Transformers – Introduction  Transferring electrical energy from one circuit to another through time-varying magnetic field ECE430  Applications: both power and communications fields  In transmission, distribution, and utilization of electrical energy: step- Power Circuits and Electromechanics up or step-down voltages at a fixed frequency (50/60 Hz), at power of hundreds of watts to hundreds of megawatts Dr Nam Nguyen-Quang Fall 2009  In communications, transformers can be used for impedance matching, DC isolation, and changing voltage levels at power of a few watts over a very wide frequency range http://www4.hcmut.edu.vn/~nqnam/lecture.php  This course concerns only power transformers Lecture Ideal transformer  i1 wound as shown Ignore losses, stray + v1 – capacitance, and leakage flux N1 i2 + v2 – N2  Magnetic permeability is infinite or zero reluctance d dt Ideal transformer (cont.)  Consider a magnetic core with two coils v1 t   N Lecture v t   N d dt  v1 t  N  a v2 t  N v1 N1  a v2 N2 i1 i1 N   i2 N1 a  i2 + v1 v2 v1 t i1 t   v2 t i2 t   – v1 N1  a v2 N + + v1 v2 – N1:N i1 i1 N   i2 N a v1 t i1 t   v t i2 t  Ideal i2 – – N1:N a is called turns ratio  Total mmf is given by Ideal +  It can be shown, for an ideal transformer mmf  N1i1  N i2  R  i1 t  N   i2 t  N1 a k 1 Lecture L2 i1 v    i2 v1 a L1 Impedance-changing property of ideal transformer  L1 N 22  L2 N 12 Lecture 4 Impedance matching  Consider an ideal transformer with resistive load accross winding  The impedance-changing property can be used for maximizing power v2  RL  By Ohm’s law, i2  Subsituting v2  v1 a and i2  ai1 transferring between to windings, or matching impedances i1 Ideal + i2 + v1 v2 – – N  v1  a R L    R L i1  N1  RL  An ideal transformer is placed between power source (impedance Zo ) and load (impedance ZL) Turns ratio is so selected that N1:N Z o  N N  Z L  Ex 3.7: Two ideal transformer (each of ratio 2:1) and one resistor R are used to maximize power transfer Find R  The discussion can easily be extended to systems with complex load It can be verified that V1  N  I  N Load resistance  together with R referred to the input side is (R +  V2  N    Z L  a Z L    I  N1  Lecture 4(2)2)(2)2 For maximum power transfer, 10  R  64  R  13.5  Lecture Power transformer More pictures  Two windings mounted on a magnetic core, minimizing leakage flux  “Primary” winding (N1 turns) connected to power supply, “secondary” winding (N2 Small power turns) connected to load circuit Small 3-phase Control Cast resin  Assuming an ideal transformer: no leakage flux, winding resistances are neglected, magnetic core has infinite permeability, and is lossless  Let v1(t) = Vm1cost is the voltage applied to the primary winding, it can be shown that Vm1  2fN 1 max or V1  4.44 fN 1 max Lecture Example 110 kV, oil 10 kV, oil 500 kV, oil Lecture Equivalent circuit of transformer with linear core  Ex 3.8: Given N1, N2, core cross-sectional area, mean core length, B-  Consider now a transformer with leakage flux and winding resistances H curve, and the applied voltage Find maximum flux density, and Equivalent directly derived from physical model is simple but somewhat useless required magnetizing current The equations on the secondary side is mulplied by a (= N1/N2) and i2 is replaced V1  4.44 fN 1 max  max Hence, where by i2/a, to derive a more useful equivalent circuit V1  230 V, f  60 Hz, N1  200 i1 230   4.32  10 3 webers 4.44  60  200 + v1 L1 – aM R1 i2 + a2R2 a L2 – aM + RL v2 + i1 v1 aM i2/a av2 a2RL 4.32 10 3 Therefore, Bm   0.864 webers/m 0.005 The required H m  0.864  300  259 At/m , the peak value of  L1 – aM is termed the leakage inductance of winding 1, a2L – aM is termed magnetizing current is (259)(0.5)/200 = 0.6475 A Hence, Irms = 0.46 A is “referred” (to side 1) leakage inductance of winding aM is the magnetizing – – – N1:N – inductance, and its associated current is called magnetizing current the magnetizing on the primary side Lecture Equivalent circuit of transformer with linear core (cont.) Lecture 10 Transformer under sinusoidal steady-state conditions  There are losses in the magnetic core due to hysteresis and eddy current  For steady-state operation, impedances and phasors can be used in These losses are very difficult to calculate analytically The sum of these losses the equivalent circuit represents the total loss in the magnetic circuit of the transformer, and depends + be placed in parallel with the magnetizing inductance aM to account for them i1 R1 L1 – aM a2R2 a L2 – aM + v1 V1 Ideal + Rc1 (aM)1 – v2 – – I1 ja2xl2 I2 a Rc1 jXm1 + I2 + aV V2 – – Ideal ZL N1:N where RL N1:N  Real load RL and its associated voltage and current can be retained by   L1  aM   xl1  Leakage reactance of winding  aM   X m1  Magnetizing reactance referred to winding  L2  M a   xl  Leakage reactance of winding  a L2  aM   a xl  Leakage reactance of winding referred to side referring them back to the secondary side, using an ideal transformer Lecture a2R2 – i2 + av2 jxl1 R1 only upon the value of B m They are called core or iron losses A resistance can 11 Lecture 12 Transformer under steady-state (cont.) Approximate equivalent circuit  All quantities can be referred to winding jxl1 R1 +  Magnetizing branch makes computation somewhat difficult, hence this a2R2 I1 branch is moved to the terminals of winding 1, yielding an approximate ja2xl2 V1 Rc1 equivalent circuit, with no serious numerical error introduced + I2 a a2Z jXm1 aV2 L – R1 + – ZL jXm1/a + V2 – I1 V1 Rc1 – I2 a jXm1 R1eq  R1  a R2 aV2 x1eq  xl1  a xl – 13 Open- and short-circuit tests of transformers + a2ZL – Lecture aV2 jx1eq R1eq + I2 Rc1/a2 a2ZL – jxl2 R2 aI1 V1 a + I2 a jXm1 – jxl1/a2 R1/a2 + ja2xl2 a2R2 I1 V1 Rc1  Or they can be referred to winding jxl1 Lecture 14 Open-circuit test  Parameters in equivalent circuit can be determined by two simple  The test is performed with all instrumentation on the LV side with the tests: open-circuit test and short-circuit test HV side being open-circuited Rated voltage is applied to LV side Voc,  In power transformers, the windings are called high-voltage (HV) and Ioc, and Poc are measured with the meters low-voltage (LV) windings A Rc  I oc W Voc V oc Hence, IR V Rc LV IX Voc2 Poc IR  Voc Rc I oc  I R  I X I oc I X  I oc2  I R2 Voc V X m  oc IX Xm HV IR IX Rc  Rc and Xm are the values referred to the LV side Open-circuit test Equivalent circuit Lecture 15 Short-circuit test Lecture 16 Example  All the instrumentation is on the HV side Rated current is supplied to  Ex 3.9: Given OC and SC tests’ readings Find equivalent circuit HV side Vsc, Isc, and Psc are measured with the meters parameters referred to the HV side A Xm I sc W Req From OC test Xeq Vsc Vsc Rc  220 2 50 I X  12  0.227  0.974 A V IR   968  Xm  220  0.227 A 968 220  225.9  0.974 From SC test HV Req  Psc I sc2 Z eq  LV V sc I sc Req  X eq  Z eq2  Req2 Z eq   Req and Xeq are referred to the HV side Lecture 60 17 2  0.2076  15  0.882  17 X eq  0.882  0.2076  0.8576  17 Lecture 18 Efficiency and voltage regulation In-class quiz  Efficiency is defined as the ratio of output power to input power   Problem 3.22 and 3.23 Pout Pout Pout  100%   100% Pin Pout  losses Pout  Pc  Pi Losses are copper loss Pc and iron losses Pi  Alternatively, if input power is known,  Pin  Pc  Pi  100% Pin  Voltage regulation is defined as % voltage regulation  Vno load  Vload  100% Vload Lecture 19 Lecture 20 ... jXm1 aV2 L – R1 + – ZL jXm1/a + V2 – I1 V1 Rc1 – I2 a jXm1 R1eq  R1  a R2 aV2 x1eq  xl1  a xl – 13 Open- and short-circuit tests of transformers + a2ZL – Lecture aV2 jx1eq R1eq + I2 Rc1/a2... circuit of the transformer, and depends + be placed in parallel with the magnetizing inductance aM to account for them i1 R1 L1 – aM a2R2 a L2 – aM + v1 V1 Ideal + Rc1 (aM )1 – v2 – – I1 ja2xl2... + v1 L1 – aM R1 i2 + a2R2 a L2 – aM + RL v2 + i1 v1 aM i2/a av2 a2RL 4.32 10 3 Therefore, Bm   0.864 webers/m 0.005 The required H m  0.864  300  259 At/m , the peak value of  L1 – aM