ECEN 613. Rectifier & Inverter Circuits

Transcription

1 Module8a Rectifier & Inverter Circuits Professor: Textbook: Dr. P. Enjeti with Michael T. Daniel Rm. 04, WEB Power Electronics Converters, pplications & Design (Third edition), by: Ned Mohan et al., John Wiley COURSE WEBPGE: 1

2 DCC Inverters Single Phase Halfbridge Inverter Fullbridge Inverter Dual output voltage configuration Output filter design Three Phase Inverter Deadtime effect PWM Techniques Sinusoidal PWM Selective harmonic elimination (SHE) Space vector PWM

3 Single Phase Halfbridge Inverter v d i d 1phase switch mode inverter filter v o i o v o Rectifier 3 Inverter i o 1 Inverter 4 Rectifier v o i o During intervals 1 and 3, power flow is from the dc side to ac side and during intervals and 4, power flow is from the ac side to dc side. Thus, power flow is bidirectional. 3

4 Single Phase Halfbridge Inverter o T T N i o N Output oltage O = / when T is on O = / when T is on Switch Ratings oltage rating: rate = Current rating: I rate = i peak Maximum output voltage max = / Reference voltage, control (Frequency, f control ) Carrier wave, tri (Frequency, f tri ) 0 t control ( t) = sin( ωt) m 4

5 Single Phase Sinusoidal PWM Principle:Bipolar 0 t 0 tri t 1 t T tri control o control o = t tri d verage area per cycle is 1 d d d = t1 ( t t1) ( T ) T o, av t T t1 = (control 4tri T t = (4tri 4 tri tri ) control ) tri control t Substituting t1 and t in o,av yields = m d sin( ωt) o,1 where, m a = modulation index a control ; tri = 0 m a 1 5

6 Single Phase Sinusoidal PWM Principle:Bipolar o T T i o N Fundamental of O Output Chopped voltage O 0 t 0 m f N ma ftri = f 1 1 ( d ) h / o 1 m f Harmonics h of O m = 0.8, m 15 a f = m f ( m f ) ( m ) f 3m f ( 3m f ) 6

7 Single Phase Sinusoidal PWM Principle:Bipolar h m a m f m f ± m f ± 4 m f ± 1 m f ± 3 m f ± 5 3m f m f ± m f ± 4 m f ± 6 Table1. Generalized Harmonics of O for a arger m f ( Fundamental of O Output Chopped voltage 0 t d ) / o h 1 m f Harmonics h of m = 0.8, m 15 a f = m f ( m f ) ( m f ) O O 3m f ( 3m f ) 7

8 Single Phase Sinusoidal PWM Principle:Bipolar o T i o T N N SPWM output voltage waveform analysis =100, m a =0.8, m f =333, Fundamental f 1 =60Hz Switching f s = m f *60 = 19.98kHz, from Table 1, v ao ( t) dc dc = 0.8 ( ) sin( ωt) ( ) sin( m f ωt) dc dc 0. ( ) sin(( m f ± ) ωt)) ( ) sin((m f ± 1) ωt)) 8

9 Single Phase Sinusoidal PWM Principle:Bipolar control tri 0 t ( d 4 ( = 1.78) π ) / o inear Overmodulation Squarewave 0 t 1 m Overmodulation: a Higher fundamental component Contains lower order harmonics Fundamental component does not vary linearly 0 t (for m f =15) 9

10 Single Phase Sinusoidal PWM Principle:Bipolar o T i o T N N SPWM example: Bipolar Switching = 100, m a = 0.8 m f =333, Fundamental f 1 =60Hz 10

11 Single Phase Sinusoidal PWM Principle:Bipolar o T T N i o N SPWM example: Bipolar Switching = 100, = 10mH, R = 10 ohms m a = 0.8 m f = 333, Fundamental f 1 = 60Hz Switching fs= mf *60 = 19.98kHz, from Table 1, we have: Calculate load current io v O dc dc (t) = 0.8 ( ) sin( ωt) ( ) sin(mf ωt) dc 0. ( ) sin((mf ± ) ωt)) = 40sin( ωt) 11sin(331ωt) 40.9sin(333ωt) 11sin(335ωt) 11

12 Calculate Inverter Input current idc o T i o T N N Homework: Calculate first 3 terms of i dc (DC, first harmonic, and second harmonic) (Hint: use sw 1 switching function and i o from previous slide) 1

13 13

14 Single Phase Full Bridge Inverter Reference voltage, control (Frequency, f control ) Carrier wave, tri (Frequency, f tri ) 0 t b O T D oad T B i B D B T D T B D B d Unipolar switching sequence d Fundamental of O Output voltage O T is on when control > tri T B is on when control < tri T B is on when control > tri T is on when control < tri 14

15 Single Phase Sinusoidal PWM Principle: Unipolar Reference voltage, control (Frequency, f control ) Carrier wave, tri (Frequency, f tri ) Reference voltage, control (Frequency, f control ) Carrier wave, tri (Frequency, f tri ) 0 t 0 t O BO 0 0 d B = B Output voltage B Fundamental of B d 15

16 Single Phase Sinusoidal PWM Principle: Unipolar Reference voltage, control (Frequency, f control ) Carrier wave, tri (Frequency, f tri ) 0 t b O T T D D oad T B T B i B D B D B d d Fundamental of O Output oltage O Unipolar PWM switching results in higher quality output voltage 16

17 Single Phase Sinusoidal PWM Principle: Unipolar Output Chopped voltage O ( O ) d Fundamental of O h 1 m f Harmonics h of f 1 m f 3m ( m f 1) ( m 1 ) f f 4m f 17

18 Single Phase Sinusoidal PWM Principle: Unipolar SPWM example: Unipolar Switching b O T T D D oad T B T B i B D B D B =100, m a =0.8, m f = 33 (even), Fundamental f 1 =60Hz R = 10 ohms, = 10 mh Switching f s = m f *60=19.9kHz, from Table 1, v v O BO B ( t) = 40sin( ωt) 11sin(330ωt ) 40.9sin(33ωt ) ( t) = 40sin( ωt π ) 11sin(330( ωt π )) 40.9sin(33( ωt π )) ( t) = v 11sin(334ωt ) 1.5sin(663ωt ) 1.5sin(665ωt )... 11sin(334( ωt π )) 1.5sin(663( ωt π )) 1.5sin(665( ωt π ))... O ( t) v BO ( t) = 80sin( ωt) sin(663ωt ) 15sin(665ωt )... 18

19 Sinusoidal PWM Principle: Unipolar Example Contd. Calculate load current i = 19

20 Half Bridge Inverter for 10/40 Single Phase Output I DC O B I IB inear oad N Two half bridge inverters are employed Two IGBTs per phase and a split dclink can be used Reduced cost topology = 150 µh ; = 15 µf ; 40kHz switching frequency 5kW load per phase Homework: Calculate the required dc. Simulate the performance Calculate switch voltage & current ratings Calculate RMS current rating of the DC link capacitors 0

21 I DC Inverter feeding a 5 kw/ph inear load O B I IB inear oad N 5kW/ph 1

22 Inverter feeding a 5 kw/ph inear load I DC O B I IB inear oad I rms =4 I pk =58 N

23 Inverter feeding a 5 kw/ph inear load I DC O B I IB inear oad N 10Hz 3

24 Inverter feeding a 5 kw on phase and phaseb is open (linear load) I DC O B I IB inear oad N 60Hz 10Hz 4

25 Inverter feeding a 5 kw/ph Nonlinear load I DC O B I IB Noninear oad N 5

26 Inverter feeding a 4 k on phase and phaseb is open (Nonlinear load) I DC O B I IB Noninear oad N 6

27 6Switch Inverter For 1phaseDual oltage (10/40) DC n a b Output Filter Six IGBT devices are employed. Two IGBTs are used to create the neutral phase n is same; No split dclink; dclink current is twice the frequency compared to half bridge under unbalance load 7

28 6Switch Inverter For 1phase UnBalanced oad (5kW/ph) DC n a b Output Filter 8

29 Inverter Output Filter Design Considerations 9

30 Output Filter Design Sizing C Filter: Inductor and Capacitor o Inverter f C f Nonlinear oad Output filter is to attenuate harmonics at the load terminals. Smaller size of filter components Typical load is nonlinear (ex: computer power supplies,etc.) & draws 3,5,7 harmonics ssumption: Output filter is lossless, and the third harmonic current (of the load) is 80% of its fundamental current. 30

31 Output Filter Design Sizing C Filter: Inductor and Capacitor jnx i,n jxc n o,n OD Z,n Thetransferfunctionforilteris H n = o, n i, n = nx X << X C jx jz Z C, n (, n n X X C (1) ) ssume: ; The gain at n = 1 (fundamental frequency H1) X C H C,1 1 jx jz,1 Z X C 1 () 31

32 Output Filter Design Sizing C Filter: Inductor and Capacitor lso for no load condition, H n = n XC X X C = therefore Eqn(1) is Z, n 1 ( 3) X n 1 X C InordertosatisfyTHDrequirementoflessthan5% n 1 X X C ; 1 X X C 3. n (4) 3

33 Output Filter Design Sizing C Filter: Inductor and Capacitor jhx jxc h For the Nonlinear load The load terminal voltage is jhx X X h X h I h C = I (5) h C h h I h X X h c : equivalent voltage : harmonic : current at " h" harmonic : capacitive reactance (60Hz) : inductive reactance (60Hz) 33

34 Output Filter Design Sizing C Filter: Inductor and Capacitor From(5)wehave h hx = X 1 h X C *I h (6) Here X X C is small, Therefore, h hx I h h X X C << 1 Forthe3rdharmonic,h=3, 3 1 = 3 X 1 I 3 where, THD is 3 1 = 0.03 or 3% (7) 34

35 Output Filter Design Example 10k(5K per phase) single phase inverter, 1 = 10, produce I 1 = 41.67, f s = 0kHz, f 1 = 60Hz, and n= f s / f 1 = , THD= 3%. By using equation (4) X.09* 10 X C 4 The filter resonant frequency can be found with f f r 1 = X X C n f r 4150Hz 35

36 Output Filter Design Example I 3 = I 1 *0.8 = 5.95 and from equation (7), X = X = πf1 = 13µ H Use equation (4) to find the capacitor impedance, X C = C = πf1 X = 1µ F C 36

37 Inverter feeding a 5 kw/ph inear load I DC O B I IB inear oad N 37

38 Inverter feeding a 5 kw/ph Nonlinear load I DC O B I IB Noninear oad N 38

39 Inverter feeding a 5 kw/ph Nonlinear load I DC O B I IB Noninear oad N 39

40 Inverter feeding a 5 kw/ph Nonlinear load I DC O B I IB Noninear oad N 40

41 Inverter feeding a 5 kw on phase and phaseb is open (Nonlinear load) I DC O B I IB Noninear oad N 41