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PWM Harmonics
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Chapter 4 Inverters.pdf
- 1. CHAPTER 4 INVERTERS: Converting DC to AC
- 2. CONTENTS ۩ Introduction ۩ Basic Principles of Inverter ۩ Single-phase Half-Bridge Square-Wave Inverter ۩ Single-phase Full-Bridge Square-Wave Inverter ۩ Quasi Inverter ۩ Three-phase inverter ۩ Fourier Series and Harmonics Analysis ۩ Pulse-Width Modulation (PWM)
- 3. Definition: Converts DC to AC power by switching the DC input voltage in a pre-determined sequence so as to generate AC voltage. INTRODUCTION Applications: Induction motor drives, traction, standby power supplies, and uninterruptible ac power supplies (UPS).
- 4. INTRODUCTION V V + + Vdc Vac - - General block diagram
- 5. Three types of inverter: INTRODUCTION (a) Voltage source inverter (VSI) (b) Current source inverter (CSI)
- 6. Three types of inverter: (cont.) INTRODUCTION (c) Current regulated inverter
- 7. BASIC PRINCIPLES The schematic of single-phase full-bridge square-wave inverter circuit
- 9. Single-phase Half-bridge Square-wave Inverter V0 1 . V + - 2 V0 G. VDC The basic single-phase half-bridge inverter circuit
- 10. The total RMS value of the load output voltage, The instantaneous output voltage is: (refer to slide 30/pp:88) Single-phase Half-bridge Square-wave Inverter 2 2 2 2 / 0 2 DC T DC O V dt V T V = = ∫ The fundamental rms output voltage (n=1)is 2,4,.... n for 0 sin 2 ,... 5 , 3 , 1 = = = ∑ = n DC O t n n V v ω π DC DC O V V V 45 . 0 2 1 2 1 = = π
- 11. In the case of RL load, the instantaneous load current io , where Single-phase Half-bridge Square-wave Inverter ∑ ∞ = − + = ,.. 5 , 3 , 1 2 2 ) sin( ) ( 2 n n DC o t n L n R n V i θ ω ω π ) / ( tan 1 R L nω θ − = where The fundamental output power is ) / ( tan 1 R L n n ω θ − = + = = = 2 2 2 1 1 1 1 1 ) ( 2 2 cos L R V R I I V P DC o o o o ω π θ
- 12. The total harmonic distortion (THD), Single-phase Half-bridge Square-wave Inverter = ∑ ∞ 2 1 V THD = ∑ = ,.... 7 , 5 , 3 2 1 1 n n O V V THD ( ) 2 1 2 1 1 o o O V V V THD − =
- 13. Example 3.1 The single-phase half-bridge inverter has a resistive load of R = 2.4Ω and the DC input voltage is 48V. Determine: (a) the rms output voltage at the fundamental frequency (b) the output power Single-phase Half-bridge Square-wave Inverter (c) the average and peak current of each transistor. (d) the THD
- 14. Solution VDC = 48V and R = 2.4Ω (a) The fundamental rms output voltage, Vo1 = 0.45VDC = 0.45x48 = 21.6V Single-phase Half-bridge Square-wave Inverter (b) For single-phase half-bridge inverter, the output voltage Vo = VDC/2 Thus, the output power, W R V P o o 240 4 . 2 ) 2 / 48 ( / 2 2 = = =
- 15. Solution (c) The transistor current Ip = 24/2.4 = 10 A Because each of the transistor conducts for a 50% duty cycle, the average current of each transistor is IQ = 10/2 = 5 A. Single-phase Half-bridge Square-wave Inverter (d) 2 1 2 1 1 o o O V V V THD − = ( ) % 34 . 48 45 . 0 2 45 . 0 1 2 2 = − = DC DC DC V V xV
- 16. The switching in the second leg is delayed by 180 degrees from the first leg. The maximum output voltage of this inverter is twice that of half-bridge inverter. Single-phase Full-bridge Square-wave Inverter
- 17. The output RMS voltage And the instantaneous output voltage in a Fourier series is Single-phase Full-bridge Square-wave Inverter DC DC O V dt V T V = = ∫ 2 2 The fundamental RMS output voltage In the case of RL load, the instantaneous load current ∑ = = ,... 5 , 3 , 1 sin 4 n DC O t n n V v ω π DC DC V V V 9 . 0 2 4 1 = = π ( ) ( ) n n DC o t n L n R n V i θ ω ω π − + = ∑ ∞ = sin 4 ,... 5 , 3 , 1 2 2
- 18. Example 3.2 A single-phase full-bridge inverter with VDC = 230 and consist of RLC in series. If R = 1.2Ω, ωL = 8 Ω and 1/ωC = 7 Ω, find: (a) The amplitude of fundamental rms output current, Single-phase Full-bridge Square-wave Inverter (a) The amplitude of fundamental rms output current, io1 (b) The fundamental component of output current in function of time. (c) The power delivered to the load due to the fundamental component.
- 19. Example 3.3 A single-phase full-bridge inverter has an RLC load with R = 10Ω, L = 31.5mH and C = 112µF. The inverter frequency is 60Hz and the DC input voltage is 220V. Determine: (a) Express the instantaneous load current in Fourier series. (b) Calculate the rms load current at the fundamental Single-phase Full-bridge Square-wave Inverter (b) Calculate the rms load current at the fundamental frequency. (c) the THD of load current (d) Power absorbed by the load and fundamental power. (e) The average DC supply current and (f) the rms and peak supply current of each transistor
- 21. Viewed as extensions of the single-phase bridge circuit. The switching signals for each switches of an inverter leg are displaced or delayed by 120o. With 120o conduction, the switching pattern is T6T1 – T1T2 – T2T3 – T3T4 – T4T5 – T5T6 – T6T1 for the positive A-B-C sequence. When an upper switch in an inverter leg connected with the Three-Phase Inverter When an upper switch in an inverter leg connected with the positive DC rail is turned ON, the output terminal of the leg (phase voltage) goes to potential +VDC/2. When a lower switch in an inverter leg connected with the negative DC rail is turned ON, the output terminal of that leg (phase voltage) goes to potential -VDC/2.
- 23. The line-to-neutral voltage can be expressed in Fourier series Three-Phase Inverter ∑ ∑ ∞ = ∞ = − = + = ,.. 5 , 3 , 1 ,.. 5 , 3 , 1 2 sin 3 sin 2 6 sin 3 sin 2 n dc bn n dc an t n n n V v t n n n V v π ω π π π ω π π The line voltage is vab = √3van with phase advance of 30o ∑ ∞ = − = ,.. 5 , 3 , 1 6 7 sin 3 sin 2 n dc cn t n n n V v π ω π π ( ) ∑ ∑ ∑ ∞ = ∞ = ∞ = − = − = + = ,.. 5 , 3 , 1 ,.. 5 , 3 , 1 ,.. 5 , 3 , 1 sin 3 sin 3 2 3 sin 3 sin 3 2 3 sin 3 sin 3 2 n dc ca n dc bc n dc ab t n n n V v t n n n V v t n n n V v π ω π π π ω π π π ω π π
- 24. Fourier series is a tool to analyze the wave shapes of the output voltage and current in terms of Fourier series. Fourier Series and Harmonics Analysis ( ) ∫ = π θ π 2 0 1 d v f ao Inverse Fourier ( ) ( ) ∑ ∞ + + = sin cos 1 n bn n a a v f θ θ π 0 ( ) ( ) ∫ = π θ θ π 2 0 cos 1 d n v f an ( ) ( ) ∫ = π θ θ π 2 0 sin 1 d n v f bn ( ) ( ) ∑ = + + = 1 0 sin cos 2 1 n n n bn n a a v f θ θ Where t ω θ =
- 25. If no DC component in the output, the output voltage and current are Fourier Series and Harmonics Analysis ( ) ∑ ∞ = + = 1 sin ) ( n n n o t n V t v θ ω ( ) ∑ ∞ + = sin ) ( t n I t i φ ω The rms current of the load can be determined by ( ) ∑ = + = 1 sin ) ( n n n o t n I t i φ ω 2 1 1 2 , 2 ∑ ∑ ∞ = ∞ = = = n n n rms n rms I I I Where n n n Z V I =
- 26. The total power absorbed in the load resistor can be determined by Fourier Series and Harmonics Analysis ∑ ∑ ∞ = ∞ = = = 1 2 , 1 n rms n n n R I P P = = 1 1 n n
- 27. Since the objective of the inverter is to use a DC voltage source to supply a load that requiring AC voltage, hence the quality of the non-sinusoidal AC output voltage or current can be expressed in terms of THD. Total Harmonics Distortion current can be expressed in terms of THD. The harmonics is considered to ensure that the quality of the waveform must match to the utility supply which means of power quality issues. This is due to the harmonics may cause degradation of the equipments and needs to be de-rated.
- 29. The THD of the load voltage is expressed as, Total Harmonics Distortion rms rms rms rms n rms n v V V V V V THD , 1 2 , 1 2 , 1 2 2 , ) ( − = = ∑ ∞ = The current THD can be obtained by replacing the harmonic voltage with harmonic current, rms n rms n i I I THD , 1 2 2 , ) ( ∑ ∞ = =
- 30. Harmonics of Square- wave Waveform DC V 0 1 0 2 0 = − + = ∫ ∫ π π π θ π DC DC V d V a 0 ) cos( ) cos( 2 = − = ∫ ∫ π π θ θ θ θ π d n d n V a DC n π 2 π t ω θ = DC V − 0 ) cos( ) cos( 0 = − = ∫ ∫ π θ θ θ θ π d n d n an [ ] [ ] ( ) [ ] π π π π π π π θ θ π θ θ θ θ π π π π π π π n n n n n n V n n n V d n d n V b DC DC DC n cos 1 2 ) cos 2 (cos ) cos 0 (cos ) cos( ) cos( ) sin( ) sin( 2 0 0 2 − = − + − = + − = − = ∫ ∫
- 31. When the harmonics number, n of a waveform is even number, the resultant of Therefore, Harmonics of Square-wave Waveform 1 cos = π n 0 = b Therefore, When n is odd number, Hence, 0 = n b 1 cos − = π n π n V b DC n 4 =
- 32. Spectrum of Square- wave • Harmonic decreases with a factor of (1/n). • Even harmonics are absent • Nearest harmonics is the (0.33) (0.2) (0.14) (0.11) (0.09) • Nearest harmonics is the 3rd. If fundamental is 50Hz, then nearest harmonic is 150Hz. • Due to the small separation between the fundamental and 3rd harmonics, output low- pass filter design can be very difficult.
- 33. Quasi-square wave an = 0, due to half-wave symmetry
- 34. Quasi-square wave Therefore, If n is even, bn = 0; If n is odd, α π cos 4 n V b dc n =
- 35. Example 3.4 The full-bridge inverter with DC input voltage of 100V, load resistor and inductor of 10Ω and 25mH respectively and operated at 60 Hz frequency. Determine: Determine: (a) The amplitude of the Fourier series terms for the square-wave load voltage. (b) The amplitude of the Fourier series terms for load current. (c) Power absorbed by the load. (d) The THD of the load voltage and load current for square-wave inverter.
- 36. Amplitude and Harmonics Control π π 2 t ω α α α α The output voltage of the full- bridge inverter can be controlled by adjusting the interval of on each side of the pulse as zero .
- 37. The rms value of the voltage waveform is The Fourier series of the waveform is expressed as Amplitude and Harmonics Control π α ω π α π α 2 1 ) ( 1 2 − = = ∫ − DC DC rms V t d V V The amplitude of half-wave symmetry is ∑ = odd n n O t n V t v , ) sin( ) ( ω ) cos( 4 ) ( ) sin( 2 α π ω ω π α π α n n V t d t n V V DC DC n = = ∫ −
- 38. The amplitude of the fundamental frequency is controllable by adjusting the angle of α. Amplitude and Harmonics Control α π cos 4 1 = DC V V The nth harmonic can be eliminated by proper choice of displacement angle α if α π cos 1 = V 0 cos = α n OR n o 90 = α
- 39. Pulse-width modulation provides a way to decrease the total harmonics distortion (THD). Types of PWM scheme Natural or sinusoidal sampling Pulse-Width Modulation (PWM) Natural or sinusoidal sampling Regular sampling Optimize PWM Harmonic elimination/minimization PWM SVM
- 40. Several definition in PWM (i) Amplitude Modulation, Ma Pulse-Width Modulation (PWM) tri , sin , carrier , , m e m m reference m a V V V V M = = If Ma ≤ 1, the amplitude of the fundamental frequency of the output voltage, V1 is linearly proportional to Ma. (ii) Frequency Modulation, Mf tri , carrier , m m e tri carrier f f f f f M sin reference = =
- 41. Bipolar Switching of PWM Pulse-Width Modulation (PWM) vtri (carrier) vsine (reference) VDC -VDC (a) (b) S1 and S2 ON when Vsine Vtri S3 and S4 ON when Vsine Vtri
- 42. Sinusoidal PWM Generator -Bipolar Pulse-Width Modulation (PWM) G1, G2 G3, G4 G3, G4
- 43. Unipolar Switching of PWM Pulse-Width Modulation (PWM) vtri (carrier) vsine (reference) (a) (b) (c) S4 S2 Vo Vdc 0 Vdc 0 S1 is ON when Vsine Vtri S2 is ON when –Vsine Vtri S3 is ON when –Vsine Vtri S4 is ON when Vsine Vtri
- 45. ■ Advantages of PWM switching - provides a way to decrease the THD of load current. - the amplitude of the o/p voltage can be controlled with the modulating waveform. - reduced filter requirements to decrease harmonics. Pulse-Width Modulation (PWM) - reduced filter requirements to decrease harmonics. ■ Disadvantages of PWM switching - complex control circuit for the switches - increase losses due to more frequent switching.
- 46. Harmonics of Bipolar PWM Assuming the PWM output is symmetry, the harmonics of each kth PWM pulse can be expressed PWM Harmonics Finally, the resultant of the integration is + = = ∫ ∫ ∫ + + + k k k k k t d t n -V t d t n V t d t n t v V DC DC T nk δ α α α δ α ω ω ω ω π ω ω π 1 k ) ( ) sin( ) ( ) ( ) sin( 2 ) ( ) sin( ) ( 2 0 [ ] ) ( cos 2 cos cos 2 1 k k k k DC nk n n n n V V δ α α α π + − + = +
- 47. PWM Harmonics k α k k δ α + 1 + k α k δ Symmetric sampling
- 48. Harmonics of Bipolar PWM The Fourier coefficient for the PWM waveform is the sum of Vnk for the p pulses over one period. PWM Harmonics ∑ = p V V The normalized frequency spectrum for bipolar switching for ma = 1 is shown below ∑ = = k nk n V V 1 ma = 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 n=1 1.00 0.90 0.80 0.70 0.60 0.50 o.40 0.30 0.20 0.10 n = mf 0.60 0.71 0.82 0.92 1.01 1.08 1.15 1.20 1.24 1.27 n=mf +2 0.32 0.27 0.22 0.17 0.13 0.09 0.06 0.03 0.02 0.00 Normalized Fourier Coefficients Vn/Vdc for Bipolar PWM
- 49. Example 3.5 The inverter has a resistive load of 10Ω and inductive load of 25mH connected in series with the fundamental frequency current amplitude of 9.27A. The THD of the inverter is not more than 9.27A. The THD of the inverter is not more than 10%. If at the beginning of designing the inverter, the THD of the current is 16.7% which is does not meet the specification, find the voltage amplitude at the fundamental frequency, the required DC input supply and the new THD of the current.
- 50. Example 3.6 The single-phase full-bridge inverter is used to produce a 60Hz voltage across a series R-L load using bipolar PWM. The DC input to the bridge is 100V, the amplitude modulation ratio is 0.8, and the frequency amplitude modulation ratio is 0.8, and the frequency modulation ratio is 21. The load has resistance of R = 10Ω and inductance L = 20mH. Determine: (a) The amplitude of the 60Hz component of the output voltage and load current. (b) The power absorbed by the load resistor (c) The THD of the load current