Lecture 22 - Three-phase square-wave inverters

Transcription

1 Lecture - Three-phase square-wave inverters Three-phase voltage-source inverters Three phase bridge inverters can be viewed as extensions of the single-phase bridge circuit, as shown in figure.1. The switching signals for each inverter leg are displaced by 10with respect to the adjacent legs. The output line-line voltages are determined by the potential differences between the output terminals of each leg. Symmetrical three phase voltages across a three-phase load can be produced by switching the devices ON for either 180 or 10 of the output voltage waveform. With 180 conduction, the switching sequence is T1TT3 TT3T4 T3T4T5 T4T5T6 T5T6T1 T6T1T T1TT for the positive A-B-C phase sequence and the other way round for the negative (A-C-B) phase sequence. With 10 conduction, the switching pattern is T1T TT3 T3T4 T4T5 T5T6 T6T1 T1T for the positive A-B-C sequence and the other way round for the negative (A-C-B) phase sequence. Whenever an upper switch in an inverter leg connected with the positive DC rail is turned ON, the output terminal of the leg goes to potential +V d / with respect to the center-tap of the DC supply. Whenever a lower switch in an inverter leg connected with the negative DC rail is turned ON, the output terminal of that leg goes to potential V d / with respect to the center-tap of the DC supply. Note that a center-tap of the DC supply V d has been created by connecting two equal valued capacitors across it. The center-tap is assumed to be at zero or earth Lecture - DC-AC (Inverter Circuits) 1 F. Rahman

2 potential. However, this contraption is artificial and really not essential; the center-tap may not exist in practice. We assume, henceforth, that the three-phase load connected to the output terminals of the inverter is balanced. P i d +V d / T1 D1 T3 D3 T5 D5 V d 0V A i a B i b C i c T4 D4 T6 D6 T D N V d / Phase A A Phase B Phase C R R N R C B Figure.1 Lecture - DC-AC (Inverter Circuits) F. Rahman

3 Inverter waveforms with 180 conduction angle In this case, each switch is turned ON for 180. Switches T1 and T4, which belong to the left-most inverter leg, produces the output voltage for phase A. The switching signals for T1 and T4 are complementary, as are for T3 and T6 or T5 and T. The switching signals for switches T3 and T6, (which are for phase B, belonging to the middle leg), are delayed by 10 from those for T1 and T4 respectively, for the ABC phase sequence. Similarly, for the same phase sequence, the switching signals for switches T5 and T are delayed from the switching signals for T3 and T6 by 10. The phase terminal voltages at A, B and C (sometimes called respective pole voltages) are determined by the states of the switches connected at each pole. Note that with 180 conduction (i.e., complementary switching), each pole voltage can have only two values (or discrete states), namely V d V d or. Considering that there are three poles, the number possible output voltage states from the inverter are 3 = 8. Line-line voltage waveforms The line-line voltages, v AB, v BC and v CA are determined from the switching states at the poles) and the DC source voltage, (V d ). Thus, when switches T1 and T3 are ON, v AB = 0V, when T1 and T6 are ON, v AB = +V d, and so on. The line-line voltages v AB, v BC and v CA (for the +ve or ABC phase sequence) are therefore quasi-square waveforms of 10 of ON and 60 of OFF durations, as Lecture - DC-AC (Inverter Circuits) 3 F. Rahman

4 shown in figure.. Each is phase displaced from its adjacent ones by 10. Line-neutral voltage waveforms Line-neutral voltages are determined from the switching states and the neutral point voltage of the load which can be found by assuming that the load consists of a balanced three-phase resistor bank. For instance, if T1, T3 and T are ON, the potential of the neutral point of the load is Vd 3 1 Vd 3 and therefore V AN and V BN will each be at potentials V d 3 while v CN will be at. Similarly, when T4, T and T3 are ON, the potential of the neutral point will be 1 Vd 3 V 3. As a result, the potential v BN will become d 1 V. and v AN and v CN will each be at d For other types of 3 balanced loads, such as R-L loads, line-neutral voltages from the harmonics of the line-line voltages may be used to find the phase harmonic currents first and then the potential of the neutral point is found from V di vn Ri L dt d n n (.1) where v n and i n represent the harmonic voltage and current respectively in a phase circuit. Lecture - DC-AC (Inverter Circuits) 4 F. Rahman

5 T1 T T3 T4 T5 T6 v AB +V d V d +V d v BC +V d v CA V d v AN 1 3 V d v AN 3 V d i A 1 3 V d 3 V d v BN i B v BN v CN ic v CN Figure. Lecture - DC-AC (Inverter Circuits) 5 F. Rahman

6 v N with respect to ve DC link; V d = 150V v N with respect to center-tap of DC link; V d = 150V Figure.3 Analysis of output voltage Line-line voltage The line-line output voltages are obtained by subtracting two square-wave waveforms which are 10 displaced from each other. Each of these waveforms would consist of harmonics orders 1, 3, 5, 7, 9, and so on. Because of the 10 phase shift between the waveforms, the triplen harmonics (of order which are multiples of 3) of both will of the same phase and hence these cancel in the process of subtraction. Consequently, the triplen order harmonic voltages are eliminated from the line line voltage. The remaining harmonics are at n = 6r ± 1 where r is any positive integer, the n th harmonic having an amplitude 1/n times the fundamental component. Lecture - DC-AC (Inverter Circuits) 6 F. Rahman

7 V d = V d = 10 Figure.4 60 The line-line quasi-square output voltage waveform of figure.4 has amplitude V d and duration = 10. Fourier series representation of this waveform is 4Vd n o n (.) n1,3,5,... vll sin cosn t V cos t cos 5 t cos 7 t cos 11 t = d o o o o (.3) The RMS values of the fundamental and higher order output voltages are, 6Vd 6Vd 6Vd 6Vd V ll,1 ; V ll,5 ; V ll,7 ; V ll,11 ; and so on. (.4) Thus, the fundamental RMS output, Vll,1 0.78V d and 0.78Vd Vll,h h where h = 6n 1 and n = 1,, 3,.. (.5) Lecture - DC-AC (Inverter Circuits) 7 F. Rahman

8 Line-neutral voltage The line-neutral voltage waveform for this inverter is as shown in figure.5. Fourier series representation of this waveform is given by d 3 V 1 d 3 V V 3 d V 3 d Figure.5 4Vd n vln sin cosn ot n (.6) n1,3,5,... n1,3,5,... 4Vd n180 4 n60 3n 3n Vd sin sin cosn ot (.7) = V d cos o t cos5 o t cos7 o t cos11 o t (.8) Lecture - DC-AC (Inverter Circuits) 8 F. Rahman

9 RMS values of the fundamental and higher order terms of the line-neutral voltage are: Vd Vd Vd Vd V ln,1 ; V ln,5 ; V ln,7 ; V ln,11 ; and so on. (.9) Inverter waveforms with 10 conduction angle In this switching scheme, switches T1 T6 are each turned ON for 10, (instead of 180 for the previous scheme), as shown below. Switching signals for each phase leg is displaced from the switching signals for the adjacent legs by 10. As a result, the switching signals for each phase leg have 60 of non overlap. Because of this, switches of a phase leg do not need any dead-time (which is the time each switch waits before the other completely turns OFF). Note that only two switches conduct at any one time, in contrast to three of the previous scheme. Lecture - DC-AC (Inverter Circuits) 9 F. Rahman

10 T1 T T T4 T5 T6 V AB V d / V d V d V d / V BC V CA v AN i A V d / i A v AN v BN i B V d / V d / v CN i C Figure.6 Lecture - DC-AC (Inverter Circuits) 10 F. Rahman

11 Analysis of output voltage waveforms Line-line voltage The line-line voltage waveform for this inverter is as shown in figure.7. Fourier series representation of this waveform is given by V d 1 V d V d V d Figure.7 4Vd n vll sin cosn ot n (.10) n1,3,5,... n1,3,5,... 4 Vd n180 4 Vd n60 sin sin cosn ot n n (.11) V cost cos5t cos7t cos11t = d o o o o (.1) Lecture - DC-AC (Inverter Circuits) 11 F. Rahman

12 The RMS values of the fundamental and higher order output voltages are: 3Vd 3Vd 3Vd 3Vd V ll,1 ; V ll,5 ; V ll,7 ; V ll,11 ; and so on. (.13) Line-neutral voltage The line-neutral voltage waveform of this inverter is as shown in figure.8 V d / = V d / = Figure.8 The line-line quasi-square output voltage waveform of figure.8 has amplitude V d / and duration = 10. Fourier series representation of this waveform is given by 4Vd n o n (.14) n1,3,5,... vln sin cosn t = Lecture - DC-AC (Inverter Circuits) 1 F. Rahman

13 Vd cos ot cos 5 ot cos 7 ot cos 11 ot (.15) The RMS values of the fundamental and higher order output line-neutral voltages are thus, Vd Vd Vd Vd V l n,1 ; V l n,5 ; V l n,7 ; V l n,11 ; (.16) and so on. Lecture - DC-AC (Inverter Circuits) 13 F. Rahman