CHAPTER 3 Diode Rectifiers After completing this chapter, students should be able to do the following: List the types of diode rectifiers and their advantages and disadvantages. Explain the operation and characteristics of diode rectifiers. List and calculate the performance parameters of diode rectifiers. Analyze and design diode rectifier circuits. Evaluate the performances of diode rectifiers by SPICE simulations. Determine the effects of load inductance on the load current Determine the Fourier components of rectifier outputs. Design output-side filters for diode rectifiers. Determine the effects of source inductances on the rectifier output voltage Symbols and Their Meanings Symbols Meaning Toiavys Ttems) ‘Average and rms diode currents, respectively Teas): Fors ‘Average and rms output currents, respectively Ips ly rms primary and secondary currents of an input transformer, respectively Paci Pac dc and ac output powers, respectively RF, TUF, PF (Output ripple factor, transformer utilization factor, and power factor, respectively wp (03 in() Instantaneous diode voltage and diode current, respectively (D5 to); HCO) Instantaneous input supply, output, and ripple voltages, respectively Vins Votan: Votens) Peak, average, and rms output voltages, respectively Vacopys Veco): Vices) Peak to peak, peak, and rms ripple output voltages, respectively 1 Vp Ve ‘Transformer turns ratio, rms primary voltage, and secondary voltage, respectively 103104 Chapter 3 Diode Rectifiers 3.1. INTRODUCTION Diodes are extensively used in rectifiers. A rectifier is a circuit that converts an ac signal into a unidirectional signal. A rectifier is a type of ac~de converter. A rectifier may also be considered as an absolute value converter. If vis an ac input voltage, the waveform of the output voltage v,, would have the same shape, but the negative part will appear as a positive value. That is, », = |»,|. Depending on the type of input sup- ply, the rectifiers are classified into two types: (1) single phase and (2) three phase. A single-phase rectifier can be either a half wave or a full wave. A single-phase half-wave rectifier is the simplest type, but it is not normally used in industrial applications. For the sake of simplicity the diodes are considered to be ideal. By “ideal” we mean that the reverse recovery time f,, and the forward voltage drop Vp are negligible. That is, ty, = Oand Vp = 0. 3.2. PERFORMANCE PARAMETERS Although the output voltage of a rectifier in Figure 3.la should ideally be a pure de, the output of a practical rectifier contains harmonics or ripples as shown in Figure 3.1b. A tectifier is a power processor that should give a de output voltage with a minimum amount of harmonic contents. At the same time, it should maintain the input current as sinusoidal as possible and in phase with the input voltage so that the power factor is near unity. The power-processing quality of a rectifier requires the determination of harmonic contents of the input current, the output voltage, and the output current, We can use the Fourier series expansions to find the harmonic contents of voltages and currents, The performances of a rectifier are normally evaluated in terms of the following parameters The average value of the output (load) voltage, Vic The average value of the output (load) current, [ic The output de power, Pac = Vaclac GA) output with ripple AC Tr = pe s ideal de (a) Rectifier (b) Output voltage FIGURE 3.1 Input and output relationship of a rectifier.3.2. Performance Parameters 105 ‘The root-mean-square (rms) value of the output voltage, Vins The rms value of the output current, [ims The output ac power Pao = Verslims 3.2) The efficiency (or rectification ratio) of a rectifier, which is a figure of merit and permits us to compare the effectiveness, is defined as Prag = 33 =p. G3) It should be noted that 1» is not the power efficiency. It is the conversion effi- ciency which is a measure of the quality of the output waveform. For a pure de output, the conversion efficiency would be unity. The output voltage can be considered as composed of two components: (1) the de value and (2) the ac component or ripple. The effective (rms) value of the ac component of output voltage is Voc = VV inns ~ Vie G4) The form factor, which is a measure of the shape of output voltage, is Vs FF= = 3.5) Vie The ripple factor, which is a measure of the ripple content, is defined as 66) G7) G8) where V, and J, are the rms voltage and rms current of the transformer secondary, respectively. The input power can be determined approximately by equating input power with the output ac power. That is, the power factor is related by Pac Vols Crest factor (CF), which is a measure of the peak input current [y(peai) 48 com- pared with its rms value /,, is often of interest to specify the peak current ratings of devices and components. CF of the input current is defined by PE 3.9) Lipeak) cF= — (3.10)106 Chapter 3 Diode Rectifiers Key Points of Section 3.2 + The performance of a rectifier that is measured by certain parameters is poor. The load current can be made continuous by adding an inductor and a freewheel- ing diode. The output voltage is discontinuous and contains harmonics at mul- tiples of the supply frequency. 3.3. SINGLE-PHASE FULL-WAVE RECTIFIERS A full-wave rectifier circuit with a center-tapped transformer is shown in Figure 3.2a. During the positive half-cycle of the input voltage, diode D; conducts and diode Dy is in a blocking condition. The input voltage appears across the load. During the negative half-cycle of the input voltage, diode Dy conducts while diode D; is in a blocking condi- tion. The negative portion of the input voltage appears across the load as a positive volt- age. The waveform of the output voltage over a complete cycle is shown in Figure 3.2b. Because there is no de current flowing through the transformer, there is no de saturation problem of transformer core. The average output voltage is ™ Vin Vm Sin wt dt = —* = 0.6366V), (3.11) Jo Vac = on (a) Circuit diagram. (b) Waveforms FIGURE 3.2 Full-wave rectifier with center-tapped transformer,3.3. Single-Phase Full-Wave Rectifiers 107 ; 7 AD BD Dy BD, aoe + = “epee Una (a) Cireuit diagram (©) Waveforms FIGURE 3.3 Full-wave bridge rectifier. Instead of using a center-tapped transformer, we could use four diodes, as shown in Figure 3.3a, During the positive half-cycle of the input voltage, the power is supplied to the load through diodes D, and D2. During the negative cycle, diodes D; and Ds conduct. The waveform for the output voltage is shown in Figure 3.3b and is similar to that of Figure 3.2b. The peak inverse voltage of a diode is only V,,. This circuit is known as a bridge rectifier, and it is commonly used in industrial applications (1, 2). Some of the advantages and disadvantages for the circuits in Figures 3.2 and 3.3 are listed in Table 3.1 TABLE 3.1 Advantages and Disadvantages of Center-Tapped and Bridge Rectifiers Advantages Disadvantages Center-tapped Simple, only two diodes Limited low power supply, less than 100 W ‘wansformer Ripple frequencyis twice the Tnereased cost due to the center-tapped supply frequency transformer Provides an electrical isolation De current flowing through each side of the second- ary willinctease the transformer cost and size Bridge reetier Suitable for industrial ‘The load cannot be grounded without an input-side applications up to 100 kW transformer Ripple frequency is twice the Although an input-sie transformer is not needed supply frequency forthe operation ofthe rectifier, one is normally Simple to use in commercially Connected to isolate the load electrically from the available units supply108 Chapter 3 Diode Rectifiers Example 3.1 Finding the Performance Parameters of a Full-Wave Rectifier with a Center-Tapped Transformer If the rectifier in Figure 3.2a has a purely resistive load of R, determine (a) the efficiency, (b) the FF (c) the RF, (d) the TUF (¢) the PIV of diode Dj, (f) the CF of the input current, and (g) the input power factor PF. Solution From Eq, (3.11), the average output voltage is We Vag = = = 0.63661, and the average load current is Vise _ 0:6366V Tie = R= R ‘The rms values of the output voltage and current are my, Vu = [2 [Wa sinay? ar] = 4 = 0707, 0 v2 1 es 0.707V mm TR R From Eq. (3.1) Pic = (0.6366Vj,)°/R, and from Eq, (3.2) Pye = (0.707Vp,)/R. a, From Eq, (3.3), the efficiency — (0.6366V;,)"/(0.707V;,)” = 81%. b. From Eq, (3.5), the form factor FF = 0.707Vj/0.6366V, = 1.11. ©. From Eq, (3.7), the ripple factor RF = VIE — 1 = 0.482 or 48.2%, 4d. ‘The rms voltage of the transformer secondary V, = Vj/'VZ = 0.707V,,. The tms value of transformer secondary current J, = 0.5Vq/R. The volt-ampere rating (VA) of the transformer, VA = V2V,l, = V2 X 0.707Vj, X 0.5Vq/R. From Eq. (3.8), 0.6366" TUF = ———_. V2 x 0.107 x 05 = 0.81064 = 81.06% fe. The peak reverse blocking voltage, PIV = 2Vq = VmlR and L, = 0.707Vjy/R. The CF of the input current is CF = Iypesiy/le = V2. g. The input PF for a resistive load can be found from P, 0.707% VA V2 x 0.107 x 0S PF = =10 Note: TUF = 1/0.81064 = 1.136 signifies that the input transformer, if present, must be 1.75 times larger than that when it is used to deliver power from a pure ac sinusoidal voltage. The rectilier has an RF of 48.2% and a rectification efficiency of 81%.3.4 Single-Phase Full-Wave Rectifier with RL Load 109 Example 3.2 Finding the Fourier Series of the Output Voltage for a Full-Wave Rectifier ‘The rectifier in Figure 3.3a has an RL load. Use the method of Fourier series to obtain expres- sions for output voltage v(t) Solution ‘The rectifier output voltage may be described by a Fourier series (which is reviewed in Appendix E) as ml = Vac +S) (@q 08 not + by sin net) where a 1 277 2Vm Vie = 5 f n9(0) (wt) = 5 [ Vou sin wt (wt) = —* a, = t 1% 608 net d(ot) = 2 f[ Voy Sin wot cos mat dt) Wn & =1 - S ery H =0 forn = 1,3,5, by 4 v9 sin nat d(wt) = 2 [ Vou sin wt sin reat d(oot) — 0 Substituting the values of a, and b,, the expression for the output voltage is Vin Vin Wan Way ny) = = v8 ar = TE cos dt — 2 00s Gur — (G2) 34 Note: The output of a full-wave rectifier contains only even harmonics and the second harmonic is the most dominant one and its frequency is 2f(= 120 Hz). The output voltage in Eq. (3.12) can be derived by spectrum multiplication of switching function, and this is explained in Appendix C. Key Points of Section 3.3 * There are two types of single-phase rectifiers: center-tapped transformer and bridge. Their performances are almost identical, except the secondary current of the center-tapped transformer carries unidirectional (de) current and it requires alarger VA rating. The center-tapped type is used in applications less than 100 W and the bridge rectifier is used in applications ranging from 100 W to 100 kW. The output voltage of the rectifiers contains harmonics whose frequencies are multiples of 2f (two times the supply frequency). SINGLE-PHASE FULL-WAVE RECTIFIER WITH RL LOAD With a resistive load, the load current is identical in shape to the output voltage. In practice, most loads are inductive to a certain extent and the load current depends on the values of load resistance R and load inductance L. This is shown in Figure 3.4a.110 Chapter 3 Diode Rectifiers (a) Cireuit FIGURE 3.4 Full-bridge rectifier with RL load, (©) Supply line current 9 = Vy sin (ot) (@) Discontinuous current3.4 Single-Phase Full-Wave Rectifier with RL Load 111 A battery of voltage E is added to develop generalized equations. If v, = V;, sin wt V2 V, sin wt is the input voltage, the load current ig can be found from di LD + Rip + B= |V2V;sinwt| for j= 0 which has a solution of the form v2V, E ig Zz sin(ot — | + Aye RID R (3.13) where load impedance Z = [R? + (wL)’]'”, load impedance angle 6 = tan“"(wL/R), and V, is the rms value of the input voltage. Case 1: continuous load current. This is shown in Figure 3.4b. The constant A; in Eq. (3.13) can be determined from the condition: at wt = 1, ig = Ip E v2, = (b+ R- a °) et Nas) R Substitution of A, in Eq. (3.13) yields /2V, /2V. ig < * sin(wr — 6) + ( + 5 = 4 Esina )emnine-9 = 5 (3.14) Under a steady-state condition, ig(wt = 0) = ig(wt = 7). That is, ig(wt = 0) Applying this condition, we get the value of Jy as VV, 1 + ethiliats) = 7 in6 RDew) ~ R ly for = 0 (3.15) which, after substituting J in Eq. VI, ar 3.14) and simplification, gives 2 — 9) ging iD] — {sino ®) + —Sanatin oe i for 0 = (wt — 0) = mandig = 0 (G.16) The rms diode current can be found from Eq, (3.16) as Lp qe inne) = [pu [Blot and the rms output current can then be determined by combining the rms current of each diode as Togs) = (ems) + Fipems)"” = V2 hy The average diode current can also be found from Eq. (3.16) as tt Tow) = 7 [ fig (wot)2 Chapter 3 Diode Rectifiers Case 2: discontinuous load current. This is shown in Figure 3.4d. The load cur- rent flows only during the period « = wt = . Let us define x = E/V_, = E/V2V, as the load battery (emf) constant, called the voltage ratio. The diodes start to conduct at ot = «given by a = sin” a sin (x) At of = a, iq(ot) = 0 and Eq, (3.13) gives o 2V, a= [E-Bay] eo4e g G17) At of = B, the current falls to zero, and (wt = 8) = 0. That is, v2, E_ V2V, ] E s - FNM nla = 0) | Riene-pye — Zsin(B = 0) + [z Zina — 8) |e R79 G8) Dividing Eq. (3.18) by V2V,Z, and substituting R/Z = cos and wL/R we get an 6, x , wn x caw 7 8M = 0) és - SM? B can be determined from this transcendental equation by an iterative (trial and error) method of solution, Start with 8 = 0, and increase its value by a very small amount until the left-hand side of this equation becomes zero. As an example, Mathcad was used to find the value of B for 8 = 30°, 60°, and x = Oto 1. The results are shown in Table 3.2. As x increases, B decreases. Atx = 1.0, the diodes do not conduct and no current flows. ‘The mms diode current can be found from Eq. (3.17) as wo 12 Iowa = [3 [% aa} The average diode current can also be found from Eq. (3.17) as sin(B - 0) 4 ( (G19) 1 fe Joos) = 3 { fo d(ot) TABLE 3.2 Variations of Angle 8 with the Voltage Ratio, x Voltage Ratios 0 01 020304 05 06 070809 «dO Biore=30° 210 203,S197,=s90—S«s83A7S-sG7?s1SBA7-—132 90 Bioro= 60° 244-234-225. 2IS.-20S.s19H_SsAB}ATL:SsAS7.—38 903.4 Single-Phase Full-Wave Rectifier with RL Load 113 Boundary: Discontinuous/Continuous Region 08 ost 2 x0) 04 4 o2b ° i i i i i i 00 iz 14 : = Load impedance angle, radians 2 FIGURE 3.5, Boundary of continuous and discontinuous regions for single-phase rectifier. Boundary conditions: The condition for the discontinuous current can be found by setting Jy in Eq, (3.15) to zero. ViV2 O87 SO) BS | R which can be solved for the voltage ratio x = EI(V2V,) as sin(®) cos(0) (3.20) The plot of the voltage ratio x against the load impedance angle 0 is shown in Figure 3.5. The load angle 6 cannot exceed 7/2. The value of x is 63.67% at 6 = 1.5567 rad, 43.65% at 8 = 0.52308 rad (30°), and 0% at @ = 0. Example 3.3 Finding the Performance Parameters of a Full-Wave Rectifier with an RL Load The single-phase full-wave rectifier of Figure 3.4a has L = 6.5 mH, R = 2.5 9, and E = 10V. ‘The input voltage is V, — 120 V at 60 Hz. (a) Determine (1) the steady-state load current Ig at ot = 0, (2) the average diode current Ipj.x), (3) the rms diode current [pin (4) the rms output current J,(g,), and (5) the input power factor PF. (b) Use PSpice to plot the instantaneous output current ip, Assume diode parameters IS = 2.22E — 15, BV = 1800 V.114° Chapter 3 Solution Diode Rectifiers It is not known whether the load current is continuous or discontinuous. Assume that the load current is continuous and proceed with the solution. If the assumption is not correct, the load current is zero and then moves to the case for a discontinuous current. b R=250, L [R + (LY) 6.5mH, f= 60Hz, w = 2 x 60 = 377 radis, V, = 120V,Z = = 35.0, and @ = tan" (wL/R) = 44.43° 1. The steady-state load current at wt = 0, fy = 32.8 A. Because fy > 0, the load current is continuous and the assumption is correct. 2, The numerical integration of ip in Eq. (3.16) yields the average diode current as Tpgay) = 19.61 A. 3. By numerical integration of i} between the limits wt — 0 and , we get the rms diode current a8 Ip(ms) = 28.5 A. 4, The rms output current Lo(mma = V2, = V2 X 28.50 = 40.3 A. 5. The ac load power is Py, = En R = 40.3? X 2.5 =4.06 kW. The input power factor is Pro _ 4,061 x 10 TE Ves 120X403 = 0.84 (lagging) Notes 1. ig has a minimum value of 25.2 A at wt = 25.5° and a maximum value of 51.46 A at wt = 125.25°. iy becomes 2741 A at of = 0 and 48.2.A at wf = 6 + 1. There- fore, the minimum value of ip occurs approximately at wf = 0. 2, The switching action of diodes makes the equations for currents nonlinear. A numerical method of solution for the diode currents is more efficient than the classical techniques. A Mathcad program is used to solve for J, Ingax), aNd Ip(em) by using numerical integration, Students are encouraged to verify the results of this example and to appreciate the usefulness of numerical solution, especially in solving nonlinear equations of diode circuits. FIGURE 3.6 Single-phase bridge rectifier for PSpice simulation. ‘The single-phase bridge rectifier for PSpice simulation is shown in Figure 3.6. The list of the circuit file is as follows:3.4 Single-Phase Full-Wave Rectifier with RL Load 115 Example 3.3 Singl vs R ve pL D2 D3 Da we MODEL TRAN 1 PROBE, =ND MoD us Phase Bridge Rectifier with RL load SIN (0 169.7V 60H) 6. SM 2.8 DC 10V; Voltage source to measure the output current MoD ; Diode model MOD MoD DMC ope D(IS=2.228-15 BV=1800V) ; Diode model parameters 32MS 16. 667MS vansient analysis Graphics postprocessor The PSpice plot of instantaneous output current jy is shown in Figure 3.7, which gives fy = 31.83 A, compared with the expected value of 32.8 A. A Dbreak diode was used in PSpice simulation to specify the diode parameters. 0A 40a 20 (iT (VxX) 200 100) . ov 4 =100v ++ 4 16ms —18ms —-20ms_-—-22ms 24m 26ms —-2Bms_—30ms_—_32ms eVGA) C1 = 22.747 m, 50.179 Time C2 = 16.667 m, 31'824 dif = 6.0800 m, 18.355, FIGURE 3.7 PSpice plot for Example 3.3,116 Chapter 3 Diode Rectifiers Key Points of Section 3.4 * An inductive load can make the load current continuous. There is a critical value of the load impedance angle 0 for a given value of the load emf constant x to keep the load current continuous. 3.5 SINGLE-PHASE FULL-WAVE RECTIFIER WITH A HIGHLY INDUCTIVE LOAD With a resistive load, the input current of the single-phase rectifier will be a sine wave. With an inductor load, the input current will be distorted as shown in Figure 3.4c. If the load is highly inductive, the load current will remain almost constant with a small amount of ripple content and the input current will be like a square wave. Let us con- sider the waveforms of Figure 3.8, where », is the sinusoidal input voltage, i, is the instantaneous input current, and jy is its fundamental component. If 6 is the angle between the fundamental components of the input current and voltage, «bis called the displacement angle. The displacement factor is defined as DF = cos (3.21) The harmonic factor (HF) of the input current is defined as p- By" ((¢ y I HF =(-—— =|(4) -1 (3.22) (Oa In 6m) where J, is the fundamental component of the input current /,. Both /,; and J, are ex- pressed here in rms. The input power factor (PF) is defined as ~ Vila da PF = 770s # = 7 0086 (3.23) Input voltage Fundamental component FIGURE 3.8 Waveforms for input voltage and current.3.5. Single-Phase Full-Wave Rectifier with a Highly Inductive Load 117 Notes 1. HF is a measure of the distortion of a waveform and is also known as total har- monic distortion (THD) 2, Ifthe input current j, is purely sinusoidal, /,, = 1, and the power factor PF equals the displacement factor DF. The displacement angle becomes the impedance angle @ = tan”'(wL/R) for an RL load, 3. Displacement factor DF is often known as displacement power factor (DPF). 4. An ideal rectilier should have » = 100%, Vj. = 0, RF = 0, TUF = 1, HF = THD = 0, and PF = DPF = 1 Example 3.4 Finding the Input Power Factor of a Full-Wave Rectifier A single-phase bridge rectifier that supplies a very high inductive load such as a de motor is shown in Figure 3.9a. The turns ratio of the transformer is unity. The load is such that the motor draws a ripple-free armature current of /, as shown in Figure 3.9, Determine (a) the HF of input current and (b) the input PF of the rectifier. Von Vin Fundamental =~4— component N 4 \, g re 72" BD, Rds - i —> + ° BD, Bde - 0 ot (@) Circuit diagram (b) Waveforms FIGURE 3.9 Full-wave bridge rectifier with de motor load. Solution Normally, a de motor is highly inductive and acts like a filter in reducing the ripple current of the load. a, The waveforms for the input current and input voltage of the rectifier are shown in Figure 3.9b. The input current can be expressed in a Fourier series as118 Chapter 3. Diode Rectifiers i) = he +S (acos mot + by sin nos) & where 26 ae le = zl i(0 dwt) = sf Led(ot) = 0 1f* 2 a, ==] if) cosnot d(wt) = 1, cos most d(.ot) = 1” 277 4a by == [ (0) sin not d(wt) = = | J,sin not d(wt) = — al, ah, od Substituting the values of a, and b,, the expression for the input current is — He (ae sinSor ) 24 1 The rms value of the fundamental component of input current is iD, avi Ina = 0.901, ‘The rms value of the input current is, wo matlt() +@)°G)+G)+~ Pa From Eq, (3.22), 2 -|(Y = b. The displacement angle 6 = 0 and DF = cos 6 = 1. From Eq, (3.23), the PF = (iail) cos & = 0.90 (lagging). Key Points of Section 3.5, * The input power factor of a rectifier with a resistive load is PF = 1.0 and PF = 0.9 for a highly inductive load. The power factor will depend on the load inductor and the amount of distortion of the input current 3.6 | MULTIPHASE STAR RECTIFIERS We have seen in Eq. (3.11) the average output voltage that could be obtained from single-phase full-wave rectifiers is 0.6366V,, and these rectifiers are used in applications up to a power level of 15 kW. For larger power output, three-phase and multiphase rec- tifiers are used. The Fourier series of the output voltage given by Eq. (3.12) indicates that the output contains harmonics and the frequency of the fundamental component is two times the source frequency (2/). In practice, a filter is normally used to reduce the level of harmonics in the load; the size of the filter decreases with the increase in3.6 Multiphase Star Recti frequency of the harmonics. In addition to the larger power output of multiphase recti- fiers, the fundamental frequency of the harmonics is also increased and is q times the source frequency (qf). This rectifier is also known as a star rectifier. The rectifier circuit of Figure 3.2a can be extended to multiple phases by hav- ing multiphase windings on the transformer secondary as shown in Figure 3.10a. This circuit may be considered as q single-phase rectifiers and can be considered as a half- wave type. The kth diode conducts during the period when the voltage of kth phase is higher than that of other phases. The waveforms for the voltages and currents are shown in Figure 3.10b. The conduction period of each diode is 2n/q. It can be noticed from Figure 3.10b that the current flowing through the second- ary winding is unidirectional and contains a de component. Only one secondary wind- ing carries current at a particular time, and as a result the primary must be connected Di Len = = = Sz 10x a q q q q q q (b) Waveforms: FIGURE 3.10 ‘Multiphase rectiiers.120° Chapter 3 Diode Rectifiers in delta to eliminate the de component in the input side of the transformer. This mini- mizes the harmonic content of the primary line current. Assuming a cosine wave from /q to 2z/q, the average output voltage for a q-phase rectifier is given by 2 fr Yae = sag J, Yon cos orate) = Vag sing (3.25) 2 7, 4 12 vam [pec f"Vien? wet] Inia Jy o) =v,J4(t41 any) 3. | (3 tye 626) If the load is purely resistive, the peak current through a diode is Jy = Vn R and we can find the rms value of a diode current (or transformer secondary current) as 2pm wn L= LanJ, {cos wt dw) | 1 (m1 2n)\]!? _ Vins tl ae(G eae) | =e 62) Example 3.5 Finding the Performance Parameters of a Three-Phase Star Rectifier A three-phase star rectifier has a purely resistive load with R ohms. Determine (a) the efficiency, (b) the FF, (c) the RF, (d) the TUF factor, (e) the PIV of each diode, and (f) the peak current through a diode if the rectifier delivers {y, = 30 A at an output voltage of Vig = 140 V. Solution For a three-phase rectifier q = 3 in Eqs. (3.25) to (3.27) a, From Eq, (3.25), Vic = 0.827Vq and [ye = 0.827Vp IR. From Eq. (3.26), Vins = 0.84068 Vy and Ire ~ 0.84068V,/R. From Eq. (3.1), Pie = (0.827V,)/R: from Eq. (32), Py. = (0.84068V,,)"/R; and from Eq. (3.3), the efficiency (0.827V in) (0.34068V,,.7 b. From Eq, (3.5), the FF = 0.84068/0.827 = 1.0165 or 101.65%. . From Eq, (3.7), the RF = V10165 0.1824 = 18.24% . The rms voltage of the transformer secondary, V;— Vj/'V2 ~ 0.707V Eq, (3.27), the rms current of the transformer secondary, a = 96.77% 0.4854Vj, = 0.4854Iq, = ‘The VA rating of the transformer for q = 3 is 0.4854V yp VA — 3Vihy X 0.707V pq, Xm Eq. (38), oar uF = = 0.6643 TUR = Sy 0707 x 04asa 664 PF 084068" _ 9 6544 ~ 3X 0.707 x 0.4854 e. The peak inverse voltage of each diode is equal to the peak value of the secondary line-to-line voltage. Three-phase circuits are reviewed in Appendix A. The line-to-line voltage is ‘V3 times the phase voltage and thus PIV = V3 V, f. The average current through each diode is, 2" lit Io) = 5 [ In cos ot (wt) = In sin (3.28) For q = 3, Ip(ay) = 0.2757Ihy. The average current through each diode is Ipi4y) = 30/3 = 10 A and this gives the peak current as I, = 10/0.2757 = 36.27 A, Example 3.6 Finding the Fourier Series of a q-Phase Rectifier a, Express the output voltage of a q-phase rectifier in Figure 3.10a in Fourier series. b. If q = 6,V, = 170V, and the supply frequency is f = 60 Hz, determine the rms value of the dominant harmonic and its frequency. Solution a. The waveforms for q-pulses are shown in Figure 3.10b and the frequency of the output is q times the fundamental component (qf). To find the constants of the Fourier series, ‘we integrate from —/q to a/q and the constants are by PV cosutcos mend or n-1 n+l =. od After simplification and then using the following trigonometric relationships, sin(A + B) = sin A cos B + cos A sin B and sin(A ~ B) = sin A cos B ~ cos A sin B we get 24Vm ( ne om nw *) ay = 4 (sin ™ cos = — cos sin™ 3.29) me = 1) qo 4 q4 eo”)122 Chapter 3 Diode Rectifiers For a rectifier with g-pulses per cycle, the harmonics of the output voltage are: qth, 2gth, 3qth, and 4gth, and Eq. (3.29) is valid for n = 0, 1g, 2g, 34. The term sin(nniq) = sin = 0 and Eq, (3.29) becomes (3.30) n= 24S a,cos mat v5 = VasEsin (1 - cos cosmo) (31) Gg PHT Gg b. For q = 6, the output voltage is expressed as = 09549V, (: 2 os bur - 2 cos 1208 + ++ ) 3.32) y(t) = 0.9549, (1 + 35 00s Gut ~ TF eos 12a ) (3.32) ‘The sixth harmonic is the dominant one. The rms value of a sinusoidal voltage is 1/'V2 times its peak magnitude, and the rms of the sixth harmonic is Vig = 0.9549V,, x 2/(35 x V2) = 6.56 V and its frequency is fi = 6f = 360 Hz. Key Points of Section 3.6 + A multiphase rectifier increases the amount of de component and lowers the amount of the harmonic components. The output voltage of a q-phase rectifier contains harmonics whose frequencies are multiples of q (q times the supply fre- quency), gf. 3.7. THREE-PHASE BRIDGE RECTIFIERS A three-phase bridge rectifier is commonly used in high-power applications and it is shown in Figure 3.11. This is a full-wave rectifier. It can operate with or without a trans- former and gives six-pulse ripples on the output voltage. The diodes are numbered in order of conduction sequences and each one conducts for 120°. The conduction se- quence for diodes is D, — Dz, Dy — D3, Dy — Ds, Ds — Dy, Ds — Ds, and D, — Dg. The pair of diodes which are connected between that pair of supply lines having the highest amount of instantaneous line-to-line voltage will conduct. The line-to-line volt- age is V3 times the phase voltage of a three-phase Y-connected source, The wave- forms and conduction times of diodes are shown in Figure 3.12 [4].