Inverter Circuit, Construction of Equivalent Circuit - Power Electronics | ECEN 5797, Assignments of Electrical and Electronics Engineering

Download Inverter Circuit, Construction of Equivalent Circuit - Power Electronics | ECEN 5797 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! 3-) 3 MODELING DIODE REVERSE RECOVERY (30 POINTS) In the half-bridge inverter ciccuit illustrated in Fig. 3 (see Pig. 4.11 of the text), the diodes exhibit a reverse recovery process with recovered charge Q, and reverse recovery time /,. The transistors have for- ward voltage drop V, and the diodes have forward vellage drop Vp. Fig. 3 Flalf-bridge inverter circuit of Problem 3 You may assume thal the (rausistors do net conduct reverse current. The converter operates in the contin uous conduction mode, ~ith negligible inductor current ripple. (a) For the caye when D > 0.5, sketch the waveforms of v,(z) and 1,(). Derive au averaged equiva- lent cuicuit that models the steady-state behavior oF the circuit, irclading the effects of diode reverse recovery and the semiconductor forward vallage drops (15 points). (b) Repeat part (a) for the case when D < 0.5 (15 points). kh t da Ve >O > & FO an Dr z L OAS is wch Ya onl eo art (@ pesrble wes te Atve d uy, ) Capacitor charge. balance \ V ZEy eo = 44> a Tag \nduetor’ Valt- second lealan ca. LuL> = O ® nate VL VR Ng So CV = ove? -\a-V . t wh 2vg> = D(a\,-Vyp) 7D Vo so 0 = Daw -VP) HD Vy “Ng WV é > oF (ad-i)Vq — Vy ~ DNS WV aueTag wput curreat ] : ‘ — tt. On . 4t,> = DILt+ Tet oe = Tp overage imput current #21 node eguaton : a zt, So ee nt ( = Ziy> - > «7! t, ar 2 Eig > = Sip> a DIL+ Te t+ = Tp  Lvg> = Y@yr\y) tv! V+ z le> = WD Ory tre (- <ig> = Der e+ te CT) ductor” velt- second balance Z2N,> =o 2 op > - V5 ~V © =2DW+DdVy +d! V_ =\g - V 6 ~P- DIN, +DV, + DY Vy ONerase. vmpat Curent sh 2. Ze >a tT. = p. a + Qr tr £ B (-T.) = + z (2) fy>= eo Ted, = ts Soe ty, = DEL + e+ Cn) wal ext Arta gla cna 3-7 \ dopende.st Seurees with trans tomes repiace Bs before 4