AC to AC Controllers in Power Electronics, Exercises of Electronics

Download AC to AC Controllers in Power Electronics and more Exercises Electronics in PDF only on Docsity! UNIVERSITY OF ANBAR COLLEGE OF ENGINEERING ELECTRICAL ENGINEERING DEPARTMENT Power Electronic Fourth Class Chapter 05 AC to AC Controller Dr.Hameed F. Ifech 2014-2015 Power Electronics 1 CHAPTER 5 Ac to Ac Controllers 5.0 Introduction An alternating current voltage controller or regulator converts a fixed AC voltage source to a variable AC voltage source. The output frequency is always equal to the input frequency. The simplest way to control the AC voltage to a load is by using an AC switch. This switch can be bidirectional switch like a triac or a pair of SCRs connected in antiparallel as shown in Figure 5.1. Switching devices other than thyristors can also be used to implement bidirectional switches. For most purposes, the control result is independent of the switch used. The practical limitations of the presently available triac ratings often make it necessary to use SCRs in very high power applications for which triac might otherwise be used. Figure 5.1: Basic AC Power Controller (a) an SCR Circuit, (b) a Triac Circuit The major applications of AC voltage controllers include lighting control, industrial heating, spot (resistance) welding, on-load transformer tap changing, static var compensation and speed control for induction motors ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 4 Vi = RMS value of input voltage = dVm / Because Ton can be varied only as an integer, the average value of the load power is not a continuous function but has only discrete levels. The number of steps available for regulating the average power depends on the total number of cycles included in the repeat pattern. Power conversion is the ratio of the average power output (Po(avg)) to the maximum possible power output (Po(max)). Po(avg) / Po(max) is equal to the duty cycle d = Ton/ (Ton+Toff) = Ton/T where T = time period = Ton+Toff The source current is always in time phase with the source voltage. However, this does not mean that an integral cycle control circuit operates at unity power factor- for part of the time, the source current is not present at all and therefore is not in phase with the source voltage. The power factor is given by dTTPF on == / 5.3 It is clear from equation 5.3 that a power factor of one will result when Ton = T, which would result in sinusoidal operation. A closed loop control system can be used to vary the value of Ton to maintain some variable close to a selected set point. Such a system would depend on sufficient energy storage in the controlled system to smooth variations that result from the on-off nature of the control. Integral cycle control has the advantage of fewer switching operations and low radio frequency interference (RFI) due to control during the zero crossing of the AC voltage, that is, in this method, switching occur only at zero voltage for resistive loads. The rate of change of the load current depends on the system frequency, which is small, so there is low electrical noise compared with the other control method Example 5.1 A single phase 120V AC source controls power to a 5Ω resistive load using integral cycle control. Find a) the average value of output current b) the maximum switch current c) the maximum power produced ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 5 d) the duty cycle and the value of Ton to produce 1kW power e) the power factor for part (d) Solution a) The average value of the output current over any number of complete conduction cycles is 0 b) A9.33)24(2I A245/120 m )( == ==RMSoI c) Maximum power will be produced when the switch is always on W288024*120(max) ==oP d) For W100)( =avgoP 35.0 2880 1000 (max) )( ==== o avgoon P P T T d If we choose T = 15 cycles, then Ton = 0.35*15 = 5 cycles e) 58.0 15 5 === T T PF on 5.3 AC PHASE CONTROL 5.3.1 In Circuits with a Resistive Load The basic circuit in Figure 5.1 can be used to control the power to a resistive load. As is done with a controlled rectifier, output voltage is varied by delaying conduction during each half cycle by and angle α. The delay angle α is measured from the source voltage zero. SCR1 which is forward biased during the positive half cycle, is turned on at an angle α. It conducts from α to π. Supplying power to the load. SCR2 is turned on half cycle later at π+α. It conducts up to 2π, supplying power to the load. The waveform in Figure 5.3 are identical to those of the full wave rectifier with a resistive load. The difference here is that each second half cycle has a negative current rather than a positive one. There is however no effect on the power, because power is a squared function. The equation for the RMS value of the output voltage is ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 6 2/1 )( 2 2sin1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ +−= π α π α iRMSo VV 5.4 The equation for the RMS value of the output current with a resistive load is similar to equation 5.4 2/1 )( 2 2sin1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ +−= π α π α R V I i RMSo 5.5 By varying the delay angle α, the output current of the load can be continuously adjusted between the maximum value of Vi/R at α = 0 and zero at α = 1800. The RMS current rating of the triac is given by IT(RMS) = Io(RMS) 5.6 The RMS current rating of the SCRs is given by Iscr(RMS) = I o(RMS)/√2 5.7 Output power is given by Po(avg) = I2 o(RMS) or V2 o(RMS)/R 5.8 Table 5.1: Examination of equation 5.5 and 5.8 shows that the load power can be varied by changing α over the full range from zero to 1800. Suitable trigger circuits exist to allow conduction to be adjusted essentially over the entire range. The control characteristics of a single phase AC power controller can be calculated as a function of the delay angle. If we assume Vi = 50V and load resistance R =100Ω, then at α = 00, using Equation 4 output voltage Vo(RMS) = Vi = 50V and ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 9 Example 5.3 A single phase power controller as shown in Figure 5.1(a) is supplying a resistive load. Plot the waveform of the output voltage if the delay angle is a) 300 b) 1200 Solution Figure 5.5 : AC Phase Control Waveforms for a Resistive Load, for Delay Angles Varying from 300 to 1500 Example 5.4 A single phase power controller as shown in Figure 5.1(a) is supplying a 100Ω resistive load through a 50V source. Plot the waveforms for output voltage, output current, voltages across SCR1 and SCR2 and current through SCR1 and SCR2 if the delay angle is 600 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 10 Solution The waveforms are shown in Figure 5.6 Figure 5.6 : Waveforms for resistive load and a delay angle of 600 Example 5 A 120V source control power to a 5 Ω resistive load using a phase control switch. If the load power required is 1 kW, find a) the maximum load current b) the RMS value of load current c) the delay angle α d) the RMS value of switch current if the switch is triac e) the average current in each of the two SCRs if the switch is like that in Figure5.1(a) f) the peak reverse voltage rating of the switch g) the power factor ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 11 Solution a) A34 5 170I V 170)120(2 m === == R V V m m b) for A14.14 5*I 1000 W1000 )( 2 o(RMS) )( = = = RMSo avgo I P c) using equation 5.5 to solve for α so α = 1050 d) IT(RMS) is the same as the load current , 14.14 A e) The SCR current is a half wave controlled waveform. The average value of each SCR current can be found by using A4)105cos1( 2 34)cos1( 2 0 )( =+=+= π α π m avgSCR I I f) The switch block at least the maximum source voltage 170V g) 58.0 2 105*2sin1051 2/1 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ +−= ππ PF ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 14 For an inductive load, the output current lags the voltage. If the load is purely inductive, the phase angle is 900. Therefore if the delay angle is less than 900, the current will not be symmetrical. With a delay angle of 300 the waveform shown in Figure 5.8(a) results, where the conduction in SCR1 is last for more than 1800 and SCR2 is not conducted at all because it does not experience forward voltage when it receives its firing pulse at π +300. The output current is unidirectional. To avoid this condition the firing angle should be at least 900 (figure 5.8(b)). When π is between 900 and 1800 the waveforms are of the from in Figure 5.8(c) and 5.8(d). Thus for inductive loads α is limited to the range 900 to 1800 Example 7 A single phase voltage controller with an RL load is connected to a 110V source. If R = 10Ω, L = 20mH and α = 800, find a) the RMS output current b) the SCR RMS current c) the power delivered to the load d) the power factor Solution a) A5.580cos80sin 180 6 2 180cos 180 8014 10 110 2/1 2 )(0 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=RMSI b) A9.32/)()( == RMSoRMSSCR II c) W5.302)10(5.5 22 )()( === RIP RMSoavgo d) 5.0 )( )(0 === RMSoi avg IV P S PPF ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 15 5.4 Three Phase AC Phase Control 5.4.1 In Circuits with a Resistive Load Figure 5.9: AC Phase Control, Three Phase Switch (a) Line Controlled Y- Connected , (b) Line Controlled Δ Connection, (c) Branch Δ Connection, (d) Neutral Point Switching using Six SCRs (e) Neutral Point Switching using Three SCRs The phase-control methods applied to single phase loads can also be applied to three phase systems. A three phase AC power controller consists of three single phase bidirectional connections using the phase control principle. The circuits shown in Figure 5.9 can be used to vary the power supplied to a three phase Y or ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 16 Δ- connected resistive load. As shown, the switch in each line is implemented by using two SCRs in an inverse parallel arrangement. The primary considerations in the selection of circuits shown in Figure 5.9 are: i) For a given power, circuits Figure 5.9(a) and (b) give lower voltages (√3/2 times the supply phase voltage or half the line voltage) and higher currents in the SCR. Two SCR pairs are always required in series to block voltage or conduct current. ii) Circuit 5.9(c) gives higher voltages and lower currents in the SCRs. Each SCR can conduct current independently of the other. iii) Circuit 5.9(d) is functionally similar to Figure 5.9(a). It produces identical output voltage waveforms, but since each SCR is part of only one current path instead of two, the average SCR current is halved. In addition, as in circuit 5.9 (c) each SCR can conduct current independently of other. iv) For circuit 5.9(e), control of 3Ф output voltage is also possible by using 3 SCRs instead of six. The waveform corresponds to the six SCR current of Fig. 5.9(d), however the SCR current ratings must be doubled. To illustrate the method of analysis of the three phase AC voltage controllers. We will use circuit shown in Figure 5.9(a) as an example. The SCRs turn on is delayed by the angle α beyond the normal beginning of conduction. For symmetrical operation of the circuit, the gate trigger pulse of the thyristors in the three branches must have the same sequence and phase displacement as the supply voltage. If SCR1 is trigger at α, SCR3 must be turned on at α =1200 and SCR5 at α = 2400. The inverse parallel SCRs are triggered 1800 from their partners. Therefore SCR4 (which is across SCR1) is triggered at α = 1800, SCR6 at α + 3000, and finally SCR2 at α +4200 (or α +600). The SCR conduction order is therefore SCR1, SCR2,SCR3, SCR4, SCR5, SCR6, SCR1,….., with a phase displacement of 600 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 19 Figure 5.11.(b) 300 Figure 5.11.(c) 600 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 20 Figure: 5.11.(d) 900 Figure5.11:(e) 1200 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 109 Figure 5.11 shows the phase and line voltage waveforms for the circuit in Figure 5.9(a) for different delay angles. Figure 5.11(a) shows the maximum output condition which occurs when α = 00. Note that the delay angle α for each SCR is measured from the reference point where current start to flow through a purely resistive load. When the delay angle is small, as in Figure 5.11(b), where α = 300 conduction in each phase stops 1800 after the reference point. All three lines start conducting again as each SCR is turned on. When α become 600 (Figure 5.11(c)), the turning on of one SCR causes another SCR that was previously conducting to turn off, so that only two lines are always conducting. For α> 900, the conduction period is reduced to the point where it becomes necessary to fire pairs of SCRs simultaneously to establish conducting paths. This means that each SCR must received two firing pulses separated by 600 in each cycle as shown in Figure 5.11(d) and (e). If α reaches 1500, the current in each line falls to zero, giving zero output. Thus the operating range for the delay angle is form 00 to 1500. The preceding analysis can be summarized into the following three possible modes of operation for the circuit in figure 11a. Mode I ( )0600 ≤≤ α One device in each line conducts in other words, three devices conduct simultaneously and normal three phase theory applies. Full output occurs when α = 0. When α = 600 and all three devices are in conduction, the load currents are the same as for an uncontrolled three phase resistive load. The RMS value of the output current is given by 2/1 )( 4 2sin 23 1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ +−= π α π α R V I i RMSo 5.12 Mode II ( )00 9060 ≤≤ α One device conducts in each of two AC lines, that is the total of only two SCRs are conducting and two lines act as a single phase supply to the load. During the intervals when one of the line currents is zero, the remaining two phases are effectively in series and form a single phase load connected to two of three lines of the voltage source. The phase voltage is equal to half the line voltage. The conduction pattern during any 600 is repeated during the following 600 interval with the permutation in phases and the sign of the current. For example the current variation for phase A during a given 600 interval is repeated during the next 600 for phase C, except for a ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 112 Solution Figure 5.13 Figure 5.14: Voltage Waveforms for a Δ- Connected Resistive Load ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 113 5.4.2 In Circuits with an Inductive (RL) Load With an RL load, the waveforms in Figure 5.11 are slightly different because the current is no longer continuous at the points where it switches. The voltages and currents cannot be determined easily since each depends not only on the present value but also on the previous conditions. The waveform of the load currents shown in Figure 5.15 are drawn for an inductive load with a delay angle of 1000. As shown the current waveform in one phase is identical to the current waveform of another phase, except for a 600 phase shift and reversal of sign.The voltage rating of the switches should be at least equal to the maximum line voltage of the source. Since the load is inductive, each switch is subjected to a rapid change in voltage as its current becomes zero. A snubber circuit across the switch is usually used to prevent unscheduled firing. The current rating of the switching devices is determined by the current at α = 00 Figure 5.15: Waveforms for an Inductive Load with a Delay Angle of 1000 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 114 Example 9 The three phase power controller shown in Figure 5.9(c) supplies a balanced inductive load. Plot the waveform of the output voltage, the voltage across the SCR and the phase and line currents for the following delay angles a) 900 b) 1200 c) 1500 d) 1650 Solutions Refer to Figure 5.16 (a),(b),(c),(d) Figure 5.16: Waveforms for a Balanced RL load (a) α = 900, (b) α =1200 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 117 2/1 )( 224 11 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ −= π α R V I i RMSo 5.18 Mode III ( )00 210120 ≤≤ α Only one SCR and one diode conduct and at 2100 the power delivered to the load is zero. The equation for the RMS value of the output current is 2/1 )( 16 2sin 16 2cos3 424 7 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ +−−= π α π α π α R V I i RMSo 5.19 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Ac-Ac Controllers 118 Figure 5.18 (a): Three Phase Half Wave AC Voltage Controller Phase Voltage and Line Voltage for Delay Angles of 300 Figure 5.18 (b) : 900 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech Power Electronics 119 Figure 5.18 (c) : 1500 ------------------------------------------------- Ac-Ac Controllers --------------------------------------------------------------------------------------------- Dr.Hameed F. Ifech