Chapter 15: Control of A.C. Drives

Chapter 15: Control of A.C. Drives SOLVED EXAMPLES Example 15.1 A 3-phase, star connected, 50 Hz, 4-pole induction motor has the following parameters in ohms per phase referred to the stator: R 1 = R = 0.034 and X 1 = X = 0.18 The motor is controlled by the variable frequency control with a constant (V /f ). Determine the following for an operating frequency of 15 Hz: (a) The breakdown torque as a ratio of its value at the rated frequency for motoring and braking. (b) The starting torque and rotor current in terms of their values at the rated frequency. Solution: From Eq. (15.34) of the book, K = 15 50 = 0.3 (a) From Eq. (15.5) of the book, the ratio of breakdown torques for K = 0.3 and K = 1, is T T max max ( K = 0.3) ( K = 1) = ( ) 0.034 ± 0.034 + (0.36) 0.034 Ê0.034ˆ ± (0.36) 0.3 Á + Ë 0.3 For motoring, For braking, T max Tmax T max Tmax ( K = 0.3) = 0.806 ( K = 1) ( K = 0.3) = 1.4. ( K = 1) (b) Substitution of s = 1 in Eq. (15.51) of the book gives an expression for the starting torque T. Thus, È Í 3 rated T s = Í V ( R / K) w Í s ÊR1 + Rˆ ÍÁ + ( X1 + X) ÍÎË K From Eq. (136), of the book the ratio of starting torques for K = 0.3 and K = 1 is, Ts ( K = 0.3) = T ( K = 1) s (0.034 0.3) (0.068 0.3) + (0.36) 0.034 (0.068) + (0.36) The starting rotor current is given by =.47 (i) I = V Ê R1 + Rˆ Á Ë K rated 1 X + ( X + ) (ii) 11 9/7/07, 4:4 PM

1 Power Electronics The ratio of starting currents for K = 0.3 and 1 is I ( K = 0.3) (0.068) + (0.36) = = 0.86 I ( K = 1) Ê0.068ˆ Á + (0.36) Ë 0.3 The above calculations of the ratios of starting torques and starting rotor currents show that the constant (V/f ) control provides a high starting torque with a reduced motor current. Example 15. For the motor Example 15.1, if the rated slip is 4% then determine the motor speed for rated torque and f = 5 Hz. The motor is controlled with a constant (Vf ) ratio. Solution: K = 5 50 = 0.5 From Eqs (15.49) of book for rated torque and f = 50 Hz, È Ê0.034ˆ Í Vrated 3 Á 0.04 Í Ë 3V T rated = w Í = s ÍÊ 0.034ˆ 0.034 + + (0.18 + 0.18) w ÍÁ Ë 0.04 Î and for 5 Hz from Eq. (15.51) of the book, È Ê0.034ˆ Í Vrated 3 Á 0.5S Í Ë T rated = w Í s ÍÊ0.034 0.034ˆ + + (0.36) ÍÁ 0.5 0.5S ÎË Equating Eqs (i) and (ii) gives, or rated s ( 0.034 0.5 S ) Ê0.034 0.034ˆ Á + + (0.36) Ë 0.5 0.5 S ( 0.034 0.5S ) Ê0.034ˆ Ê0.034 0.034ˆ Ê 0.34 ˆ Á + + + (0.36) Ë 0.5 Á Ë 0.5 0.5 S Á Ë0.5 S or S 0.474 S + 0.0344 = 0 (0.933) (i) = 0.933 = 0.933 0.474( ± 0.474) 4 0.0344 or S = = 0.384 or 0.089 The slip on the stable part of the speed torque curve will be 0.089. Now, synchronous speed for 5 Hz = 50 = 5 rps = 1500 rpm. 4 Therefore, the motor speed = 1500 (1 0.089) = 1366.5 rpm. (ii) 1 9/7/07, 4:4 PM

Control of A.C. Drives 13 Example 15.3 A three-phase, 460 V, 50 Hz, 4-pole, 140 rpm, star-connected induction motor has the following parameters per phase referred to the stator: R 1 = 0.66 W, R = 0.38 W, X 1 = X = 1.14 W, X m = 3 W. The motor is controlled by a variable frequency control at a constant flux of rated value. Determine the following: (a) The motor speed and the stator current at half the rated torque and 5 Hz. (b) By assuming the speed torque curves to be straight lines, solve for part (a), for S < S m. (c) The frequency, the stator current, and voltage at a rated braking torque and 100 rpm. Solution: From the rated conditions of operation: Synchronous speed, N s = w s = 1500 p 60 10 f 10 50 = = 1500 P 4 = 157.08 rad/s and \ S = 1500 140 1500 = 0.05 Rotor impedance, Z = R 0.38 + j 1.14 = + j = 7.6 + j1.14 = 7.69 8.53. S 0.05 Stator impedance, Z 1 = 0.66 + j 1.14 = 1.3 59.93 \ Machine impedance, Z Zm Z in = Z i + Z + Z (7.6 + j1.14) ( j3) = (0.66 + j 1.14) + = 6.91 + j3.01 = 7.54 3.54 W (7.6 + j1.14 + j3) \ Stator current, I 1 = 460 3 4 m = 66.4 A \ I = Zm j 3 Ê3ˆ I1 = 66.4 = Á (66.4) = 6.49 A Zm + Z j37.6 + + j1.14 Ë34 \ E = I Z = 6.49 (7.69) = 480.55 V 3 \ Rated torque = I w s R S 3 6.49 0.38 = 566.81 N-m 157.08 0.05 = ( ) (a) At 5 Hz, K = 5 50 = 0.5 Substituting the known values in Eq. (15.41) of the book yields ( ) 566.81 3 È (480.55) 0.38 0.5 S = Í = S = 0.0535 157.08 ÍÎ(0.38) (0.5 S) + (1.14) From Eq. (15.39) of the book, w r = Kw s (1 s) or N = KN s (1 s) = 0.5 1500 (1 0.0535) = 709.88 rpm At 5 Hz, E = 0.5 480.55 = 40.8 V 13 9/7/07, 4:4 PM

14 Power Electronics R Z = S + jkx = 0.38 + j 0.5 1.14 = 7.10 + j0.57 = 7.15 4.59 0.0535 Now, taking E as a reference vector, E 40.8 I = Z = = 33.7 4.59 A 7.15 4.59 E 40.8 \ I m = = = 15.0 90 A jkx m j0.5 3 \ I 1 = I + I m = 33.7 4.59 + 15.0 90 A Hence, I 1 = 38 A (b) Slip speed in rpm at the rated torque and frequency N ss = sn s = 0.05 1500 = 75 rpm. Since the speed torque curve is a straight line, slip speed at half the rated torque, N ss = 0.5 75 = 37.5 rpm. At 5 Hz, N = 5 1500 = 750 rpm. 50 Since the slip speed remains constant for a given torque, Motor speed, N = N s N ss = 750 37.5 rpm = 71.5 rpm. Now, for a constant flux, the (V /f ) ratio must be constant. Hence, at 5 Hz, Nss 37.5 E = 0.5 480.55 = 40.8 V. \ S = N = 750 = 0.05 R Z = s Now, taking E as a reference vector, + jk X = 7.6 + j0.57 = 7.6 4.9 W E 40.8 I = Z = = 31.53 4.9 A. 7.64.9 E 40.8 I m = = = 15.0 90 A. jkx m j16 \ I 1 = I + Im = 31.53 4.9 + 15.0 90 = 35.9 8.93. Hence I 1 = 35.9 A (c) Now, at the rated braking torque, the slip speed will be the negative of the slip speed at rated motoring torque. \ Therefore slip speed, N ss3 = 75 rpm. Synchronous speed π N + N ss3 = 100 75 = 115 rpm. s Frequency = Ê115ˆ Á Ë 1500 50 = 37.5 Hz. \ K = 37.5 50 = 0.75. 14 9/7/07, 4:4 PM

Control of A.C. Drives 15 Ê At 37.5 Hz, E = 37.5 ˆ Á Ë 50-75 480.55 = 360.41 V, S = = 0.067. 115 R Z = s + jkx = 5.67 + j 0.855 = 5.73 171.4 W E 360.41 Now, taking E as a reference vector, I = Z = = 6.9 171.4 A 5.73 171.4 I m remains the same as the foregoing. I 1 = 6.9 171.4 + 15.0 90 = 6. j4.4 = 66.81 158.58. \ V = E + Z1 I1 = 360.41 + (0.66 + j0.75 1.14) 66.81 158.58 or V = 340. j69.8 = 347. 11.51 Since the phase-angle between V and I 1 is more than 90, therefore, power flows from the motor to the source. Example 15.4 The motor in Example (15.3) is fed by a variable frequency current source. The motor is made to operate at the rated flux at all operating points. Determine the following: (a) Slip speed for I 1 80 A (b) The frequency and stator current for operation at 750 rpm for the following torque values: (i) 50 N-m (ii) 566.81 N-m. (c) Also, obtain the solution of b(ii) assuming speed torque curves to be straight lines in the region of interest. Solution: From Example (15.3), for 50 Hz operation, N s = 1500 rpm. w s = 157.08 rad/s. Rated torque = 566.81 N-m. Slip speed at rated torque = 75 rpm. E at rated conditions = 480.55 V E 480.55 (a) I m = X = m 3 = 15.0 A Substituting the known values in Eq. (15.66) of the book gives: È Ê0.38 ˆ Í Á (1.14) (15.0) s + = Í Ë (80) Í Ê0.38ˆ ÍÁ + (33.14) ÍÎË s \ S = 0.061 Slip speed in rpm = SN s = 0.061 1500 = 91.5 rpm. (b) (i) As the flux is constant for a given torque, the slip speed will also be constant for all frequencies. Hence, the slip speed can be obtained from the rated frequency operation. Now, T = 3 È Erated R / s Í ws ÍÎ( R/ s) + X 15 9/7/07, 4:4 PM

16 Power Electronics Substituting the given values yields 3 È(480.55) 0.38/ s 50 = Í S = 0.045 157 08 ÍÎ (0.38) + (1.14) Slip speed, N ss1 = 0.045 1500 = 67.5 rpm. Now, for operation at 750 rpm, synchronous speed, N s = N + N ss1 = 750 + 67.5 = 817.5 rpm. Ê817.5ˆ 7.5 Frequency = Á Ë 50 = 7.5 Hz K = = 0.545 1500 50 Nss 67.5 1 S = N = s 817.5 = 0.085, K s = 0.545 0.085 = 0.045 Now, substitute all the determined values in Eq. (15.68) of the book. 1/ È (0.38) /(0.045) + (31.14) + \ I 1 = 15.0 Í or I Î (0.38) /(0.045) + (1.14) 1 = 60.8 A. (ii) The given torque value is the rated braking torque. Therefore, the slip-speed will be the same as for the rated motor torque but of the opposite sign. \ N ss = 75 rpm., N s = N + N ss = 750 75 = 675 rpm. Frequency = 675.5 50 =.5 Hz., K = 1500 50 = 0.45 Nss -75 \ S = = = 0.11, k Ns 675 s = 0.05 Now, substitute all the determined values in Eq. (15.68) of the book, È ( 0.38) /(- 0.05) + (31.14) + \ I 1 = 15.0 Í = I Î (0.38) /(- 0.05) + (1.14) 1 = 66.45 A (c) As, speed torque curve for different frequencies are straight lines at a constant flux, therefore, slip speed, N ss3 = 50 75 = 68.81 rpm. 566.81 Hence, synchronous speed, N s = N + N ss3 = 750 + 68.81 = 818.81 rpm Ê818.81ˆ Frequency = Á Ë 1500 50 = 7.9 Hz, S = Nss 68.81 3 N = 818.81 = 0.084 K = 7.9 = 0.545 \ S k = 0.045 50 Now, substituting all the determined values in Eq. (15.68) of the book yields 1/ È (0.38) /(0.045) + (31.14) + I 1 = 15.0 Í Î (0.38) /(0.045) + (1.14) s 1/ = 60.8 A 16 9/7/07, 4:4 PM

Control of A.C. Drives 17 Example 15.5 An inverter supplies a 4-pole, three-phase cage induction motor rated at 0 V, 50 Hz. Determine the approximate output required of the inverter for motor speeds of (i) 900, (ii) 100, (iii) 1500, (iv) 1800 rpm. Solution: The slip may be neglected, approximately with the output inverter frequency related to synchronous speed. \ Frequency, f = (speed pairs of poles)/60 Also, at each condition, the voltage/frequency ratio = 0/50. Hence, the required inverter outputs are (i) f = 900 60 = 30 Hz, \ Voltage, V = 0 50 30 = 13 V. (ii) f = 100 60 = 40 Hz. \ V = 0 50 40 = 176 V. (iii) f = 1500 60 = 50 Hz. \ V = 0 V. (iv) f = 1800 60 = 60 Hz \ V = 0 50 60 = 64 V. Example 15.6 A three-phase, 400 V, delta connected induction motor has the following parameters at 50 Hz. R 1 = 0.5 W, R = 1.5 W, X 1 = X =.5 W, X m = 130 W. This motor is fed from a square-waved inverter. The voltage waveform is such that its fundamental is equal to the rated voltage of the motor. Determine, input current waveform corresponding to a rotor frequency of Hz when the supply frequency is 50 Hz and 10 Hz and the voltage applied is proportional to frequency. What waveform do you expect at 10 Hz if the voltage is varied to keep air gap flux constant? Solution: (i) Voltage proportional to frequency At 50 Hz, V = 400 V, slip S = /50 = 0.04. 1.5 Secondary impedance at this slip = + j.5 = (37.5 + j.5) W. 0.04 Mutual reactance = 130 W. Total impedance of the motor, j130 (37.5 + j.5) Z t = 0.5 + j.5 + = 3.65 + j5.736. 37.5 + j 13.5 400 Fundamental RMS current = = 60.61 57.71 A 3.65 + j 5.736 Instantaneous value of fundamental current = 60.61 sin (w t 57.71 ) A. = 85.7 sin (w t 57.71 ) A. 17 9/7/07, 4:4 PM

Ë Ê Á 18 Power Electronics Slip for fifth harmonic, S 5 = 5 50 48 5 50 = 1.19 Voltage applied = 400 5 Secondary impedance = 1.5 1.19 Total impedance, Z t = 0.5+ j1.5+ Fifth harmonic rms current = = 80 V + j1.5 = (1.58 + j1.5) W. j 650 (1.58 + j1.5) =1.734 + j 4.76=4.8 85.99. 1.58 + j 66.5 80 4.8 85.99 Instantaneous value of fifth harmonic current Slip for seventh harmonic, S 7 = = 3.3 85.99 = 3.3 sin (5 w t 85.99 ) = 4.557 sin (5 w t 85.99 ) A 350 48 350 = 0.863 Secondary impedance = Ê Á Ë 1.5 0.863 + j17.5 ohms Total impedance = 0.5 + j 17.5 + j910(1.738 + j17.5) 1.738 + j 97.5 =.06 + j 34.67 = 34.74 186.35 W RMS value of seventh harmonic = 57.143 86.35 = 1.645 A 34.74 Instantaneous value of current =.36 sin (7 w t 86.35 ) A Thus the total armature current, = 85.7 sin (w t 57.71 ) + 3.3 sin (5 w t 85.99 ) +.36 sin (7 w t 86.35 ) A. (ii) Current waveform at 10 Hz: Fundamental voltage at 10 Hz = 80 V. Slip = /10 = 0. Secondary impedance = 15. + j 0.5 = (7.5 + j 0.5) W 0. X m = 130/5 = 6 W \ Total impedance, Z t = 0.5 + j 0.5 + j 6(7.5 + j 0.5) W 7.5 + j 6.5 =(.96 + j 7.17) = 7.53 67.8 18 9/7/07, 4:4 PM

Control of A.C. Drives 19 Instantaneous value of fundamental current = 80 7.53 sin (10 p t 67.8 ) = 15.0 sin (10 p L 67.8 ) Fifth harmonic voltage = 16 V 5 10 8 Slip for fifth harmonic = = 1.16. 5 10 Secondary impedance = (1.93 + j.5) W X m = 13 W \ j130(1.93 + j.5) Total impedance = 0.5 + j.5 + 1.93 + j 13.5 = 5.59 70.343 W. Instantaneous value of fifth harmonic, i 5 = 16 sin (5w t 70.343) = 4.3 sin (5w t 70.343) A 5.59 Seventh harmonic: Voltage of seventh harmonic = 11.48 V Slip for seventh harmonic = 7 10 8 7 10 = 0.886, X m = 18 W j18(1.693 + j3.5) Total impedance, Z t = 0.5 + j3.5 + 1.693 + j 185.5 =.171 + j 6.934 = 7.66 7.6. Instantaneous value of seventh harmonic 11.486 = sin (7 w t 7.6) 7.66 =. sin (7 w t 7.6) The total armature current = 15.0 sin (w t 67.8 ) + 4.3 sin (5 w t 70.343) +. sin (7 w t 7.6 ) If the voltage is varied to keep air gap flux constant, we get the same current waveforms both at 50 Hz and 10 Hz for a rotor frequency of Hz. Example 15.7 A three-phase, 400V, 50 Hz, 4-pole, 1450 rpm, star-connected squirrel-cage induction motor has the following parameters per phase referred to the stator: R 1 = 0.11 W, R = 0.09 W, X 1 = 0.4 W, X = 0.6 W, and X m = 1 W. The motor is controlled by a six-step inverter. The d.c. input to the inverter is provided by a six-pulse, fully-controlled rectifier. (a) What should be the rectifier firing angle for getting the rated fundamental voltage across the motor if the rectifier is fed by an a.c. source of 400 V, 50 Hz. 19 9/7/07, 4:4 PM

130 Power Electronics (b) If the machine is operated at a constant flux then determine (i) the inverter frequency at 560 rpm. and rated torque (ii) the inverter frequency at 460 rpm and half the rated torque. Also determine the motor current Solution: The fundamental RMS line voltage of a six-step inverter is given by E 1 = 6 p E dc (i) where, E dc = 3 p E m cos a (ii) E m is the peak of a.c. source line voltage. 3 6 \ E 1 = p E m cos a, \ cos a = E E 1 m p 3 6 (iii) Here, E 1 = 400 V, E m = 400 V. p 400 cos a =, \ a = 18.5 3 6 400 (b) (i) As discussed in previous sections that for a given torque the motor operates at a fixed slip speed for all frequencies when the flux is maintained constant. \ Slip speed in rpm at the rated torque, 10 50 N ss = N s N = 1450 = 50 rpm. 4 Hence, synchronous speed at 560 rpm N s = N + N ss = 560 + 50 = 610 rpm. Ê 610 ˆ The inverter frequency = Á Ë 1500 50 = 0.33 Hz (ii) The back emf at the rated operation E rated = I [(R /S ) + X ] 1/ where I = E / 3 1 1+ + 1+ ( R R / S) ( X X ) Also, S = 1500-1450 1500 = 0.033 \ I = 400/ 3 Ê 0.09 ˆ Á0.11 + + (0.4 + Ë 0.6) 0.033 È Ê 0.09 ˆ E rated = 76.98 Í Á + (0.6) ÎË0.033 1/ = 76.98 A = 14.99 V 130 9/7/07, 4:4 PM

Control of A.C. Drives 131 Now, torque at a constant flux is given by Eq. (15.41) of the book 3 È Erated R ( KS) \ T = Í w s ÍÎR ( KS) + X Note that w s and X in this Eq. are for rated frequency 3 where T = w I R /S but w s = 1500 p = 157.08 rad/s 60 s 3 Ê 0.09 ˆ \ T = 157.08 (76.98) Á Ë 0.033 = 308.66 N-m. Equation (iv) becomes \ or or which gives 308.66 = È Ê0.09ˆ (14.99) 3 Í Á Ë KS Í 157.08 ÍÊ0.09ˆ ÍÁ + (0.6) Ë KS Î Ê ÁË 0.09 KS 1 KS = 6.75 ˆ + (0.6) = 0.514 KS Ê 1 ˆ Ê 1 ˆ Á - 63.46 Ë Á KS ËKS + 44.44 = 0 wr 460 0.31 Now, K = = = ws (1 -S) 1500(1 -S) (1 -S) Substituting from Eq. (v), for K, gives 1 \ 6.75 S = 0.31 (1 - S) which gives S = 0.048 \ From Eq. (v), k = 0.33 Thus, frequency = 0.33 50 = 16.6 Hz. Substituting the known value in Eq. (15.38) of the book yields 14.99 I = = 37.86 A (0.09) (6.75) + (0.6) Machine fundamental phase voltage, V 1 = I 1/ R ˆ 1+ ( 1+ ) ÈÊ ÍÁR K X X ÎË S 1/ È Ê 0.09 ˆ = 37.86 ÍÁ0.11 + + (0.33) (0.4 + 0.6) ÎË 0.048 = 76.19 V. (iv) (v) 131 9/7/07, 4:4 PM

13 Power Electronics Now, taking V 1 as a reference vector, 9.5 I = I tan K ( X + X ) = 37.86 tan 1 0.33(0.4 + 0.6) (0.11 + 0.09 /0.048) 1 1 ( R1 + R/ S) E1 ** = 90 = kxm =19.1 90 76.19 90 0.331 = 37.86 Now, I 1 = I + I m = 31.86 9.5 + 19.1 90 or I 1 = 45.14A RMS harmonic current is given by I n = Ê 1 1 ˆ V Á K( X 4 1 + X) Â Á Ë h h= 5,7,11,13 Neglecting harmonics higher than 13 gives 0.046 V1 I h = K ( X + X ) 1 0.046 76.19 \ I h = = 10.56 A. 0.33(0.4 + 0.6) \ The RMS input current = (I 1 + I 4 ) 1/ = [(45.14) + (10.56) ] 1/ = 46.36 A Example 15.8 A three-phase, 400 V, 50 Hz, 980 rpm, six-pole, star-connected, squirrel-cage induction motor has the following parameters per phase referred to the stator: R 1 = 0.0 W, R = 0.1 W, X 1 = 0.18 W, X = 0.4 W, and X m = 10.3 W. The current source inverter controls the motor. At the rated value, flux is maintained constant. Compute the following: (a) The stator current and d.c. link current, when the machine operates at rated torque and 50 Hz. (b) The inverter frequency and d.c. link current for a speed of 500 rpm and rated torque. (c) The motor speed, stator current, and d.c. link current for half of the rated torque and inverter frequency of 5 Hz. Solution: At the rated operations, Synchronous speed, N s = 1 10 f 10 50 = = 1000 rpm = 104.7 rad/s. P 6 (vi) Rated slip = 1000 980 1000 = 0.0 Rotor impedance, Z = 0.1 + j 0.4 = 6 + j 0.4 = 6.01 3.81 W 0.0 13 9/7/07, 4:4 PM

Control of A.C. Drives 133 Z Zm (6 + j0.4) ( j10.3) Machine impedance = Z 1 + = 0.0 + j 0.18 + Z Zm 6+ j 10.7 = 4.43 + j.94 = 5.3 33.57 W I = I m = Zm Zm + Z Zm Z + t m I 1 = I 1 = I 1 = 400 3 = 43.41 33.57 A 5.333.57 10.3 90 43.41 33.57 = 36.44 4.9 A 1.7 60.7 6.01 3.81 (43.41 33 57 ) = 1.6 90.48 1.7 60.7 Torque = 3 3. ( ) Ê0.1ˆ I R S = (36.44) ws 104.70.0 Á Ë = 8.4 N-m. (a) For the three-phase current source inverter (six-step), the fundamental RMS current is given by I 1 = 6 p I d (i) p I \ I d = d p = 43.41 = 55.68 A 6 6 Also, RMS stator current is given by the relation, I rms = ( 3) I d = ( ) 3 (55.68) = 45.46 A (ii) (b) We know that when the motor is controlled at a constant flux for a given torque, the slip speed has a constant value. Therefore, slip speed at the rated torque and frequency is N ss = S N s = 0.0 1000 = 0 rpm. Therefore, at the motor speed of 500 rpm, synchronous speed, N s = 500 + 0 = 50 rpm. Inverter frequency = (50/1000)50 = 6 Hz Also, when the motor is controlled at a constant flux, for a given torque, the stator current remains constant at all speeds. Since the stator current is constant, the d.c. link current also remains constant at 55.68 A. (c) The slip-speed is constant at all frequencies as the flux is constant for a given torque. The slip speed for 5 Hz operation at half the rated torque can be determined from 50 Hz operation. For 50 Hz operation, E rated = I m X m = 1.6 10.3 = 18.98 V Now, T = 3 w s È E R S Í Î( R S) X rated + 8.4 = 3 È(18.98) 0.1/ S Í 104.7 ÍÎ( 0.1(0.4) S ) + 133 9/7/07, 4:4 PM

134 Power Electronics or Ê0.1 ˆ Á Ë S + (0.4) = 1.44 S \ 0.0144 x 1.44 x + 0.16 = 0, where x = 1/S. \ x = 100, which gives S = 0.01. Slip speed, N ss = S N s = 0.01 1000 = 10 rpm Now, consider the operation at 5 Hz, K = 5 50 = 0.5 \ Synchronous speed, N s = 0.5 1000 = 500 rpm. Hence, motor speed = 500 10 = 490 rpm \ S = Nss 10 N = 500 KE \ I = ( R S) ( K X ) From Eq. (15.6), of the book we have s rated + = 0.5 18.98 ( ) 0.10.0 + (0.5 0.4) = 18.4 A I = 1 Im I 1 + X m or (18.4) = I1 (1.6) 0.4 1 + 10.3 or I 1 = 8.47 A Also, d.c. link current I d can be given by the formula p I1 p 8.47 I d = = = 36.51 A (iii) 6 6 The RMS stator current I rms = 3 36.51 = 9.81 A. Example 15.9 A cycloconverter is operating on 415 V, 50 Hz three-phase system and is controlling an induction motor in the speed range 5 to 0 Hz (non-reversible) using constant V /f control. Load power factor is 0.8 and the input displacement factor is 0.7. Determine (a) Range of variation of firing angle of cycloconverter (b) Worst value of input power factor (c) Highest value of distortion-factor. Solution: As the drive is operating in constant torque zone only, V /f will be constant, since the output frequency only varies from 5 to 0 Hz, i.e. 10% to 40% of rated frequency, RMS value of output voltage will also vary from 10% to 40% of rated value, i.e. 41.5 V to 166 V. \ The corresponding maximum value of output voltage is E max 1 = 41.5 and E max = 166 Now, E max = 1.35 E cos q or 1.35 415 cos q 1 = E max1 = 41.5 \ q 1 = 84. 134 9/7/07, 4:4 PM

Control of A.C. Drives 135 Similarly, 1.35 415 cos q = 166 \ q = 65.3. Hence, range of variation of firing angle is 65.3 to 84. (b) Input power factor is given by cos f = cos q cos q0 where cos f 0 is load power factor. Worst value of input power factor i.e. lowest value will occur for lowest values of cos q. Hence, cos f = cos q cos f0 Here, cos q = 0.1044, cos f 0 = 1.1044 0.8 = 0.059 (c) Distortion factor is given by cos q cos f0 cos qm cos f0 m =. \ m m = = cos f cos f i i 0.4178 0.8 = 0.339. 0.7 SOLVED EXAMPLES Example 15.10 A three-phase, 400 V, 50 Hz, 960 rpm, six-pole, star-connected, wound-rotor induction motor has the following parameter per-phase referred to the stator: R 1 = 0.3 W, R = 0.5 W, X 1 = X = 1.6 W, X m = 35 W, stator-to-rotor turns ratio is. The motor speed is controlled by the static rotor resistance control. The filter resistance is 0.01 W. The value of external resistance is chosen such that a = 0, the breakdown torque is obtained at standstill. Determine the following: (a) The value of the external resistance. (b) a for a speed of 750 rpm at 1.5 times the rated torque. (c) The speed, for a = 0.5 and 1.5 times the rated torque. Neglect friction and windage loss. Solution: N s = 10 f 10 50 = = 1000 rpm, V = 400 3 = 30.94 V. P 6 w s = 1000 p = 15.66 rad/s 50 1000 960 Full-load slip = 1000 Without rotor resistance control, T = 3 w s = 0.04 V ( R S) ( ) È Í Í Î R R S X X 1 + + ( 1 + ) 135 9/7/07, 4:4 PM

136 Power Electronics È Ê 0.5 ˆ Í (30.94) 3 Á 0.04 Í Ë Full load torque, = 15.66 Í ÍÊ 0.5 ˆ 0.3 + + (3.) ÍÁ 0.04 ÎË (a) From Fig. 15.8(b), of the book R m = [(R 1 + R k ) + (X 1 + X ) ] 1/ S when the breakdown torque occur at standstill, (R m ) = (R 1 + R k ) + (X 1 + X ) or (R m ) = R k + R k R 1 + R 1 + (X 1 + X ) From Eqs (15.89) and (15.90), = 91.43 N-m (i) Ê p ˆ R k = Á 1 9 Ë R m = 0.0966 R m (ii) Substituting from Eq. (i) and known values in Eq. (iii, Ex. 15.8) gives (R m ) 9.33 10 3 (R m ) 0.0579 R m 0.09 10.4 = 0 0.9906 R m 0.0579 R m 10.33 = 0 \ R m = 3.6 W and \ R k = 0.315 W From Eq. (15.90) of the book, R e * = R m R = 3.6 0.5 =.76 W..76 R e * = = 0.69 W a T 1 From Eq. (15.81) of the book, for a = 0, R = R e * R d = 0.69 0.01 = 1.37 W (b) With rotor resistance control, from Eq. (15.95), of the book T = V ( R S) ( R + R + R S) + ( X + X ) 3 m Í s Í 1 k m 1 w From Eq. (i), R k = 0.0966 R m È Î 1000 750 S = = 0.5 1000 \ Substituting all known values in above equation yields ( R ) 3 È (30.94) 0.5 m 1.5 91.43 = Í 15.66 ÍÎ(0.3 + 0.0966 Rm + Rm /0.5) + (3.) or 0.504 R m 0.97 R m + 0.31 = 0 136 9/7/07, 4:4 PM

Control of A.C. Drives 137 which gives R m = 1.4 W or 0.44 W The latter value is not feasible because it is less than R. Hence, R m = 1.4 W. From Eq. (15.90) of the book, 1.4 0.5 R e * = ( Rm R a T = = 0.5 W 1 () From Eq. (15.81) of the book, * Re Rd 0. 5 0.01 (1 a) = =, R 1.37 \ a = 0.68 (c) From Eq. (15.81) of the book R e * = 0.5 [0.01 + (1 0.50) 1.37] = 0.35 W R m + a T 1 R e * = 0.5 + 4 0.35 = 1.9 W R k = 0.0966 1.9 = 0.184 W Substituting all known values in torque equation, 1.5 91.43 = È Í 3 Í (30.94) (1.9/ S) 15.66 ÍÊ 1.9 ˆ Á0.3 + 0.184 + + (3.) ÍÎË S Ê 1.9ˆ or Á0.484 + + 10.4 = 17.64 Ë S S which gives S = 0.8 or 1.3. The 1.3 value is not feasible. \ N = N s (1 S) = 100 (1 0.8) = 70 rpm Example 15.11 A three-phase, four-pole, 50 Hz induction-motor has a choppercontrolled resistance in the rotor circuit for speed control. Load torque is w. When the thyristor is ON, the torque is 30 N-m at a slip of average 0.03. If T ON /T OFF = 1, compute the average torque and speed. The motor develops a torque of 80 per cent of ON torque when the thyristor is OFF. The speed variation ranges down to 100 rpm from synchronous speed. Determine the ratio T ON /T OFF to give an average torque of 5 Nm. Solution: The synchronous speed = 1500 rpm. S = 0.03. The speed of the motor at this slip = (1 0.03) 1500 = 1465 rpm. When the thyristor chopper is OFF, the complete resistance is included in the circuit. When T ON /T OFF = 1, the average torque, T av = 30 1 + 4 1 = 7. 137 9/7/07, 4:4 PM

138 Power Electronics Since T a N, therefore, 30 (1465) Speed when T ON /T OFF = 1 is, 7 = N N = 7 30 (1465) = 1390 rpm. When the speed is 100 rpm for chopper OFF, torque would be (100) (1405) 30 = 0.13 N-m Average-torque = 5 N-m 30 TON + 0.13TOFF 5 = T + T ON OFF T T \ ON OFF = 0.97 I MICROPROCESSOR CONTROLLED A.C. DRIVES Todays industry places high demands on control accuracies, flexibility, ease of operation, repeatability of parameters for many drives applications. To meet these requirements, use of microprocessors have become imperative. In a powerelectronic system, the microcomputer functions can in general be categorized as follows: Control of feedback loops Gate firing control of phase-controlled converters PWM or square-wave signal generation of inverters Optimal and adaptive control Estimation of feedback signals General sequencing control Protection and fault overriding control Signals monitoring and warning Data acquisition Diagnostics Miscellaneous computation and control. The superiority of microcomputer control over the conventional hardware based control can easily be recognized for complex drive control system. The simplification of hardware saves control electronics cost and improves the system reliability. The digital control has inherently improved noise immunity which is particularly important here because of large power switching transients in the converters. The software control algorithms can easily be altered or improved without changing the hardware. Another important feature is that the structure and parameters of the control system can be altered in real time making the control adaptive to the plant characteristics. The complex computation and decision 138 9/7/07, 4:4 PM

Control of A.C. Drives 139 taking capabilities of microcomputer make possible to apply the modern optimal and adaptive control theories to optimize the drive system performance. In addition, powerful diagnostics can be written in the software. Microcomputer technology is moving at such a fast rate that the use of efficient high level language with large hardware integration already is possible, and possibly VLSI implementation of the controller is the next goal. Unlike dedicated hardware control, a microcomputer executes control in serial fashion, i.e. multitasking operations are performed in time multiplexed method. As a result, the slow computation capability may pose serious problems in executing the fast control loops. However, the problems can be solved by multi-microprocessor control, where judicious partitioning of the task can significantly enhance the execution speed. Microcomputers are having a major impact on industrial applications, including the areas of testing, control, instrumentation, data acquisition, numerical machine control, and even robotics. The microprocessor controlled a.c. drives are widely used in the following industrial applications: Fans and pumps Compressors Travel and hoist drives in cranes and conveyors. Roller tables in rolling mills. Paper machines Write drawing machines in metal industry etc. 1 Control of Induction Motor Drive It is essential to keep ratio of voltage to frequency (V/f ) constant for any motor if constant torque is required throughout the speed range. This task is achieved through pulse-width modulation (PWM) technique. Interest has been growing in microprocessor-based PWM schemes for a.c. drive systems in recent years. Modern PWM a.c. drive systems are continually seeking improvement of performance and reliability with reduction of control and power conversion cost. In a conventional hardware modulator, the PWM waveforms are generated by comparing the sine reference wave with the triangular carrier wave by the natural sampling process. As the linear PWM region is exceeded into the transition region, the harmonic quality of the waves deteriorates seriously with the introduction of the lower order harmonics. In addition, the dropping of pulses near the middle of the wave causes a current surge problem. In a microprocessor-based modulator, the wave can be fabricated precisely in the transition region controlling the harmonics and voltage jump, and the nonlinearity problem can be easily overcome. However, precision PWM wave generation in real time, as required by the drive system operation, remains a challenge because of the time critical performance requirement of the microcomputer. A basic block diagram of the control is shown in Fig. 15.1. The microprocessor performs all the functions of closed-loop control using software. A suitable software and digital logic are used to develop the necessary PWM waveform. 139 9/7/07, 4:4 PM

140 Power Electronics The uniform sampling techniques are used to develop the necessary PWM waveforms. A microprocessor system may be evolved to provide the real time simulation besides implementing the control programs. The real time aspects of the processor ADCs, DAC, parallel interfaces and the capability of the system to provide control and other functions are taken to advantage. The system can be designed with a flexibility to provide monitoring functions to change the parameters such as sampling rates of ADCs. The functions of a microprocessor in a PWM control are: (a) Processing of speed signals obtained from a shaft encoder. Comparing this with set speed in a controller and to provide an output of current based on the speed error (speed controller program), the necessary limiting may also be accomplished. (b) The analog signal of measured signal is processed after conversion to digital signal by ADC. This is compared with the output of speed controller in current controller. The output of this controller determines the firing instants of the inverter which gives the desired voltage waveform. However, the speed of microprocessor may set a limit on the maximum frequency of operation. This may be overcome by a hard wired digital logic. Wherever the speed of a microprocessor is not sufficient to perform a given task, it can be accomplished using a suitable hardware. (c) The microprocessor controller must be capable of performing control tasks as it interacts with the system. This includes selection and Fig. 15.1 Block-diagram of microprocessor based speed control of induction motor 140 9/7/07, 4:4 PM

Control of A.C. Drives 141 sampling of signals, mathematical computation to implement control, the necessary A/D and D/A conversion. Here, if the execution speed of the microprocessor is slow, the control may not be accurate because some input measurements may be missed or false timing of signals may be generated. The multi-task requirement of a microprocessor may lead to reduced reliability. The task may be executed in parallel almost simultaneously by developing algorithms which may be performed independently by a multiprocessor system. The control tasks to be performed simultaneously may be recognized and separate processors be used to execute these. The trends in the design and development of microprocessor based controllers for induction motors operating on PWM inverters include 16-bit processors, special purpose processors with on board data conversion and additional hardware functions, faster processor. These trends in microprocessor hardware and software may be applied to power electronic controls, more efficient algorithms may be developed and applied which require computer implementation, as in the case of harmonic elimination, would become relatively simple. Control of Synchronous Motor Drive In recent years, increasing interest has been shown for the commutatorless d.c. motor (also known as self-controlled synchronous motor) which consists of a synchronous machine fed from a current-source inverter or cycloconverter. The advent of microprocessor has raised interest in the digital control of powerconverter systems and electronic motor drives since the microprocessor provides a flexible and low-cost alternative to the conventional approach. For motor drive systems, microprocessor control offers several interesting features, principally improved performance and reliability, versatility of the controller, reduced components count and reduced development and manufacturing cost. 1. Drive System A block diagram of the microprocessor-controlled synchronous motor drive is shown in Fig. 15.. The synchronous machine is fed from a current source d.c. link converter system, which consists of a thyristor bridge rectifier connected to a three-phase thyristor inverter through a smoothing inductor. The input rectifier is powered from three-phase a.c. supply lines, and its gating signals are provided by a digitally controlled firing circuit. The optical encoder which is composed of a coded disk attached to the motor shaft and four optical sensors, providing rotor speed and position signals. The inverter triggering pulses are synchronized to the rotor position reference signals with a delay angle determined by an 8-bit control input. The inverter thyristors are naturally commutated by the machine voltages during normal operation. The speed signal, which is a pulse train of frequency proportional to the motor speed, is fed to a programmable counter used for speed sensing. The machine field current is supplied by a d.c. source through a chopper which is under the microprocessor control. 141 9/7/07, 4:4 PM

14 Power Electronics The stator and field currents are detected by current sensors, and amplified by optically-isolated amplifiers. The output signals are multiplexed and converted to digital form by a high speed analog-to-digital converter. The principal functions of the microprocessor are monitoring and control of the system variables for the purpose of obtaining desired drive features. It can also perform various auxiliary tasks, such as protection, diagnosis and display. Fig. 15. Block diagram of micropocessor synchronous motor drive Since the self-controlled synchronous motor is unable to start by itself, the microprocessor has to ensure forced-commutation of the inverter from standstill up to speeds about 10 per cent of the nominal speed, when the machine voltages are sufficient to commutate the thyristors. In normal operation, commands are fetched from the input-output terminal, and system variables (the d.c.-link current, the field-current, the rotor position and speed) are sensed and fed to the CPU. After processing, the microprocessor issues control signals to the input rectifier, the machine inverter and the field chopper, so as to provide the programmed drive characteristics. 14 9/7/07, 4:4 PM

Control of A.C. Drives 143. Digital Control of Power Circuits The digital control of the power converters requires suitable interface circuits between the CPU and the power stages. Pulse generation is accomplished by hardware logic circuits, with a view to save the CPU time for monitoring and control tasks. These circuits convert the CPU commands to appropriate triggering pulses for the converters. (a) Input Rectifier Firing Angle Control The d.c. link current is controlled by varying the input rectifier firing angle. The schematic diagram of the digital firing circuit which is used as interface between the microprocessor and the thyristor bridge rectifier is shown in Fig. 15.3. Fig. 15.3 Schematic diagram of digitally control firing circuit It consists essentially of three delay control circuits and a pulse distributing circuit. Phase-locked loop principle is used to synchronize control signals to line-to-line voltages and the desired delay angle with respect to the a.c. input is provided by a digital comparator. This angle is determined by the 8-bit control input, providing a resolution of 0.7 over the range 0 180. (b) Rotor Position Sensing and Inverter Firing Angle Control In order to obtain self-control feature, the inverter firing, signals must be synchronized to the rotor position. This can be achieved by using as reference, either the machine 143 9/7/07, 4:4 PM

144 Power Electronics voltages for the signals provided by a position sensor. Here, a simple optical encoder is used to produce both speed and position reference signals necessary for starting and normal operation. Figure 15.4 shows details of the encoder construction and the output waveforms. The encoder disk has 56 teeth on the outer perimeter and two 90 slots on the inner perimeter. Each of the four sensors consists of light emitting diode and a phototransistor mounted in a moulded housing. The phase reference signals (P 1, P and P 3 ) are produced by three sensors positioned at 10 intervals. The fourth sensor produces a higher frequency square wave (S 4 ) of frequency proportion to the motor speed. As both rising and falling edges of this signal are deleted, the number of pulses generated for every 360 of rotation is equal to 56. This pulse train is used in inverter firing angle control and speed sensing. The schematic diagram of the inverter firing angle control circuit is shown in Fig. 15.5 with the waveforms for one phase. It consists of three delay control circuits, a thyristor address register, and a pulse distributing circuit. Linear digital ramp technique is applied to convert the digital control input to corresponding delay angle for the inverter triggering pulses. The 8-bit input control word contains mode operation and delay angle informations, which are defined respectively by the first bit and the remaining 7 bits. Depending on the mode bit, motoring or generation operation is selected with a delay angle equal to f or (180 f). Idealized motor voltage and current waveforms are shown in Fig. 15.5. waveforms Fig. 15.4 Optical-Encoder. (a) Encoder disk (b) Output 144 9/7/07, 4:4 PM

Control of A.C. Drives 145 Fig. 15.5 Idealized motor voltage and current waveforms (a) Motoring (b) regeneration At the beginning of each half-cycle of the phase-reference signal, the delay angle information f is loaded into the counter, and the count is incremented or decremented by the speed pulses from the encoder. When the terminal count is reached, a pulse is generated and applied to the clock input of the D-type flip-flop. The three-output signals, with a delay angle f or (180 f) with respect to the phase reference signals, are combined in the decoder to produce six modulated triggering pulse trains. These pulse trains are 10 wide with 60 of phase-angle between each other. The maximum count for each half-cycle is 17 so that the resolution in delay angle is about 1.4 electrical degrees. If a higher resolution is desired, a more elaborate encoder is required to generate a higher rate speed signal. (c) Field-chopper control The machine field current is supplied by a d.c. source through a transistor chopper which is controlled by a 0 10 V input voltage. The bit command from the microprocessor is converted to corresponding analog signal by a digital-to-analog converter. The obtained resolution for the control voltage is 40 mv. 3. System Operation The characteristic of the drive depends on the d.c. link current, the field current, and the inverter firing angle. These variables are independently controlled by the microprocessor to provide the desired features for all operation conditions. The converter system is symmetrical about the d.c. link so that power can flow in both directions, making regeneration operation possible. 145 9/7/07, 4:4 PM

146 Power Electronics (a) Motoring During motoring operation, the power flow is from the a.c. supply lines to motor, and a positive torque is developed. The firing angle of the input rectifier is between 0 and 90, resulting in a positive d.c. link voltage. The firing angle of the machine inverter is between 90 and 180, and it is naturally commutated by the motor terminal voltages. For constant stator and field currents, the average torque is a function of the inverter firing angle, the maximum value obtained for f = 180. Therefore, it is desirable that the inverter delay angle is maintained as close as possible to 180 so as to develop maximum torque. However, some margin is required to ensure reliable commutation. Fig. 15.6(a) The relation between the inverter firing angle and the stator current is stored in a look-up table in the memory. Desired torque speed characteristic is obtained by varying the field current in terms of the stator current. A separately excited d.c. motor is provided if the air-gap flux is kept constant. If the field current is 146 9/7/07, 4:4 PM

Control of A.C. Drives 147 made proportional to the stator current, a series d.c. motor characteristic results. In this case, the commutation is improved as the machine voltage increases with increased motor current. As for the delay angle, the relation between the field current and the stator current is provided by a function generator in the form of a look-up table stored in the memory. (b) Regeneration During regenerative braking, the motor operates as an alternator, supplying power to the a.c. lines. The developed torque is negative with the effect of reducing the motor speed. Since the d.c. link current diction is fixed, power flow reversal is achieved by reversing the d.c. link voltage. The firing angles must be consequently changed to be in the 90 180 range for the input rectifier, and in the 0 90 range for the machine converter. Their functions are then reversed. As in the motoring mode, the braking torque is determined by the inverter firing angle and the field current. Fig. 15.6 (b) and (c) Inverter delay angle control circuit various waveforms (c) Starting At standstill, the machine terminal voltage is not available so that the synchronous motor is unable to start itself. Consequently, the microprocessor is required to commutate the inverter to start and accelerate the motor until the 147 9/7/07, 4:4 PM

148 Power Electronics terminal voltage is sufficiently large to ensure reliable commutation. The forced commutation can be accomplished either by an auxiliary commutating circuit or by interrupting the d.c. link current prior to each commutation. The latter solution is used in the present system for simplicity. During starting operation, the delay counters are disabled, and the thyristor pairs are addressed directly by the CPU through the thyristor address register. The phase-reference signals available at standstill are used to determine the initial rotor position with a precision of ± 30 electrical degrees. An appropriate thyristor pair is triggered to create a motor torque in the desired direction. Immediately at the first change of state of the phase-reference signals, the inverter delay angle is precisely fixed at 180, providing maximum starting torque. At each change of state of the phase-reference signals, the forced commutation is executed by microprocessor. The rectifier is forced into inverter operation so as to reverse the d.c. link voltage until the current becomes zero. Triggering pulses are directed to the following thyristor pair by the CPU which writes an appropriate 3-bit address into the thyristor address register. Natural commutation begins when the motor speed has reached a value which produces sufficiently large terminal voltage to ensure reliable commutation. The delay counters are then enabled, and the thyristors are triggered with a delay angle determined by an 8-bit control word from the CPU. Fig. 15.7 Idealized relation between average torque and inverter firing angle (d) Reversing Reversal of rotation direction is achieved without any change in power connections, by reversal of the triggering sequence of the inverter thyristors as the frequency goes through zero value. This is accomplished by a simple selection logic circuit added to the inverter pulse distributing circuit. Thus, full four-quadrant operation of the drive is obtained. 4. Speed Control System Example In the proposed microprocessorbased synchronous motor drive, the feedback configuration, as well as the controller characteristics are defined by software. Therefore, they can be readily modified without change in hardware to provide various drive characteristics in order to satisfy particular applications requirements. (a) System hardware: Here, a 4-pole 10 kva synchronous motor with the parameter a mature current 7 A, field current.9 A, speed 1500 rpm, 148 9/7/07, 4:4 PM

Control of A.C. Drives 149 X ml = 3.5 W, X md = 5.4 W, R a = 0.113 W, R f = 41 W, and 8-bit microcomputer based on the 8080 CPU is used. The thyristor input rectifier is powered from a three-phase, 50 Hz, 0 V line-to-line supply providing maximum d.c. link voltage equal to 97 V. Its firing angle is determined by an 8-bit control word, resulting in 0.7 angle resolution. The smoothing inductor parameters are L d = 40 mh, and R d = 0.9 W. The inverter firing angle is determined by a 7-bit control word, which provides 1.4 of resolution. A d.c. generator supplying a resistor bank is used as a load for the motor. The load torque is, therefore, proportional to the motor speed. However, it can be considered as a constant for small speed variations. In order to provide fast current acquisition, a high speed input output module was built using an 8-bit, 6 ms analog to digital converter. Motor speed information is provided by a counter which is clocked by speed pulses from the optical encoder. (b) Control scheme and software implementation: The feedback configuration of the speed control system under consideration is depicted in Fig. 15.8. It Fig. 15.8 Microprocessor based speed control system consists of a main speed control loop including an inner current control loop, and a field current control loop. This system is a non-linear sampled data one, 149 9/7/07, 4:4 PM

150 Power Electronics with multi-rate sampling, precise study of which requires hybrid or digital simulation. Nevertheless, system behaviour can be predicted with acceptable accuracy using linearized models for system components, and classical Z- transform analysis (or state space analysis). Simplified control diagrams of the current regulating routine and speed regulating routine are shown in Fig. 15.9. These routines are executed on interrupt signals provided by the real-time clocks. The motor speed is sampled and processed every 0 ms on interrupt signals generated by a programmable timer. The motor speed is sensed and compared with the speed command provided by the CPU. Speed error is processed by the speed controller which generates link current command. Current limiter feature is provided by the limiter incorporated in the speed controller. Fig. 15.9 Simplified control diagrams The digital proportional integral algorithm executed by the speed controller is defined by the differential equation: I r(k) = I r(k 1) k p e n(k 1) + (k p + k i T sn )e n(k) (15.1) where I r(k) is the link current reference value at the k-th point, e n(k) is the speed error value at k-th point, T sn is the speed sampling period, and k p and k i are, respectively, the proportional and integral control gains. 150 9/7/07, 4:4 PM

Control of A.C. Drives 151 Inverter and field current commands are derived from the link current command by two function generators stored in the memory in the form of look up tables. An interrupt signal for the current regulating routine is generated on each zero-crossing of the three-phase a.c. input voltage. Therefore, the link current is sampled and processed every one-sixth of the a.c. input period, that is 3.33 ms for 50 Hz supply. The reason for this sampling rate is that the thyristor rectifier firing angle can be modified only at discrete instants with average period equal to one-sixth of the a.c. input period. The link current is sensed and compared to the reference value provided by the speed regulating routine. Current error is processed by the current controller, producing firing angle command for the thyristor input rectifier. The proportional integral algorithm executed by the current controller is defined by the difference equation: a(k) = a(k 1) K p e i (k 1) + (k p + k i T Si )e i (k) (15. ) Fig. 15.10 Speed regulating routine Fig. 15.11 Current regulating routine 151 9/7/07, 4:4 PM

15 Power Electronics Fig. 15.1 Starting routine where a(k) is the firing angle at k-th point, e i (k) is the current error at k-th point, and T si is the current sampling period. The flow charts of speed regulating routine, current regulating routine and starting routine, are shown respectively in Figs 15.10, 15.11 and 15.1. 15 9/7/07, 4:4 PM