the next smaller size should be selected. Step No. 8. Calculate the resistance, Rs of the secondary winding, using the data from Table 11. µΩ −6 Rs = MLT ( N s ) 10 , [ohms] cm ( ) Rs = ( 2.8 )( 500 )( 3402 ) 10 −6 , [ohms] Rs = 4.76, [ohms] Step No. 9. Calculate the secondary output power, Po. Po = (V o + 2 V d )(I s ) , [watts] Po = 0.07, [watts] ( ) Po = ( 5 + 2 ( 1 . 0 ) ) ( 0.01, ) [watts] 6 Step No. 10. Calculate the acceptable core loss, Pfe . core loss, % Pfe = Po , [watts] 100 3 Pfe = 0.07 , [watts] 100 Pfe = 0.0021, [watts] Step No 11. Calculate the watts / kilogram, WK. (Supermalloy 2 mil) ( ) ( ) B ( ) , [watts/kilogram] ( ) ( ) WK = 0.179 (10 ) (10000 ) (0.1) , [watts/kilogram] WK = 0.179 10 −3 f −3 1.48 2.15 ac 1.48 2.15 WK = 1.05, [watts/kilogram] or [milliwatts/gram] Step No. 12. Calculate the core loss, Pfe in watts. Pfe = Wtfe 10− 3 WK , [watts] Pfe = (1.8 ) 10 −3 (1.05 ) , [watts] Pfe = 0.00189, [watts] Step No. 13. Calculate the core error in, %. Core Loss, % ( error ) = Core Loss, % ( error ) = Pfe Po ( ( ) ) (100 ) , [%] [%] Core Loss, % ( error ) = 2.7, [%] This design meets the 3% maximum error requirement. Summary: ( 0.00189) (100 ) , ( 0.07) The author hopes that this article has enlightened engineers to the fact that the current transformer is a fairly straightforward design. Current transformers are very useful in low and high power circuits. The majority of the current transformers, designed by the aerospace engineers, are level detectors and do not have to be very accurate. I would like to thank Richard Ozenbaugh who has reviewed the math and the step-by-step approach of this design. The author is now serving as a consultant for the Reliability Engineering Office at the Jet Propulsion Laboratory, in Pasadena, California. He is working on reliability and robust issues for Wound Magnetics design. REFERENCES 7 William Dull, Dr. A. Kusko and Thor Knutrud, “Current and Power Transformers”, EDN, March 5, 1975. Tape Wound Cores, Design Manual TWC-500, Magnetics, Division of Spang & Company, Butler PA. BIBLIOGRAPHY Colonel William T. McLyman, Transformer and Inductor Design Handbook, Second Edition, Marcel Dekker Inc., New York, 1988. Colonel William T. McLyman, Magnetic Core Selection for Transformers and Inductors, Second Edition, Marcel Dekker Inc., 1997. Colonel William T. McLyman, Designing Magnetic Components for High Frequency, dc-dc Converters, Kg Magnetics, Inc., 1993. For information regarding the above Books and Companion Software in Windows 95', 98', NT, or to order the books and software, contact: Kg Magnetics, Inc. 38 West Sierra Madre Blvd. Suit J Sierra Madre, Ca. 91024 Phone (626) 836-7233 or Fax (626) 836-7263 See the Web Page for a pictorial software demonstration. Web Page: www.kgmagnetics.com Email:
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