vg DC to AC converter (Inverter) 1 Single phase t 0 T/2 T 0 T/2 T vg 2 0.5Vs C RL V VL s 0.5Vs C D IL Q1 Vg1 D t Q2 Dead band = 1 μs Vg2 n=∞ vL=Σn=1,3,5,.. (2 Vs) / (nΠ) × sin(nωt) VL, IL Vs/2 Vn 0 T/2 t T -Vs/2 Q1 Q2 VLrms= Vs / 2 1 2 3 4 5 6 7 8 n Harmonic contents in the output voltage VL1= (2Vs) / (√2Π) Q1 Heavily inductive load (RL → 0) 0.5Vs C RL V s LL VL 0.5Vs C D1 IL vL Vs/2 0 Q1 t T -Vs/2 Q1 Vg1 D2 T/2 Q2 Q1 iL Q2 Vg2 t R-L load D1 Q1 D 2 iL= Σ (2Vs) / [ nΠ √ (R2+n2ω2L2) ] × sin (nωt - Θn) Θn= tan-1(nωL / R) Q2 Performance Parameters HFn Harmonic Factor for the nth harmonic HFn= (VLn) / VL1 THD for n > 1 Total Harmonic Distortion The harmonic voltage Vh ∞ THD = 1 / VL1 ( Σ V2n ) 0.5 n=2, 3, 4,… ∞ Vh= ( Σ VLn2 ) 0.5 = ( VLrms2 – VL12 )0.5 n= 3, 5, 7, .. DF Distortion Factor ∞5 DF = 1 / VL1 [ Σ ( VLn / n2 )2 ] 0.5 n=2, 3, … The Distortion Factor of the nth harmonic = VLn / ( VL1 n2) for n > 1 Lowest Order Harmonic LOH is the harmonic component that is the closest to the fundamental and its amplitude Is ≥ 3% of the fundamental Calculate: a) The rms value of the load fundamental voltage. b) The output power. c) The average and peak current in the transistor. d) The THD, DF, the HF and DF of the LOH. 0.5Vs C RL V s D1 LL Vg1 VL 0.5Vs Q1 IL Q2 D2 C Vg2 a) vL1 = (2Vs) / Π × sin ( ωt) Vs= 48V R= 2.4 Ω VL1rms = ( 2 × 48 ) / ( Π × √2 ) = 21.6 V b) VLrms= 0.5 Vs = 24 V PL= (VLrms)2 / R = 242 / 2.4 = 240 W c) iQ1 iQ2 t 0 T/2 T 0 Peak current in each transistor = 24/2.4 = 10A Average current in each transistor = 5 A d) Vh= ( 242 – 21.62 )0.5 = 10.46 V THD = 10.46 / 21.6 = 0.4843 T/2 T t DF = 1/21.6 ×{ [ 7.2/32]2+ [4.32/52]2 +[3.086/72] 2}0.5 VL3= 21.6/3 = 7.2 V = 1/21.6 ×{ 0.64 + 0.02986 VL5= 21.6/5 = 4.32 V +0.004+ .. }0.5 VL7= 21.6/7 = 3.086 V = 0.038 The LOH = 3rd harmonic HF3= 1/3 = 0.3333 DF3= 0.3333/32 = 0.03703 note that VL3= 0.3333 which is > 0.03 so LOH =3 Vg1, Vg2 The H-bridge single phase inverter Q1 D IL RL D s Q4 VL D T/2 T 0 T/2 T Vg3, Vg4 Q3 Vg1 V t 0 Vg3 D Q2 Vg4 t Vg2 Dead band = 1 μs VL, IL n=∞ vL=Σn=1,3,5,.. (4 Vs) / (nΠ) × sin(nωt) Vs Vn 0 T/2 T t -Vs Q1, Q2 Q3, Q4 VLrms= Vs 1 2 3 4 5 6 7 8 n Harmonic contents in the output voltage VL1= (4Vs) / (√2Π) Q1, Q2 Calculate: a) The rms value of the load fundamental voltage. b) The output power. c) The average and peak current in the transistor. V d) The THD, DF, the HF and DF of the LOH. s D Q1 IL D RL Q3 Vg1 Vg3 VL Q4 Q2 D D Vg4 Vg2 a) vL1 = (4Vs) / Π × sin ( ωt) Vs= 48V R= 2.4 Ω VL1rms = ( 4 × 48 ) / ( Π × √2 ) = 43.2 V b) VLrms= Vs = 48 V PL= (VLrms)2 / R = 482 / 2.4 = 960 W c) iQ1, iQ2 iQ3, iQ4 t 0 T/2 T 0 Peak current in each transistor = 48/2.4 = 20A Average current in each transistor =10 A d) Vh= (482 – 43.22 )0.5 = 20.92 V THD = 20.92 / 43.2 = 0.4843 (same) T/2 T t DF = 1/43.2 ×{ [ 14.4/32]2+ [8.64/52]2 +[6.17/72] 2}0.5 VL3= 43.2/3 = 14.4 V = 1/43.2 ×{ 1.6 + .3456 VL5= 43.2/5 = 8.64 V +0.1259+ .. }0.5 VL7= 43.2/7 = 6.17 V = 0.033 (same) LOH = 3rd harmonic HF3 = 1/3 DF3= 1/(3×32) = 0.03703 (same) note that VL3= 14.4 which is > 0.03×VL1 so LOH =3 The quality of the output voltage is the same as for the 2-transistor circuit however, the H bridge inverter the output power is 4 times higher and the fundamental output Voltage is twice that of the 2-transistor circuit. The H-bridge inverter shown in figure has an RLC load with R=10Ω, L=31.5mH, C=112μF. D D Q1 L C R IL The inverter frequency is 60 Hz and the dc input V V Voltage is Vs=220V. VL s D Q a) Express the instantaneous load current in 4 D Fourrier series. V b) Calculate the rms load current at the fundamental frequency. c) Calculate the THD of the load current. d) Calculate the total power absorbed by the load as well as the fundamental power. e) Calculate the average dc current drawn from the supply. f) Calculate the rms and the peak current of each transistor. Q3 g1 g4 Vg3 Q2 Vg2 120o conduction Three-phase inverters 180o conduction 120o conduction R a Q1 V s a Vg1 Q4 Vg4 D1 Q3 b Vg3 Q6 D4 Vg6 D3 Q5 Q2 Vg2 R c b Vg5 D6 D5 D2 R c @ any time only 2 transistors are conducting: 1 in an upper leg 1 in another lower leg vG 1 60o ωt vG 2 60o ωt vG 3 60o ωt vG 4 60o ωt vG 5 60o ωt vG 6 60o ωt For 60o ≤ ωt < 120o For 0 ≤ ωt < 60o R R R b s n’ b R c R n’ V s For 180o ≤ ωt < For 240o ≤ ωt < 300o 240o c b R n’ s R R c R R a R n’ V s c n’ For 300o ≤ ωt < 360o a a b b R R V a V s R c R a a V For 120o ≤ ωt < 180o R b R c R V s n’ vab Vs CV 60o vbc 0.5Vs CV CV CV - 0.5Vs CV CV CV CV ωt -Vs ωt 60o vca ωt 60o van’ ωt 60o vbn’ 0.5Vs 60o Vcn’ 60o ωt -0.5Vs ωt 180o conduction ( 3 transistors are conducting at any time) vG1 60o ωt vG2 60o ωt vG3 vG4 60o 60o ωt ωt vG5 vG6 60o 60o ωt ωt For 60o ≤ ωt < 120o For 0 ≤ ωt < 60o a c R R R a R n’ For b a n’ b R R b b n’ R R c c R Vs Vs 180o For 120o ≤ ωt < 180o ≤ ωt < 240o n’ R Vs For 240o ≤ ωt < 300o R a R b c c For 300o ≤ ωt < 360o a n’ a R R c R R n’ R R R Vs Vs Vs b vab Vs CV 60o vbc CV ωt CV CV -Vs 60o ωt vca 60o ωt van’ ⅓Vs ⅔Vs 60o ωt vbn’ 60o ωt Vcn’ 60o ωt Voltage control techniques of single phase inverters Multiple pulse width modulation Single pulse width modulation VL VL Vs Vs δ δ 3Π/2 0 Π/2 Π δ 7Π/6 ωt 0 Π/6 Π/3 Π/2 2Π/3 5Π/6 Π 2Π αm=2 δ -Vs δ 3Π/2 4Π/3 δ 11Π/6 5Π/3 δ ωt 2Π δ -Vs P= # of pulses per half cycle P=3 ∞ vL= Σn=1,3,5,.. (4Vs / nΠ) sin(nδ/2) sin(nωt) VLrms= Vs √(δ/Π) ∞ Decreases DF significantly 2p vL= Σn=1, 3, ..Σm=1{4Vs /(nΠ) sin{ nδ/4 [ sin n(αm+3δ/4) – sin n(Π+αm+δ/4) ] }× sin(nωt) VLrms= Vs √ (pδ/Π) δ = M T/ (2p) Where M is the amplitude modulation index 0 ≤ M ≤ 1 Sinusoidal Pulse Width Modulation Ac Ar Reference waveform MA = Amplitude Modulation Index Ar Carrier waveform MA = _______ Ac MF = Frequency Modulation Index carrier frequency MF = --------------------------reference frequency 0 ≤ MA ≤ 1 If MA > 1 over-modulation (= 5) fC = carrier frequency fR = reference frequency if MF is an odd number, quarter-wave symmetry is obtained and no even harmonics are present in the output voltage. α1 α ωt 180o2- α1 For a 3-phase inverter, MF should be an odd triplen number 180o – α2 SPWM reduces greatly the DF U1 Vs <1 ωt over-modulation 0 0 1 MA -Vs