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EEEB143 Test 1 Answer Scheme

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  (a) Describe the piecewise linear model of a diode. [5 marks] T[c pieccr,r,ise lirtcar llrrclel is tn apprtiximation ol tlte diode's cLtn'ent-r'oltage clrlrircteristics Lrsing lineal rclrlionshipr or s;tlaighl liries, When the cliocle 1t&|.lllgfcl lrrascc] u,ilh /r ] lu, r.r'r assLnlc a slraight-[ine alrproxitlation ri']rosc slollc is l/r7. _-..--', .'-.--.-. .-, ;,, .x; -l'be eclr,rivzLiclll jr-il: titl .iJp:-go-ttsjt11 ol'a,cgr1 -a$-yql 4g9-191r ic ivith v{qe 111--c1t- ,; irr vollage. (, in serics r,r'ith a lolrn'alcl qp*_iq{l 11, ]:t-,ih''g1e;pictcci in ligr-rlc (a) /':. belolv. When thc cliclde is t'et'r'rsc-lrrirsed, iviLh Va 1V.,,tn.;iiis i.ssLnrie a sttarghl-li nte /; I belo''. Whe, thc diode is rerer-ycl_ l]tt;95|1_gL_11):p*S.yr,,*e *zLssrtttie ii stl'argnl-l,tr.*.. 1. f, Irltplpxinralin pet'allcl ri,ilh llic, axis. hence lhe e{lLlivxlenl circttit is an open--circttit lts tlcpictcel in ligure {b) belriir II[) (b) Using the piecewise linear model, determine the diode voltage, Ve and the diode current, Ip in the circuit shown in Figure I if R1 : 4.8 kQ and Rz : 3.7 k{>. Subsequently, determine the power Pp dissipated in,the diode. Assume piecewise linear diode parameters of Vy:0.7 V and r7:72 O. [7 ma:ks] Vp ++ Vy Vn ,,j-i -? (b) R1 t*l +Y Vo Page I of5  Ily K\i i.. it Li,s* lDRi+Vu=0 Vn - l'r, )- 11,r, - 0 7 a- l2lD .---.:1 Vr, = l/t, * l,,r',= 0 7 + (0.8gj6 x tn-t'@l /10?' I ,-) \ I it,' i1: \ ,/ .a't,1'].,, \ -.-rP11 = Vplp :0.71A7 x (0,893fr x 10-3) = 0-ti :151 rnl''l .+ QUESTION 2 (10 Marks) Consider the voltage regulator circuit shown in Figure 2. The Zener diode voltage, Vz ismeasured to be 12 V at a Zener cuffent, Is of l0 mA. The Zener resistance, ry is known to be 12 {r. : Figure 2 Given that the input resistohce, R; is 4 kCI, the nominal supply voltage, 7p5 is 100 V, and the nominal load resistance, R1 is I kO, load resistance (Rr : 1 kO) if [6 marks] (a) Determine the source regulation percentage at nominal the 7p5 varies bY +10 Percent' ll Vrt - Vz : lz* lt V, - l'r,, Vt D t7 1\/_ ;'t : .a I ; l-r' Rl ;: I ',1 \/z /-i Page 2 of5  EEEBI43; Semester 2 2012/2013 V, + Ri /1 rt I - t. ln.ni vt, vL vps . vzo 't'z RL Ri rz 1 1\ /p.s Vtu 17 R r./ Rt rz (?. ?)r' irr;rrRi) t - For l/p51rn;1,; = l/ps X Frlt' /1,51,,,s1-) = V ps x 11OVo - V4nttxleo% = eo u, /W /V ps(nttn\ Vro\ .' vt.(,,,u,, : {,ff . ; )(R, lt';llRr) - ( J -. Ig) (4oooil 1 2l 1ooo) \+oon' tz l'(r,\ = 11.9705 V /'V 41 ',I10 r/. / ly/ = (v r 'o't - ?) (Ri llr; llRr.) \ ,{i tz / rregtrlatiott. V ;(rrax) - Vqtnb.t) t_ \rz %o sou: 11. 90296110 12.4 ) It l) 1./ 55 olo q70 X 100% = 0.29 /b (tr/ps: t00 V) 14 marksl if (b) : (,'::=* t1 ^uu) (+oooll lzll rooo) \4000 12 /'r = tzoztgo, l0 Vesqmaxl - /ps(nrin) r00%=( 1 Determine the load regulation percentage at the R1 varies over the range of 1 kO < R, < For R1 = 1 l<f) (Full load). v t( rurr rourt) = (? - ?) rr ll,; llRL) ,/ . .. : (vE *b)(n,tt,;) '/ L.(notoud\ - \& t*r.r, Page 3 of5 ,w r 100 1l.BBr = [ -*- r --':-- l r+ooo1112ll1ooo) \4000 12 /' /a-\ = 12.000u , -/U/ nominal supply voltage mO. Fol Rl. = co (no load),  / 100 11.88\ EEEBI43' Semester 2 2012/2\ =(no*+ u )Uoooly\ = 12.7436, / |/ vps - vzot Jt. 100 - 11.88 = = 21..9641. mA  _I t I - tz - RiIrr. 4000+12 Vt(norond): Vz: VvL\ * l7r2 = 11'BB + (27'964L m)(12) = 1'2'L436V o/o l<tad regulailon.12.1436 * L2 Vttuo toatL) - \ /'ir- t,(f ttlt LorLcL) OUESTION 3 (8 Marks) Given that the input voltage, u1 is a triangular wave as shown in Figure 3(a) below, sketch the steady-state output voltage, ?e verSUS time, f for the diode circuits shown in Figure 3(b) and Figure 3(c). Assume diode piecewise linear parameters of Vr:0'7 V and r7:0 Q' For tlrc diode clipper circuit in Figure 3(b) the rliode is in Rrrrvarcl bias 1O1 1) when tlre irtpitt n\, r,oltage v1 ) Vs * Vy :4,0 V. 'l'he orLlpr-rt vohagc clips at ao = Vs * \ = 4'0 V. When n'rcrT,/ c'liode is in reverse bias (OFF). the outpr-rt voltage hence follou,s the inpr-rt voltage (vo : vr)' x100%=( ) V tUt u too.a t2 n) ) x 100% : 1.1964Vr/ u t7 q lI _t /t Page 4 of5
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