Author Dr.Ali Hussein Numan. Electromechanical Engineering Department 2 hrs / one week EME 401. Lecturers: Dr.Ali Hussein Numan & Dr.Shatha K.

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1 Introduction Author Dr.Ali Hussein Numan University of Technology Power Electronics and Electromechanical Engineering Department Electrical Drives 2 hrs / one week EME 41 Fall Energy- Systems branch Lecturers: Dr.Ali Hussein Numan & Dr.Shatha K. Baqir Contents 1) Definition of power electronics. 2) Power Semiconductor devices. 3) Power electronics application s. 4) Classification of power electronics. 5) Characteristics of different power semiconductor devices. Objectives 1) Identify the types of power semiconductor devices. 2) Identify the major applications of power electronics. 3) Classify the power electronics converters. 4) Understand the characteristics of power semiconductor devices References 1) N.Mohan, et al, Power Electronics, Converters, Applications, and Design, 3 rd Edition, John Wiley and Sons,23. 2) P.C.Sen, Principles of Electric Machines and Power Electronics, 3 rd Edition, John Wiley and Sons, ) B.K.Bose, Modern Power Electronics and AC Drives, Prentice Hall Inc, 22. 4) C.W.Lander, Power Electronics, 2 nd Edition, McGraw Hill, ) M.H. Rashid, Power Electronics Handbook Devices Circuits and Applications, 3 rd Edition, Elsevier Inc.,

2 Introduction Author Dr.Ali Hussein Numan 1-1 Power Electronic: Solid state electronic used to control the flow of electrical power by supplying voltages, currents, and frequencies in a form that is optimally suited for user load. Fig.1-1 shows a simple power electronic system. 1-2 Power semiconductor devices: An Electronic switches (power processor) controlled by control signals that is generated by controller (integrated circuits and/ or digital signal processor).these power semiconductor devices can be broadly classified into three groups according to their degree of controllability: 1. Uncontrolled turn on and off (Power Diode). 2. Controlled turn on uncontrolled turn off (Thyristors). 3. Controlled turn on and off (GTO, BJT, MOSFET, IGBT). 1-3 POWER ELECTRONICS APPLICATIONS: Power electronics has a wide range of applications including:- 1. Power supplies (TV, Radio, Receiver, PC and its Peripherals, UPS, mobile phone battery charger). 2. Air Conditioning and Refrigeration. 3. Elevators. 4. Electric drives. 5. Light control (dimmer). 6. High Voltage Direct Current (HVDC) systems. 7. Flexible AC Transmission (FACT) system. 8. Solar power. 9. Micro grid. 1. Wind generation. 1-4 CLASSIFICATION OF POWER ELECTRONIC CONVERTERS Power electronics converters can be divided into the following categories: 1. AC-DC converter (rectifier): Convert input AC voltage into a DC output voltage. 2. DC-DC converter (DC Chopper): Convert fixed input DC voltage into variable DC output voltage or vice versa. 3. AC-AC converter (AC voltage regulator also called cycloconverter): Convert fixed input AC voltage and frequency into variable AC output voltage and variable (lower) frequency. 4. DC-AC converter (Inverter): Convert input DC voltage into a variable AC output voltage 2

3 Introduction Author Dr.Ali Hussein Numan Finally, Figure 1.1 presents a categorization of power electronic converters into families according to their type of electrical conversion. Rectifier AC Voltage Regulator Chopper Inverter Fig.1-2 Families of solid state power converters categorized according to their conversion function CHARACTERISTICS OF POWER ELECTRONIC DEVICES Power Diode Power diode has two terminals anode (A) (positive) and cathode (K) (negative).when anode is positive with respect to the cathode, the diode is said to be forward biased and it begins to conduct with only small forward across it, which is on the order of 1v. Fig.1.3 (a) show the circuit symbol for the diode and (b) & (c) its steady state i-v and idealized characteristics respectively. On OFF Fig.1-3 Diode (a) symbol, (b) i-v characteristic, (c) idealized characteristic. Power diode is used in power electronics circuits to perform one the following functions: 1) Switches in rectifiers. 2) Freewheeling in switching regulators. 3) Charge reversal of capacitor. 4) Voltage isolation. Power diodes are similar to ordinary PN junction signal diode with slight difference. Power diode is capable to handle high power, voltage, and current. Frequency response or switching speed is low compared with that of signal diodes. 3

4 Introduction Author Dr.Ali Hussein Numan POWER DIODES TYPES Power diodes can be classified as General purpose diodes: Used in rectifier circuits with voltage rating upto 5KV and current 3.5KA. Fast and ultra fast recovery diodes: Used in high frequency circuits with voltage rating up to 3KV and current 1KA Schottky diode: Used in low voltage, high current application such as switched mode power supplies with voltage rating up to.1kv and current.3ka. (a) (b) (c) Fig.1-4 Photos for different types of diode (a) General purpose, (b) fast and ultra fast recovery, (c) schottky Thyristor or silicon controlled rectifier (SCRs) Thyristors has three terminals anode, cathode, and gate. They are available with voltage rating up to (5KV) and current rating up to (3KA). The circuit symbol for the thyristor, and its i-v characteristics along with the idealized characteristics are shown in Figs.1.4a through Figs.1.4c respectively. Thyristor can be turned on by applying a pulse of positive gate current for a short duration, and then remain conducting as a diode until the gate current falls to zero. For successful to turn off thyristor, the anode current should be goes to negative by using an external circuit. Fig.1.4 Thyristor : (a) symbol,(b) i-v characteristics,(c) idealized. 4

5 Introduction Author Dr.Ali Hussein Numan Gate Turn Off thyristor (GTO) The circuit symbol for the GTO is shown in Fig. 1.5a and its steady state i-v as well as idealized characteristics is shown in Fig. 1.5b and Fig. 1.5c respectively. The GTO belong to a thyristor family. It can be turned on by a short pulse of gate current and turned off by a reverse gate pulse. This reverse gate current must be (2 %) of the anode current which is considered very large. The GTOs are available with voltage rating (4.5 kv), current rating (3kA), and switching frequency (1 KHz).Fig 1.6 show photo of GTOs. Fig.1.5 A GTO: (a) symbol, (b) i-v characteristics, (c) idealized. Fig 1.6 photos for GTOs CLASSIFICATION OF THYRISTORS 1) Phase control thyrisors: Suitable for use in AC and DC motor drives as well as in HVDC power transmission. Available with voltage rating (5-7 KV) and current (4KA) 2) Inverter grade thyrisors: Used in inverter and chopper with voltage rating (2.5kV) and current rating (1.5KA). It can be turned-on using force-commutation method. 3) Light activated thyrisors: Similar to phase controlled, but triggered by pulse of light guide by optical fibers. It can be used in very high power applications. 4) TRIAC: Dual polarity thyristors (4KV) (3KA). 5

6 Introduction Author Dr.Ali Hussein Numan Metal Oxide Semiconductor Field Effect Transistors (MOSFETs) The circuit symbol of an n-channel MOSFET is shown in Fig.1.7 and its steady state i-v as well as idealized characteristics is shown in Fig. 1.7b and Fig. 1.7c respectively. Fig.1.8 show photos for different types of MOSFET. The MOSFET is on when the gate source voltage is below the threshold value, V GS (th).the main features of these devices are fast switching typically (1-1 KHz) and available in voltage rating up to 1KV and current rating up to.1 KA. Fig.1.7 N-channel MOSFET: (a) symbol, (b) i-v characteristics, (c) idealized. Fig.1.8 Photos for different types of MOSFET Insulated Gate Bipolar Transistors (IGBTs) The circuit symbol for an IGBT is shown in Fig.1.9a and its steady state i-v as well as idealized characteristics is shown in Fig. 1.9b and Fig. 1.9c respectively. Fig.1.9 An IGBT: (a) symbol, (b) i-v characteristics, (c) idealized. 6

7 Introduction Author Dr.Ali Hussein Numan IGBTs are voltage controlled four-layer devices that combine the characteristics of BJT and MOSFET. They are currently available with voltage rating 4.5 KV, current rating 1.2 KA and switching frequency up to 1 KHz. Fig.1.1 show photo for IGBTs. Fig.1.1 Photos for IGBTs MOS CONTROLLED THYRISTORS (MCTs) The MCT is voltage controlled device like IGBT and the MOSFET. The current density of MCT is high compared to a power MOSFET, and IGBT. The circuit symbol for an MCT is shown in Fig.1.11a and its steady state i-v as well as idealized characteristics is shown in Fig. 1.11b and Fig. 1.11c respectively. Fig.1.11 An MCT : (a) symbol, (b) i-v characteristics, (c) idealized. 7

8 Introduction Author Dr.Ali Hussein Numan TRIAC TRIAC is similar to two back to back (anti parallel) connected thyristors but with only three terminals. Unlike thyristor the TRIAC can conducts in both directions from anode to cathode and vice versa. Ratings from (2-5 A) and (2-8 V). Used in lamp dimmers, home appliances, and hand tools. Not as rugged as many other device types, but very convenient for many ac applications.. The circuit symbol for the TRIAC is shown in Fig.1.12a and its steady state i-v is shown in Fig. 1.12b. Fig.1.13 show photo for TRIACs. ( a) ( b) Fig.1.12 TRIAC: (a) symbol, (b) i-v characteristics. Fig.1.13 Photos for TRIACs. Table 1.1 Summery of the Power Semiconductor Devices. Current Uncontrollable On Controllable On and Off Controllable Uni-Direction Diode Thyristor GTO MOSFET IGBT MCT Bi-Direction TRIAC 8

9 University of Technology Power Electronics and Electromechanical Engineering Department Electrical Drives 1 hrs / Five weeks EME 41 Fall Energy- Systems branch Lecturers: Dr.Ali Hussein Numan & Dr.Shatha K. Baqir AC/DC Converter (Rectifier) Single and Three Phase AC/DC Converters Fig. (1) shows the inputs and outputs of AC to DC converters. The input is single phase or three phase AC supply normally available from the mains. The output is the controlled DC voltage and current. Fig. (1) AC to DC converters The AC to DC converters include diode rectifiers as well as controlled rectifiers. The controlled rectifiers mainly use SCRs. Since the input is AC supply, the SCRs are turned off natural commutation. Hence external commutation circuits are not required. Hence AC to DC converters are also called as line (supply) commutated converters. These converters are used for DC drives, uninterruptible power supplies ( UPS), high voltage DC transmission (HVDC) systems. 1- Single phase diode rectifiers Rectification is the process of conversion of alternating input voltage to direct output voltage. As stated before, a rectifier converts ac power to dc. In diode based rectifiers, the output voltage cannot be controlled. In this section, uncontrolled single phase rectifiers are studied. The diode is assumed ideal has no forward voltage drop (V d =). 9

10 1-1 Single phase half wave uncontrolled rectifier This is the simplest type of uncontrolled rectifier. It is never used in industrial applications because of its poor performance. Its study is, however, useful in understanding the principle of rectifier operation. In a single phase half wave rectifier, for one cycle of supply voltage, there is one half cycle of output, or load, voltage. The load on the output side of rectifier may be R, RL or RL with a freewheeling diode.these are now discussed briefly. a- R load The circuit diagram of a single half wave rectifier is shown in fig. (2-a). The waveforms of v s, v o, i o and v d are sketched in fig.(2-b). For a resistive load, output current i o has the same waveform as that of the output voltage v o. Diode voltage is zero when diode conducts. Fig. (2) Single phase half wave diode rectifier with R load (a) circuit diagram and (b) waveforms v ss = v D v When diode is forward biased,it is therefore turned on v DD = (ideal) v ss = v oo VV mmmmmmmm = VV oo = 1 2ππ VV mm sin θθθθθθ = VVVV 2ππ cos θθ ππ = VVVV ππ ππ.(1) 1

11 II mmmmmmmm = II oo = VV oo RR = VV mm ππππ..(2) PP oo mmmmmmmm = VV oo II oo..(3) VV oo rrrrrr = 1 (VV 2ππ mm sin θθ) 2 ππ dd θθ 1 2 = VV mm 2 2ππ ππ 2θθ 1 cos 2 dddd 1 2 = VV mm 2 4ππ θθ sin 2θθ 2 ππ 1 2 = VV mm 2 (4) II rrrrrr = II oo = VV oo rr RR PP oo rrrrrr = VV oo II oo = VV mm 2RR (5) (6) The efficiency of a rectifier ηη = PP mmmmmmmm PP rrrrrr (7) The form factor (FF), which is a measure of the shape of output voltage FF = VV rrrrrr VV mmmmmmmm (8) The ripple factor (RF), which is a measure of the ripple content RF = VV rrrrrr VV mmmmmmmm 2 = VV rrrrrr 2 VV 2 mmmmmmmm VV mmmmmmmm 2 = VV rrrrrr = FFFF 2 1 VV mmmmmmmm...(9) Ex: The rectifier in fig.(1) has a purely resistive load of R. Determine a) the efficiency, b) the form factor, c) the ripple factor. Solution: VV mmmmmmmm = VV mm ππ =.318 V mm II mmmmmmmm = VV mmmmmmmm RR =.318VV mm RR 11

12 P mmmmmmmm = (.318 VV mm ) 2 RR VV rrrrrr = VV mm 2 =.5 VV mm II rrrrrr = VV rrrrrr RR =.5 VV mm RR PP rrrrrr = (.5 VV mm ) 2 RR a) ηη = (.318 VV mm ) 2 = 4.5 % ( Low ) (.5 VV mm ) 2.5 VV b) FF = mm = 1.57 = 157 % ( high ).318 VV mm 2 c) RF = (1.57) 2 1 = 1.21 = 121% ( high ) b- RL load A single phase half wave rectifier feeding RL load is shown in fig. ( 3-a ). Current continues to flow-even after source voltage v ss has become negative; this is because of the presence of inductance L in the load circuit. Voltage v RR = i oo RR has the same wave shape as i oo. Inductor voltage v LL = v ss v RR is also shown. The current i oo flows till the two areas A and B are equal. Area A represents the energy stored by L and area B the energy released by L. It must be noted that mean value of voltage v LL across inductor L is zero. 12

13 Fig.(3) Single phase half wave diode rectifier with RL load (a) Circuit diagram and (b) waveforms When i oo = at = ββ, v LL =, v RR = and voltage v ss appears as reverse bias across diode D as shown. At ββ, voltage v DD across diode jumps from zero to VV mm sin ββ where > ππ. Here ββ = γγ is also the conduction angle of the diode. ββ VV mmmmmmmm = 1 VV 2ππ mm sin θθ dddd = VV mm (1 cos ββ) ( 1) 2ππ II mmmmmmmm = VV mmmmmmmm RR = VV mm (1 cos ββ) (11) 2ππππ A general expression for output current i oo for < θθ < ββ can be obtained as under: When diode is conducting, the circuit of fig.(3-a) gives RRi oo LL ddi oo dddd = VV mm sin θθ The load current i oo consists of two components, one steady state component i ss and the other transient component i tt. Here i ss is given by i ss = VV mm 2 sin(θθ ) RR 2 XX 2 Where = tan 1 XX RR and XX = ωωωω.here is the angle by which rms current I s lage V s. The transient component i tt can be obtained from equation RRi tt LL ddi tt dddd = Its solution gives i tt = AAee RR LL tt Total solution for current i oo is, therefore, gives by i oo = i ss i tt = VV mm ZZ sin(θθ ) AA ee RR LL tt..(12) 2 Where ZZ = RR 2 XX 2 Constant A can be obtained from the boundary condition at ωωωω =. 13

14 At ωωωω =, or at tt =, i oo =. Thus, from eq. (12) = VV mm ZZ sin AA AA = VV mm ZZ sin Substitution of A in eq. (12) gives i oo = VV mm ZZ sin(θθ ) sin ee RR LL tt.(13) c- RL load with freewheeling diode Performance of single phase half wave diode rectifier with RL load can be improved by connecting a freewheeling diode across the load as shown in fig.(4-a). Output voltage is v oo = v ss for θθ ππ. At ωωωω = ππ, source voltage v ss is zero, but output current i oo is not zero because of L in the load circuit. Just after ωωωω = ππ, as v ss tends to reverse, negative polarity of v ss reaches cathode of FD through conducting diode D,whereas positive polarity of v ss reaches anode of FD direct. Freewheeling diode FD, therefore, gets forward biased. As a result, load current i oo is immediately transferred from D to FD as v ss tends to reverse. After = ππ, diode current i ss = and it is subjected to reverse voltage with PIV equal to VV mm. Fig.(4) Single phase half wave diode rectifier with RL load and freewheeling diode (a) circuit diagram and (b) waveforms 14

15 After ωt = ππ, current freewheels through circuit RL and FD. The energy stored in L is now dissipated in R. When energy stored in L = energy dissipated in R, current fails to zero at ωωωω = ββ < 2ππ. The effects of using freewheeling diode are as under: 1- It prevents the output voltage from becoming negative. 2- As the energy stored in L is transferred to load R through FD, the system efficiency is improved. 3- The load current waveform is more smooth, the load performance is therefore improved. The waveforms for the v ss, v oo, i oo, v DD, i ss and i fd are drawn in fig.(4-b). ππ VV mmmmmmmm = 1 VV 2ππ mm sin θθ dddd = VV mm ππ.( 14 ) II mmmmmmmm = VV mm ππππ.( 15 ) d- RE load Single phase half wave rectifier with load resistance R and load E is shown in fig.(5-a). Fig.(5) Single phase half wave diode rectifier with RE load 15

16 The diode would not conduct at θθ = because diode is reverse biased until source voltage v ss equals E. When VV mm sin θθ 1 = EE, diode D starts conducting and turn-on angle θθ 1 is given by v ss = EE VV mm sin θθ 1 = EE 1 EE θθ 1 = sin VV mm (16) The diode now conducts from θθ = θθ 1 to θθ = (ππ θθ 1 ), i.e. conduction angle for diode is (ππ 2θθ 1 ) as shown in fig.(5-b). During the conduction period of diode, the voltage equation for the circuit is VV mm sin θθ = EE ii oo i oo = VV mm sin θθ EE RR.(17) II mmmmmmmm = 1 ππ θθ 1 2ππππ (VV mm sin θθ EE) dddd θθ 1 = 1 2ππππ [2VV mm cos θθ 1 EE(ππ 2θθ 1 )] (18) II rrrrrr = 1 2ππ = 1 ππ θθ 1 θθ 1 2 VV mm sin θθ EE RR dddd 1 2 ππ θθ 1 (VV 2 2ππRR 2 mm sin 2 θ EE 2 2VV mm EE sin θθ)dddd 1 2 θθ 1 = 1 {(VV 2ππRR 2 ss 2 EE 2 )(ππ 2θθ 1 ) VV 2 ss sin 2θθ 1 4VV mm EE cos θθ 1 } 1 2..(19) 2 PP oo = EEII oo I rrrrrr RR (watts) (2) From fig.(5-b) of v DD : At θθ = οο then v DD = EE, at θθ = θθ 1 then v DD = During period diode conducts, v DD =. When θθ = 3ππ 2, vv ss = VV mm and v DD = (VV mm EE). Thus PIV for diode is (VV mm EE) 16

17 1-2 Single phase bridge uncontrolled rectifier a- R load A single phase bridge rectifier employing diodes is shown in fig.(6-a).when point (a) is positive with respect to point (b), diodes D 1, D 2 conduct together so that output voltage is v aaaa. Each of the diodes D 3 and D 4 is subjected to a reverse voltage of v ss as shown in fig.(6-b), When point (b) is positive with respect to point (a), diodes D 3, D 4 conduct together and output voltage is v bbbb. Each of the two diodes D 1 and D 2 experience a reverse voltage of v ss as shown. Fig.(6) Single phase bridge rectifier (a) circuit diagram and (b) waveforms VV mmmmmmmm = VV oo = 2 VV 2ππ mm ππ sin θθθθθθ = 2VVVV ππ. (21) II mmmmmmmm = II oo = VV oo = 2VV mm RR ππππ..(22) VV oo rrrrrr = 2 (VV 2ππ mm sin θθ) 2 ππ dd θθ 1 2 = VV mm 2.(23) II rrrrrr = II oo = VV oo rr RR = VV mm 2RR (24) 17

18 Ex: The rectifier in fig.(6) has a purely resistive load of R. Determine a) the efficiency, b) the form factor, c) the ripple factor. Solution: VV mmmmmmmm = 2222 mm ππ =.6366 V mm II mmmmmmmm = VV mmmmmmmm RR =.6366VV mm RR P mmmmmmmm = (.6366 VV mm ) 2 RR VV rrrrrr = VV mm 2 =.77 VV mm II rrrrrr = VV rrrrrr RR =.77 VV mm RR PP rrrrrr = (.77 VV mm ) 2 RR a) ηη = (.6366 VV mm ) 2 = 81 % (.77 VV mm ) 2 b) FF =.77 VV mm = 1.11 = 111 %.6366 VV mm 2 c) RF = (1.11) 2 1 =.482 = 48.2% The performance of a bridge rectifier is improved compared to that a half wave rectifier 18

19 b- RL load Fig.(7) Single phase bridge diode rectifier with RL load (a) Circuit diagram and (b) waveforms When L is very high the change of current and the ripple can be assumed to be zero. The load current can be assumed to be pure dc or the load current will be approximately pure dc. II DD1(rrrrrr ) & II DD2(rrrrrr ) = 1 II 2 2ππ mmmmmmmm dddd ππ 1 2 = II mmmmmmmm 2.(25) II ss rrrrrr = 2 II 2 2ππ mmmmmmmm dddd ππ = II mmmmmmmm (26)

20 2- Three phase half bridge uncontrolled rectifier Fig. (8) Three phase half bridge rectifier (a) circuit diagram (b) waveform In this case only one D will be at any instant this diode is the one connected to the branch having the highest voltage (e.g D on if V, it the highest voltage).when D 1 is on then when V 2 become higher than V 1, D 2 will be on and commutates D 1. For usual case the load is inductive with L very high therefore it can be assumed to be continues level. VV mmmmmmmm = 1 2ππ 3 II DD1 rrrrrr = 1 5ππ 6 ππ 6 VV mm sin θθ dddd = 3 3 2ππ VV mm pph 5ππ 6 II 2 2ππ mmmmmmmm dddd 1 2 = II mmmmmmmm ππ 6 3.(27).(28) 2

21 3- Three phase bridge uncontrolled rectifier Fig. (9) Three phase bridge rectifier (a) circuit diagram (b)waveform A three phase bridge rectifier is commonly used in high-power applications and it is shown in fig. This is a full wave rectifier.it can operate with or without a transformer and gives six-pulse ripples on the output voltage. The diodes are numbered in order of conduction sequences and each one conducts for (12 ). The conduction sequence for diodes is 12, 23, 34, 45, 56 and 61. 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. When v aa is the most positive phase diode D 1 conducts and during this period first v bb is the most negative 21

22 with diode D 6 conducting until v cc becomes more negative when the current in diode D 6 commutates to diode D 2. The load voltage follows in turn six sinusoidal voltages during one cycle, these being (v aa v bb ), (v aa v cc ), (v bb v cc ), (v bb v aa ), (v cc v aa ), (v cc v bb ),all having the maximum value of the line voltage that is 3 times the phase voltage. The line voltage can be found by V XN, V YN. VV oo = VV XXXX VV YYYY = VV XXXX ( VV YYYY ) = VV XXXX VV YYYY VV mmmmmmmm = 1 2ππ 6 2ππ 3 VV ππ 3 mm llllllll sin θθ dd θθ = 3VV mm llllllll ππ..(29) II aa rrrrrr = 2 2ππ 5ππ 6 ππ 6 2 dddd II mmmmmmmm 1 2 oooo 2 2ππ 2ππ 3 2 dddd II mmmmmmmm 1 2 VV rrrrrr = 1 2ππ 6 = 2 3 II mmmmmmmm..(3) 2ππ 3 ππ 3 (VV mm llllllll ) (sin θθ) 2 dddd = VV mm.(31) Ex: A three phase bridge rectifier has a purely resistive load of R. Determine a) the efficiency, b) the form factor, c) the ripple factor. Solution: VV mmmmmmmm = 3333 mm llllllll ππ = 3 3V m π = V mm II mmmmmmmm = VV mmmmmmmm RR = 1.654VV mm RR P mmmmmmmm = (1.654 VV mm ) 2 RR VV rrrrrr = VV mm II rrrrrr = VV rrrrrr RR = VV mm RR 22

23 PP rrrrrr = ( VV mm ) 2 RR a) ηη = (1.654 VV mm ) 2 = % ( VV mm ) 2 b) FF = VV mm = 1.8 = 1.8 % VV mm 2 c) RF = (1.8) 2 1 =.4 = 4% Single phase half wave rectifier Single phase bridge rectifier Three phase bridge rectifier η 4.5% 81% 99.83% FF 157% 111% 1.8% RF 121% 48.2% 4% 23

24 3- Single phase thyristor rectifiers (controlled rectifier) Controlled rectifiers are basically AC to DC converters. The power transferred to the load is controlled by controlling triggering angle of the devices. Fig.(1) shows this operation. Fig. (1) Principle of operation of a controlled rectifier The triggering angle (α) of the devices is controlled by control circuit. The input to the controlled rectifier is normally AC mains. The output of the controlled rectifier is adjustable DC voltage. Hence the power transferred across the load is regulated. Use thyristors (or one of its family traic, power transistor) as the main components. 3-1 Single phase half wave controlled rectifier a- R load Fig.(11) Single phase half wave thyristor circuit with R load 24

25 An SCR can conduct only when anode voltage is positive and a gating signal is applied. The SCR will be off until a pulse is supplied to the gates which turn the SCR on. For resistive load, current i o is in phase with v o. Firing angle is therefore measured between the supply zero and the instant of pulse. ππ VV mmmmmmmm = 1 V 2π m sin θθ dddd α = VV mm (1 cos α).(32) 2π The maximum value of V mean occurs at α = VV mmmmmmmm = VV mm 2 = VV mm 2ππ ππ.(33) II mmmmmmmm = VV mmmmmmmm RR = VV mm (1 cos αα) (34) 2πR ππ VV rrrrrr = 1 (VV 2ππ mm sin θθ) 2 dddd αα 1 2 = VV mm (ππ αα) 1 sin 2 2αα 1 (35) 2 ππ 2 II rrrrrr = VV rrrrrr RR.(36) b- RL load A single phase half wave thyristor circuit with RL load is shown in fig.(12-a). At ωωωω = αα, thyristor is turned on by gating signal. The load voltage v oo at once becomes equal to source voltage v ss as shown. But the inductance L forces the load, or output, current i oo to rise gradually. After some time, i oo reaches maximum value and then begins to decrease. At ωωωω = ππ, v oo is zero but i oo is not zero because of the load inductance L. After ωωωω = ππ, SCR is subjected to reverse anode voltage but it will not be turned off as load current i oo is not less than the holding current. At some angle ββ > ππ, i oo reduces to zero and SCR is turned off as it is already reverse biased. After ωωωω = ββ, v oo = and i =. At ωωωω = 2ππ αα, SCR is triggered again, v oo is applied to the load and load current develops as before. The voltage equation for the circuit is VV mm sin θθ = RRi oo LL ddi oo dddd.(37) The load current i oo consists of two components: 25

26 Fig. (12) Single phase half wave circuit with RL load 1- The steady sate component i ss VV mm i ss = sin(ωωωω ) RR 2 (ωωωω) 2 1 ωωωω Where = tan. Here is the angle by which rms current II RR ss and VV ss. 2- The transient component i tt RRi tt LL ddi tt dddd = (RR LL It solution gives, i tt = AAee )tt i oo = i ss i tt = VV mm ZZ LL sin(ωωωω ) AAee (RR )tt.(38) Where ZZ = RR 2 (ωωωω) 2 Constant A can be obtained from the boundary condition at ωωωω = αα. At this time tt = αα ωω, i oo =. Thus, from eq.(38), = VV mm ZZ LLLL sin(αα ) AAee RRRR 26

27 AA = VV mm ZZ ωωωω sin(αα )ee RRRR Substitution of A in eq.(38) gives i oo = VV mm sin(ωωωω ) VV mm sin(αα ) eeeeee. RR (ωωωω αα)...(39) ZZ ZZ ωωωω For αα < θθ < ββ ββ VV mmmmmmmm = 1 VV 2ππ αα mm sin θθθθθθ = VV mm (cos αα cos ββ)...(4) 2ππ II mmmmmmmm = VV mm (cos αα cos ββ) (41) 2ππππ ββ VV rrrrrr = 1 2ππ (VV mm sin θθ) 2 dddd αα 1 2 = VV mm (ββ αα) 1 (sin 2ββ sin 2 2αα) 1 (42) 2 ππ 2 rms load current can be obtained from eq.(39) if required. c- RL load with freewheeling diode Fig.(13) Single phase half wave circuit with RL load and a freewheeling diode 27

28 ππ VV mmmmmmmm = 1 VV 2ππ αα mm sin θθ dddd = VV mm 2ππ (1 cos αα) (43) II mmmmmmmm = VV mmmmmmmm RR = VV mm (1 cos αα).(44) 2ππππ To remove the negative part form the load voltage wave form freewheeling diode is used as shown in the circuit diagram. Here, at zero crossing of v ss to negative half cycle, LL ddi will cause f d to forward biased. The thyristor will be reverse biased dddd by the supply voltage and turned off. Load current will flow through the loop ( L- R - f d ). As seen from the load current waveform that, during the positive half cycle of v ss, f d is reversed biased and the load current equation is found from VV oo = RRi LL ddi. dddd During the negative half cycle of v ss, f d will be forward biased due to LL ddi, hence i dddd oo is found from the equation = RRi LL ddi. It can be seen from the load current dddd waveform that increasing L to a very high value cause the load current waveform to approach a dc shape. 3-2 Single phase bridge controlled rectifier a- Full controlled single phase bridge rectifier Fig.(14) Single phase full controlled bridge rectifier 28

29 VV mmmmmmmm = 1 ππ ααππ αα VV mm sin θθ dddd = 2VV mm ππ cos αα.(45) The induced voltage LL dddd in the inductance of the load keeps pairs of thyristors (T 1 dddd & T 2 or T 3 & T 4 ) ON after the zero crossing point. Therefore negative parts appears at the load voltage. However for example, when T 1 and T 2 are ON, then switching ON T 3 will cause T 1 to be reversed biased (can be seen from the circuit diagram and its voltages distributions) and hence switched off. The same can be said for T 4 and T 2. The negative parts can be removed by adding freewheeling diode across the load. Also these negative parts can be removed by changing T 4 and T 2 to diodes. Using f d is preferred since it gives the required turn off time of the thyristors. b- Half controlled single phase bridge rectifier Fig. (15) Half controlled single phase bridge rectifier 29

30 VV mmmmmmmm = 1 ππ VV ππ αα mm sin θθ dddd = VV mm ππ (1 cos αα) (46) II fd rrrrrr = 2 (II 2ππ mmmmmmmm ) 2 αα dddd 1 2 = II mmmmmmmm αα ππ..(47) Compared to the full controlled circuit, the half controlled circuit is cheaper, but ac supply current is more distorted due to its zero periods. 4- Three phase half wave controlled rectifier Fig. (16) Three phase half wave controlled circuit thyristors 3

31 a- From fig.(16-b) for ( < αα < 3 ) VV mmmmmmmm = 1 2ππ 3 5ππ ππ VV mm pph sin θθ dddd = 3 3 VV 6 αα 2ππ mm pph cos αα 6 αα..(48) b- From fig.(16-c) for (3 < αα < 9 ) VV mmmmmmmm = 1 2ππ 3 5ππ ππ VV mm pph sin θθ dddd = 3 3 VV 6 αα 2ππ mm pph cos αα 6 αα (49) c- Form fig.(16-d) for (9 < αα < 15 ) VV mmmmmmmm = 1 2ππ 3 ππ VV mm pph ππ 6 αα sin θθ dddd = 3VV mm pp h 2ππ 5- Three phase bridge controlled rectifier 1 cos αα ππ (5) 6 Fig.(17) Three phase bridge controlled circuit using thyristors 31

32 VV mmmmmmmm = 1 2ππ 6 2ππ 3 αα VV mm llllllll ππ sin θθ dddd = 3VV mm llllllll 3 αα ππ cos αα (51) 6- Converter operation 6-1 Overlap In previous, the assumption was made that the transfer or commutation of the current from one diode (or thyristor) to the next took place instantaneously. In practice, inductance and resistance must be present in the supply source, and time is required for a current change to take place. The net result is that the current commutation is delayed, as it takes a finite time for the current to decay to zero in the outgoing diode (or thyristor), while the current will rise at the same rate in the incoming diode. The inductive reactance of the ac supply is normally much greater than its resistance and, as it is the inductance which delays the current change, it is reasonable to neglect the supply resistance. The ac supply may be represented by its Thevenin equivalent circuit, each phase being a voltage source in series with its inductance. The major contributor to the supply impedance is the transformer leakage reactance. To explain the phenomenon associated with the current transfer, the three phase half wave rectifier connection will be used, as once the explanation with this circuit has been understood, it can be readily transferred to the other connections Three phase half bridge uncontrolled rectifier Fig.(18-a) shows the three phase supply to be three voltages, each in series with an inductance (L). Reference to the waveforms in fig.(18-b) shows that at commutation there is an angular period (γγ) during which both the outgoing diode and incoming diode are conducting. This period is known as the overlap period, and (γγ) is defined as the commutation angle or alternatively the angle of overlap. The change of i 1 and i 2 during this period will be as shown in the current waveforms, (assuming a heavily inductive load), i 1 i 2 = i o at any instant. The load voltage during this period equal to v 1v 2 during (γγ). 2 32

33 Fig.(18) Overlap in the three phase half wave uncontrolled rectifier Fig. (19) Conditions during the overlap period 33

34 To determine the factors on which the overlap depends, and to derive an expression for the diode current, a circulating current i can be considered to flow in the closed path formed by the two conducting diodes D 1 and D 2 as shown in fig.(19). v 2 v 1 = LL ddi ddi LL dddd dddd..(52) The voltage v 2 v 1 is the difference between the two phases, having a zero value at t=, the time at which commutation commences. The voltage difference between two phases is the line voltage having a maximum value 3VV mm where V m is of the phase voltage. VV llllllll = 2LL ddi dddd VV mm llllllll sin θθ = 2LL ddi dddd ddi = VV mm llllllll 2LL sin ωωωω dddd Integrating both sides, i = 3VV mm pph 2LL cos ωωωω CC ωω At tt =, i =, CC = 3VV mm pp h 2ωωωω i = 3VV mm pp h 2ωωωω (1 cos ωωωω)..(53) The overlap is complete when i = II LL = II oo = II mmmmmmmm, at which instant θθ = ωωωω = γγ, the overlap angle. Also ωωωω = XX, the supply source reactance. Hence, II mmmmmmmm = 3VV mm pp h 2XX (1 cos γγ)...(54) cos γγ = 1 2II mmmmmmmm XX 3VV mm pp h (55) From eq. (53), the current change in the diodes during overlap is consinusoidal, as illustrated in fig.(18-b). To determine the mean voltage of the waveform shown in fig.(18-a), one can use calculus to find the area under the two sections of the curve, one based on the sine wave shape after overlap is complete and the other during overlap. 34

35 During overlap, the load voltage is the mean between two sine waves, that is, the shape is sinusoidal, but if we consider the curve as a cosine wave, then the integration limits will be to γ on a cosine wave of peak value VV mm ph sin ππ, giving 6 VV mmmmmmmm = 1 2ππ 3 5ππ 6 = 3 3VV mm pp h 4ππ VV ππ mm pph sin θθ dddd VV mm pph sin 6 γγ γγ ππ 6 cos dd (1 cos γγ) (56) Three phase half bridge controlled rectifier Fig.(2) Overlap in the three phase half wave controlled rectifier 35

36 In the controlled 3-pulse circuit, the overlap will lead to the waveform shown in fig.(2), where it can be seen that with a firing delay angle (α) a finite voltage is present form the start of commutation. v 2 v 1 = LL ddi ddi LL dddd dddd VV llllllll = 2LL ddi dddd VV mm llllllll sin(ωt α) = 2LL ddi dddd when i =....(57), where t is the time from the start of commutation, ddi = VV mm llllllll 2LL sin(ωωωω αα) dddd Integrating both sides, i = 3VV mm pph 2LL cos(ωωωω αα) CC ωω At tt =, i =, CC = 3VV mm pp h 2ωωωω cos αα i = 3VV mm pp h 2ωωωω (cos αα cos(ωωωω αα))..(58) The overlap is complete when i = II LL = II oo = II mmmmmmmm, at which instant θθ = ωωωω = γγ, the overlap angle. Also ωωωω = XX, the supply source reactance. Hence, II mmmmmmmm = 3VV mm pp h 2XX [cos αα cos(γγ αα)]...(59) Compared to the uncontrolled case (α=), the overlap angle γ will be shorter and the current change during commutation will be towards a linear variation. The mean load voltage is given by VV mmmmmmmm = 1 2ππ 3 5ππ ππ VV mm pph sin θθ dddd VV 6 ααγγ mm sin ππ cos dd αα 6 6 αα ααγγ = 3 3VV mm pp h 4ππ [cos αα cos(αα γγ)] (6) 36

37 6-2 Power factor The power factor of a load fed from an ac supply is defined as: pppppppppp ffffffffffff = mmmmmmmm pppppppppp VV rrrrrr II rrrrrr (61) In the usual ac system where the current is sinusoidal, the power factor is the cosine of the angle between current and voltage. The rectifier circuit, however, draws non-sinusoidal current from the ac system, hence the power factor cannot be defined simply as the cosine of the displacement angle. The waveforms of the various controlled rectifiers in previous shows that firing delay has the effect of delaying the supply current relative to its phase voltage. The current dose contain harmonic components which result in its overall rms value being higher than the rms value of its fundamental component, therefore the power factor is less than that calculated from the cosine of its displacement angle. Normally, the supply phase voltage can be taken as being sinusoidal, hence there will be no power associated with the harmonic current, which therefore result in pppppppppp = VV 1 rrrrrr II 1 rrrrrr cos 1.(62) Where the suffix 1 relates to the fundamental component of the input to rectifier current, 1 being the phase angle between the voltage and the fundamental component of the current. For a sinusoidal voltage supply, substituting eq.(62) into eq.(61) yields pppppppppp ffffffffffff = VV 1 rrrrrr II 1 rrrrrr cos 1 VV rrrr ss II rrrrrr Where VV 1 rrrrrr = VV rrrrrr pppppppppp ffffffffffff = II 1 rrrrrr II rrrrrr cos 1..(63) Where and II 1 rrrrrr II rrrrrr = iiiiiiiiii dddddddddddddddddddd ffffffffffff.(64) cos 1 = iiiiiiiiii ddiiiiiiiiiiiiiiiiiiiiii ffffffffffff...(65) 1 will equal the firing delay angle αα in the fully-controlled rectifier connection that have a continuous level load current.the power factor will always be less than unity when there are harmonic components in the supply current, even when the current is in phase with the voltage, as in the diode case. 37

38 DC-AC Inverters Author Dr. Ali Hussein Numan University of Technology Power Electronics and Electromechanical Engineering Department Electrical Drives 8 hrs / Four weeks EME 41 Fall Energy- Systems branch Lecturers: Dr.Ali Hussein Numan & Dr.Shatha K. Baqir Contents 1- Voltage Source Inverter Vs Current Source Inverters 2- Single phase bridge inverter with R load 3- Single phase bridge inverter with RL load 4- Quasi-square wave output RL load 5- Three phase bridge inverter six step( 18 o ) 6- Three phase bridge inverter six step ( 12 o ) 7- Single phase pulse width modulation (PWM) inverter 8- Three phase bridge PWM inverter 9- Classification of PWM techniques 38

39 DC-AC Inverters Author Dr. Ali Hussein Numan Definition: Inverters are static circuits that convert power from a DC to AC power at specified output voltage and frequency. Typical Applications:Invertersare used in the following industrial applications: 1) Variable speed AC motor drivers. 2) Induction heating. 3) Aircraft power supply. 4) Uninterruptable power supplies for computers. 5) Traction. 6) High Voltage DC (HVDC). Classification Single phase (1Ф) and three phase (3Ф) inverters can be classified into: 1) Voltage source inverters (VSI). The VSI convert the DC input voltage source into a square wave AC output voltage source as shown in Fig.1.The DC input voltage supply to the VSI can be a battery or the output of a controlled rectifier.a shunt capacitor in the input circuit of the VSI is used to provide the stiff voltage source. The VSI is commonly used for low and medium power application. Vi InputDC voltage Source C - V S I R Output AC S 3Ф Voltage source T Fig.1 Voltage source inverter (VSI). V -V 2) Current source inverters (CSI). The CSI convert the DC input current source into a square wave AC output current source as shown in Fig.2.A series inductor in the input circuit of the CSI is used to provide the stiff current source.the CSI is usually used for high power application. Ii InputDC currentsource - L C S I R S T Output AC 3Ф current source I -I Fig.2 Current source inverter(csi). 39

40 DC-AC Inverters Author Dr. Ali Hussein Numan Voltage Source Inverter (VSI) 1) Single Phase Bridge Inverter (square wave) (a) Resistive Load (R): The circuit of single phase bridge inverter fed resistive load is shown in Fig.3. i s V S _ ig 1 A T 1 D 1 i L R V L - D 3 T 3 B ig 3 ig 4 T 4 D 4 D 2 T 2 ig 2 Fig.3 Single phase bridge inverter with R load. Circuit Operation D 1,D 2,D 3,D 4 are eliminated from the circuit (always OFF) due to resistive load. From KVL T 1 &T 4 and T 3 &T 2 cannot be ON at the same time (short circuit). This circuit is operates in two modes as shown in table below. Mode Period Conducting Devices Equivalent Circuit V L i L i s T 1 T 3 I T/2 T 1 &T 2 V s - A i L - R V L B Vs Vs/R T 4 T 2 i s T 1 T 3 II T T 3 &T 4 V s - A - i L R V L B -Vs -Vs/R T 4 T 2 4

41 DC-AC Inverters The inverter waveforms are shown below: ig 1 & ig 2 (a) Author Dr. Ali Hussein Numan t ig 3 & ig 4 (b) t it 1 & it 2 V s /R (c) t it 3 & it 4 V s /R (d) t V L Vs (e) t -Vs i L V s /R (f) i L =it 1 -it 4 T/2 T 3T/2 t -V s /R T 1 and T 2 ON T 3 and T 4 OFF T 3 and T 4 ON T 1 and T 2 OFF T 1 and T 2 ON T 3 and T 4 OFF Fig.4 Waveforms (a) & (b) triggering, (c) &(d) transistors currents, (e) load voltage, and (f) load current. Mathematical Analysis: 1) The rmsvalue of load voltage is given as, VV LL(rrrrrr ) = 1 TT 2 TT/2 (VV ss) 2 dddd = VV SS 41

42 DC-AC Inverters Author Dr. Ali Hussein Numan 2) The instantaneous rmsvalue of load voltage in terms of Fourier series is: Where, CC nn = aa nn 2 bb nn 2 ππ vv LL(ωωωω ) = VV LL(aaaaaa ) CC nn sin(nnnnnn φφ nn ) nn=1, φφ nn = tan 1 aa nn bb nn and T=period=2π aa nn = 2 TT TT vv LL(ωωωω ) cos(nnnnnn)dddddd 2ππ aa nn = 2 (VV 2ππ ss )cccccc(nnnnnn) dddddd ( VV ss )cccccc(nnnnnn) dddddd = for all values of n ππ Now,the value of b n can be calculated as: ππ bb nn = 2 TT TT vv LL(ωωωω ) sin (nnnnnn)dddddd 2ππ bb nn = 2 2ππ (VV ss) ssssss (nnnnnn) dddddd ( VV ss ) ssssss(nnnnnn) dddddd = 2 VV ss (1 cccccccccc) nnnn ππ 4VV ss ffffff oooooo vvvvvvvvvvvv oooo nn bb nn = nnnn ffffff eeeeeeee vvvvvvvvvvvv oooo nn CC nn = bb nn 2 = bb nn andφφ nn = tan 1 = CC nn = 4VV ss nnnn ffffff oooooo vvvvvvvvvvvv oooo nn Since the waveform (e) of Fig.4 has symmetric positive and negative cycles. Hence average value of such waveform is zero i.e. VV LL(aaaaaa ) = Therefore, the Fourier series for bridge inverter can be written as: Where,n is the harmonic order vv LL(ωωωω ) = 4VV ss sin nnnnnn nnnn nn=1,3,5 forn= 2,4,6, vv LL(ωωωω ) =, thus, output waveform contain only odd harmonics 3) Therms value of fundamental component of load voltage (n=1) : 42

43 DC-AC Inverters Author Dr. Ali Hussein Numan vv LL1 (ωωωω ) = 4VV ss ππ sin ωωωω This has the same frequency(ωt) as that of square wave, the peak value of fundamental is, CC 1 = 4VV ss ππ The rms value of fundamental component is, vv LL11 (rrrrrr) = CC = 44VV ss 22 ππ =. 99VV ss The rms values of n th harmonics can be obtained as, vv LLnn (rrrrrr ) = 4VV ss 2 nnnn =.9VV ss nn 4)Thermsvalue of load currentii LL(rrrrrr ) = VV LL(rrrrrr ) RR 5)The value of output powerpp oo = VV LL 2 (rrrrrr ) RR (b)heavily Inductive Load for n = 1,3,5, 1- Square wave output: The circuit of single phase bridge inverter fed heavily inductive load is shown in Fig.5. i s V S _ ig 1 T 1 i L D 1 R L V L - D 3 T 3 ig 3 ig 4 T 4 D 4 D 2 T 2 ig 2 Fig.5 Single phase bridge inverter with RL load. Circuit Operation D 1,D 2,D 3,D 4 are used to protect circuit against Ldi/dt. From KVL T 1 &T 4 and T 3 &T 2 cannot be ON at the same time (short circuit). This circuit is operates in four modes as shown in table below. 43

44 DC-AC Inverters Author Dr. Ali Hussein Numan Mode Period Conducting Devices Equivalent Circuit V L i L i s I -t 1 D 1 & D 2 D 1 T 1 T 3 V L R L V s A - i L - D 3 B Vs -Imin T 4 D 4 T 2 D 2 i s T 1 D 1 T 3 V L D 3 II t 1 -T/2 T 1 &T 2 R L V s A - i L - B Vs Imax T 4 D 4 T 2 D 2 i s III T/2-t 2 D 3 & D 4 T 1 T 3 D 1 V L R L V s A - i L - D 3 B -Vs Imax T 4 D 4 T 2 D 2 i s T 1 D 1 T 3 V L D 3 IV t 2 -T T 3 &T 4 R L V s A - i L - B -Vs -Imin T 4 D 4 T 2 D 2 The inverter waveforms are shown in Fig.6. 44

45 DC-AC Inverters ig 1 & ig 2 (a) Author Dr. Ali Hussein Numan t ig 3 & ig 4 (b) t V L Vs (c) t -Vs it 1 & it 2 (d) t it 3 & it 4 (e) t id 1 & id 2 (f) id 3 & id 4 (g) t t i L (h) i L1 i L2 I max I min t i s i s =it 1 -id 1 (-T/2) i s =it 3 -id 4 (T/2-T) D 1 & D 2 ON t 1 T/2 t 2 T t 3 T 1 and T 2 ON D 3 & D 4 ON T 3 and T 4 ON T 1 and T 2 ON Fig.6 Waveforms (a) & (b) triggering, (c) load voltage,(d) & (e) transistors currents,(f) &(g) diodes currents,(h) load current, and (i) supply current. 45 (i) D 1 & D 2 ON 3T/2 t

46 DC-AC Inverters Author Dr. Ali Hussein Numan Mathematical Analysis: 1) Since the load voltagewaveform in inductive load shown in Fig.6(c) has similar shape to the load voltagewaveform in resistive load shown in Fig.4(e). Therefore, the load voltage equations for resistive load can be applied for inductive load. 2) The load current can be expressed as: ii LL1 (tt) = VV SS RR tt 1 ee ( ττ ) tt ( II mmmmmm ee ττ ) t T/2 ii LL2 (tt) = VV SS RR TT 1 ee tt 2 /ττ TT tt II mmmmmm ee 2 /ττ T/2 t T Where:- ττ:l/r Time constant I min : minimum value of load current I max :maximum value of load current The maximum value of load current equal to minimum value but with negative sign as: I max = -I min II mmmmmm = VV SS RR ( ee TT 2222 ) 1 1 ee ( TT ) 2222 II mmmmmm = VV SS RR 1 ee ( TT 2222 ) 1 ee ( TT 2222 ) 46

47 DC-AC Inverters Author Dr. Ali Hussein Numan Example (1):A single phase bridge inverter shown in Fig.7used to supply inductive load with R=1Ω, L=25mH,the DC supplyvoltage V S =1v, and the outputfrequency 5 Hz. Write an equations for instantaneous loadcurrent. T 1 D 1 D 3 T 3 V S 1 V _ i L 1Ω 25mH V L - T 4 D 4 D 2 T 2 Fig.7 Single phase bridge inverter. Solution:The period (T) of the load voltage and time constant (ττ): TT = 1 ff = 1 =.2 ssssss 5 ττ = LL RR =.25 1 =.25 ssssss Now, the minimum value and the maximum value of load currentsare: II mmmmmm = II mmmmmm = 1 1 The expressionsinstantaneous load currentsare: ii LL1 (tt) = 1 1 = 1 2ee ( tt.25 ) t.1 ( ) ee 1 1 ee (.2 = 1 A ) 2.25 tt 1 ee (.25 ) 1ee ( tt.25 ) ii LL2 (tt) = ee (tt.1)/.25 1ee (tt.1)/.25 = 1 2ee (tt.1)/.25.1 t.2 47

48 DC-AC Inverters Author Dr. Ali Hussein Numan Example (2):Inthe single phase bridge inverter of Fig.8, the load current isi L = 22 sin (ωt-45 o ) and the DC supply voltage V S =1v. (a) Draw waveforms of V L,i L, and i s. Indicate on the waveforms of i L, and i s the device that are conducting during various intervals of time. (b)determine the average value of the supply current i s,and the power from the supplyp S.(c) Determine the power delivered to the load P L. i S T 1 D 1 D 3 T 3 V S i L R L _ 1 V V L - T 4 D 4 D 2 T 2 Fig.8 Single phase bridge inverter. Solution:(a)the waveforms are shown in Fig.9. V L Vs t -Vs i L t i S D 1 & D 2 ON T 1 and T 2 ON D 3 & D 4 ON T 3 and T 4 ON T/2 T t (b) D 1 & D 2 ON T 1 and T 2 ON D 3 & D 4 ON T 3 and T 4 ON Fig.9waveforms. 48

49 DC-AC Inverters Author Dr. Ali Hussein Numan ii ss aaaaaa = ii ss = 1 ππ ππ 22 sin(ωωωω 45oo )dd (ωωωω) = 22 ππ = 99 AA 2 PP ss = VV ss xx ii ss = 1 99 = 9.9KKKK (c)from the Fourier analysis of V L (square wave), the rms value of the fundamental load voltage is: vv LL1(rrrrrr ) = 4VV ππ 2 = 4 1 = 9 vv ππ 2 PP oooooo = vv LL1(rrrrrr ) II LL cos θθ = 9 22 cos 45oo 2 = KKKK 2- Quasi-square wave output (RL load) :the square waveformoutputfrom single phase bridge inverter can be improved by adding zero periods to obtain waveform called quasi-square. The circuit of quasi-square waveform is shown in Fig.5. i s V S _ ig 1 T 1 i L D 1 R L V L - D 3 T 3 ig 3 ig 4 T 4 D 4 D 2 T 2 ig 2 Fig.1 Single phase bridge inverter with RL load. Circuit Operation 49

50 DC-AC Inverters Author Dr. Ali Hussein Numan The operation of quasi square wave inverter can be summarized as shown in the following table Mode Period Conducting Devices V L i L I -t 1 D 1 D 2 V s il 4 (-) II t 1 -t 2 T 1 T 2 V s il 1 () III t 2 -t 3 T 2 D 4 il 2 () IV t 3 -t 4 D 3 D 4 -V s il 3 () V t 4 -t 5 T 3 T 4 -V s il 3 (-) VI t 5 -t 6 T 3 D 1 il 4 (-) The inverter waveforms are shown in Fig.6. ig 1 ig 2 ig 3 ig 4 (a) (b) (c) (d) t t t t V L Vs φ (e) φ t -Vs I max I min i L il 1 il 2 t 1 t 2 t 3 t 4 t 5 t 6 it 4 il 3 il 4 D 1 & D 2 ON T 1 and T 2 ON T 2 & D 4 ON D 3 & D 4 ON (f) T 3 and T 4 ON T3 & D1 ON t 5

51 DC-AC Inverters Author Dr. Ali Hussein Numan it 1 (g) t it 2 (h) t it 3 (i) t it 4 (j) t id 1 (k) t id 2 (l) t id 3 (m) t id 4 (n) D 1 & D 2 ON t 1 t 2 t 3 t 4 t 5 t 6 T 2 D 3 T3 T 1 and T 2 ON & & T 3 and T 4 ON & D 4 D 4 D1 ON ON ON Fig.11 quasi square waveforms with inductive load: (a)-(d) triggering, (e) load voltage, (f) load current, (g)-(i)transistors currents, and (k)-(n) diodes currents. t Mathematical Analysis: From Fig.11(f),the load current can be described as: iill 1 (tt) = V s R iill 3 (tt) = V s R t ( 1 e τ ) II mmmmmm e (t τ ) iill 2 (tt) = II mmmmmm e (t τ ) t ( 1 e τ ) II 1 e (t τ ) 51

52 DC-AC Inverters Author Dr. Ali Hussein Numan II mmmmmm = V s R iill 4 (t) = II 2 e (t τ ) II mmmmmm = II 2 e (T τ ) t2 ( 1 e τ ) II mmmmmm e (t 2 τ ) II 1 = II mmmmmm e ( T 2τ ) II 2 = V t s ( 5 (1 e τ ) II R 1 e (t 5 τ ) Example (3):For a single phase quasi square wave inverter feeding heavily inductive load obtain an expression for rmsload voltage. Solution: From Fig.11 (e),the rms load voltage is VV LL (rrrrrr ) = 1 TT TT (VV LL) 2 dddddd VV LL (rrrrrr ) = 1 ππ φφ 2ππ φφ 2ππ (VV ss) 2 dddddd ( VV ss ) 2 dddddd VV LL (rrrrrr ) = (VV ss) 2 2ππ [(ππ φφ) (2ππ φφ) (ππ)] = (VV ss) 2 [(ππ φφ) (ππ φφ)] 2ππ ππ VV LL (rrrrrr ) = VV ss 1 (ππ φφ) ππ For perfect quasi square wave (φφ ππ) = i.e 12o pulse width. Then therms load voltage will be, VV LL (rrrrrr ) = VV ss 1 ππ 2ππ 3 VV LL (rrrrrr ) = VV ss 2 3 =.816 VV ss 52

53 DC-AC Inverters Author Dr. Ali Hussein Numan Example (4):For a single phase quasi square wave inverter with RL load. Derive an expression for rms value of n th harmonic of load voltage. Solution: From Fig.11 (e), the general expression for Fourier series is given as vv LL(ωωωω ) = VV LL(aaaaaa ) CC nn sin(nnnnnn φφ nn ) nn=1 Where, CC nn = aa nn 2 bb nn 2 aaaaaa φφ nn = tan 1 aa nn bb nn To obtain a n ππ φφ aa nn = 2 TT TT vv LL(ωωωω ) cos(nnnnnn)dddddd 2ππ φφ = 2 2ππ (VV ss)cccccc(nnnnnn) dddddd ( VV ss )cccccc(nnnnnn) dddddd = VV ss ππ ssssss(nnnnnn) nn ππ φφ ππ ssssss(nnnnnn) nn 2ππ φφ = VV ss [ssssss nn(ππ φφ) ssssss nn(2ππ φφ) ssssss nn(ππ)] nnnn ππ Since (φφ ππ) = φφ = ππ Substitute the value ofzero periods(φφ) in the 33 above the equation will result aa nn = VV ss nnnn ssssss nn 2ππ 3 ssssss nn 5ππ ssssss nn(ππ) = for all values of n 3 To obtain b n ππ φφ bb nn = 2 TT TT vv LL(ωωωω )ssssss(nnnnnn)dddddd 2ππ φφ = 2 2ππ (VV ss) ssssss (nnnnnn) dddddd ( VV ss ) ssssss(nnnnnn) dddddd = VV ss ππ cccccc(nnnnnn) nn ππ φφ ππ cccccc(nnnnnn) nn 2ππ φφ ππ 53

54 DC-AC Inverters Author Dr. Ali Hussein Numan = VV ss [ cccccc nn(ππ φφ) 1 cccccc nn(2ππ φφ) cccccc nn(ππ)] nn ππ bb nn = VV ss nn ππ 1 cccccc nn 2ππ 3 cccccc nn 5ππ cccccc nn(ππ) 3 bb nn = for even values of n and, bb nn = VV ss nn ππ 1 cccccc nn 2ππ 3 cccccc nn 5ππ cccccc nn(ππ) 3 for odd values of n except (3, 9,15..) i.emultiples of 3 To obtain Fourier series Since a n = thus, CC nn = aa nn 2 bb nn 2 = bb nn CC nn = VV ss nn ππ 1 cccccc nn 2ππ 3 cccccc nn 5ππ cccccc nn(ππ) 3 and φφ nn = tan 1 aa nn bb nn = From the waveform of Fig.11 (e), it is clear that VV LL(aaaaaa ) =.Hence Fourier series will be vv LL(ωωωω ) = VV ss nn ππ 1 cccccc nn 2ππ 3 cccccc nn 5ππ 3 nn=1 cccccc nn(ππ) sin(nnnnnn) For odd values of n except multiples of 3 54

55 DC-AC Inverters Author Dr. Ali Hussein Numan 2) Three Phase Bridge Inverter (Six Step Inverter) Three phase bridge inverter are used in high power applicationssuch as AC motors and three phase power system. The basic circuit of three phase bridge inverter is shown in Fig.9. The circuit is consists of six transistors or thyristors,and six diodes connected in parallel with each power semiconductor switch. T 1 T 3 T 5 i g1 D 1 D 3 D 5 i g3 i g5 V s - a b c T 4 T 6 T 2 i g4 D 4 D 6 D 2 i g6 i g2 ia ib ib R va R vb R vc vab vca vbc Fig. 9 basic three phase bridge inverter with resistive load. Three phase bridge invertercan be controlled using the following conduction mode: 1) 18 o conduction 2) 12 o conduction In both cases each power semiconductor switch will conductsevery 6 o interval. 1) 18 o conduction with resistive (R) load In this type of conduction each transistor in Fig.9 will conducts 18 o. The switching sequence:561 (V 1 ) 612 (V 2 ) 123 (V 3 ) 234 (V 4 ) 345 (V 5 ) 456 (V 6 ) 561 (V 1 ) Next table summarize operation of three phase inverter with 18 o conduction angle and resistive load. 55

56 DC-AC Inverters Author Dr. Ali Hussein Numan Mode Conducting transistors Actual cct. Equivalent cct. i 1 a c Phase voltage Line voltage V a V b V c V ab V bc V ca V s - T 1 T 3 T 5 a b c V s - V s /3 R Va Vc R T 4 T 6 T 2-2V s /3 R Vb I (π/3) T 1 T 6 T 5 ia ib ic R va R vb R vc R eq =(R//R)R=3R/2 i 1 =Vs/R eq =2Vs/3R V a =V c = i 1 (R//R)= b VV ss ss 33 VV ss 33 V s - V s i 1 *(R/2)=Vs/3 V b = - i 1 *R=-2Vs/3 V ab =V a -V b V bc =V b -V c V ca =V c -V a i 2 a - V s T 1 T 3 T 5 a T 4 T 6 T 2 b c V s - 2V s /3 -V s /3 R R Vb Vc Va R II ( 2π/3) T 2 T 1 T 6 ia Va ib Vb ic Vc b R eq = (R//R)R=3R/2 i 2 =v s /R eq =2V s /3R V a = i 2 *R=2Vs/3 c 2222 ss 33 VV ss 33 VV ss 33 V s - V s V b = V c = - i 2 *(R//R) = - Vs/3 V ab =V a -V b V bc =V b -V c V ca =V c -V a i 3 b a - V s T 1 T 3 T 5 a b c - V s V s /3 R Vb Va R T 4 T 6 T 2-2V s /3 R Vc III (π) T 3 T 2 T 1 ia ib ic Va Vb Vc R eq = (R//R)R=3R/2 i 3 =Vs/R eq =2Vs/3R V a =V b = i 3 (R//R)= c VV ss 33 VV ss ss 33 V s - V s i 3 *(R/2)=Vs/3 V c = - i 3 *R=-2Vs/3 V ab =V a -V b V bc =V b -V c V ca =V c -V a 56

57 DC-AC Inverters Author Dr. Ali Hussein Numan i 4 b - V s T 1 T 3 T 5 a T 4 T 6 T 2 b c V s - 2V s /3 -V s /3 R R Vc Va Va R IV (4π/3) T 4 T 3 T 2 ia ib ic Va Vb Vc c R eq = (R//R)R=3R/2 i 4 =v s /R eq =2V s /3R a VV ss ss 33 VV ss 33 -V s V s V b = i 4 *R=2Vs/3 V a = V c = - i 4 *(R//R) = - Vs/3 V ab =V a -V b V bc =V b -V c V ca =V c -V a i 5 c b V s - T 1 T 3 T 5 a b c - V s V s /3 R Vc Vb R T 4 T 6 T 2-2V s /3 R Va V ( 5π/3) T 5 T 4 T 3 ia ib ic Va Vb Vc R eq = (R//R)R=3R/2 i 5 =Vs/R eq =2Vs/3R a 2222 ss 33 VV ss 33 VV ss 33 -V s V s V b =V c = i 5 (R//R)= i 3 *(R/2)=Vs/3 V a = - i 5 *R=-2Vs/3 V ab =V a -V b V bc =V b -V c V ca =V c -V a i 6 c V s - T 1 T 3 T 5 a T 4 T 6 T 2 b c V s - 2V s /3 -V s /3 R R Va Vb Vc R VI (2π) T 6 T 5 T 4 ia ib ic va vb vc a R eq = (R//R)R=3R/2 i 6 =v s /R eq =2V s /3R V c = i 6 *R=2Vs/3 b VV ss 33 VV ss ss 33 - V s V s V a = V b = - i 6 *(R//R) = - Vs/3 V ab =V a -V b V bc =V b -V c V ca =V c -V a 57

58 DC-AC Inverters Author Dr. Ali Hussein Numan The three phase bridge inverter waveforms with 18 o conduction angle and resistive load are shown in Fig.1. ig 1 ig 2 Mode I T 1 T 6 T 5 Mode II T 2 T 1 T 6 Mode III T 3 T 2 T 1 Mode IV T 4 T 3 T 2 Mode V T 5 T 4 T 3 Mode VI T 6 T 5 T 4 ωt Transistor ON ωt ig 3 ig 4 ig 5 ig 6 ωt ωt ωt Switching sequence it 1 2Vs/3R ωt Vs/3R ωt it 2 ωt it 3 it 4 Vs/3 ωt ωt Transistor currents it 5 ωt it 6 π 2π ωt 58

59 DC-AC Inverters Author Dr. Ali Hussein Numan Va,ia Vs/3 i a =it 1 -it 4 Vb,ib i b =it 3 -it 6-2Vs/3 Vc,ic Vs/3 i c =it 5 -it 2 Vab Vs 2Vs/3 ωt ωt ωt Line current = Phase current (Y) connected &Phase Voltages V ab =V a V b ωt Vbc Vs V bc =V b V c Vca Vs ωt Line voltages (load voltage) V ca =V c V a π/3 2π/3 π 4π/3 5π/3 2π ωt Fig.1 Three phase bridge inverter waveforms with 18 o conduction and resistive load. Mathematical Analysis: The line voltage in terms of phase voltage in a 3- phase system with phase sequence (a,b,c) are: V ab = V a V b V bc = V b V c V ca = V c V a 59

60 DC-AC Inverters Author Dr. Ali Hussein Numan Where, V ab,v bc,v ca line voltage V a,v b,v c phase voltage V ab V ca = 2V a (V b V c ) But in a balance three phase system, the sum of the three phase voltages is zero V a V b V c = V ab V ca = 3V a V a = (V ab V ca ) 3 Similarly, the (b) and (c) phase voltages are: V b = (V bc V ab ) 3 V c = (V ca V bc ) 3, From Fig. 1,the three phase output voltages are six stepped and can be described by Fourier series. V a (t) = 2V s sin nn (ωt) nπ n=6k±1 V b (t) = 2V s sin nn ωt 2π nπ 3 n=6k±1 Where k=, 1, 2, 3,.. V c (t) = 2V s cos nπ nπ 6 sin nn ωt 4π 3 n=6k±1 For n=3,cos nπ 6 =, thus, all multiples of 3rd harmonics are cancelled. 6

61 DC-AC Inverters Author Dr. Ali Hussein Numan The three phase output voltage can be rewritten as: V a (t) = 2VV ss ππ ssssssssss 1 5 ssssss5ωωωω 1 7 ssssss7ωωωω V b (t) = 2VV ss ππ sin (ωωωω 2π 3 ) 1 5 ssssss5(ωωωω 2π 3 ) 1 7 ssssss7(ωωωω 2π 3 ) V c (t) = 2VV ss ππ sin (ωωωω 4π 3 ) 1 5 ssssss5(ωωωω 4π 3 ) 1 7 ssssss7(ωωωω 4π 3 ) Also, the three line output voltages can be described by Fourier series as follows: V ab (t) = 4V s cos nπ nπ 6 sin nn ωt π 6 1,3,5 V bc (t) = 4V s cos nπ nπ 6 sin nn ωt π 2 1,3,5 V ca (t) = 4V s cos nπ nπ 6 sin nn ωt 5π 6 1,3,5 Also, all multiples of 3 rd harmonics are cancelled.thus, Note that cos π 6 = = 3 V ab (t) = 2 3 ππ VV ss ssssssssss 1 5 ssssss5ωωωω 1 7 ssssss7ωωωω V bc (tt) = 2 3 ππ VV ss ssssss ωωωω 2π ssssss 5 ωωωω 2π ssssss 7 ωωωω 2π 3 V ca (tt) = 2 3 ππ VV ss ssssss ωωωω 4π ssssss5 ωωωω 4π ssssss7 ωωωω 4π 3 61

62 DC-AC Inverters Author Dr. Ali Hussein Numan The amplitude of the fundamental line voltage can be obtained as Where, V s :DC input voltage to the inverter V ab 1 = 2 3 ππ VV ss V ab1 : The amplitude of the fundamental line voltage The RMS value of fundamental line voltage is V ab 1(rms ) = 2 3 2ππ VV ss = 6 ππ =.7797VV ss The phase voltages are shifted from the line voltages by 3 V a = 2 3 ππ VV ss / 3 = 2 ππ VV ss The RMS value of fundamental phase voltage is V a1(rms ) = 2VV ss ππ 1 2 = V ab 1(rms ) 33 =.45 VV ss It is seen from the line voltage waveform (V ab ) in Fig.1 that the line voltage is (V s ) from (-12 o ). Therefore, the rms value of line voltage (V ab (rms)) is 2ππ 3 V ab (rms ) = 1 ππ VV ss 2 dddd = VV ss =.8165VV ss Also, the rms value of phase voltage (V a (rms)) is ππ 3 V a(rms ) = 1 2 ππ VV ss 3 dddd 2VV 2 ss 3 dddd ππ/33 Orsince the load is star connected the phase voltage is 2ππ 3 62 ππ 2222/33 VV ss 3 2 dddd =.4714VV ss

63 DC-AC Inverters Author Dr. Ali Hussein Numan V a(rms ) = V ab(rms) 33 Total Harmonics Distortion (THD) The THD of the line voltage is =.8165VV ss 33 =.4714VV ss TTTTTT llllllll vvvvvvvvvvvvvv = (VV aaaa ) 2 (VV aaaa 1 ) 2 (VV aaaa 1 ) 2 = (.8165)2 (.7797) 2 (.7797) 2 = 31.8 % The THD of the phase voltage is TTTTTT pphaaaaaa vvvvvvvvvvvvvv = (.4714)2 (.45) 2 (.45) 2 = 31.1 % Note:Absence of the 3 rd harmonics is the most important advantage of a three phase inverter. 2) 12 o conduction with resistive (R) load In this type of conduction each transistor in Fig.9 will conducts 12 o. The switching sequence:61 (V 1 ) 12 (V 2 ) 23 (V 3 ) 34 (V 4 ) 45 (V 5 ) 56 (V 6 ) 61 (V 1 ) Next table summarize operation of three phase inverter with 12 o conduction angle and resistive load. 63

64 DC-AC Inverters Author Dr. Ali Hussein Numan Mode Conducting transistors Actual cct. Equivalent cct. i 1 a Phase voltage Line voltage V a V b V c V ab V bc V ca V s - T 1 T 3 T 5 a b c V s - R Va - I (π/3) T 1 T 6 T 4 T 6 T 2 ia ib ic R va R vb R vc R Vb - b R eq = RR=2R i 1 =Vs/R eq =Vs/2R V a = i 1 *R=Vs/2 VV ss 22 VV ss 22 V s VV ss 22 VV ss 22 V b = - i 1 *R=-Vs/2 V c = V ab =V a -V b V bc =V b -V c V ca =V c -V a i 2 a V s - T 1 T 3 T 5 a b c V s - R Va - II ( 2π/3) T 2 T 1 T 4 T 6 T 2 ia ib ic R va R vb R vc R R eq = RR=2R Vc - c i 2 =Vs/R eq =Vs/2R V a = i 2 *R=Vs/2 VV ss 22 VV ss 22 VV ss 22 VV ss 22 -V s V c = - i 2 *R=-Vs/2 V b = V ab =V a -V b V bc =V b -V c V ca =V c -V a i 3 b V s - T 1 T 3 T 5 a b c V s - R Vb - III (π) T 3 T 2 T 4 T 6 T 2 ia ib ic R va R vb R vc R R eq = RR=2R Vc - c i 3 =Vs/R eq =Vs/2R V b = i 3 *R=Vs/2 VV ss 22 VV ss 22 VV ss 22 V s VV ss 22 V c = - i 3 *R=-Vs/2 V a = V ab =V a -V b V bc =V b -V c V ca =V c -V a 64

65 DC-AC Inverters Author Dr. Ali Hussein Numan i 4 b V s - T 1 T 3 T 5 a b c V s - R Vb - IV (4π/3) T 4 T 3 T 4 T 6 T 2 ia ib ic R va R vb R vc R R eq = RR=2R i 4 =Vs/R eq =Vs/2R V b = i 4 *R=Vs/2 Va - a VV ss 22 VV ss 22 - V s VV ss 22 VV ss 22 V a = - i 4 *R=-Vs/2 V c = V ab =V a -V b V bc =V b -V c V ca =V c -V a i 5 c V s - T 1 T 3 T 5 a b c V s - R Vc - V ( 5π/3) T 5 T 4 T 4 T 6 T 2 ia ib ic R va R vb R vc R Va - a R eq = RR=2R i 5 =Vs/R eq =Vs/2R V c = i 5 *R=Vs/2 VV ss 22 VV ss 22 VV ss 22 VV ss 22 V s V a = - i 5 *R=-Vs/2 V b = V ab =V a -V b V bc =V b -V c V ca =V c -V a i 6 c V s - T 1 T 3 T 5 a b c V s - R Vc - VI (2π) T 6 T 5 T 4 T 6 T 2 ia ib ic R va R vb R vc R R eq = RR=2R Vb - b i 6 =Vs/R eq =Vs/2R V c = i 6 *R=Vs/2 VV ss 22 VV ss 22 VV ss 22 -V s VV ss 22 V b = - i 6 *R=-Vs/2 V a = V ab =V a -V b V bc =V b -V c V ca =V c -V a 65

66 DC-AC Inverters Author Dr. Ali Hussein Numan The three phase inverter waveforms with 12 o conduction angle and resistive load are shown in Fig.11. ig 1 Mode I T 1 T 6 Mode II T 2 T 1 Mode III T 3 T 2 Mode IV T 4 T 3 Mode V T 5 T 4 Mode VI T 6 T 5 Transistor ON ig 2 ωt ωt ig 3 ig 4 ig 5 ig 6 ωt ωt ωt Switching sequence ωt it 1 Vs/2R Vs/2R ωt it 2 ωt it 3 it 4 ωt ωt Transistor currents it 5 ωt it 6 π 2π ωt 66

67 DC-AC Inverters Author Dr. Ali Hussein Numan Va,ia Vs/2 i a =it 1 -it 4 Vb,ib i b =it 3 -it 6-2Vs/2 Vc,ic i c =it 5 -it 2 Vab Vs Vs/2 V ab =V a V b Vbc V bc =V b V c -Vs/2 Vca ωt ωt ωt ωt ωt Line current = Phase current (Y) connected &Phase Voltages Line voltages (load voltage) V ca =V c V a -Vs π/3 2π/3 π 4π/3 5π/3 2π ωt Fig.11Three phase bridge inverter waveforms with 12 o conduction and resistive load. Mathematical Analysis: The line voltage in terms of phase voltage in a 3- phase system with phase sequence (a,b,c) are: V ab = V a V b V bc = V b V c V ca = V c V a 67

68 DC-AC Inverters Author Dr. Ali Hussein Numan Where, V ab,v bc,v ca line voltage V a,v b,v c phase voltage V ab V ca = 2V a (V b V c ) But in a balance three phase system, the sum of the three phase voltages is zero V a V b V c = V ab V ca = 3V a V a = (V ab V ca ) 3 Similarly, the (b) and (c) phase voltages are: V b = (V bc V ab ) 3 V c = (V ca V bc ) 3, From Fig. 11, the three phase output voltages can be described by Fourier series. V a (t) = 2V s cos nπ nπ 6 sin nn ωt π 6 n=1,3,5 V b (t) = 2V s cos nπ nπ 6 sin nn ωt π 2 n=1,3,5 For n=3, cos nπ 6 V c (t) = 2V s cos nπ nπ 6 sin nn ωt 5π 6 n=1,3,5 =, thus, all multiples of 3rd harmonics are cancelled. The three phase output voltage can be rewritten as: 68

69 DC-AC Inverters Author Dr. Ali Hussein Numan Note that cos π 6 = = 3 V a (t) = 3VV ss ππ ssssssssss 1 5 ssssss5ωωωω 1 7 ssssss7ωωωω V b (t) = 3VV ss ππ sin (ωωωω 2π 3 ) 1 5 ssssss5(ωωωω 2π 3 ) 1 7 ssssss7(ωωωω 2π 3 ) V c (t) = 3VV ss ππ sin (ωωωω 4π 3 ) 1 5 ssssss5(ωωωω 4π 3 ) 1 7 ssssss7(ωωωω 4π 3 ) Also, the three line output voltages can be described by Fourier series as follows: Where k=,1,2,3,.. V ab (t) = 3V s sin nn ωt π nπ 3 n=6k±1 V bc (t) = 3V s sin nn ωt π nπ 3 n=6k±1 V ca (t) = 3V s sin nn ωt π nπ 3 n=6k±1 Also, all multiples of 3 rd harmonics are cancelled.thus, V ab (t) = 3 ππ VV ss ssssssssss 1 5 ssssss5ωωωω 1 7 ssssss7ωωωω V bc (tt) = 3 ππ VV ss ssssss ωωωω 2π ssssss 5 ωωωω 2π ssssss 7 ωωωω 2π 3 V ca (tt) = 3 ππ VV ss ssssss ωωωω 4π ssssss5 ωωωω 4π ssssss7 ωωωω 4π 3 69

70 DC-AC Inverters Author Dr. Ali Hussein Numan The amplitude of the fundamental line voltage can be obtained as V ab 1 = 3 ππ VV ss Where, V s :DC input voltage to the inverter V ab1 : The amplitude of the fundamental line voltage The RMS value of fundamental line voltage is V ab 1(rms ) = 3 2ππ VV ss = 3 V a1(rms ) =.6755 VV ss The phase voltages are shifted from the line voltages by 3 V a = 3 ππ VV ss / 3 =.551VV ss The RMS value of fundamental phase voltage is V a1(rms ) = 3VV ss ππ 1 2 = V ab 1(rms ) 33 =.39 VV ss It is seen from the phase voltage waveform (V a ) in Fig.11 that the phase voltage is (V s /2) from (-12 o ). Therefore, the rms value of phase voltage (V a (rms)) is V a(rms ) = 1 2 ππ VV ss 2 dddd = ππ 3 VV ss 2 = VV ss 6 =.482VV ss Also, the rms value of line voltage (V ab (rms)) is V ab (rms ) = 33 V a(rms) = VV ss =.77VV ss 7

71 DC-AC Inverters Author Dr. Ali Hussein Numan Total Harmonics Distortion (THD) The THD of the line voltage is TTTTTT llllllll vvvvvvvvvvvvvv = (VV aaaa ) 2 (VV aaaa 1 ) 2 (VV aaaa 1 ) 2 = (.77)2 (.6755) 2 (.6755) 2 = 3.89 % The THD of the phase voltage is TTTTTT pphaaaaaa vvvvvvvvvvvvvv = (.482)2 (.39) 2 (.39) 2 = 3.89 % The square waveform output from the inverter supplied to load is not good enough. But, the sinusoidal load voltage is usually the most desirable. But how do we approximate a sinusoidal output with only three states (Vdc, Vdc, )? The answer: Unipolar PWM modulation 3) Single Phase Bridge Pulse Width Inverter (PWM) The PWM is very commonlyusedmethod to control the output voltage and frequency fromthe inverter through controlling the switching instantsof the inverter transistors (ON and OFF).The PWM technique produces lower order harmonic contents in comparison to other technique. However, there are two types of single phase bridge PWM inverter. a) Bipolar PWM In this method,a triangular carrier waveform of high frequency (V c ) is compared with a low frequency sinusoidal reference waveform (V r ) and the crossover points are 71

72 DC-AC Inverters Author Dr. Ali Hussein Numan used to determine switching instantsas shown in Fig.12. The dotted curve is the desired output; also the fundamental frequency. Carrier waveform (V c ) Reference waveform (V r ) wt Load voltage (V L ) Fundamental component of Load voltage (V L1 ) V S -V S wt ig 1 wt Firing pulses ig 2 ig 3 wt wt ig 4 wt Fig.12Bipolar switching scheme. V r > V c VV LL = VV ss V r < V c VV LL = VV ss v control < vtri TA on and VAN = control > vtri TB v on and V = V BN d 72

73 DC-AC Inverters Author Dr. Ali Hussein Numan b) Unipolar PWM In this method,triangular carrier waveform of high frequency (V c ) is compared with two low frequency sinusoidal reference waveforms (V ra and V rb ),and the crossover points are used to determine switching instants as shown in Fig.13.The phase shift between these two waveforms is (18 o ). Unipolar switching schemeproduces better harmonics. But it is moredifficult to implement. Load Voltage VL Fig.13Unipolar switching scheme. Modulation Index(MI):defined as the ration of the sinusoidal reference waveforms amplitude (V r ) to the carrier waveform amplitude (V c ). Amplitude of the sinusoidal reference waveform MMMM = Amplitude of the carrier waveform MI is related to the fundamental (sinusoidalwaveform)output voltagemagnitude. If M is high, thenthe sinusoidalwaveform output is high and vice versa. If MI 1 73

74 DC-AC Inverters Author Dr. Ali Hussein Numan VV 11 = MMMM VV iiii Where,V 1 and V in are fundamental of the outputvoltage and input (DC) voltage,respectively. Fig.14a showsa maximum output, a reduction to half this value being made by simply reducing the reference sinusoidal waveform to half value as shown in Fig.14b. Fig.14c shows how a reduction in frequency of the reference sinusoidal waveform increases the number of pulses within each half cycle. Carrier waveform Reference waveform Fundamental component of output waveform (a) V s (b) (c) 74

75 DC-AC Inverters Author Dr. Ali Hussein Numan Fig.14 Switching instants forpwm waveform (a) At maximum output voltage (b) At half maximum output voltage and (c) At half voltage and half frequency. 4) Three Phase Bridge Inverter (PWM Inverter) The circuit of three phase bridge PWM inverter is shown in Fig.15. In this type, the PWM waveform generation is identical to the six step inverter, but the switching sequence is more complex. L T 1 T 3 T 5 D 1 D 3 D 5 i g1 D 1 D 3 D 5 i g3 i g5 Va 3φ Vb 5Hz Vc C a b c T 4 T 6 T 2 D 4 D 6 D 2 i g4 D 4 D 6 D 2 i g6 i g2 ia ib ib R va R vb R vc vab Fig.15 Three phase bridge PWM inverter. vca vbc The operation of PWM inverter is depending on the technique that may be used to generate the firing signal waveforms. Conventionally, these waveforms have been produced by comparing a triangular carrier waveformof high frequency (V c ) with a low frequency sinusoidal reference waveform(v r ). 3-1 Classification of PWM The PWM can be broadly classified into the following types: 1) Natural (sinusoidal) sampling 75

76 DC-AC Inverters Author Dr. Ali Hussein Numan Problems with analogue circuitry, e.g. Drift,sensitivity etc. 2) Regular sampling Simplified digital version of natural sampling appropriate for digital hardware or microprocessor implementation. Widely used in industry. 3) Optimum PWM PWM waveform is constructed based oncertain performance criteria, e.g. THD. 4) Harmonic elimination/minimisation PWM PWM waveforms are constructed to eliminatesome undesirable harmonics from the outputwaveform spectra. Highly mathematical in nature. 5) Space-vector modulation (SVM) Advanced computationally intensive technique that offers superior performance in variablespeed drives. Used to minimize harmonic content of the three-phase isolated neutral load. 1) Three Phase sinusoidal PWM In this method, a triangular carrier waveform of high frequency (V c ) is compared with a three low frequency sinusoidal reference waveforms (V ra,v rb,v rc ) to determine the switching instants of each transistor. Principle of sinusoidal PWM for three phase bridge inverteris shown Fig

77 DC-AC Inverters Author Dr. Ali Hussein Numan Reference sinewave phase A V ra Reference sinewave phaseb Reference sinewave phasec V rb V rc Triangular carrier waveform V c Fig.16Principle of sinusoidal PWM for three phase bridge inverter. 77