Question

Draw a well-labelled circuit diagram of a Buck-Boost converter and draw the equivalent circuit diagrams for its two modes of operation (no waveforms). Label all circuits. [30%] With the help of the circuits in 2(a)(i), explain the working of a Buck-Boost converter. [20%] (ii) (b) Consider the following single-phase, half-wave rectifier circuit that is used to charge a battery through a resistor R. GO v, = Vm sin ot = (i) Prove that the average value of the charging current is: 1. = [21.. cos 6, – E(T – 20,)] 2nR where, ê, is the minimum value of the firing angle for the thyristor. [15%] (ii) For an AC source voltage of 230V, 50Hz, find the value of average charging current if R= 8 92 and E= 150V. Hence compute the power supplied to the battery, [20%] (iii) Sketch the waveforms for the source voltage and the load current for one full cycle of the input source voltage. Label the waveforms clearly. [15%]

Answer 1

2. Mathematical Analysis of Sus Buck Buck Boostconvertor Boost Circuit program Basic sulten D ZL c R { Vo unipolar unidirectional switches are employed. D) when switch is on : [Oct ZOT] Diode reverse biased Copen circuired Vacat i t &L c T B & Vo in when switch is OFF : [DT CELT @ pn modo (), KUL at input loop Vdes lai — @ KLL at output nude codeco + yo no - ☺

in maholli) dt Kve : Ldü - - V. KLL: ca vo vozil dt we knew by burning - in which Cauld In DC-DC converters olp In DC-DC converters ole power power can be controlled on and turning off the solter. ratio of twin on time to total time ratio is Duty cycle ratio Tont Toff=T Top =D Ton= DT] : Toff = T-Ton & (i-DT Toff = (1-0)T] t*) volt-sec balance equation: from equations ② y ③ Voc (OT) - VO (1-0)T (01) + Ldin (I-DT - Labin (OT+T-OT) : Vdc (DT)- vo (ID)T =0 (in steady state die zo) Voz Dude (1-0) when : OLD CO.5 0.5<D<1 acts Buck converter acts as Boost convertor Voevac Vozva

6) Amper sec (ou charpe balonce equation. fem equarios ☺ , weget O(07) +IL(1-NT = ((dvot bo) 07+ de pe teren CDT Il ll-OT = TVO (OT+T-07) C: at steady state 50) IL 1-0)= Vot IL = I love circuit O Given single-phase , half-wave rectifier N) Us= Um Sinut () sont L ED 20+ Tre3 brate puse

-0, SCR Comer into foro and blocking Kino mode when azo ser starts conduction ; a= triggering polje when the load has emf and SCR is triggered with Constant De Signal then a=0 when SCRIS ON: Riot E= Um Sinut io = Um Siowe -E lo= um sinut - - soverope output current Ti-0 Io = 1 1 1 (um sincet. - Et dot ( wm [ue vot i oce (vt 310 ] Io zne [um (cos@rcos &]-5 (0-8-]] as elzo Io - Zor (2. vm Cose - E 11-20)] LA when switch is off ioso Vo=Vs E - Vm sin of w=0) (o= sint (m)

(ii) Given 230V 50HZ RMS voltage Um=23052 = 325-26u R=82, E=1500 - 6-sin' (Tom) - Son" (2003) o= sin( 15 ) 0=27.46° (or 0-479 rodians. :: overope current in output Tocuge. ZR zum Coso - El1-20)] to 121325-26) Los 27146-150111-240:479)) Ioang = 4.96 A gmist be in radian : power supplied to battetry P= E Io ave =15014.96) p=745 u w

(170) Vm-23DS- 150E ezd:2746 on Ter__-20 1-1-29 turn off time I conduction angle WEEnt20 - (1129