Điện tử công suất (AC DC convertor chapter 2 )

Chapter AC to DC CONVERSION (RECTIFIER) • Single-phase, half wave rectifier – Uncontrolled: R load, R-L load, R-C load – Controlled – Free wheeling diode • Single-phase, full wave rectifier – Uncontrolled: R load, R-L load, – Controlled – Continuous and discontinuous current mode • Three-phase rectifier – uncontrolled – controlled Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 Rectifiers • DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches • Basic block diagram AC input DC output • Input can be single or multi-phase (e.g 3phase) • Output can be made fixed or variable • Applications: DC welder, DC motor drive, Battery charger,DC power supply, HVDC Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 Single-phase, half-wave, R-load + vs _ + vo _ vs π vo ωt 2π io Output voltage (DC or average), π V Vo = Vavg = Vm sin(ωt )dωt = m = 0.318Vm 2π π Output voltage (rms), π Vm (Vm sin(ωt )dωt ) = = 0.5Vm Vo , RMS = 2π Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 Half-wave with R-L load i + vs _ + vR _ + + vL _ _ vo KVL : vs = v R + v L di (ωt ) dωt First order differential eqn Solution : Vm sin(ωt ) = i (ωt ) R + L i (ωt ) = i f (ωt ) + in (ωt ) i f : forced response; in natural response, From diagram, forced response is : i f (ωt ) = Vm ⋅ sin(ωt − θ ) Z where : Z = R + (ωL) θ = tan −1 ωL R Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 R-L load Natural response is when source = 0, di (ωt ) i (ωt ) R + L =0 dωt which results in : in (ωt ) = Ae −ωt ωτ ; τ = L R Hence Vm i (ωt ) = i f (ωt ) + in (ωt ) = ⋅ sin(ωt − θ ) + Ae −ωt ωτ Z A can be solved by realising inductor current is zero before the diode starts conducting, i.e : Vm ⋅ sin(0 − θ ) + Ae −0 ωτ Z V V A = m ⋅ sin(−θ ) = m ⋅ sin(θ ) Z Z i ( 0) = Therefore the current is given as, [ Vm i (ωt ) = ⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ Z Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 ] R-L waveform vs, io β vo vR vL 2π π 3π 4π ωt Note : v L is negative because the current is decreasing, i.e : di vL = L dt Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 Extinction angle Note that the diode remains in forward biased longer than π radians (although the source is negative during that duration)The point when current reaches zero is whendiode turns OFF This point is known as theextinction angle, β [ ] Vm ⋅ sin( β − θ ) + sin(θ )e − β ωτ = Z which reduces to : i(β ) = sin( β − θ ) + sin(θ )e − β ωτ = β can only be solved numerically Therefore, the diode conducts between and β To summarise the rectfier with R - L load, [ Vm ⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ Z i (ωt ) = for ≤ ωt ≤ β ] otherwise Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 RMS current, Power The average (DC) current is : β 2π Io = i (ωt ) dωt = i (ωt )dωt 2π 2π The RMS current is : β 2π 2 I RMS = i (ωt ) dωt = i (ωt )dωt 2π 2π POWER CALCULATION Power absorbed by the load is : Po = ( I RMS )2 ⋅ R Power Factor is computed from definition : P S where P is the real power supplied by the source, which equal to the power absorbed by the load pf = S is the apparent power supplied by the source, i.e S = (Vs, RMS ).( I RMS ) pf = P (Vs,RMS ).(I RMS ) Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 Half wave rectifier, R-C Load + vs _ iD + vo _ vs Vm π /2 2π 3π /2 π 3π 4π vo Vmax Vmin ∆Vo iD α θ Vm sin(ωt ) when diode is ON vo = V e −(ωt −θ ) / ωRC when diode is OFF θ vθ = Vm sin θ Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 Operation • Let C initially uncharged Circuit is energised at ωt=0 • Diode becomes forward biased as the source become positive • When diode is ON the output is the same as source voltage C charges until Vm • After ωt=π/2, C discharges into load (R) • The source becomes less than the output voltage • Diode reverse biased; isolating the load from source • The output voltage decays exponentially Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 10 Center-tapped waveforms is iD1 D1 + vD1 − − vo + vs1 _ + vs _ + vs2 _ iD2 Center-tapped Vm + vD2 − + io D2 vs 2π π Vm 3π 4π vo vD1 -2Vm vD2 -2Vm io iD1 iD2 is Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 27 Full wave bridge, R-L load iD1 io + vR _ + vL _ is + vs _ + vo _ vs π iD1 , iD2 2π ωt iD3 ,iD4 io vo is Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 28 Approximation with large L Using Fourier Series, vo (ωt ) = Vo + ∞ Vn cos(nωt + π ) n = 2, where the DC term Vo = 2Vm π and the harmonics terms 2Vm 1 − π n −1 n +1 The DC curent Vo Io = R The harmonic currents : V Vn In = n = Z n R + jnωL Vn = As n increases, Vn harmonic decreases Thus I n decreases rapidly very increasing n If ωL is large enough, it is possible to drop all the harmonic terms, i.e : V 2V i (ωt ) ≈ I o = o = m , for ωL >> R, R πR Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 29 R-L load approximation Approximate current V 2V Io = o = m , R πR ( ) I RMS = I o + I n, RMS = I o Power delivered to the load : Po = I RMS R vs π iD1 , iD2 2π ωt iD3 ,iD4 io vo is Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 30 Examples Given a bridge rectifier has an AC source Vm=100V at 50Hz, and R-L load with R=100ohm, L=10mH a) determine the average current in the load b) determine the first two higher order harmonics of the load current c) determine the power absorbed by the load Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 31 Controlled full wave, R load is T3 io iD1 T1 + vs _ + vo _ T2 T4 Average (DC) voltage : Vo = 1π πα V Vm sin (ωt )dωt = m [1 + cos α ] π RMS Voltage Vo, RMS = π πα [Vm sin (ωt )] dωt α sin (2α ) − + 2π 4π The power absorbed by the R load is : = Vm VRMS Po = R Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 32 Controlled, R-L load iD1 io + vR _ is + vs _ + vL _ + vo _ io α π β π+α 2π vo Discontinuous mode π+α io α π β 2π vo Continuous mode Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 33 Discontinuous mode Analysis similar to controlled half wave with R - L load : i (ωt ) = [ Vm ⋅ sin(ωt − θ ) − sin(α − θ )e −(ωt −α ) ωτ Z for α ≤ ωt ≤ β Z = R + (ωL) ωL L R R For discontinous mode, need to ensure : and θ = tan −1 ;τ = β < (α + π ) Note that β is the extinction angle and must be solved numerically with condition : io ( β ) = The boundary between continous and discontinous current mode is when β in the output current expression is (π + α ) For continous operation current at ωt = (π + α ) must be greater than zero Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 34 ] Continuous mode i (π + α ) ≥ sin(π + α − θ ) − sin(π + α − θ )e −(π +α −α ) ωτ ≥ Using Trigonometry identity : sin(π + α − θ ) = sin(θ − α ), [ sin(θ − α ) − e −(π ωτ ) ] ≥ 0, Solving for α α = tan −1 ωL R Thus for continuous current mode, −1 ωL α ≤ tan R Average (DC) output voltage is given as : 2V α +π Vo = Vm sin (ωt )dωt = m cos α π α π Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 35 Single-phase diode groups D1 io + vs _ D3 vp + vo _ D4 D2 vo =vp −vn • In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential Therefore, the diode with its anode (+) at the highest potential will conduct (carry) id • For example, when vs is ( +), D1 conducts id and D3 reverses (by taking loop around vs, D1 and D3) When vs is (-), D3 conducts, D1 reverses • In the bottom group, the anodes of the two diodes are at common potential Therefore the diode with its cathode at the lowest potential conducts id • For example, when vs (+), D2 carry id D4 reverses When vs is (-), D4 carry id D2 reverses Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 36 Three-phase rectifiers D1 + van - io D3 n + vbn + vcn - D5 - vpn D2 D6 vnn + vo _ vo =vp −vn D4 van Vm vbn vcn vp Vm vo =vp - π 2π Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 3π 4π 37 Three-phase waveforms • Top group: diode with its anode at the highest potential will conduct The other two will be reversed • Bottom group: diode with the its cathode at the lowest potential will conduct The other two will be reversed • For example, if D1 (of the top group) conducts, vp is connected to van If D6 (of the bottom group) conducts, connects to vbn All other diodes are off • The resulting output waveform is given as: vo=vp-vn • For peak of the output voltage is equal to the peak of the line to line voltage vab Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 38 Three-phase, average voltage vo vo π/3 Vm, L-L π/3 2π/3 Considers only one of the six segments Obtain its average over 60 degrees or π radians Average voltage : Vo = = = 2π π 3π Vm, L − L sin(ωt )dωt 3Vm, L − L π 3Vm, L − L π [cos(ωt )]π2π33 = 0.955Vm, L− L Note that the output DC voltage component of a three - phase rectifier is much higher than of a single - phase Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 39 Controlled, three-phase T1 + van n - + vbn + vcn io T3 T5 - vpn + vo _ T2 vnn T6 T4 α Vm van vbn vcn vo Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 40 Output voltage of controlled three phase rectifier From the previous Figure, let α be the delay angle of the SCR Average voltage can be computed as : Vo = = • 2π 3+α π 3π Vm, L − L sin(ωt )dωt 3+α 3Vm, L − L π ⋅ cos α EXAMPLE: A three-phase controlled rectifier has an input voltage of 415V RMS at 50Hz The load R=10 ohm Determine the delay angle required to produce current of 50A Power Electronics and Drives (Version 3-2003), Dr Zainal Salam, 2003 41 ... (Version 3 -20 0 3), Dr Zainal Salam, 20 03 RMS current, Power The average (DC) current is : β 2 Io = i (ωt ) dωt = i (ωt )dωt 2 2 The RMS current is : β 2 2 I RMS = i (ωt ) dωt = i (ωt )dωt 2 2 POWER... 100u ∆Vo = Vm (c) Capacitor current : ωCVm cos(ωt ) ic (ωt ) = Vm sin(θ ) −(ωt −θ ) /(ωRC ) − R ⋅e 6.4 cos(ωt ) A = − 0.339 ⋅ e −(ωt −1. 62 ) /(18.8 5) (ON) (OFF) (ON) A (OFF) (d) Peak diode current... Vm sin(ωt ) = 169.7 sin(ωt ) vo (ωt ) = V sin θ ⋅ e −(ωt −θ ) / ωRC (ON) (OFF) m = 169.7 sin(ωt ) 169.5e −(ωt −1. 62 ) /(18.8 5) (ON) (OFF) Power Electronics and Drives (Version 3 -20 0 3), Dr Zainal