chool of Electricl Engineering & Telecommunictions, UNW olution of Tutoril Converter driven DC motor drive Question 1. T V s D V I L E V 50 V,.5, I 0 A rted rted f 400 Hz, 0 rev/ min s rted (i) 0 6.8 rd / sec 50 I A KE 50 0.5 KE 6.8 50 50 KE 3.18 V / rd / sec 6.8 300 rev /min 31. 4 rd /sec V I K 05. 318. 314. 150V E 150 D 06. 50 At constnt field, when the lod torque reduces to 50%, the rmture current is lso reduced by 50%. olution of Tutoril 1 F. hmn/aug 013
I 10 A 3 V 10. 5 3. 18 31. 4 15 V 15 D 05. 50 (ii) The pek-pek ripple cn be derived in the following procedure: I V 1 e E 1 e DT / mx T / I V E e 1 DT / e 1 min T / The pek-pek ripple is given by: DT / DT / V 1e V e 1 I Imx Imin T / T / 1e e 1 Multiplying both numertor nd denomintor of the first term by e T / T / ( 1D) T / DT / T / ( 1D) T / DT / V e e e 1 V e e e 1 I T / T / T / e 1 e 1 e 1 Mximum ripple occurs for D 05. T / 0. 5T / 0. 5T / T / 05T. / V e e e 1 V e e 1 I T / T / e 1 e 1 Let e x T / V x x 1 I x 1 For pek-pek ripple of 10%, 01I. V x x 1 x 1 01. 05. ( x1) 50 ( x 1)( x 1) or ( x 1) 0. 10. 5 ( x 1) 50 olution of Tutoril F. hmn/aug 013
00x. 00. x 1 098x. 10. 10. x 1. 040816 098. so tht 05T. / e 1. 040816 0.5 Ts 0.04 05T. 0. 0315 004. L 0. 0315 L 0. 0315. 5 78 mh Question T E I L V s D V I V E e 1 DT / e 1 min T / For discontinuous conduction, I min 0 T 1/ f 1000 sec D 005., T DT 50 sec ON V V, 0. 035, T 10 E D I 13 13 0. 035 1. 83 V K 1. 966 T olution of Tutoril 3 F. hmn/aug 013
Where K T is found from 100 I KE 400 0. 035 KE 6. 8 K K 1. 966 V / rd /sec or Nm / A E T 50 / e 1 1. 83 1000 / 0. 035 e 1 0. 035, where is in msec. 50 / 1000 / e 1. 8 e 1. 83 0 50 / 1000 / or 1. 05e 0. 05e 1 olving this eqution itertively: 50 / in µsec 105e. 1000 1 0 05e /. 50 500 1000 000 3000 4000 5000 1.06 1.063 00 5000 sec is the solution 6 L 5000 10 0. 035 0. 165 mh olution of Tutoril 4 F. hmn/aug 013
Question 3 V V V, I 400 A, 3 0. 035, L 0. 163 mh, L / 5 10 sec 1 KE 1. 966 V / rd /sec, rted 0 rev /min T 1. 966 400 786. 4 Nm, tted 1 T tted 393. Nm 393. I 00 A 1. 966 Assuming continuous conduction, 0 V DV I KE 00 0. 035 1. 966 6. 5 13. 47 19. 96 V 19. 96 D 05. Whether I is continuous cn be determined in two wys: 800 1. For 800 rev /min, E 1. 966 164. 6 V V 19. 96 V with continuous conduction V E 19. 96 164. 6 I 1066. 5A which is ve 0. 035. I DT / V e 1 E min T / e 1 which is ve. ince I cn not be ve, when the motor is driven from 1-Q converter, it must be discontinuous. olution of Tutoril 5 F. hmn/aug 013
3(i) V s v E T on = DT s T off =(1-D)T s t I i t t T s V E t lne 1 1e E DT / DT / 3 3 3 05. 10 05. 10 0. 163 10 3 164. 6 3 510 510 3 ln e 1 1e 0. 77 10 sec 0. 035 164. 6 05. V ( 1077. ) E V 167. 8 V 1 167. 8 164. 6 I 99. 77 A 0. 035 T 1. 966 99. 77 196. 15 Nm 3(ii) t V DV ( 1 ) E T rms 0. 5 ( 1 0. 77) 164. 6 33800 633 00 V insted of V 05. V 184V for continuous conduction. rms s olution of Tutoril 6 F. hmn/aug 013
Question 4 V 00 V, 0. 33, L 11 mh, 100 rev /min, when D 1 I 0 A, 800 rev /min f 500Hz; T m sec. 00 I E 0 0. 33 K 0 6. 6 15. 6 K E E 00 6. 6 KE 1. 54 V / rd /sec. 15. 6 800 @ 800 rpm E 1. 54 6. 8 18. 95 V V I E 6. 6 18. 95 135. 55 V D 0. 678 I V e 1 E 00 e 1 18. 95 e 1 0. 33 e 1 0. 33 3 3 DT / 0. 678 10 /( 33. 310 ) min T / 3 3 10 /( 33310. ) I 1. 04156 1 6 410. 75 1. 0619 1 406. 87 390. 76 16. 11 A is continuous. Question 5 I 5 A, V 50 V, 0. 7, L mh, f 1000Hz; T 1 msec. For D 1, 1000 rev /min V DV 50 I KE 1000 or, 50 5 0. 7 KE 17. 5 104. 67 K E olution of Tutoril 7 F. hmn/aug 013
50 17. 5 KE. V / rd /sec 104. 67 For just discontinuous conduction V DT / e 1 I min 0 T / e 1 E DT s/ e 1 E Vs e T/ s 1 Given D 0., 5 V 50 V, 0. 7, 3 3 L / 10 / 0. 7. 857 10 sec 05T. s/ e 1 E 50 114. 1 V Ts/ e 1 114.1 114.1 rd/sec K. E 114.1 491 rev/min. DV E 0.5 50 114.1 A s I 15.57 0.7 T K I. 15.57 34.57 Nm. T olution of Tutoril 8 F. hmn/aug 013
Question 6 00Vdc T 1 D i L V s T D 1 E (i) 1 005. T 100sec, 0. 08 sec 10000 1. 8 15 Torque 15 Nm, I 15A, 1 150 peed 150 rev /min 130. 8 rd /sec E 130. 8 K 130. 8 V E V E I 130. 8 15 1. 8 157. 83 V 157. 83 0. 789 00 T 78. 9 sec, T 1. 1 sec ON OFF (ii) I V 1 e E 00 1 e 130. 8 1 e 1. 8 1. 8 1 e 0. 0000789 TON / 008. mx T / 0. 0001 0. 08 1 0. 997186109 0. 008319 111. 1 7. 66 111. 1 7. 66 15. 039 A 1 0. 996434941 0. 003559 0. 0000789 TON / 008. V e 1 E e 1 I min 111. 1 7. 66 T / 0. 0001 e 1 008. e 1 87. 568 7. 66 14. 965 A Imx Imin 1. 5039 14. 965 0. 074 A olution of Tutoril 9 F. hmn/aug 013
Note: ipple current clcultion t switching frequencies of 10 khz or so involves subtrction of two quntities which re nerly the sme. The switching times, so found, re lso smll, often few tens of microseconds. Gret cre thus must be tken in your clcultions. Up to six or seven deciml plces must be retined before rounding off. (iii) For brking D E I L + V s T V emember tht for the brking mode of opertion, T ON is the on time for D nd OFF time for T. For pek brking current of 3 A, TON TON / Vs e 1 E e 1 3 I min 7. 66 T 0. 0001 0. 08 e 1 e 1 TON / e 1 111. 1 7. 66 1. 0035778 1 /.. T ON 40 66 0 0035778 e 1 1. 309391 10 3 0. 001309391 111. 1 T ON / or e 1. 001309391 3 or T / 1. 3085344 10 0. 0013085344 ON 5 or T 3. 6638965 10 36. 6 sec ON 36. 6 T 1 0. 664 (i.e., T is ON for 100 36.6 = 63.4 sec) 100 Alterntive solution for (iii) Note tht the pek brking current is bout the sme s the DC current I (ssuming tht ripple is smll). With this ssumption we my lso write, olution of Tutoril 10 F. hmn/aug 013
I s DVs E 3 DV 3 1.8 130.8 73.V s DVs 130.8 1.8 73. D 0.366 00 Diode D ON time = DT 0.366 100 sec 36.6 sec s Thus, T ON time = 100 36.6 = 63.4 sec, s found previously. (iv) When motor speed flls to 750 rev/min, E 750 KE 1 78.5 V If the sme breking current is mintined, I s DVs E DVs 78.5 3 1.8 DV 3 1.8 78.5 5.9 V s 5.9 D 0.195 00 Diode D ON time = DT 0.195 100 sec 1.95 sec s At this speed, T ON time increses to: 100 1.95 = 87.05 sec. Using this procedure, you should be ble to find the speed t which the current controller comes out of limiting mode (i.e., becomes unble to mintin the breking current t -I rted ). olution of Tutoril 11 F. hmn/aug 013
Question 7 V~ V = 0.3 10hp 30V 100rev/min E Figure 1 K E = 0.18 V/rev/min = 0.18 V/rev/sec = A. I = 38 A T = K t I = 1.7438 = 66 Nm 0.18 rd/sec = 1.74 V/rd/sec B. E ; E = V I K E V mx cos I cos30 38 0.3 = 0.79 11.4 = 191.39 V 191.39 N 105 rev/min 0.18 olution of Tutoril 1 F. hmn/aug 013
C. DC output power of converter, Pdc 0.79 38 7706 Wtts Input voltge, v s, V Output voltge, v o, V Input current, I p, A The input current wveform to the converter is the rms input current = 1 id 38 A 0 The input Volt-Ampere (VA) = rms input voltge rms input current = 38 = 9880 The input Power Fctor = Output Pdc of Converter Input VA 7706 0.78 lgging. 9880 olution of Tutoril 13 F. hmn/aug 013
Question 8. While brking regenertively, (i.e., opertion in the second qudrnt), the polrity of the motor voltge nd motor-converter connections re s indicted in the figure Q. Note tht the motor rmture connections hve been reversed, so tht it is now ble to source current into the converter, llowing the motor overhuling energy to be returned to the AC source. V~ I V 0. 3 + + 191.39V Figure Q A. I 38 A The KVL round the rmture-converter circuit is V 191.39 38 A V ( 0.3 38 191.39 ) 179.99 V V mx cos 179.99 cos 0.7686, so tht 140.3 Power fed bck immeditely t the strt of breking V I 179.99 38 6839.6 Wtts B. I = 76 A V ( 0.3 76 191.39 ) 168.59 V olution of Tutoril 14 F. hmn/aug 013
V mx cos 168.59 cos 0.7199, so tht 136. Power fed bck immeditely t the strt of regenertive brking V I 168.58 76 1,81 Wtts Question 9 V~ I L E A. T = K t I = 1.74 38 = 66 Nm B. E V I Vmx 1cosI 1 cos30 0.3 38 18.47 0.3 38 07.07 Volts 07.07 1138 ev/min 0.18 DC power to the motor = P dc = 18.4738 = 8,301.86 Wtts olution of Tutoril 15 F. hmn/aug 013
Input voltge v s, V Output voltge v o, V Input current I p, A The rms input current to the converter is Input rms current, I rms = 1 180 30 id = 1 180 180 30 38 d 180 30 38 34.69 A 180 Input Power fctor, PF = Pdc 8,301.86 V I 34.69 rms rms 0.9 lgging Question 10 () A hlf-controlled converter will not llow regenertive brking, s opposed to the fullycontrolled converter which would. (b) 0V = I KE 100 0.3 150 KE 45 15.6 K 0 45 KE 1.3933 V/rd/sec 15.6 Vmx 40 339.36 V For = 1000 rev/min, with full lod 1000 6.8 Vdc 0.3 150 1.3933 190.83 V V mx Vdc cos 190.83 E olution of Tutoril 16 F. hmn/aug 013
1 190.83 cos 8 339.36 (c) Brking torque = KT I 417.99 Nm I 150 300 A V I E ~ V dc I + 100 E KE 1.3933 + + E = 175V V mx cos 16.15 cos 300 0.3 175 85V 1 85 cos 113 16.15 (d) J Tm = 1.5 kg-m ; T Lm = 375 37.5 N Nm D Tm = 0.1 Nm/rd/sec; J D T T 150 1. 3933 417. 99 Nm Tm m Tm m Lm m 417. 99 DTm TLm 417. 99 0. 1 15. 6 37. 5 m 31 rd/sec J 1. 5 Tm 31. /sec L 31 rd 10 (e) V I E 0. 3I K 0. 3I 1. 3933 47. 1 0. 3I 65. 6 E V 16. 15cos 16. 15cos 108. 075V I ( 108. 073 65. 6)/ 0. 3 141. 5A which is more thn 5% of I rted the lod current is continuous. 900 (f) At 900 rev/min, E 1. 3933 131. 5 Volts olution of Tutoril 17 F. hmn/aug 013
which is more thn the converter output voltge, V = V mx cos 108. 1 V V E 108.1 131.5 I 77.17 A, which is negtive. 0.3 The rmture current will be discontinuous. Question 11 Motor rtings re: 500W; 1000 rev/min Motor prmeters re: = 0.15, L = 0.005 H T rted Prted 500 4.775 rted 1000 Nm We will ssume tht the motor runs t the rted speed when the firing ngle 0. For opertion t rted speed, nd ssuming continuous conduction, the rted rmture voltge is V V 40 mx cos 0 16.074 V We need to find the bck emf in order to find I min nd to determine whether the rmture current will be continuous. The motor voltge constnt K E nd the rted current hve not been given. I V E 16.074 E.(1) 0.15 Now EI 500 () By solving equtions (1) nd () simultneously I.3 A nd E 15.7 V The condition for opertion t the cont/disc conduction boundry is given by eqution 4.3 which is: olution of Tutoril 18 F. hmn/aug 013
L For this problem, L e 1 E sin V L e 1 mx 1L 1314 0.005 tn tn 79.18 0.15 [For the inequlity, LH = 0.153.14 3140.005 0.15 e 1 sin 0 79.10 0.15 0.15 3.14 314 0.005 3140.005 e 1 = 0.635 H = V E mx 15.7 0.634 40 Both sides of the inequlity re nerly equl so tht conduction (even with = 0) is on the verge of being discontinuous. At 500 rev/min (hlf the rted speed), the firing ngle is expected to be bout, so tht conduction is very likely to be discontinuous. Therefore, dditionl inductnce in the rmture circuit will thus be needed]. At 500 rev/min, becuse of the constnt lod torque, P T 50 W ince the lod torque is unchnged, rmture current I remins unchnged (for the. E. motor) nd the bck emf is hlved, so tht, E 15.7 107.86 V V I E.3 0.15 107.86 108.1 V V V cos 108.1 mx E 107.86 The H of the inequlity = 0.318 V 340 mx olution of Tutoril 19 F. hmn/aug 013
By iterting the LH of the inequlity with vrious vlues of L, it cn be found tht L 54.7 mh Thus, the dditionl externl inductnce required = 54.7.5 = 5. mh. Question 1 = 0.045 ; L = 0.73 10 3 H E = 30 V @ 3500 rev/min I rted = 89 A ; f = 50 Hz 3-phse, fully-controlled converter v o v n i L i T 1 T 3 T 5 = 0.045 v bn i b V d L = 0.073 H i c + T 4 T 6 T E V v cn (c) 3V mx Vdc 30 V cos For = 0, V dc = mximum = 30 V 30 V 40.7 V 3 mx V M, supply = V mx 170.5 V (d) 6V cos7 cos 5 sin7 sin 5 14 10 14 10 mx v6 sin6 t cos6 t The mplitude of v 6 is mximum when = 90. v 6 6Vmx 1 1 cos6t 14 10 for = 90 (worst cse). 6Vmx 1 1 V6,M 0.3V mx 55.78 14 10 V olution of Tutoril 0 F. hmn/aug 013
I 6 V6 Impednce to the 6th hrmonic voltge 55.78 0.045 j 6 50 0.73 10 55.78 55.78 I6 40.56 0.045 j1.375 1.37688 A This is the dominnt ripple current in the rmture. ince the rted rmture current is 89 A, the motor current must be derted to: 3 new I rting = 89 40.56 79. A (c) E 30 KE 0.68 3500 V/rd/sec = K T Nm/Amp 0 o 0 rev/min 6.8 rd/sec Vdc E I 0.68 6.8 0.045 79. 43 V Also, 3V V mx Vdc cos 43 43 cos 79. 3V mx olution of Tutoril 1 F. hmn/aug 013
Question 13 0. 0, L 0. 00H, K. V / rd /sec., I 500A E rted 1500 () 0 1500rev /min 157 rd /sec. E K. 5 157 353. 5 Volts E 0 3Vmx Vdc E I cos 353. 5 500 0. 0 363. 5 V Now Vmx 300 44. V V cos dc 0. 8963 3V mx cos 1 0. 8963 6. 3 ( b ) 3 3Vmx 3 L 3 314 1 10 Vdc cos I 363. 5 500 363. 5 150. 36 1. 88 Volts E V I 1. 88 500 0. 0 0. 88 Volts dc 0. 88 rd /sec 90. 17rd /sec 861. 5 rev /min 5. Question 14 For the fstest ccelertion, T mx = T rted Tmx KTI 40Nm Tmx T 40 5 mx 500 rd /sec J 0. 014 mt f Noting tht the friction torque ssists the decelertion, the mximum decelertion is Tmx T 40 5 mx 314 rd /sec J 0. 014 mt f olution of Tutoril F. hmn/aug 013