ELEC Power Electronics
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1 School of Elecrical Engineering & Telecommunicaions Universiy of New Souh Wales ELEC464 - Power Elecronics Power Elecronics Soluion of Tuorial - Basic Conceps in Power Elecronics Quesion. (a) V A R During urn-on ime, v T ( ) V it ( ) ( ) ri ri W on ri vt ( ) it ( ) V 6 ri During urn-off ime, vt ( ) V ( ) fi it ( ) fi W off ri vt ( ) it ( ) V 6 fi Woal Won Woff V ( ri fi ) 3mJ 6 (b) Poal Woal fs 3W (c) Afer urn-on, on-sae volage rop V R. V on s 4 W on on V on V on on Soluion of Tuorial
2 Power Elecronics Pon Won fs 4W () P P P loss( oal ) oal on 7 W Quesion (a) Assume ha he loa curren a urn off is V /R = A. i 3/.5 vs V L Ri V L 3.,,3V MV 9 ri (b) No pracical semiconucor swich can wihsan MV volage across i s s erminals. The evice will mos likely be esroye. This overvolage is ue o he inucance in he circui. Even for a purely resisive loa, here is always some inucance in he connecing wires (he so-calle parasiic inucance). So, some form of clamping is always necessary. Soluion of Tuorial
3 Power Elecronics Quesion 3. Circui A Vs, V orf V i V = 4V i f R f = 5 o Highly nucive Loa v s i T +f Von off V v s fi i T V +V f = 4 V V + o xr f v s i T o + 4/5 ri Swich-off ransien Swich-on ransien (b) Turn-off swiching loss A urn-off, he iniial curren hrough he swich is f = V /R f + o. This curren falls o zero in ime fi. The swich volage is V s = V f + V where V f = - f R f is he volage across he clamping resisor R f. Afer fi, he loa curren circulaes in R f an falls exponenially wih ime. Because of high frequency swiching, he off perio is shor compare o he ime consan of he loa an he clamping resisor, so he exponenial fall of volage across he swich afer fi may be neglece. Wih ime origin saring from off, Soluion of Tuorial 3
4 o f it o f fi Power Elecronics v s V R o f fi p i v s, off s W fi fi 4 soff, i.48 T v s fi fi fi 3 fi =.699 mj fi (c) Turn-on swiching loss A urn-on, he iniial curren hrough R f is assume o be o. Prior o urn-on, he loa was circulaing hrough R f an was ecaying exponenially. f he swiching frequency is high, i can be assume ha he ecay in curren uring off is negligible. Swich curren it rises from zero o o + V /R f in ime ri. Noe ha fv is smaller han ri, so ri eermines volage rise also. i T V / R o f ri vs V orf ri o V / Rf pson, V orf ri ri ri /, o V R f Wson V orf ri ri ri ri.35 3 mj 3 ri ri ri (c) Toal Swiching Power Loss P W W f s s, on s, off s W Soluion of Tuorial 4
5 Power Elecronics Circui B (a) V, V 4 V s V = 4V D Highly nucive Loa v s i o V V o rv fi ri fv Swich-off ransien Swich-on ransien (b) Turn-off swiching loss W s,off = Area of he riangle A (b) Turn-on swiching loss rv fi Vo coffvo = 7 µj 9 = W s,on = Area of riangle B = () Toal swiching power loss 6 3 P s = = 5.4 W.5 V µj ri fv o con Soluion of Tuorial 5
6 Power Elecronics Circui C V, V V 4 5 = 74V s z 4V V z = 5V Highly nucive Loa V + V z o rv fi Swich-off ransien Swich-on ransien (b) Turn-off swiching loss W s,off = Area of he riangle A = µj (c) Turn-on swiching loss rv fi V Vz o coff V Vz o =.5 V V.5 74 W s,on = Area of riangle B = = µj () Toal swiching power loss 6 3 P s = 5 = 6.65 W ri fv z o con Soluion of Tuorial 6
7 Power Elecronics Circui D (a) V, V 4V s 4V Highly nucive Loa o V V o rv fi ri fv Swich-off ransien Swich-on ransien (b) Turn-off swiching loss per swich W s,off = Area of he riangle A rv fi Vo coffvo = 7 µj 9 = (c) Turn-on swiching loss per swich W s,on = Area of riangle B = () Toal swiching power loss 6 3 P s = =.8 W.5 V µj ri fv o con (e) Leas P s Highes P s Circui B Circui D Circui C Circui A Leas P s (e) Circui A Circui C Circui D Circui B Soluion of Tuorial 7
8 Power Elecronics 4. S 5.5 Q fi rr 5nC 5ns i A/ s i Qrr. S rr 4 A The peak curren raing of he MOSFET mus be higher han A. i, i D T L rr i D () () i T rr Soluion of Tuorial 8
9 Power Elecronics Quesion 5 Waveform of Figure 5(a) ime / / = 3/ 4 i( ) cos n n,3,5... n where / 4 n an cos( n ) ( )cos( n ) sin n,,3,4... / n, a b n 4 i cos 4. 9, 4. 3, 4. 8 ( 3) ( 5) 3 5 The frequency specrum of he waveform is shown below: n n Soluion of Tuorial 9
10 Power Elecronics Waveform of Figure 5(b) /3 /3 / = a b n /3 n /3 a [ cos( n ) ( )cos( n ) 4 sin( n / 3 ) ; n,3,5... n 4 i( ) sin( n / 3 )cos n n,3,5... n 4 sin( / 3 ).7797 ; 4 3 sin(3 /3) (3 ) 4 sin( 5 / 3 ).559 (5 ) 5 4 sin(7 / 3 )..64 (7 ) 7 9 Soluion of Tuorial
11 Power Elecronics Waveform of Figure 5(c) / / / / a a n 4 bn sin(n) / 8 ; n,3,5... n ;.637 ; Waveform of Figure 5() V Vc a / Vsin( ), / V Vcos( ) / n a V sin( )cos( n ) V sin( )cos( n ) 4V for n,4,6... ; a n for n,3,5... (n ) bn 4V 4V V.3V ; V4.6V ; ( ) (4 ) 4V V6.57V... (6 ) Soluion of Tuorial
12 Power Elecronics Waveform of Figure 5(e) V c DT T V a / DV, c V n c an sindn ; n,,3... Waveform of Figure 5(f) V c /3 = T ime V 3 V cos( ) /3 3 = 3 3V Noice he perio of he waveform, we can escribe i as: f () Vcos( ) 3 b n an Vcos( )cos(n) 3 3V sin( 3n ) / 3 sin( 3n ) / 3 3n 3n n,,3,4,... V3.93V ; V6.668V... Soluion of Tuorial
13 Power Elecronics Waveform of Figure 5(g) V /3 /3 /6 V Vcos( ) /3/6 3V Noice he perio of he waveform, we can escribe i as: f () Vcos( ) 6 b n 6V sin( 6n ) / 6 sin( 6n ) / 6 6 6n 6n ; an Vcos( )cos( n ) n,,3... V6.386V ; V.945V... Soluion of Tuorial 3
14 Power Elecronics Quesion 6 For he waveform of figure 5(a), / rms / Accoring o he answer of (a), 4 4 ( ).9 THD.48 48% 4.9 For he waveform of Figure 5(b) /3 rms / sin( / 3 ) (.86 ) ( sin( / 3 )) THD.3 3% sin( / 3 ) For he waveform of Figure 5(c) ( /) [ ( ) ] [( ) ] ( ) ( ) / rms Accoring o (e) resul: (.577 ) (.573 ) THD. % Soluion of Tuorial 4
15 Power Elecronics Quesion 7 For he waveform of Figure 5() V V [ (V sin ) V ( ).77V rms Accoring o (f), V c a / V S in( ) V.6366V (.77V ) (.637V ) V.637 RF.48 48% For he waveform of Figure 5(e) DT rms c c c V [ (V ) V D V D T T Accoring o (g), V a / DV c, DVc D Vc D D RF DV D D c For he waveform of Figure 5(f) 3 /3 3 3 rms /3 V [ (V cos ) V ( ).84V 3 3 /3 3 3 c /3 V V cos( ) V.87V (.84V ) (.87V ).84V.87V RF.8 8%.87V.87V For he waveform of Figure 5(g) 3 /6 3 3 rms /6 V [ (V cos ) V ( ).9558V /6 3 c /6 V V cos( ) V.9549V Soluion of Tuorial 5
16 Power Elecronics (.9557V ) (.9549V ) V.9549 RF.4 4% Quesion 8. A 5 V/ (5 )/ (5 )/ (5 )/ R j L 5 j34*.5 5 j A( rms ) V / (3 )/ (3 )/ (3 )/ R j L 5 j68*.5 5 j a( rms ) DCPower W Power from Source : V cos.9w where Power from Source : V cos 9.746W where 6 PToal W Quesion P cos3 8 W VSrms 34 / 4V Srms 8 (5/ ) (6/ ) (/ ) 4.A 5 /.6A DF / s.6/ Soluion of Tuorial 6
17 Power Elecronics THD Srms rms P DF Cos.76 Cos3.658 F Quesion. 3 3 ime v 34sin ; V 34 / 4.45 Vol 4 an sin( n / 3) n 4 i( ) sin( n / 3 )cos n n,3,5... n 4 sin( / 3 ) A DF npu Displacemen Facor cos cos npu Power P Vcos Was Noe! Oher harmonics o no raw any ne power. 5 / 6 / 6 srms A npu Disorion Facor npu Power Facor DF s s Soluion of Tuorial 7
18 Power Elecronics S THD 3. 95% 7. 8 Soluion of Tuorial 8
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