ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Problem PS7.1 Erickson, Problem 5.1 Soluion (a) Firs, recall he operaion of he buck-boos converer in he coninuous conducion mode. The plo of he inducor curren i() in he coninuous conducion mode is illusraed in Figure 1. i() PSfrag replacemens I g DT s D T s Figure 1: Buck-boos converer: inducor curren i() in he coninuous conducion mode. From our previous analysis of he buck-boos converer and looking a Figure 1, we noe ha in he coninuous conducion mode I = D = D g (D ) and = DT s g The converer operaes in he coninuous conducion mode (CCM) when I > and in he disconinuous conducion mode (DCM) when I <. Therefore, he converer operaes in DCM when D g (D ) < DT s g }{{} K < (D ) }{{} K cri(d) K < K cri (D) This is in he sandard form for operaing in DCM: K < K cri where K = and K cri (D) = (D ). (b) The inducor curren v () and inducor curren i() in DCM are shown in Figure. Noe ha he duy raio D of he ransisor shown in Figure 1 for CCM is idenical in value o he duy raio D 1 shown in Figure. Since he duy raio D and D 1 is a known quaniy since we usually se his value independenly. Applying vol-second balance on he inducor volage yields v = 0 = D 1 g + D + D 3 0 = D 1 D g (1) Noe ha we don know D ye so we sill need o do more analysis o deermine he converer dc gain. In effec, we have one expression and wo unknowns ( and D ). The exra equaion we need will be formed by applying charge balance o he capacior curren. Charge balance can be applied by examining he circui srucure shown in Figure 3.. Caliskan Handed ou: November 1, 00
PSfrag replacemens DT s D T s ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions I v () PSfrag replacemens DT s g D T s D 1 T s g D T s D 3 T s i() g i pk = gd1ts i() D 1 T s (D 1 +D )T s T s I Figure : v Buck-boos () converer: inducor volage v () and inducor curren i() in he disconinuous conducion g mode. i D () D T s D 3 T s i D i pk = gd1ts i C + C v Area = 1 i pkd T s D 1 T s (D 1 +D )T s T s Figure 3: Buck-boos converer: diode curren i D () in he disconinuous conducion mode. Wriing a node equaion a he posiive node yields i D + i C + v = 0 Taking he average of boh sides of he equaion resuls in i D + i C + v = 0 Charge balance on he capacior requires ha i C = 0 and due o he small ripple approximaion on he capacior volage v. These simplificaions give i D = In oher words, he average value of he diode curren is equal o he negaive of he dc load curren. The average of he diode curren can be calculaed by examining he skech of he diode curren shown in Figure 3. The average value of he diode curren is given by he area under he curve divided by he period i D = 1 T s (Area) = 1 T s ( 1 i pkd T s ) = gd 1 D T s. Caliskan Handed ou: November 1, 00
ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Since i D = /, we now have a second equaion in erms of and D = gd 1 D T s () We now have wo equaions ((1) and ()) wih wo unknowns ( and D ). Eliminaing from he (1) and () yields D = K D = ± K Since D canno be negaive, we ake he posiive roo; herefore, D = K. Now we can calculae he gain of he converer by using equaion (1) M(D 1, K) = g = D 1 K (c) Wih K = 0.1 and K cri = (D ), we are in DCM if K < K cri and in CCM if K > K cri. In erms of duy raio D, we are in DCM if D < D cri and in CCM if D > D cri where he criical duy raio D cri = 1 K = 1 0.1 = 0.683. Therefore, when D < D cri, he gain / g is given by D/ K = D/ 0.1 = 3.16D. When D > D cri, he gain / g is given by D/D. The plo of he gain versus he duy raio is shown in Figure 4. PSfrag replacemens 1 3 4 5 g DCM D cri 0.5 0.5 0.75 1.0 CCM D Figure 4: Buck-boos converer: converer gain versus duy raio. (d) If D = D 1 = 0.3, K cri = (1 D 1 ) = 0.49. Since K = 0.1 and K < K cri, we are in DCM. Given K = 0.1, D = K = 0.3 and D 3 = 1 D 1 D = 0.38. Typical waveforms were already illusraed in Figure. (e) A no load ( ), he value of K = 0. The gain M(D 1, K) = / g of he converer approaches D, since lim K 0 M(D 1, K) = lim 1 K 0 K =. As a resul, he oupu volage approaches. Physically, he inducor keeps charging he oupu capacior every cycle, bu here is no load resisor o consume he energy; herefore, he capacior volage will end o have an infinie volage.. Caliskan Handed ou: November 1, 00
ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Problem PS7. Erickson, Problem 5.3 Soluion (a) In CCM, D = g wih = 8 and 35 g 70. In addiion, P load = wih 10W P load 1000W. The converer will operae in CCM provided ha K > K cri. K cri = 1 D for a buck converer. Therefore, he converer will operae in CCM if he following inequaliy is saisfied: > 1 D Subsiuing = P load, D = g and T s = 1 f s and simplifying gives P load > ) (1 g P cri (3) f s Therefore, if P load > P cri hen we are in CCM; oherwise we are in DCM. Given our range of he inpu volage g and values for, and f s, we can deermine he P cri from equaion (3). From his informaion, we can deermine he criical load curren (I cri = P cri / ). Similar o P > P cri, if I > I cri, we will be in CCM. The able below summarizes he values load power and curren above which we will be in CCM over he range of inpu volage g. g P cri I cri 35 47.5W 1.7A 50 104.5W 3.73A 70 14.5W 5.09A (b) The gain of he buck converer is given by M = D = g 1+ 1+ 4K D in CCM in DCM Therefore he ransisor duy raio in each mode is given by g in CCM D = K g in DCM 1 (4) For a given inpu volage g and oupu load power P load, we can use our resuls from par (a) o deermine if we are in CCM or DCM. Once he mode of operaion is deermined, we can use (4) o deermine he ransisor duy raio. The resuls are summarized in he able shown below where D min = 0.106 and D max = 0.8. g P load mode D 35 1kW CCM 0.8 D max 35 10W DCM 0.367 70 1kW CCM 0.4 70 10W DCM 0.106 D min. Caliskan Handed ou: November 1, 00
ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Problem PS7.3 Erickson, Problem 5.14 Soluion (a) The wors case condiion for he DCM boos converer is when i is closes o he CCM boundary. This wors case condiion occurs a minimum inpu volage and maximum load curren (or load power). Therefore, we will use his condiion o calculae, C, and i pk. Once hese are calculaed we can use he resuls o answer pars (d) and (e). From he soluion of he DCM boos converer (see Tables 5.1 & 5. on pages 11 & 14), we know ha he converer gain M and K cri are given by K cri = D(D ) (5) M = 1 + 1 + 4D K (6) where K =. For our design, we need K = αk cri where α = 0.75. A he wors case condiion, he converer gain M = g,min = 48 18 =.67. If we eliminae K from equaions (5) and (6) and solve for he duy raio D, we ge ( ) D 4 D + α [(M 1) + 1 = 0 (7) 1] Solving for D yields D = 1 + ( ) α [(M 1) 1] ± 1 + α [(M 1) 1 (8) 1] We ake he negaive sign in equaion (8) (o obain D < 1) and evaluae D using M =.67 and α = 0.75; herefore, he duy raio D and K are given by A he wors case condiion, he load resisance is Therefore, he inducance is given by D = 0.581 and K = αk cri = αd(d ) = 0.0763 = P load,max = 48 100 = 3Ω = K = (0.076)(3Ω)(6.67µs) = 5.83µH (c) The peak inducor curren a wors case condiion is given by i pk = g,mindt s = (18)(0.581)(6.67µs) 5.83µH = 1A (b) We can deermine he capacior value using a similar procedure o he one ha we have used for converers in he coninuous conducion mode. The oupu secion of he converer is illusraed in Figure 5. The diode curren i D () is equal o he capacior curren plus he load curren. If he capacior is large enough, nearly. Caliskan Handed ou: November 1, 00
ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions all of he average diode curren goes ino he load while all of he ac curren goes hrough he capacior as illusraed in Figure 5. Given he waveform for he capacior curren i C, he capacior volage is roughly given by he waveform also illusraed in he figure. Noe ha during he inerval T, he capacior curren is posiive and he capacior volage goes from is minimum value o is maximum value. The difference in he exreme values is he peak-peak capacior volage ripple and is denoed as v C. As before, he ripple can be calculaed using q = C( v C ) = 1 ( T i pk ) (9) }{{} area q where q is he area under he i C () waveform over he inerval T. Examining he figure, we can calculae he ime inerval T as ( T = 1 ) D T s (10) i pk Subsiuing he value of T from equaion (10) ino equaion (9), solving for capacior C and using componen values yields C = q = D ( T s i pk 1 v C 4 v C i pk ) = (0.350)(6.67µs)(1A) 4(1) Noe ha we have used v C = 1 and D (see Table 5. on page 14) as given by D = K D M = 0.0763.67 = 0.350 0.581 ( ) 48 1 = 4.78µF (1A)(3Ω) (d) & (e) Given our analysis from pars (a)-(c), we can simulaneously answer he quesions in pars (d) and (e). Firs of all, he duy raio D of he ransisor can be calculaed by evaluaing equaion (8). The value of K cri = D(D ) hen be calculaed once we have deermined D. The value of K is given by / where is he value calculaed in par (a) and = /P load. Once K and K cri are calculaed, we can deermine he raio K/K cri o make sure ha i is a leas 0.75, or 75%. The able below summarizes he resuls of such calculaions. case (i) case (ii) case (iii) case (iv) g 18 36 18 36 48 48 48 48 P load 5W 5W 100W 100W 5.86µH 5.86µH 5.86µH 5.86µH T s 6.67µs 6.67µs 6.67µs 6.67µs M = / g.67 1.33.67 1.33 = /P load 460.8Ω 460.8Ω 3Ω 3Ω D, from eqn. (8) 0.581 0.087 0.581 0.087 K cri = D(D ) 0.1017 0.1307 0.1017 0.1307 K = / 0.0038 0.0038 0.0763 0.0763 K/K cri (%) 3.75.9 75.06 58.39. Caliskan Handed ou: November 1, 00
ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions From he able, we see ha duy raio varies over he following range 0.087 D 0.581 herefore, he maximum ransisor duy raios are D min = 0.087 and D max = 0.581. The values of K and K cri vary over he following ranges 0.0038 K 0.0763 0.1017 K cri 0.1307 PSfrag replacemens Noe ha he wors DT s case condiion corresponds o case (iii) where he raio of K/K cri is he highes a a value of 75% as D expeced. T s g i D () i() I v () g i C C i D + v i C () D 1 T s i pk = gd1ts (D 1 +D )T s T s D 3 T s i pk q = 1 T (i pk ) T T s Area = 1 i pkd T s v() D T s v C T T s Figure 5: Boos converer: diode curren i D () and capacior curren i C () and capacior volage v() in he disconinuous conducion mode.. Caliskan Handed ou: November 1, 00