Điện tử công suất (inverter chapter 4 )

Chapter DC to AC Conversion (INVERTER) • • • • • General concept Single-phase inverter Harmonics Modulation Three-phase inverter Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB DC to AC Converter (Inverter) • DEFINITION: Converts DC to AC power by switching the DC input voltage (or current) in a pre-determined sequence so as to generate AC voltage (or current) output • General block diagram IDC Iac Vac VDC − − • TYPICAL APPLICATIONS: – Un-interruptible power supply (UPS), Industrial (induction motor) drives, Traction, HVDC Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB Simple square-wave inverter (1) • To illustrate the concept of AC waveform generation S1 S3 S4 S2 Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB AC Waveform Generation S1,S2 ON; S3,S4 OFF vO S1 VDC for t1 < t < t2 VDC S3 + vO − t1 S4 t t2 S2 S3,S4 ON ; S1,S2 OFF for t2 < t < t3 vO S1 VDC S3 t2 + vO − S4 t3 t S2 -VDC Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB AC Waveforms INVERTER OUTPUT VOLTAGE Vdc π 2π -Vdc FUNDAMENTAL COMPONENT V1 4VDC π V1 V1 3RD HARMONIC 5RD HARMONIC Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB Harmonics Filtering DC SUPPLY INVERTER (LOW PASS) FILTER LOAD L + vO C + vO − BEFORE FILTERING vO − AFTER FILTERING vO • Output of the inverter is “chopped AC voltage with zero DC component” It contain harmonics • An LC section low-pass filter is normally fitted at the inverter output to reduce the high frequency harmonics • In some applications such as UPS, “high purity” sine wave output is required Good filtering is a must • In some applications such as AC motor drive, filtering is not required Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB Variable Voltage Variable Frequency Capability Vdc2 Higher input voltage Higher frequency Vdc1 Lower input voltage Lower frequency t • Output voltage frequency can be varied by “period” of the square-wave pulse • Output voltage amplitude can be varied by varying the “magnitude” of the DC input voltage • Very useful: e.g variable speed induction motor drive Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB Output voltage harmonics/ distortion • Harmonics cause distortion on the output voltage • Lower order harmonics (3rd, 5th etc) are very difficult to filter, due to the filter size and high filter order They can cause serious voltage distortion • Why need to consider harmonics? – Sinusoidal waveform quality must match TNB supply – “Power Quality” issue – Harmonics may cause degradation of equipment Equipment need to be “de-rated” • Total Harmonic Distortion (THD) is a measure to determine the “quality” of a given waveform Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB Total Harmonics Distortion (THD) Voltage THD : If Vn is the nth harmonic voltage, ∞ (Vn, RMS )2 THDv = n= V1, RMS = V2, RMS + V3, RMS + + V2, RMS V1, RMS If the rms voltage for the vaveform is known, ∞ (VRMS )2 − (V1, RMS )2 THDv = n= V1, RMS Current THD : ∞ (I n, RMS )2 THDi = n =2 I1, RMS V In = n Zn Z n is the impedance at harmonic frequency Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB Fourier Series • Study of harmonics requires understanding of wave shapes Fourier Series is a tool to analyse wave shapes Fourier Series ao = an = bn = 2π π π π 2π 2π f (v )dθ (" DC" term) f (v) cos(nθ )dθ (" cos" term) f (v) sin (nθ )dθ ("sin" term) Inverse Fourier ∞ f (v) = ao + (an cos nθ + bn sin nθ ) n =1 where θ = ωt Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 10 Determination of switching angles for kth PWM pulse (1) AS2 AS1 v Vmsin( θ ) + Vdc Ap1 Ap2 V − dc Equating the volt - second, As1 = Ap1 As = Ap Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 32 PWM Switching angles (2) The Volt - second during the first half cycle of the PWM pulse is given as : Vdc Vdc (δ1k ) − (2δ o − δ1k ) A p1 = 2 = (Vdc )(δ1k − δ o ) Similarly for the second half, Vdc (δ 2k ) − Vdc (2δ o − δ 2k ) 2 = (Vdc )(δ 2k − δ o ) Ap2 = The volt - second supplied by the sinusoid, As1 = αk Vm sin θdθ = Vm [cos(α k − 2δ o ) − cos α k ] α k − 2δ o = 2Vm sin δ o sin(α k − δ o ) Similarly, As = 2δ oVm sin(α k + δ o ) Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 33 Switching angles (3) For small angle δ o sin δ o → δ o , As1 = 2δ oVm sin(α k − δ o ) As = 2δ oVm sin(α k − δ o ) To derive the modulation strategy, A p1 = As1; A p = As Hence, for the the first half cycle of PWM pulse, (Vdc )(δ1k − δ o ) = 2δ oVm sin(α k − δ o ) (δ1k − δ o ) = 2Vm (δ o sin(α k − δ o ) Vdc By definition, the Modulation Ratio, MI = Vm (Vdc ) is known as modulation Thus, the pulse width for the first half cycle of the PWM waveform is given by : δ1k = δ o [1 + M I sin(α k − δ o )] Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 34 PWM switching angles (4) Thus the leading edge switching angle of the kth pulse is : α k − δ1k Using similar method, pulse width of the second half cycle of PWM waveform : δ 2k = δ o [1 + M I sin(α k + δ o )] And the trailing edge angle : α k + δ 2k The above equation is valid for Asymmetric Modulation, i.eδ1k and δ 2k are different For Symmetric Modulation, δ1k = δ 2k = δ k Hence δ k = δ o [1 + M I sin α k ] Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 35 Example • For the PWM shown below, calculate the switching angles pulses no carrier waveform 2V 1.5V π 2π modulating waveform π t1 t2 t3 t4 t5 t6 t13 t15 t17 t7 t8 t9 t10 t11 t12 t14 t16 t18 2π α1 Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 36 Harmonics of bipolar PWM ∆ Assuming the PWM waveform is half δ0 δ0 + Vdc δ1k wave symmetry, harmonic δ0 δ0 δ 2k content of each (kth) PWM pulse can be computed as : bnk = = + + 1T π α k −δ1k α k +δ k π α −δ k 1k αk f (v) sin nθdθ π α −2δ k o + Vdc α k + 2δ o π α +δ k 2k V − dc sin nθdθ Vdc sin nθdθ V − dc sin nθdθ Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 37 Harmonics of Bipolar PWM Which can be reduced to : Vdc {cos n(α k − 2δ o ) − cos n(α k − δ1k ) bnk = − nπ + cos n(α k + δ k ) − cos n(α k − δ1k ) + cos n(α k + δ k ) − cos n(α k + 2δ o )} Yeilding, 2Vdc [cos n(α k − δ1k ) − cos n(α k − 21k ) bnk = nπ + cos nα k cos n 2δ o ] This equation cannot be simplified productively.The Fourier coefficent for the PWM waveform isthe sum of bnk for the p pulses over one period, i.e : bn = p bnk k =1 Next slide shows the computation of this equation Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 38 PWM Spectra M I = 0.2 Amplitude M I = 0.4 0 M I = 0.6 0.6 0.4 M I = Modulation Index 0.2 M I = p 2p 3p 4p Fundamental NORMALISED HARMONIC AMPLITUDES FOR SINUSOIDAL PULSE-WITDH MODULATION Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 39 PWM spectra observations • • The harmonics appear in “clusters” at multiple of the carrier frequencies Main harmonics located at : f = kp (fm); k=1,2,3 where fm is the frequency of the modulation (sine) waveform • There also exist “side-bands” around the main harmonic frequencies • Amplitude of the fundamental is proportional to the modulation index The relation ship is given as: V1= MIVin • The amplitude of the harmonic changes with MI Its incidence (location on spectra) is not • When p>10, or so, the harmonics can be normalised For lower values of p, the side-bands clusters overlap-normalised results no longer apply Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 40 Tabulated Bipolar PWM Harmonics 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0 MR 1.242 1.15 1.006 0.818 0.601 MR +2 0.016 0.061 0.131 0.220 0.318 n MI MR +4 2MR +1 0.018 0.190 2MR +3 0.326 0.370 0.314 0.181 0.024 0.071 0.139 0.212 0.013 0.033 2MR +5 3MR 0.335 0.123 0.083 0.171 0.113 3MR +2 0.044 0.139 0.203 0.716 0.062 0.012 0.047 0.104 0.157 0.016 0.044 3MR +4 3MR +6 4MR +1 0.163 0.157 0.008 0.105 0.068 4MR +3 0.012 0.070 0.132 0.115 0.009 0.034 0.084 0.017 0.119 0.050 4MR+5 4MR +7 Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 41 Three-phase harmonics • For three-phase inverters, there is significant advantage if MR is chosen to be: – Odd: All even harmonic will be eliminated from the pole-switching waveform – triplens (multiple of three (e.g 3,9,15,21, 27 ): All triplens harmonics will be eliminated from the line-to-line output voltage • By observing the waveform, it can be seen that with odd MR, the line-to-line voltage shape looks more “sinusoidal” • As can be noted from the spectra, the phase voltage amplitude is 0.8 (normalised) This is because the modulation index is 0.8 The line voltage amplitude is square root three of phase voltage due to the three-phase relationship Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 42 Effect of odd and “triplens” π Vdc − − 2π V RG Vdc Vdc VYG Vdc Vdc V RY − Vdc Vdc − − p = 8, M = 0.6 V RG Vdc Vdc VYG Vdc Vdc V RY − Vdc p = 9, M = 0.6 ILLUSTRATION OF BENEFITS OF USING A FREQUENCY RATIO THAT IS A MULTIPLE OF THREE IN A THREE PHASE INVERTER Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 43 Spectra: effect of “triplens” Amplitude 1.8 0.8 (Line to line voltage) 1.6 1.4 1.2 1.0 0.8 0.6 B 0.4 19 0.2 37 23 41 43 47 59 61 65 67 79 83 85 89 21 19 Fundamental A 63 23 37 39 41 43 45 47 57 59 61 83 81 65 79 67 69 77 85 87 89 91 Harmonic Order COMPARISON OF INVERTER PHASE VOLTAGE (A) & INVERTER LINE VOLTAGE (B) HARMONIC (P=21, M=0.8) Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 44 Comments on PWM scheme • It is desirable to have MR as large as possible • This will push the harmonic at higher frequencies on the spectrum Thus filtering requirement is reduced • Although the voltage THD improvement is not significant, but the current THD will improve greatly because the load normally has some current filtering effect • However, higher MR has side effects: – Higher switching frequency: More losses – Pulse width may be too small to be constructed “Pulse dropping” may be required Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 45 Example The amplitudes of the pole switching waveform harmonics of the red phase of a three-phase inverter is shown in Table below The inverter uses a symmetric regular sampling PWM scheme The carrier frequency is 1050Hz and the modulating frequency is 50Hz The modulation index is 0.8 Calculate the harmonic amplitudes of the line-to-voltage (i.e red to blue phase) and complete the table Harmonic number Amplitude (pole switching waveform) 19 0.3 21 0.8 23 0.3 37 0.1 39 0.2 41 0.25 43 0.25 45 0.2 47 0.1 57 0.05 59 0.1 61 0.15 63 0.2 65 0.15 67 0.1 69 0.05 Amplitude (line-to line voltage) Power Electronics and Drives (Version 3-2003): Dr Zainal Salam UTM-JB 46 ... an = bn = 2π π π π 2π 2π f (v )dθ (" DC" term) f (v) cos(nθ )dθ (" cos" term) f (v) sin (nθ )dθ ("sin" term) Inverse Fourier ∞ f (v) = ao + (an cos nθ + bn sin nθ ) n =1 where θ = ωt Power Electronics... Drives (Version 3-200 3): Dr Zainal Salam UTM-JB 12 Spectra of square wave Normalised Fundamental 1st 3rd (0.3 3) 5th (0. 2) 7th (0. 1 4) 9th (0.1 1) 11th (0.0 9) n 11 • Spectra (harmonics) characteristics:... 2nπ − cos nπ )] = nπ Vdc [(1 − cos nπ ) + (1 − cos nπ )] = nπ 2V = dc [(1 − cos nπ )] nπ [ ] When n is even, cos nπ = bn = (i.e even harmonics not exist) When n is odd, cos nπ = −1 4Vdc bn = nπ