A presentation on inverter by manoj

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1. Submitted to :- submitted by :- (Coordinator) manoj Kumar swami Ujjwal kalla A presentation on INVERTER

2. Introduction Inverter is a device which convert a DC input supply voltage into symmetric AC voltage of desired magnitude and frequency at the output side. It is also know as DC-AC converter. Ideal and practical inverter have sinusoidal and no-sinusoidal waveforms at output respectively. If the input dc is a voltage source, the inverter is called a Voltage Source Inverter (VSI). One can similarly think of a Current Source Inverter (CSI), where the input to the circuit is a current source. The VSI circuit has direct control over ‘output (ac) voltage’ whereas the CSI directly controls ‘output (ac) current.

3. Applications of inverter For low and medium power applications, square- wave or quasi-square wave voltages may be acceptable For high –power applications ,low distorted sinusoidal waveforms are required. Using high speed power semiconductor devices the harmonic contents at output can be reduced by PWM techniques. Industrial applications:- variable ac motor, induction heating , standby power supply ,UPS (uninterrupted power supply) Inputs are (battery, fuel cell, solar cell, other dc sources)

4. Classifications of inverter Inverter can be mainly classified into two types- Single-phase inverter Three-phase inverter Turn-ON and turn-OFF controlling devices are- Bipolar junction transistor[BJTs] Metal oxide semiconductor field-effect transistor[MOSFETs] Insulated-gate bipolar transistor[IGBTs] Gate turn-OFF thyristor[GTOs]

5. Classifications of inverter ( Cant’d) Voltage fed inverter[for constant input voltage] Current fed inverter[for constant input current] Variable DC-link inverter[for controllable input voltage] Resonant-pulse inverter :- If the output voltage or current of inverter is forced to pass through zero by creating an LC resonant circuit.

6. Single phase inverter:- Principle of operation: -The inverter circuit consists of two choppers. when only transistor Q 1 is turned ON for a time T /2 then instantaneous voltage across the load is (V =V s /2). If the transistor Q ­2 only is turned ON for a time T /2,-V S /2 appears across the load. This type of inverter is called Half bridge inverter. Root-mean square (rms) output voltage

7. Single phase inverter:- ( Cant’d)

8. Single phase inverter ( Cant’d) Instantaneous output voltage in the form of Fourier series is- Due to quarter wave symmetry along the X-axis, both a and a n are zero (or even Harmonic voltages are absent ). then b n is Output voltage- v = 0 for n=2,4…..

9. Single phase inverter ( Cant’d) For an inductive load ,the load current can not change immediately with the output voltage. diode D 1 and D 2 are known as feedback diodes .Transistor can be replaced by any switching device. There must be a minimum delay time between the outgoing device and triggering of the next incoming device, otherwise there would occur a short-circuit between devices. Maximum conduction time of a device would be

10. Single phase inverter ( Cant’d) For an RL load instantaneous load current i is Note-in most applications the output power due to the fundamental current is generally the useful power ,and the power due to harmonic current is dissipated as heat and increase the load temperature.

11. EXAMPALE - 1. A single phase half bridge inverter has a resistive load of R=2.4  and the DC input voltage is 48V,Determine- Rms value of output voltage at the fundamental frequency V 01 output power Average and peak current of each transisto The peak reverse blocking voltage VBR. The THD The DF and The HF and DF of the LOH . Solution:- V s =48V, R=2.4  .    V 01 =

12. The output power is The peak transistor current is I p =24/2.4=10A. because each transistor conducts for a 50% duty cycle ,the average current of each transistor is I Q =0.5*10=5A. The peak reverse blocking voltage V BR =2*24=48V.   V 01 = And the RMS harmonic voltage So THD = The LOH is the third ,V 03 = V 01 /3 HF 3 =V 03 /V 01 = 1/3=33.33% and DF 3 =(V 03 /3 2 )V 01 =1/27=3.704%. Because V 03 /V­ 01 =33.33% which is greater than 3% , LOH=V 03 .

13. Single phase Bridge inverter It consists of four choppers. Full bridge converter is also basic circuit to convert dc to ac. An ac output is synthesized from a dc input by closing and opening switches in an appropriate sequence. There are also four different states depending on which switches are closed. When transistors Q 1 and Q 2 are turned on simultaneously, the input voltage appears across the load. If transistor and are turned on at the same time, the voltage across the is reversed and is –V s . Transistor Q 1 and Q 2 acts as switches S 1 and S 2. , respectively.

14. Single phase Bridge inverter(cont’d) The fundamental RMS output voltage obtained from RMS output voltage is Fourier series of output voltage = 0 for n=2,4…. For an RL load instantaneous load current i is

15. Single phase Bridge inverter(cont’d)

16. Single phase Bridge inverter(cont’d) The output load voltage alternates between + Vs  when Q1 and Q2 are on and - Vs  when Q3 and Q4 are on, irrespective of the direction of current flow. It is assumed that the load current does not become discontinuous at any time. In the following analysis we assume that the load current does not become discontinuous at any time, same as for the half-bridge circuit. Bridge inverters are preferred over other arrangements in higher power ratings. With the same dc input voltage, output voltage is twice that of the half-bridge inverter.

17. Three phase inverter :- (VSI) Three - phase inverters  are used for variable-frequency drive applications and for high power applications such as HVDC power transmission. A basic three-phase inverter consists of three single-phase inverter switches each connected to one of the three load terminals. Gating signal of 1-phase inverter should be delayed by 120 with respect to each other to obtain 3-phase balanced voltages. Two types of control signals can be applied to the transistors: 180 conduction and 120 conduction.

18. EXAMPLE -2. Q.2 A single phase full bridge inverter has a resistive load of R=2.4  and the DC input voltage is 48V,Determine- Rms value of output voltage at the fundamental frequency V 01 output power Average and peak current of each transistor The peak reverse blocking voltage VBR. The THD The DF and The HF and DF of the LOH . Solution:- V s =48V, R=2.4  . V 1 = V 1 =

19. The output power is The peak transistor current is I p =48/2.4=20A. because each transistor conducts for a 50% duty cycle ,the average current of each transistor is I Q =0.5*20=10A. The peak reverse blocking voltage V BR =48V. V 1 = And the RMS harmonic voltage So THD = The LOH is the third ,V 3 = V 1 /3 HF 3 =V 3 /V 1 = 1/3=33.33% and DF 3 =(V 3 /3 2 )V 1 =1/27=3.704%. .

20. 180 Conduction In this type of conduction each transistor conducts for 180 .3-transistor remain on at any instant of time. When transistor Q 1 is switched ON ,terminal a is connected to + ve terminal of supply and if Q 4 is switched ON then terminal a is connected to – ve terminal of supply. There are 6-mode of operation in a cycle and the duration of each mode is 60 . The load may be connected in star or delta form. Inverter switches of any leg can not be switched ON and OFF simultaneously.

21. 180 Conduction(cont’d) Circuit diagram Figure: Three Phase Inverter circuit diagram

22. 180 Conduction(cont’d) Circuit diagram and waveforms

23. 180 Conduction(cont’d) Transistor Q 1 ,Q 6 act as the switching devices S 1, S 6 respectively. For states 1to 6 output voltage are non-zero and states7 to 8 produce zero line voltage.line currents freewheel through freewheeling diodes. A modulating technique is used to select the valid states only.

24. 3-mode of operation of inverter 180 Conduction(cont’d) Mode-1 .For transistors Q 1, Q 5, and Q 6 conducts. Mode-2- For transistors Q 1, Q 2, and Q 6 conducts. Mode-3- For transistors Q 1, Q 2, and Q 3 conducts.

25. 180 Conduction(cont’d) In the form of Fourier series :-there is a and a n are absent so Instantaneous L-L voltages are given as:- Triple harmonics (n=1,3,5…) are absent in L-L voltage. RMS L-L voltage is

26. 180 Conduction(cont’d) Fundamental line voltage is RMS L-N voltage is The diode across the transistors have no function with R-load . For an inductive load ,current of each arm would be delayed to its voltage. The transistors must be continuously gated ,because the conduction time of transistor and diode depends on load power factor.

27. 120 Conduction In this conduction ,each transistor conducts for 120 . Two transistor remain ON at any instant of time. Conduction sequence of transistors is as 61,12,23,34,45,56,61.

28. 3-mode of operation of inverter 120 Conduction(cont’d) Mode-1 .For transistors 1 and 6 conducts. Mode-2- For transistors 1 and 2 conducts. Mode-3- For transistors 2 and 3 conducts.

29. 120 Conduction(cont’d) L-N ,phase voltages are as Instantaneous L-L voltages are given as:- There is delay of 30 between turning ON and OFF Q 1 and turning of Q 4. At any time ,two terminals are connected to DC supply and third one remains open. Potential of open terminal depends on load characteristics.

30. Why 180 conduction is preferred compared to 120 ? In 120 conduction ,one transistor conducts for 120 ,the transistors are less utilized as compared with that of 180 conduction for the same load condition.

31. EXAMPLE-3. A three-phase bridge inverter delivers power to a resistive load from a 450V dc source. For a star-connected load of 10  per phase,determine for both (a)180 mode and (b) 120 mode, RMS value of load current RMS value of thyristor current Load power . Solution :- For a resistive load ,the waveform of load current is same as that of the applied voltage . 180 Mode :- Rms value of per-phase load current is   Rms value of thyristor current is   Power delivered to load is

32. 120 Mode :- Rms value of per-phase load current is Rms value of thyristor current is   Power delivered to load is

33. Voltage control of Inverter VCI For 1-phase inverter To control of output voltage of inverter is often necessary . To cope with the variation of DC input voltage. To regulate voltage of inverter . To satisfy constant voltage and frequency control requirement.

34. Control techniques are Single –pulse width modulation. Multiple pulse width modulation. Sinusoidal pulse width modulation. Modified sinusoidal PWM.

35. Single –pulse width modulation Single-PWM In single PWM control ,there is only one pulse per half cycle and width of pulse is varied to control inverter output voltage . The gating signals are generated by comparing rectangular reference signal of amplitude A r with a triangular carrier wave of amplitude A c . AM index : The ratio of A r to A c is the control variable ,called AM index. M= A r /A c

36. Single –pulse width modulation(cont’d) RMS output voltage is Fourier series output voltage is Time and angle of intersection is Pulse width d (or pulse angle )as Where T S = T/2

37. Single –PWM Waveform

38. Multiple pulse width modulation (M-PWM) The harmonic contents can be reduced by using several pules in each half cycle of output voltage . The frequency of reference signal sets the output frequency f , and carrier frequency determine the number of pulses per half cycle p. The modulation index control the output voltage. Number of pulses per half-cycle is P= f c /2f =m f /2 Where m f = f c /f = frequency modulation ratio.

39. M-PWM(cont’d)

40. M-PWM(cont’d) Instantaneous output voltage is V = V S (g 1 -g 4 ). RMS output voltage is The variation of M from 0 to 1 varies the pulse width d from 0 to T/2p(0 to  /p) and the rms output voltage v from 0 to v s . a , a n and even harmonics are absent . so b n =

41. M-PWM(cont’d) The coefficient B n is Time and angle of intersection is for m=1,3,….2p for m=2,4….2p Pulse width d (or pulse angle )as

42. Sinusoidal pulse width modulation (Sinusoidal -PWM) In this type of modulation the width of each pulse is varied in proportion to the amplitude of a sine wave evaluated at the center of the same pulse. The DF and LOH are reduced significantly. Comparing the bidirectional carrier signal v cr with two sinusoidal reference signals v r and - v r produces gating signals g 1 and g 4 ,respectively . The output voltage is V =V s (g 1 -g 4 ) g 1 and g 4 can not be released at the same time.

43. Sinusoidal –PWM Wave-form

44. Sinusoidal –PWM(cont’d) The number of pulses per half cycle depends on the carrier frequency. Gating signals can also be generated by using unidirectional triangular carrier wave. RMS output voltage is The coefficient B n is Time and angle of intersection is for m=1,3,….2p for m=2,4….2p Pulse width d m (or pulse angle )as

45. Sinusoidal –PWM(cont’d) The output voltage of an inverter contain harmonics .the PWM pushes the harmonics into a high –frequency range around the switching frequency f c and its multiples ,that is, around harmonics m f ,2m f ,3m f and so on. The frequency at which voltage harmonics occur can f n = peak fundamental output voltage for PWM and SPWM control is V m1 = dv s for For d=1 ,V m1(max) = V s . The operation beyond d=1.0 is called over-modulation. Over-modulation is normally avoided in UPSs

46. Modified (Sinusoidal -PWM) In this PWM the widths of pulses nearer the peak of the sine wave do not change significantly with variation of modulation index. In this PWM the carrier wave is applied during ther first and last 60 intervals per half cycle. The fundamental component is increased and its harmonic characteristics are improved.reduces switching losses .

47. Modified Sinusoidal –PWM(cont’d) Time and angle of intersection is for m=1,2,3,….p for m=1,3,….p for m=2,4….p Intersection during the last 60 interval is for m =p,p+1…..,2p-1 Pulse width d m (or pulse angle )as Instantaneous output voltage is V = V S (g 1 -g 4 ).

48. Modified Sinusoidal –PWM(cont’d)

49. Voltage control of 3-phase inverter The voltage control techniques are Sinusoidal PWM Third –harmonic PWM 60 PWM Space vector modulation.

50. Harmonic reduction The output voltage control of inverter requires varying both the number of pulses per half –cycle and the pulse widths that are generated by modulating techniques.the output voltage contain even harmonics over a frequency spectrum. Some harmonic reduction techniques are as:- Phase displacement. Bipolar output voltage notches. Unipolar output voltage notches. Transformer connections .

51. Phase displacement In this technique the n th harmonic can be eliminated by a proper choice of displacement angle  if Cos(n  ) = 0 Or  = 90 /n And third harmonic is eliminated if  =90 /3=30 .

52. Bipolar output voltage notches A pair of unwanted harmonics at the output of single –phase inverter can be eliminated by introducing a pair of symmetrically placed bipolar voltage notches . the Fourier series output voltage is B n =

53. Bipolar output voltage notches wave-form

54. Unipolar output voltage notches Similarly to bipolar notches symmetrical unipolar notches can also be introduced. ∞  v o ( t ) = ∑ B n sin ( nωt )   n =1,3,5,.  

55. Transformer connections The output of two or more inverters may be connected in series through a transformer to reduce certain unwanted harmonics. The second inverter is phase shifted by 60 . So effective output of inverter is reduced by 13.4%.

56. Transformer connections circuit diagram and wave-form

57. Transformer connections(cont’d) v o 1 ( t ) = A 1 sin ω t + A 3 sin 3 ω t + A 5 sin 5 ωt + Total output voltage is :-

58. Current source inverter (CSI) In the CSI, the current is nearly constant. The voltage changes here, as the load is changed. In an Induction motor, the developed torque changes with the change in the load torque, the speed being constant, with no acceleration/deceleration. The input current in the motor also changes, with the input voltage being constant. So, the CSI, where current, but not the voltage, is the main point of interest, is used to drive such motors, with the load torque changing

59. Single-phase Current Source Inverter The type of operation is termed as Auto-Sequential Commutated Inverter (ASCI). A constant current source is assumed here, which may be realized by using an inductance of suitable value, which must be high, in series with the current limited dc voltage source. The thyristor pairs, Th 1 & Th 3 , and Th 2 & Th 4 , are alternatively turned ON to obtain a nearly square wave current waveform. Two commutating capacitors − C 1 in the upper half, and C 2 in the lower half, are used. Four diodes, D 1 –D 4 are connected in series with each thyristor to prevent the commutating capacitors from discharging into the load. The output frequency of the inverter is controlled in the usual way, i.e., by varying the half time period, (T/2).

60. Single-phase CSI Circuit diagram

61. Single-phase CSI Wave-form

62. Operation of Single-phase CSI There are Two mode of operation:- Mode1:- The Starting from the instant, t = 0 − , the thyristor pair, Th 2 & Th 4 , is conducting (ON), and the current (I) flows through the path, Th 2 , D 2 , load (L), D 4 , Th 4 , and source, I. The commutating capacitors are initially charged equally with the polarity as given, i.e., v C 1 = v C 2 = − V C . At time, t = 0, thyristor pair, Th 1 & Th 3 , is triggered by pulses at the gates. The conducting thyristor pair, Th 2 & Th 4 , is turned OFF by application of reverse capacitor voltages. The voltage, v D 1 is obtained by going through the closed path, abcda as v D 1 + V co − (1 /( C / 2))⋅ ∫ I ⋅ dt = 0 It may be noted the voltage across load inductance, L is zero.

63. Operation of Single-phase CSI(cont’d) The value of v C is v C 1 = v C 2 = v C = − V co + (2 / C ) ⋅ ∫ I ⋅ dt , which, if computed at t = t 1 , v C 1 = v C 2 = v C ( t 1 ) = − V co + (( 2 ⋅ I ⋅ t 1 ) / C ) = − V co + ((2 ⋅ I ) / C )⋅ ( C /(2 ⋅ I ))⋅ V C = 0

64. Operation of Single-phase CSI(cont’d) Mode2:- Diodes, D 2 & D 4 , are already conducting, but at t = t 1 , diodes, D 1 & D 3 , get forward biased, and start conducting. Thus, at the end of time t 1 , all four diodes, D 1 –D 4 conduct. As a result, the commutating capacitors now get connected in parallel with the load (L). natural frequency:- f = 1/((2 ⋅ π ) ⋅( L ⋅ C ) ), ω = (2 ⋅ π ) ⋅ f = 1/( L ⋅ C ) time period:- T = 1/ f = (2 ⋅ π ) / ω = (2 ⋅ π ) ⋅ ( L ⋅ C )

65. Operation of Single-phase CSI(cont’d) The voltage across capacitor is v C = v L = L ⋅ ( di / dt ) = (( 2 ⋅ I ) ⋅ ( ω ⋅ L ))⋅ sin ( ω ⋅ t ) The total commutation interval is, t c = t 1 + t 2 = (1+ ( π / 2))/ ω = (1 + ( π / 2))⋅ ( L ⋅ C ) . The procedure remains nearly same, if the load consists of resistance, R only. The procedure in mode I, is same, but in mode II, the load resistance, R is connected in parallel with the two commutating capacitors. The direction of the current, I remains same, a part of which flows in the two capacitors, charging them in the reverse direction

66. Three-phase Current Source Inverter In this circuit, six thyristors, two in each of three arms, are used, as in a three-phase VSI. Also, six diodes, each one in series with the respective thyristor, are needed here, as used for single-phase CSI. Six capacitors, three each in two (top and bottom) halves, are used for commutation. It may be noted that six capacitors are equal, i.e. C 1 = C 2 = = C 6 = C . The numbering scheme for the thyristors and diodes are same, as used in a three-phase VSI.

67. Three-phase Current Source Inverter circuit diagram

68. Three-phase Current Source Inverter Wave-form

69. Three-phase CSI(cont’d) There are two mode of operation :- Mode:- The commutation process starts, when the thyristor, Th 3 in the top half, is triggered, i.e. pulse is fed at its gate. Immediately after this, the conducting thyristor, Th 1 turns off by the application of reverse voltage of the equivalent capacitor. It may be noted the equivalent capacitor is the parallel combination of the capacitor, C 1 and the other part, being the series combination of the capacitors, C 3 & C 5 ( C ′ = C / 2 ). value is C eq = C / 3 , When the voltage across the capacitor, C 1 (and also the other two) decreases to zero, the mode I ends.

70. Three-phase CSI(cont’d) After the end of mode I, the voltage across the diode, D 3 goes positive, as the voltage across the equivalent capacitor goes negative, assuming that initially (start of mode I) the voltage was positive. when the mode II ends. The diode, D 1 turns off, as the current goes to zero. So, at the end of mode II, the thyristor, Th 3 & the diode, D 3 conduct. This is needed to turn off the outgoing (conducting) thyristor, Th 3 , when the incoming thyristor, Th 5 is triggered. The complete commutation process as described will be repeated. The diodes in the circuit prevent the voltage across the capacitors discharging through the load.

71. Variable DC Link Inverter Varying the modulation index (or pulse width) and maintaining the dc input voltage constant has shown that a range of harmonics would be present on the output voltage. The pulse width can be fixed to eliminate or reduce certain harmonics and the output voltage can be controlled by varying the level of the dc input voltage.

72. Variable DC Link Inverter (Cont’d) Circuit diagram

73. Advantages of CSI The circuit for CSI, using only converter grade thyristor, which should have reverse blocking capability, and also able to withstand high voltage spikes during commutation, is simple.   An output short circuit or simultaneous conduction in an inverter arm is controlled by the ‘controlled current source’ used here, i.e., a current limited voltage source in series with a large inductance. The converter-inverter combined configuration has inherent four-quadrant operation capability without any extra power component.

74. disadvantages of CSI A minimum load at the output is required, and the commutation capability is dependant upon load current. This limits the operating frequency, and also puts a limitation on its use for UPS systems.   At light loads, and high frequency, these inverters have sluggish performance and stability problems.

75. References M. Kojima, K. Hirabayashi, Y. Kawabata, E. C. Ejiogu and T. Kawabata, “Novel Vector Control System Using Deadbeat Controlled PWM Inverter With Output LC Filter”, IEEE Transactions on Industrial Applications, Vol. 40, No. 1, January/February 2004, pp. 132-169. J. Nazarzadeh, M. Razzaghi, and K. Y. Nikravesh, “Harmonic Elimination in Pulse-Width Modulated Inverters using Piecewise Constant Orthogonal Functions”, Electric Power Systems Research, Vol. 40, 1997, pp. 45-49. Coker, J.O. (2004). Solar energy and its applications in Nigeria. Global Journal of Pure and Applied Sciences 10: 223 – 225. Duncan, T. (1997). Electronics for today and tomorrow 2nd Edition Hodder Education