Solution of Tutorial 2 Converter driven DC motor drive

Transcription

1 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) rd / sec 50 I A KE KE KE 3.18 V / rd / sec rev /min rd /sec V I K V E 150 D 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

2 I 10 A 3 V V 15 D (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 ( x1) 50 ( x 1)( x 1) or ( x 1) ( x 1) 50 olution of Tutoril F. hmn/aug 013

3 00x. 00. x 1 098x x so tht 05T. / e Ts T L L 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, , T 10 E D I V K T olution of Tutoril 3 F. hmn/aug 013

4 Where K T is found from 100 I KE KE 6. 8 K K V / rd /sec or Nm / A E T 50 / e / e , where is in msec. 50 / 1000 / e 1. 8 e / 1000 / or 1. 05e 0. 05e 1 olving this eqution itertively: 50 / in µsec 105e e / sec is the solution 6 L mh olution of Tutoril 4 F. hmn/aug 013

5 Question 3 V V V, I 400 A, , L mh, L / 5 10 sec 1 KE V / rd /sec, rted 0 rev /min T Nm, tted 1 T tted 393. Nm 393. I 00 A Assuming continuous conduction, 0 V DV I KE V D 05. Whether I is continuous cn be determined in two wys: For 800 rev /min, E V V V with continuous conduction V E I A which is ve 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

6 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 / ln e 1 1e sec V ( ) E V V I A T Nm 3(ii) t V DV ( 1 ) E T rms 0. 5 ( ) V insted of V 05. V 184V for continuous conduction. rms s olution of Tutoril 6 F. hmn/aug 013

7 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 K K E E KE V / rd /sec rpm E V V I E V D I V e 1 E 00 e e e DT / /( ) min T / /( ) I 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, KE K E olution of Tutoril 7 F. hmn/aug 013

8 KE. V / rd /sec 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 / sec 05T. s/ e 1 E V Ts/ e rd/sec K. E rev/min. DV E A s I T K I Nm. T olution of Tutoril 8 F. hmn/aug 013

9 Question 6 00Vdc T 1 D i L V s T D 1 E (i) T 100sec, sec Torque 15 Nm, I 15A, peed 150 rev /min rd /sec E K V E V E I V T sec, T 1. 1 sec ON OFF (ii) I V 1 e E 00 1 e e e TON / 008. mx T / A TON / 008. V e 1 E e 1 I min T / e e A Imx Imin A olution of Tutoril 9 F. hmn/aug 013

10 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 T e 1 e 1 TON / e /.. T ON e T ON / or e or T / ON 5 or T sec ON T (i.e., T is ON for = 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

11 I s DVs E 3 DV V s DVs D Diode D ON time = DT sec 36.6 sec s Thus, T ON time = = 63.4 sec, s found previously. (iv) When motor speed flls to 750 rev/min, E 750 KE V If the sme breking current is mintined, I s DVs E DVs DV V s 5.9 D Diode D ON time = DT sec 1.95 sec s At this speed, T ON time increses to: = 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

12 Question 7 V~ V = hp 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 = = 66 Nm 0.18 rd/sec = 1.74 V/rd/sec B. E ; E = V I K E V mx cos I cos = = V N 105 rev/min 0.18 olution of Tutoril 1 F. hmn/aug 013

13 C. DC output power of converter, Pdc 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 lgging olution of Tutoril 13 F. hmn/aug 013

14 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 V Figure Q A. I 38 A The KVL round the rmture-converter circuit is V A V ( ) V V mx cos cos , so tht Power fed bck immeditely t the strt of breking V I Wtts B. I = 76 A V ( ) V olution of Tutoril 14 F. hmn/aug 013

15 V mx cos cos , so tht 136. Power fed bck immeditely t the strt of regenertive brking V I ,81 Wtts Question 9 V~ I L E A. T = K t I = = 66 Nm B. E V I Vmx 1cosI 1 cos Volts ev/min 0.18 DC power to the motor = P dc = = 8, Wtts olution of Tutoril 15 F. hmn/aug 013

16 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 = id = d A 180 Input Power fctor, PF = Pdc 8, V I 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 KE K 0 45 KE V/rd/sec 15.6 Vmx V For = 1000 rev/min, with full lod Vdc V V mx Vdc cos E olution of Tutoril 16 F. hmn/aug 013

17 cos (c) Brking torque = KT I Nm I A V I E ~ V dc I E KE E = 175V V mx cos cos V 1 85 cos (d) J Tm = 1.5 kg-m ; T Lm = N Nm D Tm = 0.1 Nm/rd/sec; J D T T Nm Tm m Tm m Lm m DTm TLm m 31 rd/sec J 1. 5 Tm 31. /sec L 31 rd 10 (e) V I E 0. 3I K 0. 3I I E V cos cos V I ( )/ A which is more thn 5% of I rted the lod current is continuous. 900 (f) At 900 rev/min, E Volts olution of Tutoril 17 F. hmn/aug 013

18 which is more thn the converter output voltge, V = V mx cos V V E I 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 = H T rted Prted 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 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 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

19 L For this problem, L e 1 E sin V L e 1 mx 1L tn tn [For the inequlity, LH = e 1 sin e 1 = H = V E mx 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 V V I E V V V cos mx E The H of the inequlity = V 340 mx olution of Tutoril 19 F. hmn/aug 013

20 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 = = 5. mh. Question 1 = ; L = H E = 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 = v bn i b V d L = 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 V (d) 6V cos7 cos 5 sin7 sin mx v6 sin6 t cos6 t The mplitude of v 6 is mximum when = 90. v 6 6Vmx 1 1 cos6t for = 90 (worst cse). 6Vmx 1 1 V6,M 0.3V mx V olution of Tutoril 0 F. hmn/aug 013

21 I 6 V6 Impednce to the 6th hrmonic voltge j I j 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 = A (c) E 30 KE V/rd/sec = K T Nm/Amp 0 o 0 rev/min 6.8 rd/sec Vdc E I V Also, 3V V mx Vdc cos cos 79. 3V mx olution of Tutoril 1 F. hmn/aug 013

22 Question , L 0. 00H, K. V / rd /sec., I 500A E rted 1500 () rev /min 157 rd /sec. E K Volts E 0 3Vmx Vdc E I cos V Now Vmx V V cos dc V mx cos ( b ) 3 3Vmx 3 L Vdc cos I Volts E V I Volts dc rd /sec rd /sec 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 mt f Noting tht the friction torque ssists the decelertion, the mximum decelertion is Tmx T 40 5 mx 314 rd /sec J mt f olution of Tutoril F. hmn/aug 013