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1 10. Magnetically coupled networks EMLAB Transformer, Inductor 2 1. Transformer ① Used for changing AC voltage levels. ② Transmission line : high voltage levels are used to decrease power loss due to resistance of copper wires. The smaller magnitude of current, the less power loss, when transmitting the same power. ③ Used for Impedance matching. Transformers are used to change magnitudes of impedances to achieve maximum power transfer condition. 2. Inductors or transformers are difficult to integrate in an IC. (occupies large areas) EMLAB 3 Two important laws on magnetic field Current B-field Current generates magnetic field (Biot-Savart Law) Current Time-varying magnetic field generates induced electric field that opposes the variation. (Faraday’s law) Vinduced B-field Top view Electric field EMLAB 4 Magnetic flux Current N B i llength B llength Magnetic flux : NS B da BS i S llength Ni EMLAB 5 Self inductance Current Vinduced Vinduced d N dt B, Vinduced d N 2 S di N dt llength dt Vinduced L di dt N 2S L llength L N2 EMLAB 6 Inductor circuit (S : cross-section area of a coil, μ : permeability) Magnetic field Ni S Total magnetic flux linked by N-turn coil Ampere’s Law (linear model) N d dt Faraday’s Induction Law N d S di di N2 L dt l dt dt Assumes constant L and linear models! L Ideal Inductor di dt The current flowing through a circuit induces magnetic field (Ampere’s law). A sudden change of a magnetic field induces electric field that opposes the change of a magnetic field (Faraday’s law), which appear as voltage drops across an inductor terminals. EMLAB 7 Mutual Inductance (1) When the secondary circuit is open 2 N 2 d d L N di di N 2 1 i1 2 L1 1 L21 1 dt dt N1 N1 dt dt The current flowing through the primary circuit generates magnetic flux, which influences the secondary circuit. Due to the magnetic flux, a repulsive voltage is induced on the secondary circuit. EMLAB 8 Nomenclatures Primary circuit Secondary circuit Primary coil Secondary coil EMLAB Secondary voltage and current with different coil winding directions 9 EMLAB Two-coil system 10 (both currents contribute to flux) (2) Current flowing in secondary circuit d 1 di1 di2 1 L1 L12 dt dt dt d 2 di2 di1 2 L2 L21 dt dt dt Self-inductance Self-inductance Mutual-inductance L12 L21 M Mutual-inductance (From reciprocity) EMLAB 11 The ‘DOT’ Convention 1 d 1 di di L1 1 M 2 dt dt dt 2 d 2 di di L2 2 M 1 dt dt dt Dots mark reference polarity for voltages induced by each flux EMLAB 12 Example 10.2 Mesh 1 Voltage terms EMLAB 13 Mesh 2 Voltage Terms EMLAB 14 Example 10.4 FIND THE VOLTAGE V0 I2 I1 1. Coupled inductors. Define their voltages and currents V1 V2 KVL : 2430 2 I1 V1 VS KVL : - V2 j 2 I 2 2 I 2 0 Mutual inductance circuit V1 j 4 I1 j 2( I 2 ) V0 2I 2 V2 j 2 I1 j 6( I 2 ) / j2 0 j 2 I1 (2 j 2 j 6) I 2 / 2 j 4 VS (2 j 4) I1 j 2 I 2 j 2VS 4 (2 j 4)2 I 2 j 2VS j 2VS I2 8 j16 j 16 8 j V 2430 5.373.42 V0 2 I 2 S 4 2 j 4.4726.57 2. Write loop equations in terms of coupled inductor voltages 3. Write equations for coupled inductors 4. Replace into loop equations and do the algebra EMLAB 15 Example E10.3 WRITE THE KVL EQUATIONS I2 I1 Va Vb R1 R2 jL1 I1 R2 jM I 2 V1 R2 jM I1 R2 R3 jL2 I 2 V1 1. Define variables for coupled inductors 2. Loop equations in terms of inductor voltages Va R2 ( I1 I 2 ) V1 R1 I1 0 Vb R3 I 2 V1 R2 ( I 2 I1 ) 0 3. Equations for coupled inductors Va jL1 I1 jM ( I 2 ) Vb jMI 1 jL2 ( I 2 ) 4. Replace into loop equations and rearrange EMLAB 16 Example 10.6 Z S 3 j1() I1 DETERMINE IMPEDANCE SEEN BY THE SOURCE Z L 1 j1() V1 V 2 Zi I2 jL1 j 2() jL2 j 2() jM j1() 1. Variables for coupled inductors 2. Loop equations in terms of coupled inductors voltages Z S I1 V1 VS V2 Z L I 2 0 3. Equations for coupled inductors V1 jL1I1 jM ( I 2 ) V2 jMI1 jL2 ( I 2 ) VS ? I1 Z S jL1 I1 ( jM ) I 2 VS /( Z L jL2 ) ( jM ) I1 ( Z L jL2 ) I 2 0 / jM ( Z S jL1 )( Z L jL2 ) ( jM ) 2 I1 ( Z L jL2 )VS VS ( j M ) 2 Zi ( Z S jL1 ) I1 Z L jL2 ( j1) 2 1 1 j Z i 3 j3 3 j3 1 j1 1 j 1 j Z i 3 j3 1 j 3.5 j 2.5() 2 Zi 4.3035.54() 4. Replace and do the algebra EMLAB 17 10.2 Energy analysis di (t ) di (t ) p1 (t ) 1 (t )i1 (t ) L1 1 M 2 i1 (t ) dt dt t1 t1 0 0 di (t ) di (t ) p2 (t ) 2 (t )i2 (t ) L2 2 M 1 i2 (t ) dt dt W p1 (t ) p2 (t )dt 1 (t )i1 (t ) 2 (t )i2 (t )dt di (t ) di (t ) di (t ) di (t ) L1 1 M 2 i1 (t ) L2 2 M 1 i2 (t )dt dt dt dt dt 0 t1 t1 d 1 L1i12 (t ) L2i22 (t ) 2M i1 (t )i2 (t ) dt dt 2 0 1 L1 I12 (t1 ) L2 I 22 (t1 ) 2 M I1 (t1 ) I 2 (t1 ) 0 2 EMLAB 18 Coupling coefficient L I 2 1 1 (t1 ) L2 I 22 (t1 ) 2 M I1 (t1 ) I 2 (t1 ) 0 2 M M2 2 I 2 0 L1 I1 I 2 L2 L1 L1 M L1 L2 M k L1 L2 k M L1 L2 ; Coefficient of coupling (0 k 1) EMLAB 19 Compute the energy stored in the mutually coupled inductors Example 10.7 t 5ms Assume steady state operation We can use frequency domain techniques Merge the writing of the loop and coupled inductor equations in one step k 1 L1 2.653mH , L2 10.61mH 1 1 w (t ) L1i12 (t ) Mi1 (t )i2 (t ) L2 i22 (t ) 2 2 MUST COMPUTE M , i1 (t ), i2 (t ) L1, L2 , k M k L1L2 M 5.31mH L1 377 2.653 103 1 L2 4, M 2 I1 I2 2 I1 ( j1I1 j 2 I 2 ) 240 4 I 2 ( j 2 I1 j 4 I 2 ) 0 SOLVE TO GET I1 9.41 11.31( A), I 2 3.3333.69( A) i1 (t ) 9.41cos(377t 11.31)( A) i2 (t ) 3.33 cos(377t 33.69)( A) WARNING : The term 377t is in radians! t 0.005s 377t 1.885(rad ) 108 i1 (0.005) 1.10( A), i2 (0.005) 2.61( A) w(0.005) 0.5 2.653 10 3 (1.10) 2 5.3110 3 (1.10) (2.61) 0.5 101.6110 3 (2.61) 2 ( J ) w (0.005) 22.5mJ Circuit in frequency domain EMLAB 20 10.3 The ideal transformer 1 N1 d v1 (t ) N1 (t ) v N dt 1 1 First ideal transformer d v2 (t ) N 2 (t ) v2 N 2 equation dt v1 (t )i1 (t ) v2 (t )i2 (t ) 0 Ideal transformer is lossless 2 N 2 Insures that ‘no magnetic flux goes astray’ Since the equations are algebraic, they are unchanged for Phasors. Just be careful with signs i1 N 2 Second ideal transformer i2 N1 equations Circuit Representations v1 N1 ; v2 N 2 i1 N 2 i 2 N1 EMLAB Reflecting Impedances 21 For future reference * N N S1 V1I1* V2 1 I 2 2 V2 I 2* S2 N 2 N1 N n 2 turns ratio N1 V1 N1 (both signs at dots) V2 N 2 Phasor equations for ideal transformer I1 N 2 (Current I 2 leaving transform er) ( I 2 ) N1 V2 Z L I 2 (Ohm' s Law) N N V1 2 Z L I1 1 N1 N2 V2 n I1 nI 2 V1 ZL n2 S1 S 2 Z1 2 N V1 1 Z L I1 N2 2 N V1 Z1 1 Z L I1 N2 Z1 impedance, Z L , reflected into the primary side EMLAB 22 Non-ideal transformer M k L1 L2 V1 jL1 I1 jMI 2 Z L I 2 V2 0, (V2 jL2 I 2 jMI1 ) ( Z L jL2 ) I 2 jMI1 I 2 j M I1 Z L jL2 (M ) 2 V1 jL1 I1 I1 Z L jL2 V1 (M ) 2 (M ) 2 2 L1 L2 jL1Z L (1 k 2 ) 2 L1 L2 jL1Z L Z in jL1 I1 Z L jL2 Z L jL2 Z L jL2 (1) k 1 V jL1 Z L Z in 1 I1 Z L jL2 (2) Z L jL2 2 L N Z in 1 Z L 1 Z L L2 N2 To build ideal transformers, following two conditions are needed. (1) k=1; (2) ZL<<jωL; EMLAB Example 10.8 23 Determine all indicated voltages and currents n 1/ 4 0.25 I1 1200 1200 2.33 13.5 50 j12 51.4213.5 V1 Z1I1 Z1 1200 Z1 Z 2 Z1I1 (32 j16) 2.33 13.5 Strategy: reflect impedance into the primary side and make transformer “transparent to user.” ZL Z1 SAME COMPLEXITY Z1 32 j16 1200 120 Z1 Z 2 51.4213.5 V1 35.7826.57 2.33 13.5 83.3613.07 n2 ZL CAREFUL WITH POLARITIES AND CURRENT DIRECTIONS! I2 I1 4I1 (current into dot) n V2 nV1 0.25V1 ( is opposite to dot) Z1 32 j16 EMLAB 24 Thevenin’s equivalents with ideal transformers Replace this circuit with its Thevenin equivalent I2 0 I1 0 V1 VS1 I1 nI 2 V1 VS1 VOC nVS1 V2 nV1 To determine the Thevenin impedance... Reflect impedance into secondary ZTH ZTH n2 Z1 Equivalent circuit with transformer “made transparent.” One can also determine the Thevenin equivalent at 1 - 1’ EMLAB Thevenin’s equivalents from primary ZTH Z2 n2 In open circuit I1 0 and I 2 0 25 Equivalent circuit reflecting into primary VOC VS 2 n Thevenin impedance will be the the secondary mpedance reflected into the primary circuit Equivalent circuit reflecting into secondary EMLAB 26 Example 10.9 Draw the two equivalent circuits n2 Equivalent circuit reflecting into secondary Equivalent circuit reflecting into primary EMLAB Find I1 Example E10.8 27 n2 2 j 2 () I1 Equivalent circuit reflecting into primary 0.5 360 120 2 Notice the position of the dot marks I1 360 60 2 j1.5 I1 360 2.5 36.86 EMLAB Find Vo Example E10.9 28 n2 8 j8 240 Transfer to secondary VO 2 40 VO 2 200 (8 j8) 2 VO 400 12.81 38.66 EMLAB 29 Safety considerations Houses fed from different distribution transformers Braker X-Y opens, house B is powered down When technician resets the braker he finds 7200V between points X-Z when he did not expect to find any Good neighbor runs an extension and powers house B EMLAB