BWW 11 Naturally Commutating Converters The converter circuits considered in this chapter have in common an ac voltage supply input and a dc loadoutput.Thefunctionoftheconvertercircuitistoconverttheacsourceenergyintocontrolleddc loadpower,mainlyforhighlyinductiveloads.Turn-offofconvertersemiconductordevicesisbrought about by the ac supply voltage reversal, a process called line commutation or natural commutation. Converter circuits employing only diodes are termed uncontrolled (or rectifiers) while the incorporation of only thyristors results in a (fully) controlled converter. The functional difference is that the diode conducts when forward-biased whereas the turn-on of the forward-biased thyristor can be controlled from its gate. An uncontrolled converter provides a fixed output voltage for a given ac supply and load. Convertersemployingacombinationofbothdiodesandthyristorsaregenerallytermedhalf-controlled (orsemi-controlled).Bothfullycontrolledandhalf-controlledconvertersallowanadjustableoutput voltage by controlling the phase angle at which the forward biased thyristors are turned on. The polarity oftheoutput(load)voltageofafullycontrolledconvertercanreverse(butthecurrentflowdirectionis notreversible),allowingpowerflowintothesupply,aprocesscalledinversion.Thusafullycontrolled converter can be described as a bidirectional converter as it facilitates power flow in either direction. The half-controlled converter, as well as the uncontrolled converter, contains diodes which prevent the output voltage from going negative. Such converters only allow power flow from the ac supply to the dc load, termed rectification, and can therefore be described as unidirectional converters. Although all these converter types provide a dc output, they differ in characteristics such as output ripple and mean voltage as well as efficiency and ac supply harmonics. An important converter characteristic isthatofpulsenumber,whichisdefinedastherepetitionrateinthedirectoutputvoltageduringone complete cycle of the input ac supply. Thegeneralanalysisinthischapterisconcernedwithsingleandthreephaseacsuppliesfeeding inductive dc loads. A load back emf is used in modelling the dc machine. Generally, uncontrolled rectifier equationscanbederivedfromthecorrespondingcontrolledconvertercircuitequationsbysettingthe controlleddelayangletozero.Alsopurelyresistiveloadequationscanbederivedbysetting inductanceLtozerointheL-RloadequationsandR-LloadequationscanbederivedfromR-L-E equations by setting E, the back emf, to zero. 11.1Single-phase uncontrolled converter circuits 11.1.1Half-wave rectifier circuit with an R-L load A simple half-wave diode rectifying circuit is shown in figure 11.1a, while various circuit electrical wave-forms are shown in figure 11.1b. Load current commences when the supply voltage goes positive at t = 0. It will be seen that load current flows not only during the positive half of the ac supply voltage, 0 t ,butalsoduringaportionofthenegativesupplyvoltage,t.Theloadinductorstored energymaintainstheloadcurrentandtheinductorsterminalvoltagereversessoastoovercomethe negative supply and keep the diode forward-biased and conducting.This current continues until all the inductor energy, Li2, is released (i = 0) at the current extinction angle (or cut-off angle), t = . During diode conduction the circuit is defined by the Kirchhoff voltage equation 2sin (V)diL Ri V tdt + = (11.1) where V is the rms ac supply voltage.Solving equation (11.1) yields the load (and diode) current { }- / tan2( ) sin( - )sin e (A) 0(rad) = + tVZi t tt(11.2) Naturally commutating converters 214 Figure 11.1.Half-wave rectifier with an R-L load: (a) circuit diagram and (b) waveforms, illustrating the equal area and zero current slope criteria. where Z = (R2 + 2 L2) (ohms) tan /cos = = and L R R Z ( ) 0(A)2(rad)i tt = (11.3) ThecurrentextinctionangleisdeterminedsolelybytheloadimpedanceZandcanbesolvedfrom equation (11.2) when the current, i = 0 with t = , such that > , that is - / tan sin( - )sin0 e + = (11.4) This is a transcendental equation which can be solved by iterative techniques. Figure 11.2a can be used to determine the extinction angle , given any load impedance (power factor) angle 1tan / L R = . The mean value of the rectified current,oI , is given by integration of equation (11.2) 01222( ) (A)(1 cos ) (A) == ooVRI i t d tI(11.5) vL = 0 = Ldi/dt current slope = 0 R current extinction angle -VR R X = L Z = R2+2L2 q=1r =1s =1 p= qx r x s p= 1 i

Power Electronics 215 2/ = 0.45 while the mean output voltage Vo is given by 02 12 22sin (1 cos ) (V) = = = o oVV V t d t I R (11.6) Since the mean voltage across the load inductance is zero,= ooV I R (see the equal area criterion below).Figure 11.2b shows the normalised output voltage/oV V as a function of L / R. The rms output (load) voltage and current are given by ( ) { }( ) ( )220 12122 sin sin 2sin cos sin cos cos 1 12 cos 2 cos ( ( = = ( (( + + = = (( ` ` ( ( ) ) rmsrmsV V t d t VV ViR Z(11.7) The power delivered to the load resistance is 2L rmsP i R = (11.8) The supply power factor, using the rms current in equation (11.7), is ( )2power, apparent powersin cos1cos2 cosLR rms rms rmsrmsPpfV i R i Ri V V V == = = ( + ( = ` ( ) (11.9) Figure 11.2.Single-phase half-wave converter characteristics: (a) load impedance angle versus current extinction angle and (b) variation in normalised mean output voltage Vo / V versus L/R. Naturally commutating converters 216 11.1.1i Inductor equal voltage area criterion The average output voltage Vo, given by equation (11.6), is based on the fact that the average voltage across the load inductance, in steady state, is zero. The inductor voltage is given by / (V)Lv Ldi dt =which for the circuit in figure 11.1a can be expressed as /00( ) ( )oiLit v dt L di Li i = = (11.10) If the load current is in steady state then oi i = , which is zero here, and in general 0 (Vs)Lv dt =(11.11) The inductor voltage waveform for the circuit in figure 11.1a is shown in the last plot in figure 11.1b. The inductorequalvoltageareacriterionimpliesthattheshadedpositiveareamustequaltheshaded negative area, in order to satisfy equation (11.11). The net inductor energy at the end of the cycle is zero (specifically, unchanged since oi i= ), that is, the energy in equals the energy out. This is a useful aid in predicting and drawing the load current waveform. Itisusefultosuperimposethesupplyvoltagev,theloadvoltagevo,andtheresistorvoltagevR waveformsonthesametimeaxis,t.Theloadresistorvoltage,vR = Ri,isdirectlyrelatedtotheload current,i.TheinductorvoltagevLwillbethedifferencebetweentheloadvoltageandtheresistor voltage, and this bounded net area must be zero. Thus the average output voltage o oV I R = . The equal voltage areas associated with the load inductance are shown shaded in two plots in figure 11.1b. 11.1.1ii - Load current zero slope criterion The load inductance voltage polarity changes from positive to negative as energy initially transferred into theinductor,isreleased.Thestoredenergyintheinductorallowscurrenttoflowaftertheinputac voltage has reversed.At the instant when the inductor voltage reverses, its terminal voltage is zero, and / 0/ 0Lv Ldi dtdi dt = == that is(11.12) Thecurrentslopechangesfrompositivetonegative,whencethevoltageacrosstheloadresistance ceases to increase and starts to decrease, as shown in figure 11.1b.That is, the Ri waveform crosses thesupplyvoltagewaveformwithzeroslope,whencewhentheinductorvoltageiszero,thecurrent beginstodecrease.ThefactthattheresistorvoltageslopeiszerowhenvL = 0,aidspredictionand sketching of the various circuit waveforms in figure 11.1b, and subsequent waveforms in this chapter. 11.1.2Half-wave circuit with an R-L load and freewheel diode The circuit in figure 11.1a, which has an R-L load, is characterised by discontinuous current (i = 0) and highripplecurrent.ContinuousloadcurrentcanresultwhenadiodeDfisaddedacrosstheloadas shown in figure 11.3a. This freewheel diode prevents the voltage across the load from reversing during thenegativehalf-cycleoftheacsupplyvoltage.Theinductorenergyisnotreturnedtotheacsupply, ratherisretainedintheloadcircuit.Thestoredenergyintheinductorcannotreducetozero instantaneously,sothecurrentisforcedtofindanalternativepathwhilstdecreasingtowardszero. When the rectifier diode D1 ceases to conduct at zero volts it blocks, and diode Df provides an alternative load current freewheeling path, as indicated by the waveforms in figure 11.3b. The output voltage is the positive half of the sinusoidal input voltage. The mean output voltage (thence mean output current) is 01222sin0.45 (V) = == = =o oo oVV I R V t d tV V I R(11.13) The rms value of the load circuit voltage v0 is given by ( )20122 sin0.71 (V)2 == = rmsVV V t d tV(11.14) The output ripple (ac) voltage is defined as 2 22 22 220.545| | | | ||\ . \ .= =

Ri rms oV VV V VV(11.15) hence the load voltage form and ripple factors are defined as Power Electronics 217 q=1r =1s =1 p= qx r x s p= 1 ( )22 2/ / 1 1 1 1.211 = = = = = =

rmsv rms o v Ri oovVVFF V V RF V VFF(11.16) After a large number of ac supply cycles, steady-state load current conditions are established, and from Kirchhoffs voltage law, the load current is defined by 2 sin (A) 0diL Ri V t tdt + = (11.17) and when the freewheel diode conducts 0 (A) 2diL Ri tdt + = (11.18) Figure 11.3.Half-wave rectifier with a load freewheel diode and an R-L load:(a) circuit diagram and parameters and (b) circuit waveforms. Duringtheperiod 0t,whenthefreewheeldiode currentisgivenbyiDf =0,thesupplycurrent, which is the load current, are given by 2/ tan( ) ( )2 2sin( ) ( sin ) (A)0o ott tV Vi i t I eZ Zt = = + + (11.19) for 2/ tan/ tan / tan2 1sin (A) +=oV eIZ e e Naturally commutating converters 218 2 2( )(ohms)tan /Z R LL R = +=where Duringtheperiodt2,whenthesupplycurrenti=0,thefreewheeldiodecurrentandhence load current is given by 1( ) / tan( ) ( ) (A) 2o Df ott t i i I e t = = (11.20) for / tan1 2(A)o oI I e =For discontinuous load current (the freewheel diode current iDf falls to zero before the rectifying diode D1 recommences conduction), the appropriate integration gives the average diode currents as ( ) ( )( )/ tan 21/ tan 212 1 sin21 sin2 = + = = + DDf o DVI eRVI I I eR(11.21) Infigure11.3bitwillbeseenthatalthoughtheloadcurrentcanbecontinuous,thesupplycurrentis discontinuous and therefore has a high harmonic content. The output voltage Fourier series (Vo + V1 + Vn = 2, 4, 6..) is ( )( ). . .22,4,62 2 2 2sin cos2 1 == + onV V Vv t t n tn(11.22) Dividingeachharmonicoutputvoltagecomponentbythecorrespondingloadimpedanceatthat frequency gives the harmonic output current, whence rms current. That is ( )22n n nnnV V VIZ R jn LR n L= = =++(11.23) and 2 21, 2, 4, 6..== + rms o nnI I I (11.24) Example 11.1:Half-wave rectifier Inthecircuitoffigure11.3,thesourcevoltageis2402sin(250t)V,R=10ohms,andL=50mH. Calculate i.the mean and rms values of the load voltage, Vo and Vrms ii.the mean value of the load current, oIiii. the current boundary conditions, namely Io1 and Io2. iv. the average freewheel diode current, hence average rectifier diode current v. the rms load current, hence load power and supply rms current vi. the power factor If the freewheel diode is removed from across the load, determine vii. an expression for the current hence the current extinction angle viii.the average load voltage hence average load current ix.the rms load voltage and currentx. the power delivered to the load and supply power factor From the rms and average output voltages and currents, determine the load form and ripple factors. Solution i.From equation (11.13), the mean output voltage is given by 2 2 240V108V = = =oVVFrom equation (11.14) the load rms voltage is/ 2 240V/ 2 169.7V = = =rmsV V ii. The mean output current, equation (11.5), is 2 2240V= = 10.8A10 = =ooV VIR R iii.The load impedance is characterised by Power Electronics 219 2 22 2( )10 (2 50Hz 0.05) 18.62tan /2 50Hz 0.05H/10 1.57or 57.5 1rad = += + = == = = Z R LL R From section 11.1.2, equation (11.19) 212/ tan/ tan / tan/1.57/1.57 /1.5721sin12 240Vsin(tan 1.57) = 3.41A18.62 +=+= ooeVIZe eeIe e Hence, from equation (11.20) 1 2/ tan /1.573.41 = 25.22Ao oI I e e = = Since 23.41A 0 = >oI , continuous load current flows. iv. Integration of the diode current given in equation (11.20) yields the average freewheel diode current. 10 01.570/ tan/1.57rad( )1 12 21 25.22A25.22A 1.57rad 1 5.46A2 2Df Df ottt I i d t I e d te d t e = = (= = = ( The average input current, which is the rectifying diode mean current, is given by 110.8A 5.46A 5.34A = = = = Df s D oI I I I v. The load voltage harmonics given by equation (11.23) can be used to evaluate the load current at the load impedance for that frequency harmonic. ( )( )( ). . .22,4,62 2 2 2sin cos2 1 == + nV V Vvt t n tn The following table shows the calculations for each frequency component. harmonic n ( ).22 21 =nVVn(V) ( )22 = +nZ R n L() =nnnVIZ(A) 2nI0(108.04)10.0010.80(116.72) 1(169.71)18.629.1141.53 272.0332.972.182.39 414.4163.620.230.03 66.1794.780.070.00 83.43126.060.030.00 2 2o nI I + =160.67 The rms load current is .2 21, 2, 4.. 160.7 12.68A== + = = rms o nnI I IThe power dissipated in the load resistance is therefore 2 21012.68A 10 1606.7WrmsP I R = = = The freewheel diode rms current is ( )( )..21020( ) / tan( ) /1.57rad12125.22A 8.83A2 == =Df ottI I e d te d t Naturally commutating converters 220 Thus the input (and rectifying diode) rms current is given by 2 212 212.68 8.83 9.09A= = = =rms rms D s rms Df rmsI I I I vi. The input ac supply power factor is 1606.7W0.74240V 9.09Aoutrms rmsPpfV I= = = vii. If the freewheel diode Df is removed, the current is given by equation (11.2), that is { }{ }{ }- / tan- /1.57- /1.5722 V18.62( ) sin( - )sin e240 = sin( -1.0) 0.841 e= 18.23 sin( -1.0) 0.841 e (A) 0(rad) = ++ + tttVZi t ttt t The current extinction angle is found by setting i = 0 and solving iteratively for . Figure 11.2a gives an initial estimate of 240 (4.19 rad) when = 57.5 (1 rad). That is - /1.570 =sin( -1.0) 0.841 e + gives = 4.08 rad or 233.8, after iteration. viii. The average load voltage from equation (11.6) is 2 2 240V2 2(1 cos ) (1 cos 4.08) 86.0V = = =oVVThe average load current is /= 86.0V/10 = 8.60A = o oI V R ix.The load rms voltage is 169.7V with the freewheel diode and increases without the diode to, as given by equation (11.7), { }{ }1212sin 2240V 4.08 sin 2 4.08 181.6V ( = ( = = rmsV V The rms load current from equation (11.7) is decreased to ( )( )sin cos 12 cossin 4.08cos 4.08 1.57 240V 14.08 9.68A18.62 2 cos1.57 ( + + = ( ` ( ) ( + = =( ` ( ) rmsViZ Removal of the freewheel diode decreases the rms load current from 12.68A to 9.68A. x.The load power is reduced without a load freewheel diode, from 1606.7W with a load freewheel diode, to 2 2109.68 10 937W = = =rmsP i RThe supply power factor is also reduced, from 0.74 to 937W0.40240V 9.68A= = =outrms rmsPpfV I circuit with freewheel diodecircuit without freewheel diode Load factor form factorripple factorform factorripple factor FF = rms/aveRF = FF2-1FF = rms/aveRF = FF2-1 Voltage factor169.7V/108V = 1.571.21181.6V/86V = 2.11.86 Current factor12.68A/10.8A = 1.170.6159.68A/8.60A = 1.120.517 Power Electronics 221 q=2r =1s =1 p= qx r x s p= 2 11.1.3Single-phase, full-wave bridge rectifier circuit Single-phaseuncontrolledfull-wavebridgecircuitsareshowninfigures11.4aand11.4b.Bothcircuits appear identical as far as the load is concerned. It will be seen in part b that two fewer diodes can be employedbutthiscircuitrequiresacentre-tappedsecondarytransformerwhereeachsecondaryhas onlya50%copperutilisationfactor.Forthesameoutputvoltage,eachofthesecondarywindingsin figure 11.4b must have the same rms voltage rating as the single secondary winding of the transformer infigure11.4a.Therectifyingdiodesinfigure11.4bexperiencetwicethereversevoltage,(22 V),as that experienced by each of the four diodes in the circuit of figure 11.4a, (2 V). Figure 11.4.Single-phase full-wave rectifier bridge: (a) circuit with four rectifying diodes; (b) circuit with two rectifying diodes; and (c) circuit waveforms. Figure11.4cshowsbridgecircuitvoltageandcurrentwaveforms.Withaninductivepassiveload,(no back emf) continuous load current flows, which is given by( ) ( )/ tan/ tan2 2sinsin 01 (= + ( toVi t t e tZ e(11.25) Appropriate integration of the load current squared, gives the rms load (and supply) current: Naturally commutating converters 222 2 / tan2/ tan11 4sin tan1 ( = + = ( rms sV eI IZ e(11.26) Theloadexperiencesthetransformersecondaryrectifiedvoltagewhichhasameanvoltage(thence mean load current) of 0122 2sin 0.90 (V) = = = =o oVV V t d t I R V (11.27) Since the average inductor voltage is zero, the average resistor voltage equals the average R-L voltage.The rms value of the load circuit voltage v0 is ( )220122 sin (V) = =rmsV V t d t V (11.28) From the load voltage definitions in section 11.10, the load voltage form factor is1.112 2= = =rmsvoV VFFV V(11.29) The load ripple voltage is ( )2 222 22 2= = 0.435 (V)RI rms oV V VV V V

(11.30) hence the load voltage ripple factor is ( )222 2 2/ 121 0.483 / = = =

v RI o vvRF V V FFRF(11.31) which is significantly less (better) than the half-wave rectified value of 1.211 from equation (11.16). The output voltages and currents (rms and average) can be derived from the voltage Fourier expansion form for a half-sine wave: ( )22,4,62 2 2 2 2cos1 == + onV Vv t n tn(11.32) Thefirsttermistheaverageoutputvoltage,asgivenbyequation(11.27).Notetheharmonic magnitudesdecreaserapidlywithincreasedorder,namely 2 2 2 2: : : : ...15 3 35 63Theoutputvoltageis therefore dominated by the dc component and the harmonic at 2. Theoutputcurrentcanbederivedbydividingeachvoltagecomponentbytheappropriateload impedance at that frequency. That is ( )2222 222 212, 4, 6..= == = =+foroonnnV VIR RV VnI nZR n L(11.33) The load rms current whence load power and power factor, are given by 2 222,4== + == = rms o nL rmsrms LrmsnI I IP I RI R PpfV I V(11.34) Each diode rms current is/ 2rmsI . For the circuit in figure 11.4a, the transformer secondary winding rms current is Irms, while for the centre-tapped transformer, for the same load voltage, each winding has an rms current rating of Irms / 2. The primary current rating is the same for both transformers and is related to the secondary rms current rating by the turns ratio. 11.1.3iSingle-phase full-wave bridge rectifier circuit with an L-C filter A with an L-C filter and continuous load current Table11.1showsthreetypicalsingle-phase,full-waverectifieroutputstages,wherepartcisatypical output filtering stage used to obtain a near constant dc output voltage. Power Electronics 223 If it is assumed that the load inductance is large and the load resistance small such that continuous load current flows, then the bridge average output voltage oV is the same as the average voltage across the loadresistorsincetheaveragevoltageacrossthefilterinductoriszero.Fromequation(11.32),the dominant load voltage harmonic is due to the second harmonic therefore the ac current is predominately thesecondharmoniccurrent, , , 2oac oI I .Byneglectingthehigherorderharmonics,thevariouscircuit currents and voltages can be readily obtained as shown in table 11.1. From equation (11.32) the output voltage is given by ( ), 22cos 22 2 2 2 2cos 212 2 2 2 2cos 230.90 0.60 cos 2 = += + == + = + o o ov t V V tV Vn t nnV VtV V tfor(11.35) Withthefiltercapacitoracrosstheloadresistor,theaverageinductorcurrentisequaltotheaverage resistor current, since the average capacitor current is zero. Table 11.1.Single-phase full-wave uncontrolled rectifier circuits continuous inductor current 2nd harmonic current average output current output power Io,2oIPR+PE Full-wave rectifier circuit Loadcircuit (A)(A)(W) (a) R-L see section 11.1.3 and 11.3.3 = 0 ( ),2222 +oVR L 2 2=oVRVR 2,o rmsI R(b) R-L-E see section 11.3.4 = 0 ( ),2222 +oVR L 1 2 2| |= | |\ .oV ERVER 2, +o o rmsI R I E(c) L-R//C see section 11.1.3i ,22oVL 2 2=oVRVR 22, =o o rmsI I R R With continuous inductor current, the inductor current is ( )( ), 2222cos 22 0.90 0.60cos 2 cos 232 = += + = + +o o oo oi t I I tV V V Vt tR Z RR L(11.36) Vs Is Io VoIo,dc Io,acL CR Vs Is Io VoE + L R VsIs Io Vo L R Naturally commutating converters 224 From equation (11.36) for continuous inductor current, the average current must be larger than the peak second harmonic current magnitude, that is , 2223>>o oo oI IV VR Z(11.37) Since the load resistor must be small enough to ensure continuous inductor current, then2 > L R such that( )2 222 2 = + Z L R L . Equation (11.37) therefore gives the following load identity for continuous inductor current 21 2 1 1133 3> = >LRR Z Lthat is(11.38) The load and supply (peak) ac currents are ,,,2= =o ac s ac oI I I . The output and supply rms currents are 2 22 2,,,,2 = = + = +o oo rms s rms o ac oI I I I I I (11.39) and the power delivered to resistance R in the load is 2,=R ormsP I R (11.40) B with an L-C filter and discontinuous load current Iftheinductorcurrentreducestozero,atangle,alltheloadcurrentisprovidedbythecapacitor.Its voltage falls to Vo ( . The load power is given by 2o L rmsP I R I R = + (11.93) while the supply power factor is given by 2o rms Lrms rmsI R I R PpfV I V I+= = (11.94) The solution for the uncontrolled converter (a half-wave rectifier) is found by setting = . Example 11.4:Half-wave controlled rectifier The ac supply of the half-wave controlled single-phase converter in figure 11.8a is v = 2 240 sint. For the following loads Load #1: R = 10, L = 0 Load #2: R = 0 , L = 10 Load #3: R = 7.1, L = 7.1 Determine in each load case, for a firing delay angle = /6 the conduction angle = - , hence the current extinction angle the dc output voltage and the average output current thermsloadcurrentandvoltage,loadcurrentandvoltageripplefactor,andpower dissipated in the load the supply power factor Solution Load #1: Z = R = 10, L = 0 From equation (11.76), Z = 10 and0 = . From equation (11.81), = for all , thus for = /6, = - = 5/6. From equation (11.82) 2222(1 cos )(1 cos / 6) 100.9Vo oVVV I R= = += + = The average load current is 22/ (1 cos ) 100.9V/10 =10.1A. = = + = o oVRI V RPower Electronics 237 The rms load voltage is given by equation (11.83), that is ( ) { }( ) { }1212sin 2 )240V / 6 sin / 3 167.2V (= + (= + = rmsV V Since the load is purely resistive, the power delivered to the load is 2 2 2/ 167.2V /10 2797.0W/ 167.9V/10 16.8A= = = == = =o rms rmsrms rmsP I R V RI V R For a purely resistive load, the voltage and current factors are equal: 2167.2V 16.8A1.68100.9V 10.1A1 1.32= = = == = =i vi vFF FFRF RF FF The power factor is 2797W0.70240V 16.7A= =pfAlternatively, use of equation (11.84) gives /6 sin /6- 0.702 4 = + = pf Load #2: R = 0 , Z = X = L = 10 From equation (11.76), Z = X =10 and = . From equation (11.85), which is based on the equal area criterion, = 2 - , thus for = /6, = 11/6 whence the conduction period is = = 5/3. From equation (11.86) the average output voltage is 0VoV =The average load current is ( )( )2240 210cos sin5 / 6 cos / 6 sin / 614.9A ( = + ( = + = oVLI Using equations (11.88) and (11.89), the load rms voltage and current are { }( )( ){ }321240V sin 236.5V6 3240V 12 cos 2 sin 37.9A6 310 (= + = ( (= + + = ( rmsrmsVI Since the load is purely inductive, the power delivered to the load is zero, as is the power factor, and the output voltage ripple factor is undefined. The output current ripple factor is 237.9A2.54 2.54 1 2.3414.9A= = = = = whencermsi ioIFF RFI Load #3: R = 7.1, L = 7.1 From equation (11.76), Z = 10 and = . From figure 11.9a, for / 6 = = and , = =195 whence = 225.Iteration of equation (11.77) gives = 225.5 = 3.936 rad From equation (11.78) 2222(cos cos )240= (cos30 cos 225 ) 85.0Vo oVV I R = = = The average load current is /= 85.0V/7.1 = 12.0Ao oI V R = Alternatively,theaveragecurrentcanbeextractedfromfigure11.9b,whichfor = and Naturally commutating converters 238 / 6 = gives the normalised current as 0.35, thus 20.352240V 0.35 = 11.9A10oVIZ= = From equation (11.80), the rms current is ( )( )240V10sin cos( ) 1( )2 cossin 3.93 cos( 3.93)1 6 6(3.93 ) 18.18A62 cos ( | | + += (| | (\ . ( | | + + (|= = (| | (\ . rmsVZIThe power delivered to the load resistor is 2 218.18A 7.1 2346W = = =o rmsP I RThe load rms voltage, from equation (11.79), is ( ) { }( ) ( ) ( ) { }1 16 61212(sin 2 sin 2 )240V 3.94 (sin 2 3.94 sin 2 ) 175.1V (= (= = rmsV V The load current and voltage ripple factors are 2218.18A1.515 1 1.13812.0A175.1V2.06 1 1.885V= = = == = = =i i iv v vFF RF FFFF RF FF The supply power factor is 2346W0.54240V 18.18A= =pf 11.3.2Half-wave half-controlled Thehalf-wavecontrolledconverterwaveforminfigure11.8bshowsthatwhen , the average output voltage is negative, resulting in a net energy transfer from the load to the supply.The normalised mean output voltage Vn is / coso n oV V V = = (11.111) The rms output voltage is equal to the rms input supply voltage and is given by ( )22 21sinrmsV V t d t V += =(11.112) The ac component harmonic magnitudes in the load are given by ( ) ( ) ( )( )2 22 1 1 2cos 22 1 1 1 1| |= + | | + +\ .nVVn n n n(11.113) for n even, namely n = 2, 4, 6 The load voltage form factor, (thence ripple factor), is 2 2 cos=vFF (11.114) Thecurrentharmonicsareobtainedbydivisionofthevoltageharmonicbyitsloadimpedanceatthat frequency, that is ( )22n nnnV VIZR n L = =+(11.115) Integration of equation (11.110), squared, yields the load rms current (or equation (11.26) for = 0) ( )( )( )( )22 / tan / tan/ tan / tan2sin 2sin 1tan 1 4 sin sin 11 1 ( | | | | (= + || ` || ( \ . \ . ) rmsVI e eZ e e(11.116) Thyristor average current isoI , while thyristor rms current rating is / 2rmsI . The same thyristor current rating expressions are valid for both continuous and discontinuous load current conditions. For a highly inductive load, constant load current, the supply power factor is ( )21 cos pf = +. Critical load inductance The critical load inductance, to prevent the load current falling to zero, is given by ( )2 2cos sin cos 2coscritLR | |= + + + |\ .(11.117) Power Electronics 243 for where 1 12cossin sin2oVV = = (11.118) The minimum current occurs at the angle , where the mean output voltage Vo equals the instantaneous load voltage, vo. When the phase delay angle is greater than the critical angle , substituting = in equation (11.117) gives tancritLR = (11.119) Example 11.5:Controlled full-wave converter continuous conduction The fully controlled full-wave, single-phase converter in figure 11.11a has a source of 240V rms, 50Hz, and a 10 50mH series load. If the delay angle is 45, determine i.the average output voltage and current, hence thyristor mean current ii.the rms load voltage and current, hence thyristor rms current and load ripple factors iii. the power absorbed by the load and the supply power factor If the delay angle is increased to 75 determine iv.the load current in the time domain v. numerically solve the load current equation for , the current extinction angle vi.the load average current and voltage vii. the load rms voltage and current hence load ripple factors and power dissipated viii.the supply power factor Solution The load natural power factor angle is given by ( )-1 -1tan / tan 2 50 50mH/10 57.5 =1 rad = = = L RSince( ) 45 57.5 < < , continuous load current flows, which is given by equation (11.110). { } (1.31 -) / 1.56/ 1.56-/ 1.562 240V 2 sin(1.31-1)( ) [sin( -1)- ]18.62 118.2 [sin( -1)-1.62 ] = = tti t t eet e i.The average output current and voltage are given by equation (11.109) 2 2 2 2cos cos 45 152.8V/ 152.8V/10 15.3Ao oo oV VV I RI V R = = = == = = Each thyristor conducts for 180, hence thyristor mean current is of 15.3A = 7.65A. ii.The rms load current is determined by harmonic analysis. The voltage harmonics (peak magnitude) are given by equation (11.113) ( ) ( ) ( )( )2 22 1 1 2cos 2 2, 4, 6,..2 1 1 1 1| |= + =| | + +\ .fornVV nn n n n and the corresponding current is given from equation (11.115) ( )22n nnnV VIZR n L = =+ The dc output voltage component is given by equation (11.109). harmonic n Vn( )22nZ R n L = +nnnVIZ= 2nI0(152.79)10.0015.28(233.44) 255.6532.971.691.42 48.1663.620.130.01 63.0394.780.070.00 2 2o nI I + =234.4 Naturally commutating converters 244 From the calculations in the table, the rms load current is 2 2+ 234.4 15.3A = = = rms o nI I ISince each thyristor conducts for 180, the thyristor rms current is 12of 15.3A = 10.8A The rms load voltage is given by equation (11.112), that is 240V. 2 22 215.3A1.0 1 1.00 1 0.015.3A240V1.57 1 1.57 1 1.21152.8V= = = = = == = = = = =rmsi i iormsv v voIFF RF FFIVFF RF FFV iii.The power absorbed by the load is 2 215.3A 10 2344W = = =L rmsP I RThe supply power factor is 2344W0.64240V 15.3A= = =Lrms rmsPpfV I iv.Whenthedelayangleisincreasedto75(1.31rad),discontinuousloadcurrentflowssincethe naturalpowerfactorangle ( )-1 -1tan / tan 2 50 50mH/10 57.5 1rad = = = L R isexceeded.Theload current is given by equation (11.104) { } (1.31 -) / 1.56- / 1.562 240V( ) [sin( -1)- sin(1.31-1)] 18.6218.2 [sin( -1)-0.71 ] == tti t t et e v.Solving the equation in part iv for t = and zero current, that is -/ 1.560 sin( -1)-0.71 = egives = 4.09 rad or 234.3. vi.The average load voltage from equation (11.105) is 2 240V (cos75 - cos234.5 ) 90.8V90.89.08A10= == = =oooVV VIR vii. The rms load voltage is given by equation (11.106) { }1240V (4.09 1.31) (sin8.18 sin 2.62) 216.46V (= = ( rmsVThe rms current from equation (11.107) is ( )( )V18.62sin 4.09 1.31 cos(1.31 1 4.09) 240 14.09 1.31 13.55Acos1 ( | | + += =(| | (\ . rmsIThe load voltage and current form and ripple factors are 2 22 213.55A1.49 1 1.49 1 1.119.08A216.46V2.38 1 2.38 1 2.1690.8V= = = = = == = = = = =rmsi i iormsv v voIFF RF FFIVFF RF FFV The power dissipated in the 10 load resistor is 2 213.55 10 1836W = = =rmsP I R viii.The supply power factor is 1836W0.56240V 13.55A= = =Lrms rmsPpfV I Power Electronics 245 11.3.4Full-wave, fully-controlled circuit with R-L and emf load, E AnemfsourceandR-Lloadcanbeencounteredindcmachinemodelling.Theemfrepresentsthe machinespeedbackemf,definedbyE k = .DCmachinescanbecontrolledbyafullycontrolled converter configuration as shown in figure 11.12a, where T1-T4 and T2-T3 are triggered alternately. If in each half sine period the thyristor firing delay angle occurs after the rectified sine supply has fallen belowtheemflevelE,thennoloadcurrentflowssincethebridgethyristorswillalwaysbereverse-biased. Thus the zero current firing angle

is:

( )1sin / 2 (rad) < = < for E V (11.120) where it has been assumed the emf has the polarity shown in figure 11.12a. With discontinuous output current, load current cannot flow until the supply voltage exceeds the back emf E. That is ( )1sin / 2 (rad)0< = < for E V (11.121) Load current can always flow with a firing angle defined by

(rad) (11.122) The load circuit current can be evaluated by solving 2 sin (V)diV t L Ri Edt = + + (11.123) The load voltage and current ripple are both at twice the supply frequency. 11.3.4i - Discontinuous load current The load current is given by { }{ } (-) / tan 2E E2 2( ) [cos sin( - ) - - cos sin( - ) ](rad) = + < +t VV VRi t t et(11.124) Fordiscontinuousloadcurrentconduction,thecurrentextinctionangle,shownonfigure11.12b,is solved by iterative techniques for i(t=) = 0 in equation (11.124). { }{ } (-) / tan E E2 2cos sin( - ) - - cos sin( - ) 0 + =V Ve (11.125) The mean output voltage can be obtained from equation (11.105), which is valid for E = 0.For any E, including E = 0 ( ) 1222 sincos cos ( ) (V)0 (rad) = +| |= + + |\ .< . The relationship between the mean output voltage and current is now given by o oV E I R = + (11.140) That is, the emf term E in equations (11.120) to (11.139) is appropriately changed to - E. The load current flows from the emf source and if > , the average load voltage is negative. Power is beingdeliveredtotheacsupplyfromtheemfsourceintheload,whichisanenergytransferprocess called power inversion. In general 0 90 0 0 090 180 0 0 0< < > > > < < < < >rectificationinversiono o oo o oV P iV P i Naturally commutating converters 248 Figure 11.13.A full-wave controlled converter with an inductive load and negative emf source: (a) circuit diagram; (b) voltage waveforms for discontinuous load current; and (c) continuous load current. Example 11.6:Controlled converter continuous conduction and back emf Thefullycontrolledfull-waveconverterinfigure11.11ahasasourceof240Vrms,50Hz,anda10, 50mH, 50V emf opposing series load. The delay angle is 45. Determine i. the average output voltage and current ii.the rms load voltage and the rms voltage across the R-L part of the load iii. the power absorbed by the 50V load back emf iv. the rms load current hence power dissipated in the resistive part of the load v.the load efficiency, that is percentage of energy into the back emf and power factor vi. the load voltage and current form and ripple factors Solution From example 11.5, continuous conduction is possible since( ) 45 57.5 < < . i.The average output voltage is given by equation (11.133) 22 2402 cos2cos45 152.8VoVV == = The average current, from equation (11.134) is T1 T3 T2 T4 q=2r =1s =1 p= qx r x s p= 2

Power Electronics 249 152.8V 50V10.28A10ooV ERI = = = ii.From equation (11.112) the rms load voltage is 240V. The rms voltage across the R-L part of the load is 2 22 2240V 50V 234.7VRLrms rmsV V E = = = iii. The power absorbed by the 50V back emf load is 10.28A 50V 514WoP I E = = = iv. The R-L load voltage harmonics (which are even) are given by equations (11.135) and (11.136): ( ) ( ) ( )( ).2 22 2cos2 1 1 2cos 22 1 1 1 1 = | |= + | | + +\ .o R LnVV EVVn n n n The harmonic currents and voltages are shown in the table to follow. harmonic n Vn ( )( )22 = +nZ R n L ( )=nnnVIZA 2nI0102.7910.0010.28105.66 260.0232.971.821.66 48.1663.620.130.01 63.2694.780.040.00 2 2o nI I + =107.33 From the table the rms load current is given by 2 2+ 107.33 10.36A = = = rms o nI I IThe power absorbed by the 10 load resistor is 2 210.36A 10 1073.3W = = =L rmsP I R v. The load efficiency, that is, percentage energy into the back emf E is o oo o514W100 32.4514W 1073.3W = =+ The power factor is 514W+1073.3W0.64240V 10.36A= = =Lrms rmsPpfV I vi.The output performance factors are 2 22 210.36A1.011 1 1.023 1 0.12510.28A240V1.57 1 1.57 1 1.211152.8V= = = = = == = = = = =rmsi i iormsv v voIFF RF FFIVFF RF FFV Note that the voltage form factor agrees with that obtained by substitution into equation (11.138), 1.57. Naturally commutating converters 250 11.4Three-phase uncontrolled converter circuits Single-phase supply circuits are adequate below a few kilowatts. At higher power levels, restrictions on unbalancedloading,lineharmonics,currentsurgevoltagedips,andfilteringrequiretheuseofthree-phase (or higher - polyphase) converter circuits. Generally it will be assumed that the output current is both continuous and smooth. This assumption is based on the dc load being highly inductive. 11.4.1Half-wave rectifier circuit with an inductive R-L load Figure 11.14 shows a half-wave, three-phase diode rectifier circuit along with various circuit voltage and current waveforms. A transformer having a star connected secondary is required for neutral access, N. The diode with the highest potential with respect to the neutral conducts a rectangular current pulse. As thepotentialofanotherdiodebecomesthehighest,loadcurrentistransferredtothatdevice,andthe previouslyconductingdeviceisreverse-biasedandnaturally(line)commutated.Notethattheload voltage, hence current never reaches zero, when the load is passive (no opposing back emf). Figure 11.14.Three-phase half-wave rectifier:(a) circuit diagram and (b) circuit voltage and current waveforms. 232323N a c b t q=3r =1s =1 p= qx r x s p= 3 Power Electronics 251 In general terms, for an n-phase p-pulse system, the mean output voltage is given by //22cos (V)2 /sin( / )= (V)/ =popVV t d tppVp(11.141) For a three-phase, half-wave circuit (p = 3)the mean output voltage, (thence average current) is 5 / 62/ 6 312 sin 32 1.17 (V)/ 3 = == = o oV I R V t d tV V(11.142) The rms load voltage is ( )25 / 622/ 6 321sin 1.19 = = rmsV V t d t V (11.143) The load form factor is 1.19 /1.17 1.01rmsFFVV V VV= = = (11.144) ac voltage across the loadRF = ripple factor=dc voltage across the load 1 0.185 = =rms VV(11.145) Thediodeconductionangleis2/n,namely.Thepeakdiodereversevoltageisgivenbythe maximum voltage between any two phases, 32 V = 6 V. Fromequations(11.44),(11.45),and(11.46),foraconstantoutputcurrent, o ormsI I = ,themeandiode current is 1 13(A) D o on I I I = = (11.146) and the rms diode current is 31 1 1(A) = =o o D orms n nI I I I (11.147) The diode current form factor is / 3 = =D ID DFF I I (11.148) Figure 11.15.Three-phase zig-zag interconnected star winding: (a)transformer connection and (b) phasor diagram of transformer voltages. R-L q=3r =1s =1 p= qx r x s p= 3 Naturally commutating converters 252 q=3r =1s =2 p= qx r x s p= 6 abac bcbacaca 6,1 1,22,33,44,5 5,6 i a = iD1- i D4 i b = iD3- i D6 i c = iD5- i D2 source voltages output voltage abcIfneutralisavailable,atransformerisnotnecessary.Thefullloadcurrentisreturnedviatheneutral supply.Thisneutralsupplycurrentisgenerallynotacceptableotherthanatlowpowerlevels.The simpledelta-starconnectionofthesupplyinfigure11.14aisnotappropriatesincetheunidirectional currentineachphaseistransferredfromthesupplytothetransformer.Thismayresultinincreased magnetising current and iron losses if dc magnetisation occurs. This problem is avoided in most cases by the special interconnected star winding, called zig-zag, shown in figure 11.15a. Each transformer limb has two equal voltage secondaries which are connected such that the magnetising forces balance. The resultant phasor diagram is shown in figure 11.15b. Asthenumberofphasesincreases,thewindingsbecomelessutilisedpercyclesincethediode conduction angle decreases, from for a single-phase circuit, to for the three-phase case. Figure 11.16.Three-phase full-wave bridge rectifier:(a) circuit connection and (b) voltage and current waveforms. Power Electronics 253 11.4.2Full-wave rectifier circuit with an inductive load Figure 11.16a shows a three-phase full-wave rectifier circuit where no neutral is necessary and it will be seenthattwoseriesdiodesarealwaysconducting.Onediode(oneofD1,D3,orD5,atthehighest potential)canbeconsideredasbeinginthefeedcircuit,whiletheother(oneofD2,D4,orD6,atthe lowestpotential)isinthereturncircuit.Assuch,theline-to-linevoltageisimpressedacrosstheload. Given no two series diodes conduct simultaneously, there are six possible diode pair combinations. The rectifiercircuitwaveformsinfigure11.16bshowthattheloadripplefrequencyissixtimesthesupply. Each diode conducts for and experiences a reverse voltage of the peak line voltage, 2 VL. The mean load voltage is given by twice equation (11.142), that is 2 / 3/ 3 332 212 sin (V)31.35/ 3o o LL LV IR V t d tV V V = == = =(11.149) where VL is the line-to-line rms voltage (VL =3V). Generally the peak-to-peak ripple voltage for n-phases is 2 2 cos Vn . Table 11.3.Three-phase full-wave uncontrolled rectifier circuits 6th harmonic current average output current output power Io, 6o I PR+PE Full-wave rectifier circuit loadcircuit (A)(A)(W) (a) R-L see section 11.4.2i ( ), 6226 +oVR LoVR 2,o rmsI R(b) R-L-E ( ), 6226 +oVR L oV ER 2, +o o rmsI R I E(c) R-L-C , 66oVL oVR 22, =o o rmsI R I R Vs Is Io VoIo,dc Io,acL CR Vs Is Io VoE + L R Vs Is Io Vo L R Naturally commutating converters 254 The output harmonics of a p-pulse voltage output are 222 2sin11211npannpoVVpnpVn= ( = ( (11.150) where n = mp and m = 1, 2, 3, and Vo is the mean output voltage given by equation (11.141). The output voltage harmonics for p = 6 are given by

( )261Lo nVVn =(11.151) for n = 6, 12, 18, .. The rms output voltage is given by . 2 / 32/ 312 sin2 / 63 31 1.3522rms LL LV V t d tV V | |=|\ .= + =(11.152) Generally, for a p-pulse rectifier output, the rms output voltage is 21 sin2rms LpV Vp= + (11.153) The load voltage form factor = 1.352/1.35 = 1.001 and the ripple factor = form factor -1 = 0.06. 11.4.2iThree-phase full-wave bridge rectifier circuit with continuous load current Ifitisassumedthattheloadinductanceislarge,then(evenwithaloadbackemf),continuousload currentflowsandthedominateloadcurrentharmonicisduetothesixthharmoniccurrent,thatis let, ,6=o ac oI I . By neglecting the higher order harmonics, the various circuit currents and voltages can be readilyobtainedasshownintable11.3.Fromequations(11.149)and(11.151)theoutputvoltageis given by ( )( ), 622 22 2cos 63 3 2cos 613 3 2cos 2351.35 0.077 cos 2 = += + == + = +o o oL LL LL Lv t V V tV V n t nnV V tV V tfor(11.154) The fundamental voltage, hence current, Vo /R, is therefore much larger than the sixth harmonic current, Vo,6 / Z6,thatis , 6> o oI I .Theloadandsupplyaccurrentsare ,,, 6= =o ac s ac oI I I .Theoutputandsupply rms currents are 2 22 2,,,, 6= = + = +o o o rms s rms o ac oI I I I II(11.155) and the power delivered to resistance R in the load is 2,=R ormsP I R (11.156) 11.4.2iiThree-phase full-wave bridge circuit with highly inductive loads constant load current For a highly inductive load, that is a constant load current: the mean diode current is 1 13(A) D o on I I I = = (11.157) and the rms diode current is 31 1 1(A) = =o o D rms orms n nI I I I (11.158) and the power factor for a constant load current is 30.955 pf= = (11.159) Power Electronics 255 The rms input line currents are 23=L rms o rmsI I (11.160) The diode current form factor is / 3 = =D ID D rmsFF I I (11.161) The diode current ripple factor is 21 2ID IDRF FF = = (11.162) A phase voltage and current are given by 2 sinav V t = (11.163) ( ) ( ) 2 3sin 1 sin 1sin 6, 12, 18, ..1 1o an t n ti I t nn n ( += + + =( + ( (11.164) with phases b and c shifted by . That is substitute t in equations (11.163) and (11.164) with t. Each load current harmonic n produces harmonics n+1 and n-1 on the input current. The total load instantaneous power is given by ( ) 2cos3 21on tp t V In| |= |\ .(11.165) The apparent power is 3L s rmsS V I = (11.166) Example 11.7:Three-phase full-wave rectifier The full-wave three-phase rectifier in figure 11.16a has a three-phase 415V 50Hz source (240V phase), and a 10, 50mH, series load. During the problem solution, verify that the only harmonic that need be considered is the sixth. Determine i. the average output voltage and current ii.the rms load voltage and the ac output voltage iii. the rms load current hence power dissipated and power factor iv. the load power percentage error in assuming a constant load currentv.the diode average and rms current requirements Solution i. From equation (11.149) the average output voltage and current are 1.35 1.35 415V 560.447V560.4556.045A10= = = == = =o o LooV IR VV VIR ii.The rms load voltage is given by equation (11.152) 1.352 1.352 415V = 560.94Vrms LV V = = The ac component across the load is 2 22 2560.94V 560.447V 23.52V= = =ac rms oV V V iii. The rms load current is calculated from the harmonic currents, which are calculated from the harmonic voltages given by equation (11.151). harmonic n

( )261LnVVn = ( )22nZ R n L = +nnnVIZ= 2nI0(560.45)10.0056.04(3141.01) 632.0394.780.340.06 127.84188.760.040.00 Note the 12th harmonic current is not significant 2 2o nI I + =3141.07 Naturally commutating converters 256 The rms load current is .2 23141.07 56.05Arms o nI I I = += = The power absorbed by the 10 load resistor is 2 256.05A 10 31410.7WL rmsP I R = = =The power factor is 31410.7W0.9553 23 415V 56.05A3= = = = L Lrms rms L LP PpfV I VI This power factor 0.955 is as predicted by equation (11.159), 3, for a constant current load. iv.The percentage output power error in assuming the load current is constant is given by

2 2oo2 256.045A 10 31410.1W1 1 1 1 056.05A 10 31410.7W = = =

o LL rmsI R PP I R v. The diode average and rms currents are given by equations (11.157) and (11.158) 1 13 31 13 356.045 18.7A56.05 23.4A= = == = =DoD rms ormsI II I 11.5Three-phase half-controlled converter Figure 11.17a illustrates a half-controlled converter where half the devices are thyristors, the remainder being diodes. As in the single-phase case, a freewheeling diode can be added across the load so as to allowthebridgethyristorstocommutateanddecreasefreewheelinglosses.Theoutputvoltage expression consists of 2V 33/2 due to the uncontrolled half of the bridge and 2V 33 cos /2 due to the controlled half which is phase-controlled. The half-controlled bridge mean output is given by the sum, that is 2 23 3 3(1 cos ) (1 cos )2 22.34 (1 cos ) (V)0 (rad) = + = += + =o Lo oV V VVV I R(11.167) At = 0,

o V = 2 V 33/ = 1.35 VL, as in equation (11.46). The normalised mean output voltage Vn is

/ (1 cos ) on oV V V = = + (11.168) Thediodespreventanynegativeoutput,henceinversioncannotoccur.Typicaloutputvoltageand current waveforms for a highly inductive load are shown in figure 11.17b. 11.5i - When the delay angle is less than the output waveform contains six pulses per cycle, of alternating controlled and uncontrolled phases, as shown in figure 11.17b. The output current is always continuous (even for a resistive load) since no output voltage zeros occur. The rms output voltage is given by ( ) ( ) { }( )2 2 2 / 3 2 / 32 2/ 3 / 332 sin 2 sin23 31 1 cos 24/ 3 ++= +| |= + +| |\ . forrms L LLV V t d t V t d tV (11.169) 11.5ii - For delay angles greater than the output voltage waveform is made up of three controlled pulses per cycle, as shown in figure 11.17c. Although output voltage zeros result, continuous load current can flow throughadiodeandtheconductingthyristor,orthroughthecommutatingdiodeifemployed.Therms output voltage is given by Power Electronics 257 ( )( )2223sin23sin 22/ 3 =| |= + |\ .rms LLV V t d tVfor(11.170) Figure 11.17.Three-phase half-controlled bridge converter: (a) circuit connection; (b) voltage and current waveforms for a small firing delay angle ; and (c) waveforms for large. R-L q=3r =1s =2 p= qx r x s p= 6 Naturally commutating converters 258 q=3r =1s =1 p= qx r x s p= 3 The Fourier coefficients of the p-pulse output voltage are given by ( ) ( )( ) ( )2cos 1 cos 12 22 1 1 1sin 1 sin 122 1 1nnn nVan n npn nVbn np ( + = +( + ( ( + = (+ ( (11.171) wheren = mpandm = 1,2,3,..Forthethree-phase,full-wave,half-controlledcase,p = 6,thusthe output voltage harmonics occur at n = 6, 12, 11.6Three-phase controlled thyristor converter circuits 11.6.1Half-wave circuit with an inductive load Whenthediodesinthecircuitoffigure11.14arereplacedbythyristors,asinfigure11.18a,athree-phase fully controlled half-wave converter results. The output voltage is controlled by the delay angle . Thisangleisspecifiedfromthethyristorcommutationangle,whichistheearliestpointtheassociated thyristor becomes forward-biased, as shown in parts b, c, and d of figure 11.18. (The reference is not the phasezerovoltagecross-overpoint).Thethyristorwiththehighestinstantaneousanodepotentialwill conductwhenfiredandinturningonwillreversebiasandturnoffanypreviouslyconductingthyristor. The output voltage ripple is three times the supply frequency and the supply currents contain dc compo-nents.Each phase progressively conducts for periods of , displaced by , as shown in figure 11.18b. Figure 11.18.Three-phase half-wave controlled converter:(a) circuit connection; (b) voltage and current waveforms for a small firing delay angle ; (c) and (d) load voltage waveforms for progressively larger delay angles. Power Electronics 259 The mean output voltage for an n-phase half-wave controlled converter is given by (see example 11.8) //22cos2 /sin( / )cos (V)/nonVV t d tnnVn +==(11.172) which for the three-phase circuit considered with continuous load current gives 323cos 1.17 cos 0 / 62o oV I R V V = = = (11.173) For discontinuous conduction, the mean output voltage is ( ) ( )3231 cos / 6 / 6 5 / 62o oV I R V = = + + (11.174) The mean output voltage is zero for = .For 0 < < , the instantaneous output voltage is always greaterthanzero.Negativeaverageoutputvoltageoccurswhen>asshowninfigure11.18d.Sincetheloadcurrentdirectionisunchanged,for >,powerreversaloccurs, withenergyfeeding from the load into the ac supply.Power inversion assumes a load with an emf to assist the current flow, as in figure 11.13. If > no reverse bias exists for natural commutation and continuous load current will freewheel. Themaximummeanoutputvoltage

o V =2V33/2occursat=0.Thenormalisedmeanoutput voltage Vn is / cosn o oV V V = = (11.175) With an R-L load, at Vo = 0, the load current falls to zero. Thus for > , continuous load current does not flow for an R-L load. The rms output voltage is given by ( )2 / 32/ 31 36 823 23sin ( )2sin2rmsV V t d tV +=| |= + |\ .(11.176) 11.6.2Half-wave converter with freewheel diode Figure 11.19 shows a three-phase, half-wave controlled rectifier converter circuit with a load freewheel diode, Df. This diode prevents the load voltage from going negative, thus inversion is not possible. Figure 11.19.A half-wave fully controlled three-phase converter with a load freewheel diode. q=3r =1s =1 p= qx r x s p= 3 Df Naturally commutating converters 260 11.6.2i - < /6. The output is as in figure 11.18b, with no voltage zeros occurring. The mean output voltage (and current) is given by equation (11.173), that is 3 32 cos 1.17 cos (V)20 / 6 (rad)o oV I R V V = = = (11.177) The maximum mean output Vo = 2V 33/2 occurs at = 0. The normalised mean output voltage, Vn is given by

/ cos on oV V V = = (11.178) TheFouriercoefficientsof the3-pulse outputvoltagearegivenby(11.171).Forthethree-phase,half-wave, half-controlled case, p = 3, thus the output voltage harmonics occur at n = 3, 6, 9, 11.6.2ii - > /6. Voltage zeros occur and the negative portions in the waveforms in parts c and d of figure 11.18 do not occur. The mean output voltage is given by ( ) / 622sin2 / 31 cos( / 6 (V)2 / 3/ 6 5 / 6o oVV I R t d tV = == + + (11.179) The normalised mean output voltage Vn is

/[1cos( / 6)]/ 3 on oV V V = = + + (11.180) The average load current (with an emf E in the load) is given by ooV EIR= (11.181) These equations assume continuous load current. 11.6.2iii - > 5/6. A delay angle of greater than 5/6 would imply a negative output voltage, clearly not possible with a freewheel load diode. 11.6.3Full-wave, fully-controlled circuit with an inductive load Athree-phasebridgeisfullycontrolledwhenallsixbridgedevicesarethyristors,asshowninfigure 11.20a. The frequency of the output ripple voltage is six times the supply frequency and each thyristor alwaysconductsfor.Circuitwaveformsareshowninfigure11.20b.Themeanoutputvoltagefor both inductive and resistive loads is given by ( )/ 63 2323sin / 63cos 2.34 cos (V)0 2 / 3 ++= += = oV V t d tV V (11.182) which is twice the voltage given by equation (11.173) for the half-wave circuit, but for a purely resistive load, equation (11.182) becomes ( )3231 cos / 6 (V)/ 3 2 / 3 ( = + + oV V(11.183) Theaverageoutputcurrentisgivenby / o oI V R = ineachcase.Ifaloadbackemfexiststheaverage current becomes ooV EIR= (11.184) The maximum mean output voltage

o V = 2V 33/ occurs at = 0. The normalised mean output Vn is

/ cos on oV V V = = (11.185) Fordelayanglesupto,theoutputvoltageisatallinstancesnon-zero,hencetheloadcurrentis continuousforanypassiveload(bothresistiveandinductive).Beyondtheloadcurrentmaybe discontinuous(alwaysdiscontinuousforaresistiveload).For>thecurrentisalways discontinuous for passive loads and the average output voltage is less than zero. Power Electronics 261 Figure 11.20.A three-phase fully controlled converter:(a) circuit connection and (b) load voltage waveform for four delay angles. For continuous load current, the load current is given by ( )( )( )6tan3tan3 2 3 2sin sin61tV E V ei t tZ R Z e + += + + (11.186) The maximum and minimum ripple current magnitudes are ( )( )6tan3tan3 2 3 2sin sin21V E V eIZ R Ze = + +

(11.187) at t = +n for n = 0, 6, 12, .. ( )( )3tan3 2 3 2 1sin sin31V E VIZ R Ze = + + (11.188) at t = - +n for n = 0, 6, 12, .. With a load back emf the critical inductance for continuous load current must satisfy0 I=in equation (11.188), that is q=3r =1s =2 p= qx r x s p= 6 Naturally commutating converters 262 ( )( )/ 3tan13sinsin1 3 2R EZ e V ( + + ( (11.189) wheretan / L R = . The rms value of the output voltage for a purely resistive load is given by ( )2 / 22/ 63 3223 233 sin ( )1 cos 2rmsV V t d tV +| |=|\ .= + 0 / 3 (11.190) and( )3 32 43 2 1 sin 2 / 3rmsV V = / 3 2 / 3 (11.191) Thenormalisevoltageharmonicpeakmagnitudesintheoutputvoltage,withcontinuousloadcurrent, are ( ) ( ) ( )( )..122 23 3 1 1 2cos 221 1 1 1| |= + | | + +\ .L nV Vn n n n(11.192) for n = 6, 12, 18 The harmonics occur at multiples of six times the fundamental frequency. For discontinuous load current, at high delay angles, when the output current becomes discontinuous with an inductive load, the output current is given by ( )( ) ( )6 6tan tan3 2sin sin 16 3t tcV Ei t t e eZ Rt + += + + + (( (( (11.193) where c is the conduction period, which is found by solving the transcendental equation formed when in equation (11.193), i(t = ++c) = 0. The average output voltage can then be found from ( ) ( )3 3 2 3cos cos3 3 3o c cV EV = + + + (( (11.194) 11.6.3i Resistive load For a resistive load, the load voltage harmonics for p pulses per cycle, are given by ( ) ( )( ) ( )2cos 1 cos 12 22 1 1 1sin 1 sin 122 1 1nnn nVan n npn nVbn np ( + = (+ + ( ( + = (+ ( (11.195) for n = pm and m = 1, 2, 3, .. The harmonics occur at multiples of six times the fundamental frequency, for a 6 - pulse (p = 6) per cycle output voltage. 11.6.3iiHighly inductive load constant load current Aswithacontinuousloadcurrent,withaconstantloadcurrenttheinputcurrentcomprises alternatingpolarityblocksofcurrent,witheachphasedisplacedrelativetotheothersby, independent of the thyristor triggering delay angle.At maximum voltage hence maximum power output, thedelayangleiszeroandthephasevoltageandcurrentfundamentalareinphase.Asthephase angleisincreased,theinverteroutputvoltage,hencepoweroutputisdecreased,andthelinecurrent blockofcurrent(fundamental)shiftsbywithrespecttothelinevoltage.Reactiveinputpower increases as the real power decreases. At = , the output voltage reduces to zero, the output power iszero,andthecurrentblocksintheacinputareshiftedwithrespecttothelinevoltage, producingonlyVArsfromtheacinput.Whenthedelayangleisincreasedabove,theinverterdc output reverses polarity and energy transfers back into the ac supply (inversion), with maximum inverted power reached at = , where the reactive VAr is reduced to zero, from a maximum at = . For a highly inductive load, that is a constant load current: the mean diode current is Power Electronics 263 1 13(A) = = Tho on I I I (11.196) and the rms diode current is 31 1 1(A) = =o o Th rms orms n nI I I I (11.197) The diode current form factor is / 3 = = ThITh Th rmsFF I I (11.198) The diode current ripple factor is 21 2ITh IThRF FF = = (11.199) The rms input line currents are 23=L rms o rmsI I (11.200) A phase voltage is given by 2 sinav V t = (11.201) with phases b and c shifted by . That is substitute t with t. From equation (14.34), the line current harmonics are ( ) 134 1 cos for oddoi I n nn=i(11.202) The rms fundamental input current is 123o rmsI I= (11.203) The supply fundamental apparent power, S1, active power P and reactive power Q, are given by 1 11 11 13cossinrmsS VIP P SQ Q S== == =(11.204) Thesupplyapparentpowerisconstantforagivenconstantloadcurrent,independentofthethyristor turn-on delay angle. The power factor for a constant load current is 123 3 cos3 cos 3cos2 23 33 3= = = o rmsrms rmso o rms rmsV IV IpfV I V I(11.205) Converter shut down is best achieved regeneratively by increasing (and controlling) the delay angle to greater than such that the output voltage goes negative, which results in controlled power inversion backintotheacsupply.Iftriggeringpulsestoallthethyristorsareremoved,thedccurrentdecays slowly and uncontrolled to zero through the last pair of thyristors that were triggered. Example 11.8:Three-phase full-wave controlled rectifier with constant output current Thefull-wavethree-phasecontrolledrectifierinfigure11.20ahasathree-phase415V50Hzsource (240V phase), and provides a 100A constant current load.Determine: i.the average and rms thyristor current ii.the rms and fundamental line current iii.the apparent fundamental power S1 If 25kW is delivered to the dc load, calculate: iv. the supply power factoriv.the dc output voltage, hence the converter phase delay angle v.the real active and reactive Q1 ac supply power vi.the delay angle range if the ac supply varies by 5% (with 25kW and 100A dc). Solution i.From equations (11.196) and (11.197) the thyristor average and rms currents are 1 1 13 3 31 13 3100A 33 A100A 57.7Ao Tho ThrmsI II I= = == = = Naturally commutating converters 264 ii.The rms and fundamental line currents are 12 2100A = 81.6A3 32 23 3 100A = 78.0A = = = = L rms ormso rmsI II I iii.The apparent power is 1 13 3 415V 78A = 56.1kVArmsS VI = = iv.The supply power factor, from equation (11.205), is 25kW0.4263 3 415V 81.6A= = = Lrms rmsPpfV I v.The output voltage is power25kW250V dc100AooVI= = =Thyristor delay angle is given by equation (11.182), that is 2.34 cos415V250Vdc 2.34 cos3== oV V which yields a delay angle of = 1.11rad = 63.5 vi.For a constant output power at 100A dc, the output voltage must be maintained at 250V dc independent of the ac input voltage magnitude, thus for equation (11.182) ( )( )

( )111250Vdccos2.34 415 5%3250Vdccos 1.08rad 61.92.34 415 5%3250Vdccos 1.13rad 64.92.34 415 5%3= = = = = = = + Figure 11.21.A full-wave three-phase controlled converter with a load freewheeling diode (half-controlled). q=3r =1s =2 p= qx r x s p= 6 Power Electronics 265 2 s i n V nn2 s i n V nn(a) (b) p0 + p rectify invert 2p 2sinVpp+2sinVpp11.6.4Full-wave converter with freewheel diode Bothhalf-controlledandfullycontrolled converterscanemployadiscreteloadfreewheeldiode.These circuitshavethevoltageoutputcharacteristicthattheoutputvoltagecannevergonegative,hence powerinversionisnotpossible.Figure11.21showsafullycontrolledthree-phaseconverterwitha freewheel diode D. Thefreewheeldiodeisactivefor>.Theoutputisasinfigure11.20bfor90. V = (v1 + v2) 0 + + 0 v3-1 2 VLL recovery angle = tq VT3 (b) (e) (c) (d) commutation voltage commutation voltage Naturally commutating converters 268 The mean output voltage oVis therefore 5 / 6 / 62/ 6 / 612 / 3o ov V Vv d t v d t + + ++ += (= ( where v = (v1 + v2) ( )( ) { }5 / 6/ 6/ 6/ 62322sin32sin sinoV t d tVV t t d t +++ ++ (+ (= ( + + ( ( ) ( )33 2 23 3cos cos cos2 2 2oV V V = (11.213) ( )323cos cos4oV V ( = + + (11.214) which reduces to equation (11.173) when = 0.Substituting cos - cos ( + ) from equation (11.212) into equation (11.213) yields 323 3cos2 2o oV V LI = (11.215) 32= o o oV V LI that is (11.216) ThemeanoutputvoltageVoisreducedorregulatedbythecommutationreactanceXc=Landthis regulationvarieswithloadcurrentmagnitudeIo.Convertersemiconductorvoltagedropsalsoregulate (decrease) the output voltage.The component 3L/2 is called the equivalent internal resistance. Being an inductive phenomenon, it does not represent a power loss component. As shown in figure 11.24, the overlap occurs immediately after the delay .The commutation voltage, v2 - v1, is 3 2 V sin . The commutation time is inversely proportional to the commutation voltage v2 - v1. For rectification, as increases from zero to , the commutation voltage increases from a minimum of zero volts to a maximum of 3 2 V at , whence the overlap angle decreases from a maximum of at = 0 to a minimum of at .[For inversion, the overlap angle decreases from a minimum of at to a maximum of at , as the commutation voltage reduces back to zero volts.] From equation (11.212), with = 2 3 sin(2 / )oarc LI V =Thegeneralexpressionsforthemeanloadvoltage oVofann-pulse,fully-controlledconverter,with underlap, are given by ( )2sin cos cos2 / ( = + + o nVVn(11.217) and 2sin cos / 2/ = o c o nVV nXIn(11.218) where V is the line voltage for a full-wave converter and the phase voltage for a half-wave converter and the plus sign in equation (11.218) accounts for inversion operation.Effectively, as shown in figure 11.25, overlap reduces the mean output voltage by nfLI0 or as if were increased.Thesupplyvoltageiseffectivelydistortedandtheharmoniccontentoftheoutputis increased.Equating equations (11.217) and (11.218) gives the mean output current ( ) ( )2sin cos cos (A)ocnVIX = + (11.219) which reduces to equation (11.212) when n = 3. Harmonic input current magnitudes are decreased by a factor ( ) sin / n n . The single-phase, full-wave, converter voltage loss is 2LIo / and the overlap output voltage is zero. Power Electronics 269 Figure 11.24.Overlap for current commutation from thyristor 1 to thyristor 2, at delay angle . Figure 11.25.Overlap regulation model: (a) equivalent circuit and (b) load plot of overlap model. Io Vo n 2 V/sin/ncos nX/2 Io=0, = 0 from equation 11.180 Vo = n 2 V/sin/ncos Vo=0, = from equation 11.184 Io = 22V/Xsin/ncos Vo Io slope =-nX/2 0 (a) (b) Mean output voltage Load currrent equation (11.211) = 90 i1 i1 i2 i1 i1 i2 i2 i2 rectifying Vo > 0

= 0 = 60 = 135 v v1 v2 (v1+v2 ) vo Inverting Vo < 0 2 3VL 2 32VL i(t) t Io Io Io Io Io Io Io Io v2-v1 Naturally commutating converters 270 11.8Overlap - inversion A fully controlled converter operates in the inversion mode when > 90 and the mean output voltage is negative and less than the load back emf shown in figure 11.23a. Since the direction of the load current Io is from the supply and the output voltage is negative, energy is being returned, regenerated into the supply from the load. Figure 11.26 shows the power flow differences between rectification and inversion. As increases, the returned energy magnitude increases. If plus the necessary overlap exceeds t = , commutation failure occurs. The output goes positive and the load current builds up uncontrolled.The last commutation with in figures 11.23c and d results in a commutation failure of thyristor T1. Beforethecirculatinginductorcurrentihasreducedtozero,theincomingthyristorT2 experiencesan anodepotentialwhichislesspositivethanthatofthethyristortobecommutatedT1,v1-v2

0 vs > 0 0 < < i vs + + + i > 0 vs < 0 < < i vs + (a)(b) power out power in power in power out rectification inversion Power Electronics 271 62sin cos / 22 /2 415sin cos 0 3V 557.44V2 / 6o c o nVV nXIn = = = From equation (11.217) ( )| |2sin / cos cos2 /2 415557.44 sin / 6 1 cos2 / 68.4oVV nn ( = + + = += that is ii. = 60 622sin cos / 22 /415sin cos 60 3V 277.22V2 / 6o c o nVV nXIn = += = ( )( )22sin / cos cos2 /415277.22 cos 60 cos 602 / 60.71oVV nn ( = + + ( = + + = that is Equation (11.220) gives the maximum allowable delay angle as 1-4 2-122cos 1sin /25010 10= cos -1415171.56 and 557.41VooX IV nV = ` ) ` )= = 11.9Summary General expressions for n-phase converter mean output voltage, Vo (i) Half-wave and full-wave, fully-controlled converter 2sin( / )cos/onV Vn=where V isthe rms line voltage for a full-wave converter orthe rms phase voltage for a half-wave converter. cos cos = , the supply displacement factor From LHopitals rule,for n, Vo = 2 V cos (ii) Full-wave, half-controlled converter ( ) 2sin( / )1 cos/onV Vn= +where V is the rms line voltage. (iii) Half-wave and full-wave controlled converter with load freewheel diode 2sin( / )cos 0 //onV V nn = < < ( )21 cos / / /2 /onV V n nn + + = < < + the output rms voltage is given by Naturally commutating converters 272 ( )cos 2 sin 2 /1 / 2 /cos 2 2 / / 4 2 / 4 / = + + = + + >rmsrmsnV V nnn nV V nn n where V isthe rms line voltage for a full-wave converter orthe rms phase voltage for a half-wave converter. n = 0 for single-phase and three-phase half-controlled converters = for three-phase half-wave converters = for three-phase fully controlled converters Thesevoltageoutputcharacteristicsareshowninfigure11.27andthemainconvertercircuit characteristics are shown in table 11.4. Figure 11.27. Converter normalised output voltage characteristics as a function of firing delay angle . Power Electronics 273 11.10Definitions

average output voltage average output currentrms output voltage rms output currentpeak output voltage peak output currento orms rmsV IV IV I

rmsv vo rmsrmsi irmsoVVFF CFV VIIFF CFII= = = == = = == =

Load voltage form factor Load voltage crest factorLoad current form factor Load current crest factordc load powerRectification efficiencyac lrectifierLosso orms rmsV IV I=+oad power + rectifier losses ( )2 2222 2 1211 == = == = (= + ( effective values of ac (or)Waveform smoothness =Ripple factoraverage value of (or)wheresimilarly the current ripple factor isRivorms ovoRi an bnnV V IRFV I VV VFFVV v vR21 = = = for a resistive loadRii ioi vIF FFIRF RF 11.11Output pulse number Outputpulsenumberpisthenumberofpulsesintheoutputvoltagethatoccurduringoneacinput cycle, of frequency fs. The pulse number p therefore specifies the output harmonics, which occur at p x fs, and multiples of that frequency, mpfs, for m = 1, 2, 3, ... The pulse number p is specified in terms ofqthe number of elements in the commutation group rthe number of parallel connected commutation groups sthe number of series connected (phase displaced) commutating groups Parallelconnectedcommutationgroups,r,areusuallyassociatedwith(andidentifiedby)intergroup reactors (to reduce circulating current), with transformers where at least one secondary is effectively star connectedwhileanotherisdeltaconnected.Therectifiedoutputvoltagesassociatedwitheach transformer secondary, are connected in parallel. Seriesconnectedcommutationgroups,s,areusuallyassociatedwith(andidentifiedby)transformers where at least one secondary is effectively star while another is delta connected, with the rectified output associated with each transformer secondary, connected in series. The mean converter output voltage Vo can be specified by2 sin cosoqV s Vq= (11.221) For a full-wave fully-controlled single-phase converter, r = 1, q = 2, and s = 1, whence p = 2 q=3r =2s =2 p= qx r x s p= 12 Naturally commutating converters 274 2 221 2 sin cos cos2oVV V = = For a full-wave, fully-controlled, three-phase converter, r = 1, q = 3, and s = 2, whence p = 6 3 232 2 sin cos cos3oVV V = = Reading list Dewan, S. B. and Straughen, A., Power Semiconductor Circuits, John Wiley and Sons, New York, 1975. Sen, P.C., Power Electronics, McGraw-Hill, 5th reprint, 1992. Shepherd, W et al. Power Electronics and motor control, Cambridge University Press, 2nd Edition 1995. http://www.ipes.ethz.ch/ Power Electronics 275 Table 11.4.Main characteristics of controllable converter circuits 11.8a11.11a11.10b11.711.18a 11.19a11.17a 11.10a11.711.1911.2111.17a Naturally commutating converters 276 Problems 11.1.For the circuit shown in figure 11.28, if the thyristor is fired at = i.derive an expression for the load current, i ii.determine the current extinction angle, iii.determine the peak value and the time at which it occurs iv.sketch to scale on the same t axis the supply voltage, load voltage, thyristor voltage, and load current. Figure 11.28.Problem 11.1. 11.2.For the circuit shown in figure 11.29, if the thyristor is fired at = determine i.the current extinction angle, ii.the mean and rms values of the output current iii.the power delivered to the source E. Sketch the load current and load voltage vo. Figure 11.29.Problem 11.2. 11.3.Derive equations (11.19) and (11.20) for the circuit in figure 11.3. 11.4.Assumingaconstantloadcurrentderiveanexpressionforthemeanandrmsdevicecurrent and the device form factor, for the circuits in figures 11.4 and 11.7. 11.5.PlotloadripplevoltageKRIandloadvoltageripplefactorKv,againstthethyristorphasedelay angle for the circuit in figure 11.7. 11.6.Showthattheaverageoutputvoltageofan-phasehalf-wavecontrolledconverterwitha freewheel diode is characterised by 11 1n n22sin( / )cos (V)/0 - /1 cos (V)2 / oonnV VnnV Vn =< > L ii.R = L iii. R