Semiconductor Diodes Notes: Efficiency of a Half-Wave Rectifier

Advantages:mmmmn(i)The a.c input voltage can stepped up or down(ii)It isolates Q.C power line from the outputcircuit,ThusthereisnochanceofanelectricShock.Efficiency of a Half- wave rectifier:-To find efficiency,1We know mean value of DC DC,2 IT2TTIdc=I,do=12Th2ThIm sin (wt) x dutm0IT2TT12 = I m Sin wot=12TCIn m sin cot doot + 1 ImP = I2R0II= 12TTIm Sin (wt) (dut) + 0mT=ImSin wt . dut2TI070= Im x [-cos(wt)2TCO= Im [ cos TC - (-waso)at= Im [1+1] = Im2>Im12TT210Th 11Power, P dc = I2 d.c xR,2={I'mRL(2)rIIIThe rms value of current is girlen by,2TT1/2Irms=rms2THI2do1/22TTIT=27S2I'msin'wt xd(ws)+2thSI2 sin'wt d(wt)m0T=I'mmI21101/221/2I2=TIm=m=Imm3x2422TWPower, Pa, QC = I a ( R + L rj)rfresstancermsof diodeSo, Pac ac = I rms 2 R1atvery small2=I'mmR24)2VRectification efficiency = PacPac=I2R24T2=2In2mR2L0.406= 40.6% Ripple factor of Half wave rectifier:-We know,rIrms2=-1Idefor half wave rectifier,I = Imrms 2I dc = ImTL2r =Im22-1=1.21=4ImTC(ii) full wave rectifier.In full wave we get output during +ve & -ve halfcycle.a) Centre Tap full wave rectifierS,D,*S0Aa.conD2R LB Input3W5TMoltage 0II2TTAITwtIT23T4T5I22aoutputVoltagewt(ort)oTC2T3TT4IT5TTIT3T2T5I222Working :-During poschie half cycle of ac voltage diode D1becomes forward biased and D2 becomes reversebiased . So D1 conducts through the path S,D,ABOS,In the negative half cycle D1 becomes reversebiased and Da becomes forward biased So D2conducts through the path S2 D2ABOS2In both cases output current flows from A to BThat means the current becomes unidirectional .Efficiency of full wave rectifier = 81.2%Advantages1)The centre tap of transformer or used. 11) output is twice as that of half wave rectifierIII) PIV is one half as that of half wave rectifier.Disadvantage:-in1)It is difficult to locate the centre tap of thesecondary11)D.C output is small as each dwde utilize asone half of the transformer secondaryIII)Didde used must have high peak voltageSo full wave - bridge rethful is preferred.b)Full - wave bridge rectifierS,PD3SADP4R25?Bwt-working:-During +ve half cycle of a.c , thewtdiodes D2 and D3 become forward biasedand D1 and D4 becomes reverse biased So thecurrent flows through the path SD.ABD,S 12 32 SIn the -ve half cycle D1 and D4 becomesForcoard biased and D2 and D3 becomes reversebiased. So circuit flows through the path SDABDSS. In both the cycle current flows from A to B inthe outputAdvantage :-There is no need of a centre tapped transformer.1)11)Efficiency is highiii) Repple factor is low.10) Small transformer can be usedv)output is twice as that of centre-tapped rectifier forthe same secondary voltagevi)PIVisonehalfofthecentretapedrectified*Disadvantage:-1)four diodes are required1)This rectifier cannot be used at low voltages(1)During each 1/2 cycle , the hoo diodes are in serieshence resestance of the corcut increases. This objectionalin the secondary voltage is low.The voltage regulation is very poorEfficiency of full - wave rectifier:-(1)In full wave rectifier, as output dc flows for bothhalf cycles,2II2TI = 2TC I IL do = 2th S Insin (wt) d (wtdcO02TT= 1II Im Sin cot dwt + S Im son(wt II wt2thTC2 TTS.IT= II-cos(wt)+f-cos (cwt-tc)]2TmTC14 as-In m +2TT- Im -1-1 -1-1==Im4=2 Im2rTOI=2ImdcTC11output power, Pac = I2 dc RL2It Im RL(2)111The rms value of current is given by,2 ITI rms =S1/2ITI22do0IT21171/2=x2TI m 2 sin (cut)-dut + S I'm dutor=I'm21/2m[KIRt+TCR/T2TT21/21/2==Inm42I.Im3rmsS22(Iv)Input power,P.Fundacrms(RLXY)IonRLxof)(4)2 2VRectification efficiency n =Pdc2 Inm=RLPacnacaIm2(R, + rf)V2L=4 I'm RLItIm's (RL+ rf )2=8R,= 8TC CR, +rf)A2= 81.2 %Ripple factor of full wave rechfier:I2we know,r=rmsfrequency output:-Idcfor half wave = fFor full wave, 2fFor full wave rectifier,IIIm=rms 52.Idc = 2.ImILI22,r =TL0.48- I=S2=I82 ImTHalf waveFull wave1)only one diode usedTwo diodes are used (4 diode also)2)The output is discontinues $ pulseting output is continuous $ pulsating dic)n= 40.6%n=81.2%)only one half of a.c input wave comesBoth half cycle of a.c input waveout as d.c outputcomes out as d.c output)n is low.n is high 5. Problem:1.A av voltage regulated supply is required to run acar stereo System from 12 V battery. A zener diodewith V2 = av , Pmax max = 0.25W. is used as a voltage regularFind the value of the series resestor.?Ans:Vz av, , Pmax max = 0.25wP=I.V.maxI= Z Pmax = 0.25 . = 0.2519Vz9When current is maximum, Iz is maximum and IL=0.I = I + I = I/2 Z +0 = IzLZ.Potential difference, V = IR= = 3VR = V = 3 = 108-2I0.25192.A 10V Lener diode along with a Series resistoris connected across a 40V supply Calculate the maximumValue of the resistor required / if the maximumLener current is 50 mA.Ans,Given, V2= =P.d , V = Vi-Vz = 40 - 10 = 30VI = 50m A , ( Daaxim , So L==0):I = I + IL = 50m AV= IRR=V = 30 = 600eI 50x10-3 - The internal resistance of a diode used in ahalf wave rectifier is 25.2 and the load resistanceis 975 2. off the input a.c voltage is V = 50 Sin wtFind , , a) a.c input powerb) output powerc) Efficiency of rectifier.Given, , r f = 25 s 2 , R1 = 915s.Peak unput voltage , Vm = 50v.Im = Vm=5050 X10-3A=rf+R125+975I trms If Im = 50x10-3 = 25 x10-3A22a)a.c input power, Pas = I2 [RL+7]acrms14(25x103) 2 (25+975) = 0-625 wattb)I = Im 50x103=dcrTC= 15.92.11.3P de = , 2=(15 92x10-3 2 x 975 = 0.2471 waltc)Efficiency , I = Pdc = 0.2471 = 0.395 = 39.5%Pac 0.625