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Chapter 7 Uncontrolled Diode Rectifier Circuits • Introduction • Single-phase Rectifier Circuit – Resistive Load – Inductive Load – Capacitive Load – Voltage Source in the dc side • The effect of ac-side inductance – Half-wave rectifier with inductive load – Half-wave rectifier with capacitive load – Full-wave rectifier with inductive load • Three-phase rectifier circuits – Three-phase half-wave rectifier – Three-phase full-wave rectifier • AC-side inductance in three-phase rectifier circuits – Half-wave rectifier – Full-wave bridge rectifiers Introduction • The word rectification is used not because these circuits produce dc, but rather because the current flows in one direction. • These circuits allow power to flow only from the source to the load, unidirectional converters. • This dc voltage maintained across the output capacitor is known as raw dc or uncontrolled dc. • Circuits are said to be uncontrolled since the output voltage and current are a function only of the applied excitation, with no mechanism for varying the output level. • To simplify the analysis it will be assumed throughout this chapter that the diodes are ideal. • The analyses of rectifier circuits will include resistive, inductive, and capacitive loads, as well as loads with dc sources. • Rectification is a term generally used for converting ac to dc with power flowing unidirectionally to the load • Inversion is a term used when dc power is converted to ac power. • The circuits in this chapter are also known as naturally commuting or line- commutating converters since the diodes are naturally turned off and on by being connected directly to a single-or three-phase ac line supply. Block diagram representation for (a) single-phase and (b) three-phase rectifier circuits. SINGLE-PHASE RECTIFIER CIRCUITS Resistive Load Figure 7.2(a) shows the simplest single-phase, half-wave rectifier circuit under a resistive load. The diode conducts only when the source voltage vs is positive as shown in Fig. 7.2(b). The average value of the output voltage, Vo, which represents the dc component of the output signal, vo, 1 T Vo= vo (t) dt T o 1 T/2 = o V s sin t dt (7.1) T V = s and the average load current is given by, Vo Io= R (7.2) Vs = R We observe that the output voltage waveform contains an ac component (ripple) at the same frequency as the ac source Example 7.1 Consider a half-wave rectifier circuit of Fig. 7.2(a) with a resistive load of 25 W and a 60-Hz ac source of 110V rms. a) Calculate the average values of vo and io, b) Calculate the rms values of vo and io, c) Calculate the average power delivered to the load, d) Repeat part (c) by assuming that the source has a resistance of 60W. Solution a) The average values of vo and io are given by, Vs Vo= Vo Io= R 2 (110V) = = 1.98A = 49.52V b) The rms value of the output voltage is expressed as, 1 T 2 V o,rms = v o dt T 0 1 T/2 2 (7.3) 0 V s sin t dt 2 = T Substituting for sin2 t=(1-cos2t)/2, Eq. (7.3) gives Vo,rms=77.78V. The rms current is given by, V o,rms I o,rms = R = 3.11A c) The average power delivered to the load over one cycle is obtained from the following relation: 1 T Po = io vo dt T 0 1 T/2 2 = vo dt TR o 2 V o,rms = R = 242 W d) With a 60 W source resistance, Rs, the average power is given by, 2 RV o,rms P0 = (R + R s )2 = 20.93W Inductive Load An increase in the conduction period of the load current can be achieved by adding an inductor in series with the load resistance The load current flows not only during vs > 0, but also for a portion of vs < 0. (a) (b) Fig. 7.4 (a) Half-wave rectifier circuit with inductive load. (b) Current and voltage waveforms. The equivalent circuit mode for t>t2 and t<t2 are shown in Fig. 7.4(c) and (d), respectively, where t2 is the time when D stops conducting. (a) (b) Fig 7.4 (c) Equivalent circuit for 0 t t2 (d) Equivalent circuit for t2 t T. In steady state, the average value of the inductor voltage is zero. This is indicated by the equal areas of vL in Fig. 7.4(b). The average value of vo is calculated from Fig. 7.4(b) as follows, t 1 2 V o = Vs sin t dt T 0 (7.4) Vs = (1 - cos t 2) 2 The times t1 and t2 and can be determined from the output current equation to be derived next. The first order differential equation that mathematically represents the above circuit is given by, d io R 1 (7.5) + io = V s sin t dt L L Solving Eq. (7.5) for io requires obtaining both the transient, io,tr(t), and the steady-state, io,ss(t) solutions, as given by, io (t)= io,tr (t)+ io,ss (t) The transient current response, also known as the natural response, is obtained by setting the forced excitation, vs, to zero, resulting in -t/ i o,tr (t)= I o,i e (7.6) Where, Io,i initial value of io,tr(t) at t=0. circuit time constant equals L/R. The steady state or forced response is obtained by using the equivalent phasor circuit shown in Fig. 7.5. The phasor current is given by, V s 0 o V I o,ss = = s - (7.7) | Z | | Z | Where, | Z |= ( L )2 + R 2 L = tan 1 R In the time domain, we obtain the steady state output current response as follows, Vs io,ss (t)= sin ( t - ) (7.8) |Z | The total solution for io is obtained by adding Eqs. (7.6) and (7.8) to yield, Vs io (t)= I o,i e -t/ + sin( t - ) (7.9) |Z | The initial value, Io,i, is obtained from evaluating Eq. (7.9) at t = 0+ with io(0+) = 0 to give, V s L (7.10) I o ,i = | Z |2 Fig 7.5 Equivalent phasor circuit for Fig. 7.4(c). Evaluating Eq. (7.9) at t = t2, with io (t2) =0, we obtain the following expression in terms of t2, R 2 t 2 = - ln 1+ ( ) sin( t 2 - ) (7.11) L By normalizing the average output voltage as Vno=Vo/Vs L Rn = (7.12) R In terms of Rn, the average output voltage is expressed by, 1 - cos t 2 V no = (7.13a) 2 Where, 1 (7.13b) t 2 = - R n ln 1 sin(t 2 ) 2 Rn = tan-1 Rn (7.13c) One important design criterion in power converters is what is known as load regulation. If we assume the output voltage takes on a minimum value at low-load, Vol, and a maximum value at a full-load, Vof, then the load regulation (LRload) is defined by, V of V ol LRload = | | 100% (7.14) V of Another important parameter is known as line regulation, LRline, which gives a measure of the output variation when the line input voltage changes. The line regulation is defined as, V o,m ax - V o,m in LR line = 100% (7.15) V o,m in Where Vo,max and Vo,min are the average output voltages at maximum and minimum line voltages, respectively. Example 7.2 Consider the inductive load half-wave rectifier circuit of Fig. 7.4(a) with the vs=120 sin377t, L = 60mH and R=12W. Determine: (a) Vo, (b) Io, (c) LRline if Vs is changed by ± 20%, and (d) LRload if R decreases by 50%. Solution: a) The normalized average output voltage for L/R 1.9 is obtained from equations (7.11) and (7.12b). Vno = 0.213 Then we obtain the average output voltage as, V o = V s V no,ave = 25.55V b) The average output current is, Vo Io= R = 2.13 A c) The minimum and maximum peak input voltages are given as, Vs,min=96V Vs,max=144V The corresponding average outputs are, V0,min=20.44V V0,max=30.67V The line regulation is obtained from Eq. (7.14) as LRline=50%. d) With the load resistance decreasing by 50%, the new value is R=6W, resulting in L/R = 2.5 and Vno = 0.156. This gives a new value of the average output voltage equal to 18.72V. The resultant load regulation is LRload=26.73%. A more useful half-wave rectifier circuit is one in which an additional diode is added across the output as shown in Fig. 7.6(a). A careful investigation of Fig. 7.6(a) shows that D1 and D2 cannot conduct simultaneously. Mode 1: D1 is on and D2 is off: This mode occurs when vs is positive, causing D1 to become forward biased and D2 reverse biased. d io R 1 + io = vs (t) (7.16) dt L L The solution of Eq. (7.15) is similar to Eq. (7.5), which has the following complete solution. Vs i o1 (t)= sin( t - ) + I 1 e-t/ (7.17) |Z| Where, | Z |= ( L )2 + R2 = L/R L = tan-1 R and I1 is constant and to be determined from the initial conditions. The subscript of the load current io1, denotes the current is of Mode 1. Fig 7.6 (a) Half-wave rectifier with a flyback diode. (b) Mode 1: D1 is on and D2 is off. (c) Mode 2: D1 is off and D2 is on. (d) Voltage and current waveforms. Mode 2: D1 is off and D2 is on: Mode 2 occurs for ns < 0 at which D1 becomes reverse biased and D2 turns ON to pick up the load current. The equivalent circuit is shown in Fig. 7.6(c), and the governing equation for this topology is given by, d io R + io = 0 dt L The solution of Eq. (7.17) consists of only transient response as shown in Eq. (7.18). t -T/2 - io2 (t)= I 2 e (7.18) Where I2 is constant and can be solved for, along with I1, from the following relations that must hold in the steady state: io1 (0)= io2 (T) (7.19a) io1 (T/2)= io2 (T/2) (7.19b) The left side currents of Eqs. (7.19a) and (7.19b) are obtained from Eq. (7.16) and the right side currents are obtained from Eq. (7.18). Consequently, we obtain the solution for I 1 and I2 as follows, Vs 1 + e-T/2 I1= sin (7.20) |Z | 1 - e-T/ Vs 1 + e-T/2 (7.21) I2= sin |Z | 1 - e-T/ The load resistance according to the following relation limits the value of the dc output current, Vo Io= R V = s R The above circuit is also known as a Current-doubler since the average output current is twice the average input current. Capacitive Load In some applications in which a constant output voltage is more desired, a series inductor is replaced by a parallel capacitor as shown in the half-wave rectifier circuit of Fig. 7.8(a). Let Mode 1 correspond to the time interval in which the diode conducts, resulting in the equivalent circuit shown in Fig. 7.8(b), whereas, Fig. 7.8(c) shows the equivalent circuit when the diode is OFF, representing Mode 2. (a) (b) (c) Fig 7.8 (a) Half-wave rectifier with capacitive load. (b) Equivalent circuit mode when the diode is on. (c) Equivalent circuit mode when the diode is off. Figure 7.9 shows the steady-state waveforms for the output voltage and the source current. Fig 7.9 Steady-state waveforms for Fig. 7.8 (a). The analytical expressions for no and is are given as follows, For t1 < t < T/4, vo = V s sin t (7.23) Vs i s (t)= sin t + C V s cos t (7.24) R For T/4 < t < T+t1, -(t-T/4)/ vo (t)= V s e (7.25) is(t) = 0 (7.26) where, the time constant is given by = RC. The value of t = t1 can be determined by equating Eqs. (7.23) at t = t1 and (7.25) at t = t1+T. This yields the following relation, 3T sin t 1 - e( t1 + 4 )/ = 0 (7.27) The average value of the output voltage is determined from the following, T +t 1 1 Vo= T vo (t) dt t1 T/ 4 t1 T+ 1 = [ V s Sin t dt + V s e-(t-T/4)/ dt ] T t1 T/ 4 Vs = [cos t 1 + (1 - e-( t 1+3T/4 )/ )] 2 V = s [cost1 (1 e (t1 3T / 4 ) / )] 2 Example 7.3 Consider the half-wave rectifier circuit of Fig. 7.8(a) with vs(t) = 20sin260t , R =10kW, and C=47mF. (a) Sketch the waveforms for vo, iR, iD, and iC, and (b) determine the output voltage ripple. Solution: (a)Using t=RC=470ms and T=16.67ms, through several iteration, Eq. (7.27) give t1»3.48ms. At this time, the output voltage is 19.3V and the resistor current is 1.93mA. At t=T/4=4.17ms, vo=20V and iR = 2mA. When the diode turns OFF, iC = -iR = -2mA. However, while the diode is conducting iC is given by, iC (t)= C V s cos t = 354 cos t mA At t13.48ms, iC=354cos377(3.48ms)=90.6 mA, hence, iD= iR + iC= 1.93 +90.6 = 92.53mA. Finally, the output voltage at T or at t=0, is obtained from Eq. (7.25) as follows, -(t - T/4)/ vo =V s e = 20 e -3T/ 4 = 20 e -3(16.67)/ (4)( 470 ) = 19.5V (b) The output voltage ripple, Vr, is given by, V r = vo,m ax - vo,m in = vo (T/4) - vo ( t 1 ) = 20 - 19.3 = 0.7V Voltage Source in the DC-Side Fig 7.11 (a) Half-wave rectifier with voltage Fig 7.10 Steady-state waveforms for source in the dc side. (b) Voltage and current Example 7.3. waveforms. THE EFFECT OF THE AC-SIDE INDUCTANCE Half-Wave Rectifier with Inductive Load Let us consider the half-wave diode rectifier circuit by including an AC-side inductance, which is inevitable in all ac systems. Figure 7.12(a) shows the circuit for the half-wave rectifier including Ls and under an inductive load. Figure 7.12 (a) Half-wave rectifier with ac-side inductance.(b) Mode 1: D1 is on and D2 is off during vs(t) > 0. (c) Mode 2 (communication mode) : D1 and D2 are on while vs(t) < 0. (d) Mode 3 : D1 is off and D2 is on While vs(t) < 0. (e) The waveforms for vs,v0,iD1,iD2,and vLs. The behavior of the circuit can be easily discussed by assuming that one of the diodes, D1, is conducting for some time during the positive cycle of ns(t), while D2 is off. Since the current in D1 is constant, then the voltage across Ls is zero, and the voltage across D2 is positive. During this mode, we have the following current and voltage values: i D1 = i s (t)= I o , iD2 = 0 , vLs = 0 , vo = v s . At t=T/2, vs(t) starts to become negative, causing D1 to stop conducting. However, since the current in D1 is the same as the inductance current, which is not allowed to change instantaneously, D2 turns on in order to maintain the inductor current’s continuity. This is known as a commutation process during which currents are moved from one branch of the circuit to another branch. This is why the AC-side inductance, Ls, is known as the commutation inductance. During Mode 2, the following equations hold: di s v Ls = v s (t)= L s dt i D 1 = i s (t) iD 2 = I o - iD 1 vo = 0 The initial condition for is(t) at t=T/2 is Io. t 1 i s (t)= Ls T 2 v Ls (t )dt I o (7.28) Substituting for vLs(t)=vs(t)= Vs sint in the above integral, and solving for is(t), we obtain, Vs i s (t)= I o - (1+ cos t) T/2 < t < t 1 + T/2 (7.29) Ls At the end of commutation period, t=t1+T/2, is(t) becomes zero, forcing D1 to turn OFF at zero current and D2 remains forward biased i D1 = i s (t)= 0 , iD2 = I o , vLs = 0 , and vo = 0. To solve for the commutation time, t1, or the commutation angle, u=t1, we evaluate is(t) at t=T/2+t1 and set it to zero. The load current and t1 are given by, Vs Io= (1 cos t 1) (7.30) Ls 1 -1 Ls I o ) t 1 = cos (1 - (7.31) Vs The average output voltage is expressed by, T/2 1 Vo T v (t )dt s t1 (7.32) Vs L s I 0 (1 ) 2Vs If we normalize the output voltage by Vs, and the load current by Zo, where Zo=Ls is the characteristic impedance of the AC-line, we have the following normalized output quantities, Vo V no = Vs 1 I = (1 - no ) 2 (7.33) Zo Io I no = Vs = 1 cosu (7.34) or u is given by, u = cos-1 (1 - I no) (7.35) The maximum voltage across Ls, VLs,max, occurs at t=t1 which is given by, V Ls,m ax= v s (t1) = V s sin t1 Normalizing VLs,max by Vs and substituting the normalized current Ino into the above relation, we obtain, (7.36) V nLs,max= (2 - I no) I no Figure 7.13 shows the curve for VnLs,max vs. Ino, and Fig. 14(a) and (b) show Ino vs. u and Ino vs. Vno, respectively. Figure 7.13 Normalized maximum Figure 7.14 (a) Normalized output current versus inductor voltage versus normalized communication angle. (b) Normalized output output current voltage versus normalized output current. Half-Wave Rectifier with Capacitive Load Consider the half-wave rectifier circuit with a capacitive load including an AC-side inductance as shown in Fig. 7.15(a). Assume RC>>T/2. Its equivalent circuit is given in Fig. 7.15(b). Fig 7.15 Half-wave capacitive-load circuit with commutative inductance. (b) Equivalent circuit with RC >> T/2. (c) Current and voltage waveforms. Let us assume first that the diode is off and while the input voltage is negative. As shown in Fig. 7.15(a). At t=t1, vs=Vo and the diode turns ON. Voltage across Ls is given by, dis v LS = L s dt = v s (t) V o Since the initial inductor current value at t=t1 is zero, the solution for is(t) is given by, Vs V i s (t ) = (cos t1 cos t ) o (t - t1) (7.37) Ls Ls The peak inductor current occurs when vLs=0, i.e. when vs=Vo. Let us assume this occurs at t=t2, I s, peak = i s (t 2) Vs V = (cos t1 cos t 2) o (t 2 - t1) (7.38) Ls Ls where t2 is obtained from, V o = V s sin t 2 (7.39) Also at t = t1, when the diode first turns on, we have, V o = V s sin t 1 (7.40) where, t1+t2=T/2. As long as is(t) is positive, D remains ON, and at t=t3, is reaches zero, causing D to turn OFF. The current at t=t3 is given by, Vs V (7.41) i s (t 3) = (cos t1 cos t 3) o (t 3 - t1) Ls Ls =0 Equations (7.39), (7.40), and (7.41) can be used to solve for t1, t2, and t3 numerically. The average output current is given by, t 1 3 I o = is (t )dt Tt 1 Vs sin t 3 sin t1 V0 cost1 (t 3 t1 ) 2 = + (t 3 t1 ) 2 L s 2Ls T Normalizing Io and Vo, and using the relation (7.39) for t=t2, we obtain, Ls I o I no = Vs 1 2 = cos t 1 sin t 3 + sin t 1 Vno (7.42) 2 2 where, =(t3-t1) which is the diode conducting time. Figure 7.16 shows Ino vs. and Fig. 7.17 shows Vno vs. . Figures 7.16 Normalized load current versus conduction angle. Figure 7.17 Normalized load voltage versus conduction angle. Figure 7.18 Normalized load voltage Figure 7.19 Normalized peak inductor versus normalized load current. current versus conduction angle. The effect of Ls on the variation of the average output voltage is shown in Fig. 7.18. The normalized inductor peak current is given by, Ls I Ls, peak I nLs, peak = Vs = (cost1 cost 2 ) Vno A plot of the normalized peak inductor current vs. under different load conditions is shown in Fig. 7.19. Example 7.4 Consider the half-wave rectifier of Fig. 7.15(b) with vs(t)=25sin377t, Vo=20V and Ls = 2mH. Determine: Vno, Ino, and InLs,peak. Solution: First find the value of t1 1 V t1 sin 1 o 2.46ms t 2 T t1 5.875 ms V 2 S Through some numerical iterations we can find that t3=7.66ms. The normalized average output current can be obtained theoretically as follows, Vs sin t 3 sin t1 V0 Io = (t 3 t1 ) cos t1 + (t 3 t1 ) 2 2 L s 2Ls T 0.99 A Ls I o I no = 0.03 Vs (t 3 t1 ) 1.96 rad / s Vo Vno 0.8V VS I nLs, peak= (cos t1 cos t 2 ) V no 0.17 Full-Wave Rectifiers with an Inductive Load The effect of the commutation inductance on the waveforms and parameters of the full-wave rectifier shown in Fig. 7.20(a) is investigated next. As before, we will assume L/R>>T/2, resulting in a constant load current, Io. The typical waveforms are shown in Fig. 7.20(b). Fig 7.20 (a) Full-wave rectifier with ac-side inductance. (b) Current and voltage waveforms. Assume that D1 and D4 are conducting during the positive cycle of vs(t). Since Io is constant, the voltage across Ls is zero, and vo(t)=vs(t) The voltages across D2 and D3 are negative and equal to -vs(t). As long as vs(t) is positive, D2 and D3 remain off. The second topological mode starts at t=T/2 when vs(t) becomes negative. At this time, D2 and D3 start conducting while D1 and D4 remain on in order to maintain the continuity of the inductor current. Figure 7.21(b) shows this mode when all diodes are conducting. The following current and voltage relations are given in this mode: Figure 7.21 Circuit modes of operation for Fig. 7.20 (a) Mode 1 : equivalent circuit for vs(t) > 0. (b) Mode 2: equivalent circuit for vs(t) < 0 and during diode communication period. (c) Mode 3: equivalent circuit for vs(t) > 0 and during diode communication period. Voltage relations are given in this mode: d is v Ls = L s dt vo = 0 I o + is (7.43) iD 1 = iD 3 = 2 I o - is iD 2 = iD 4 = 2 Using the initial inductor value of is(T/2)=+Io in the above equation for is(t), we obtain: Vs i s (t ) = I o - (1 + cos t ) (7.44) Ls At t=t1+T/2, we have is(t1+T/2) = -Io. This gives the following relation for the commutation period, cos t 1 = 1 - 2 L s I o (7.45) Vs In terms of normalized values, u=t1 is given by, u = cos-1 (1 - 2 I no) Figure 7.22 shows the curves for the normalized output current as a function of the commutation angle. Fig 7.22 Normalized output current vs. communication angle. The larger the commutation inductor, the less the average output current we have. At t=T/2+t1, the currents in D1 and D4 become zero and the currents in D2 and D3 reach Io as shown in the topology of Mode 3 given in Fig. 7.21(c). The output voltage is equal to vs(t). This mode remains until t=T when the voltage becomes positive again. Since vs(t)>0, D1 and D4 turn on and D2 and D3 remain on until the current commutates between D2 and D3 to D1 and D4 at the end of the commutation period u. The current and voltage equations are the same as those in Mode 2. The initial condition for is(t) is is(T)=-Io, is(t) is given by, vs i s (t ) = - I o + (cos t - 1) (7.46) Ls The average output voltage, inductor peak current and rms value of is(t) are directly related to the normalized load current. To illustrate the source of Ls in the full-wave rectifier we refer to Fig. 7.23. This is a center-tap transformer including a leakage inductance Lk reflected to the primary side. The voltage relations are given by, N2 v1 = vs N1 N3 v2 = vs N1 Where Ls1=(N1/N2)2 Lk and Ls2=(N1/N3)2 Lk. Figure 7.23 (a) Full-wave center-tap transformer with leackage inductance. (b) Simplified equivalent circuit. THREE-PHASE RECTIFIER CIRCUITS Three-phase Half-wave Rectifier General configuration for an m-phase half-wave rectifier, connected to a single load. All diode cathodes are connected to the same point, creating a diode-OR arrangement. At any given time, the highest anode voltage will cause its corresponding diode to conduct, with all other diodes in the reverse bias state. In other words, the output voltage will ride on the peak voltages at all times. Figure 7.25 shows four random sine functions and the output voltage. Figure 7.24 m-phase half-wave rectifier Figure 7.25 Example of random four-phase connected to a single load. sinusoidal input voltages and the output voltage. The half-wave three-phase resistive load rectifier circuit is shown in Fig. 7.26(a). We assume that the three-phase voltage source is a D-configuration with the three balance voltages are given by, v1 = V s sin t v2 = V s sin( t - 120 ) v3 = V s sin( t - 240 ) o o This circuit is also known as a three-pulse rectifier circuit. Here, the number of pulses refers to the number of voltage peaks in a given cycle. Figure 7.26 (a) Half-wave three-phase resistive-load rectifier. (b) Waveforms for the output voltage. The circuit works as follows: At t=t1, we have v1=v3. This occurs when t1=T/12, and at t2=5T/12 we have v2=v1. The average output voltage is given by, T 1 V0 = T v 0 0 (t ) dt t2 3 = T v (t )dt t1 1 5T 12 3 T T 12 v (t )dt 1 (7.47) 3 3 = Vs 2 The source voltage arrangement of v1, v2 and v3 can also be of the Delta (D) or the Wye (Y) connection as shown in Fig. 7.27(a) and 7.27(b) Figure 7.27 Three-phase connections : (a) Y and (b) A connections. The voltages va, vb and vc are the balanced, positive sequence phase voltages given by, va = V s sin t vb = V s sin(t - 120 ) vc = V s sin(t - 240 ) o o Whereas, the voltage set v1, v2 and v3 are line voltages. In the D-configuration, the line-to-line voltages are the same as the phase-voltages, whereas, the line-to-line voltages in Y-configuration are given by, v12 = v1 - v 2 = v a - vb = v ab = 3 V s sin( t + 30o ) v 23 = v 2 - v3 = vb - v c = vbc = 3 V s sin( t - 90o ) v31 = v3 - v1 = vc - v a = vca Figure 7.28 Phase or diagram representation = 3 V s sin( t 150 ) o for phase and line-to-line three-phase voltages. The equivalent circuit for a half-wave circuit under a highly inductive load, and its waveforms Figure 7.29 (a) Half-wave three-phase rectifier with infinite load inductance. (b) Current and voltage waveforms. The half-wave circuit with a capacitive load is shown in Fig. 7.30(a) and its output waveform is shown in Fig. 7.30(b). Figure 7.30 (a) Three-phase half-wave capacitive-load rectifier. (b) Output voltage waveform. Assumption that the load time constant, RC, is much larger than T/3. Let t=t1 be the time at which the diode D1 turns ON when na is the most positive. At t=T/4, va=Vs and the diode D1 stops conducting, since the capacitor voltage at this instant is equal to the peak voltage Vs, all diodes remain OFF and the capacitor starts discharging through R. At t=t1+T/3, v0 becomes equal to vb and D2 turns ON. Once again, the capacitor voltage increases with vb until it equals Vs at which the cycle repeats. For T/4<t<T/3+t1, the output voltage is given by, -(t-T/4)/RC v0 = V s e At t=t1+T/3, the output voltage equals vb, i.e. vb ( t 1 + T/3) = V s sin( ( t 1 + T/3) - 120 ) 0 = V s e-( t 1+T/12)/RC The above equation can be rewritten as follows, sin t 1 = e-( t 1+T/12)/RC Like the case of the single-phase rectifier circuit, we can solve for t1 numerically. The ripple voltage, Vr, is obtained from the following relation, V r = V s - vb (t1 + T/ 3) = V s (1 - sin t1) (7.48) Using the approximation t1 ≈ T/4, the above equation becomes, T T - ( + ) /RC vo (t1 + T/ 3) = V s e 4 12 T = V s e- 3 RC T V s (1 - ) 3RC so, the ripple is given by, vr = V s - vo ( t 1 + T/3) T = Vs 3RC (7.49) 1 = Vs 3fRC It can be shown that the diode conduction time can be approximated by, T 2V r - t1 2 Vp (7.50) Three-Phase Full-wave Rectifiers Full bridge rectifier circuit including the commutation inductance. Average Output Voltage : T 4 1 3 3 Vo T Va dt T Vs 12 6 Figure 7.31 Full-bridge Figure 7.32 Output voltage rectifier circuit under waveforms for Fig. 7.31 resistive load. Full-Bridge Rectifier-Highly Inductive Load Bridge rectifier under a highly inductive load using the Y-connected voltage source with L/R>>T/6 is shown in Fig. 7.34 (a). Figure 7.34 (a) Three-phase bridge rectifier with highly inductive load. (b) Phase current waveforms. Modes of Operation – 3-phase bridge Converter Table 7.1 shows all six Modes of operation with the corresponding diode conduction angles and currents, where =t. AC-SIDE INDUCTANCE THREE-PHASE RECTIFIER CIRCUITS Half-Wave Rectifiers Three-phase, half-wave rectifier circuit including an AC-Side commutating inductance under a highly inductive load as shown in Fig 7.35. Figure 7.35 (a) Three-phase, half-wave rectifier with constant load current. Equivalent Circuit Modes and Typical Waveforms Figure 7.35 (b)Equivalent circuit with Vc(t) most positive. (c) Equivalent circuit during current communication from D3 to D1. Figure 7.36 Waveforms for vo, ia , ib ,and ic. Steady-State Analysis • Assuming equal AC-Side inductances, the following relations are obtained dia (7.52a) va = Ls + vo dt dic (7.52b) vc = L s + vo dt • Since ia+ic=Io, then dic /dt = - dia /dt, vc + v a (7.53) vo = 2 d ia 1 = (v a - v c ) dt 2 L s • Substitute for va =Vs sin(t) and vc =Vs sin(t-240o) and use the initial values of ic(T/12)=Io, ia(T/12)=0. The solution for ia is then given by, Vs 3 i a (t ) = [1 - cos( t - 30o)] 2 Ls Commutation Angle •The commutation angle is obtained by evaluating ia(t) at t1 when the current reaches Io. •The commutating waveform from phase "c" to phase "a", in Fig. 7.37. V 3 (7.54) ia(t ) = s [1 - cos( - T/12)]= t1 Io 2 Ls Vs 3 i c (t ) = I o - i a = I o - [1 - cos( t - 30 o)] (7.55) 2 L s 2 u = cos-1 (1 - L s I o ) (7.56) Vs 3 Figure 7.37 Communication waveforms from phase c to phase a in Fig. 7.35(a). Average Output Voltage The output tvoltage is given by the following integral: 3 1 vt a + v c 5T /12 Vo = 3 1 + dt + 5T /12dt = T/12 2v0 dt vadt va T T Vo (7.57) t1 v dt - v - v dt+ v dt T/12 t1 t1 t1 5T /12 3 2 Substituting for Vo from the above equation, a c = T a a T/12 T /12 t1 3 5T /12 t1 va - vc 3 t 1 di a = v adt - o Vdt= V o - Ls dt (7.58) T T/12 T /12 2 Ls =0 T T/12 dt the The first integral is the average output voltage for di a half-wave three-phase converter with 3 /6+u T Ls=0. The second integral is the average voltageLloss d to the commutation. =V o - dt due s Ls =0 /6 3 t 1 di a V o =V o - Ls dt =V o - 3 LsI o (7.59) Ls =0 T T/12 dt Ls =0 T Recall that the average value of the voltage when Ls=0 /6+u di a 3 =V o Ls =0 - L s d 3T 3V s dt /6 Vo = Ls =0 32 =V o - LsI o Ls =0 T Then V0 of Eq. (7.59) may be written as 3 3 Ls I o V0 = V s 1 - 2 3V s Normalized 3 3 I no V no = 1 - 2 3 (7.60) Full-Wave Bridge Rectifiers • Full-wave three-phase rectifier circuit by including an AC-Side inductance is shown in Fig. 7.38. Figure 7.38 Three-phase bridge rectifier with ac-side inductance. Full-Wave Bridge Rectifiers - Analysis • The three source line-to-line voltages are shown by, v ab = 3 V s sin( t + 30 ) vbc = 3 V s sin( t - 90 ) vca = 3 V s sin( t + 150 ) o o o • The operation of this circuit can be discussed in a similar way to that of the case with Ls=0, except in this case the presence of the commutation inductance will present an additional commutation angle, u, during which three diodes are on simultaneously. • For t</6, |vbc| is the maximum voltage with D3 and D5 conducting as shown in Figure 7.39(a). ia=0 ib=-I0 ic=I0 v0=-vbc vLs=0. • From Mode I to Mode II the currents commutate from zero to +Io in ia, D3 remains on for a short time until its current goes to zero at t=t1. dic dia - vb - v o - L s + vc = 0 vb - v o - L s + va = 0 dt dt (7.61) Since ia+ic=i0, then dic /dt = - dia /dt. Hence, the above relation becomes, vab + vcb vab + vbc (7.62) vo = = 2 2 Full-Wave Bridge Rectifiers – Analysis (cont’d) First order differential equations in terms of the line-to-line voltages and currents, dic - vcb - vo - Ls =0 (7.63a) dt dia vab - vo - Ls =0 (7.63b) dt Substituting for v0 in the above equations, vcb - v ab dic di = =- a 2 Ls dt dt v ab - v cb di a = 2 Ls dt Substituting for vcb-vab=vca=Vssin(t+150o), and since at t=0, ic(/6)=Io, we have t 1 2 Ls ic (t ) = 3 V s sin(t +150o)dt + I 0 /6 At t=u+/6, ic(t1+/6)=0, hence, u is given by, 2 u = cos-1 (1 - L s I o ) (7.64) Vs 3 Full-Wave Bridge Rectifiers – Equivalent Circuits Mode Figure 7.39 Equivalent circuit modes for mode 1 to mode V1. (a) Mode1 : D3 and D5 are conducting. (b) Mode I and mode II Communication. (c)Mode II: D1 and D5 are conducting. (d) Mode II and mode III commutation. e) Mode III : D1 and D6 are conducting .(f) Mode III and mode IV commutation . Full-Wave Bridge Rectifiers – Equivalent Circuits Mode (cont’d) (g) Mode IV : D2 and D6 are conducting. (h) Mode IV and mode V commutation. Mode V : D2 and D4 are conducting. (j) Mode V and mode VI communication. (k) Mode VI : D3 and D4 are conducting. (I) Mode VI and mode I communication. Typical Current Waveforms Figure 7.40 Current waveforms for ia, ib, and ic for Fig. 7.38

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