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Chapter 2 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 – R load – R-L load, – Controlled R, R-L load – continuous and discontinuous current mode • Three-phase rectifier – uncontrolled – controlled Power Electronics and 1 Drives (Version 2), Dr. Zainal Salam, 2002 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. 3- phase). • Output can be made fixed or variable • Applications: DC welder, DC motor drive, Battery charger,DC power supply, HVDC Power Electronics and 2 Drives (Version 2), Dr. Zainal Salam, 2002 Single-phase, half-wave with R-load + + vs vo _ _ vs ωt vo io ωt Output voltage (average), π Vo = Vavg = ∫ Vm sin(ωt )dωt 0 Vm = = 0.318Vm π Power Electronics and 3 Drives (Version 2), Dr. Zainal Salam, 2002 RMS voltage Output voltage (RMS) π 1 Vm ∫ (Vm sin(ω t ) ) dω t = 2 2 Vo , RMS = π 0 Output current (DC), Vo 0.318Vm Io = = R R • DC voltage is fixed at 0.318 or 31.8% of the peak value • RMS voltage is reduced from 0.707 (normal sinusoidal RMS) to 0.5 or 50% of peak value. • Half wave is not practical because of high distortion supply current. The supply current contains DC component that may saturate the input transformer Power Electronics and 4 Drives (Version 2), Dr. Zainal Salam, 2002 Half-wave with R-L load i + vR + + _ vTNB vo _ + vL _ _ vs = vR + v L di (t ) Vm sin( ω t ) = i (t ) R + L dt This is a first order differenti al equation. Solution is in the form of : i ( t ) = i f ( t ) + in ( t ) where : i f , in are known as " forced" and " natural" response, respective ly. From diagram, forced response is : Vm i f (t ) = ⋅ sin( ω t − θ ) Z where : ωL Z= R 2 + (ω L ) 2 and θ = tan −1 R Power Electronics and 5 Drives (Version 2), Dr. Zainal Salam, 2002 R-L load Natural response is when source = 0, di (t ) i (t ) R + L =0 dt which results in : in (t ) = Aet τ ; τ = L R Hence V i (t ) = i f (t ) + in (t ) = m ⋅ sin(ωt − θ ) + Ae −t τ Z A can be solved by realising inductor current is zero before the diode starts conducting, i.e : Vm i (0) = ⋅ sin(0 − θ ) + Ae −0 τ Z Vm Vm ⇒ A = ⋅ sin( −θ ) = ⋅ sin(θ ) Z Z Therefore the current is given as, V Z [ i (t ) = m ⋅ sin(ωt − θ ) + sin(θ )e −t τ ] or V Z [ i (ωt ) = m ⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ ] Power Electronics and 6 Drives (Version 2), Dr. Zainal Salam, 2002 R-L waveform vs, io β vo vR vL 0 2π ωt π 3π 4π Note : vL is negative because the current is decreasing, i.e : di vL = L dt Power Electronics and 7 Drives (Version 2), Dr. Zainal Salam, 2002 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 when diode turns OFF. This point is known as the extinction angle, β . V Z [ i ( β ) = m ⋅ sin( β − θ ) + sin(θ )e − β ωτ ]= 0 which reduces to : sin( β − θ ) + sin(θ )e − β ωτ =0 β can only be solved numerically. Therefore, the diode conducts between 0 and β To summarise the rectfier with R - L load, Vm Z [ ⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ ] i (ωt ) = for 0 ≤ ωt ≤ β 0 otherwise Power Electronics and 8 Drives (Version 2), Dr. Zainal Salam, 2002 RMS current, Power Factor The average (DC) current is : β 1 2π 1 Io = 2π 0∫ i(ωt )dωt = 2π ∫ i(ωt )dωt 0 The RMS current is : β 1 2π 2 1 2 I RMS = 2π 0∫ i (ωt )dωt = 2π ∫ i (ωt )dωt 0 Power absorbed by the load is : Po = ( I RMS )2 ⋅ R Power Factor is computed from definition : P pf = S where P is the real power supplied by the source, which equal to the power absorbed by the load. S is the apparent power supplied by the source, i.e S = (Vs , RMS ).( I RMS ) P ⇒ pf = (Vs,RMS ).(I RMS ) Power Electronics and 9 Drives (Version 2), Dr. Zainal Salam, 2002 Half wave rectifier, R-C Load + iD + vs vo _ _ vs Vm π /2 π 2π 3π /2 3π 4π Vmax vo Vmin ∆Vo iD α θ Vm sin(ωt ) when diode is ON vo = −(ωt −θ ) / ωRC Vθ e when diode is OFF vθ = Vm sin θ Power Electronics and 10 Drives (Version 2), Dr. Zainal Salam, 2002 Operation • Let C initially • After ωt=π/2, C uncharged. Circuit discharges into is energised at load (R). ωt=0 • The source • Diode becomes becomes less than forward biased as the output voltage the source become positive • Diode reverse biased; isolating • When diode is ON the load from the output is the source. same as source voltage. C charges • The output voltage until Vm decays exponentially. Power Electronics and 11 Drives (Version 2), Dr. Zainal Salam, 2002 Estimation of θ The slope of the functions are : d (Vm sin ωt ) = Vm cos ωt d (ωt ) and ( d Vm sin θ ⋅ e −(ωt −θ ) / ωRC ) d (ωt ) − 1 ⋅ −(ωt −θ ) / ωRC = Vm sin θ ⋅ e ωRC At ωt = θ , the slopes are equal, − 1 ⋅ e −(θ −θ ) / ωRC Vm cosθ = Vm sin θ ⋅ ωRC Vm cosθ 1 ⇒ =− Vm sin θ ⋅ ωRC 1 1 = tan θ − ωRC θ = tan −1 (− ωRC ) = − tan −1 (ωRC ) + π Power Electronics and 12 Drives (Version 2), Dr. Zainal Salam, 2002 Estimation of α For practical circuits, ωRC is large, then : π π θ = -tan(∞ ) + π = − +π = 2 2 and Vm sin θ = Vm At ωt = 2π + α , Vm sin( 2π + α ) = (Vm sin θ )e −( 2π +α −θ ) ωRC or sin(α − (sin θ )e −( 2π +α −θ ) ωRC = 0 This equation must be solved numerically for α Power Electronics and 13 Drives (Version 2), Dr. Zainal Salam, 2002 Ripple Voltage Max output voltage is Vmax . Min output voltage occurs at ωt = 2π + α Reffering to diagram, the ripple is : ∆Vo = Vmax − Vmin = Vm − Vm sin( 2π + α ) = Vm − Vm sin α If Vθ = Vm and θ = π 2, and C is large such that DC output voltage is constant, then α ≈ π 2. The output voltage evaluated at ωt = 2π + α is : 2π +π 2−π 2 2π − − ωRC ωRC vo (2π + α ) = Vm e = Vm e The ripple voltage is approximated as : 2π 2π − − ωRC ωRC ∆Vo ≈ Vm − Vm e = Vm 1 − e Power Electronics and 14 Drives (Version 2), Dr. Zainal Salam, 2002 Voltage ripple-cont’d Approximation of exponent term yields: −2π ω RC 2π e ≈ 1− ω RC Substituting, 2π Vm ∆Vo ≈ Vm = ω RC fRC • The output voltage ripple is reduced by increasing C. •As C is increased, the conduction interval for diode decreases. •Therefore, reduction in output voltage ripple results in larger peak diode current. • EXAMPLE: The half wave rectifier has 120V RMS source at 60Hz. R=500 Ohm and C=100uF. Determine (a) the expression for output voltage, (b) voltage ripple. Power Electronics and 15 Drives (Version 2), Dr. Zainal Salam, 2002 Controlled half-wave ig ia vo ia + + α ωt vs vo vs _ _ ig α ωt Average voltage : 1 π Vm Vo = ∫ Vm sin (ωt )dωt = 2π [1 + cos α ] 2π α RMS volatge π 2 1 VRMS = ∫ [Vm sin (ωt )] dωt 2π α Vm π 2 Vm α sin (2α ) = ∫ [1 − cos(2ω t ] dωt = 2 1 − π + 2π 4π α Power Electronics and 16 Drives (Version 2), Dr. Zainal Salam, 2002 Controlled h/w, R-L load i + vR + + _ vs vo _ + vL _ _ vs π 2π ωt vo io π β 2π ωt α −ωt V i (ωt ) = i f (ωt ) + in (ωt ) = m ⋅ sin (ωt − θ ) + Ae ωτ Z Initial condition : i (α ) = 0, −α V i (α ) = 0 = m ⋅ sin (α − θ ) + Aeωτ Z −α V A = − m ⋅ sin (α − θ ) e ωτ Z Power Electronics and 17 Drives (Version 2), Dr. Zainal Salam, 2002 Extinction angle Substituting for A and simplifying, V −(α −ωt ) m ⋅ sin (ωt − θ ) − sin (α − θ )e ωτ Z i (ωt ) = for α ≤ ωt ≤ β 0 otherwise Extinction angle, β is defined when i = 0, (α − β V i( β ) = 0 = m sin ( β − θ ) − sin ( β − θ )e ωτ Z which can only be solved numerically. Angle ( β − θ ) is called the conduction angel. i.e the diode conducts for γ degrees. Power Electronics and 18 Drives (Version 2), Dr. Zainal Salam, 2002 RMS voltage and current voltage Average : β 1 Vm Vo = ∫Vm sin(ωt )dωt = [cosα − cosβ ] 2π α 2π current Average RMScurrent β β 1 1 2 Io = ∫ i(ωt )dω I RMS = ∫ i (ωt )dω 2π α 2π α by Thepowerabsorbed theloadis : P = I RMS2 ⋅ R o • EXAMPLES • 1. Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply. Determine the power factor absorbed by the resistance. • 2. A half wave rectifier has a source of 120V RMS at 60Hz. R=20 ohm, L=0.04H, and the delay angle is 45 degrees. Determine: (a) the expression for i(ωt), (b) average current, (c) the power absorbed by the load. Power Electronics and 19 Drives (Version 2), Dr. Zainal Salam, 2002 Freewheeling diode (FWD) • Note that for single-phase, half wave rectifier with R-L load, the load (output) current is NOT continuos. • A FWD (sometimes known as commutation diode) can be placed as shown below to make it continuos io + vR + + _ vs vo _ + vL _ _ io io vo= 0 + vo= vs + + vs vo vo io _ _ _ D1 is on, D2 is off D2 is on, D1 is off Power Electronics and 20 Drives (Version 2), Dr. Zainal Salam, 2002 Operation of FWD • Note that both D1 and D2 cannot be turned on at the same time. • For a positive cycle voltage source, – D1 is on, D2 is off – The equivalent circuit is shown in Figure (b) – The voltage across the R-L load is the same as the source voltage. • For a negative cycle voltage source, – D1 is off, D2 is on – The equivalent circuit is shown in Figure (c) – The voltage across the R-L load is zero. – However, the inductor contains energy from positive cycle. The load current still circulates through the R-L path. – But in contrast with the normal half wave rectifier, the circuit in Figure (c) does not consist of supply voltage in its loop. – Hence the “negative part” of vo as shown in the normal half-wave disappear. Power Electronics and 21 Drives (Version 2), Dr. Zainal Salam, 2002 FWD- Continuous load current • The inclusion of FWD results in continuos load current, as shown below. • Note also the output voltage has no negative part. output vo io iD1 ωt Diode current iD2 0 π 2π 3π 4π Power Electronics and 22 Drives (Version 2), Dr. Zainal Salam, 2002 Full wave rectifier with R load iD1 io is + + vs vo _ _ Bridge circuit is iD1 + vs1 + _ − vo + vs _ + io vs2 _ iD2 Center-tapped circuit Power Electronics and 23 Drives (Version 2), Dr. Zainal Salam, 2002 Notes on full-wave • Center-tapped rectifier requires center-tap transformer. Bridge does not. • Center tap requires only two diodes, compared to four for bridge. Hence, per half-cycle only one diode volt-drop is experienced. Conduction losses is half of bridge. • However, the diodes ratings for center- tapped is twice than bridge. For both circuits, vo = {V Vsinsintωt − m m ω 0 ≤ ωt ≤ π π ≤ ωt ≤ 2π DC voltage : 1π 2Vm Vo = ∫ Vm sin (ωt )dωt = = 0.637Vm π0 π Power Electronics and 24 Drives (Version 2), Dr. Zainal Salam, 2002 Bridge waveforms vs Vm 2π 3π 4π π vo Vm vD1 vD2 -Vm vD3 vD4 -Vm io iD1 iD2 iD3 iD4 is Power Electronics and 25 Drives (Version 2), Dr. Zainal Salam, 2002 Center-tapped waveforms vs Vm 2π 3π 4π π vo Vm vD1 -2Vm vD2 -2Vm io iD1 iD2 is Power Electronics and 26 Drives (Version 2), Dr. Zainal Salam, 2002 Full wave bridge, R-L load io iD1 is + vR + + vs _ _ vo + _ vL _ iD1 , iD2 π 2π 3π 4π iD3 ,iD4 vo output io vs is supply Power Electronics and 27 Drives (Version 2), Dr. Zainal Salam, 2002 R-L load analysis: approximation with large L Using Fourier Series, output voltage is described as : ∞ vo (ωt ) = Vo + ∑ Vn cos(nωt + π ) n = 2, 4... where 2Vm Vo = π 2V 1 1 Vn = m − π n − 1 n + 1 The DC and harmonic currents are : Vo Vn Vn Io = In = = R Z n R + jnωL Note that as n increases, voltage amplitude for the nth harmonic decreases. This makes I n decreases rapidly for increasing n. Only a few terms sufficient to approximate output. Power Electronics and 28 Drives (Version 2), Dr. Zainal Salam, 2002 R-L load analysis IfωLis large enough, it is possible to drop all the harmonic terms, i.e. : Vo 2Vm i(ωt ) ≈ I o = = , for ωL >> R, R R The approximation with large L is shown below. iD1 , iD2 exact approx. π 2π 3π 4π iD3 ,iD4 vo output io 2Vm/R is supply Power Electronics and 29 Drives (Version 2), Dr. Zainal Salam, 2002 Examples ( I RMS = I o 2 + ∑ I n, RMS 2 = I o) Power delivered to the load : Po = I RMS 2 R • EXAMPLE: Given a bridge rectifier has an AC source Vm=100V at 50Hz, and R-L load with R=10ohm, 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 30 Drives (Version 2), Dr. Zainal Salam, 2002 Controlled full wave, R load iD1 io is + + vs vo _ _ 1π V Vo = ∫ Vm sin (ωt )dωt = m [1 + cos α ] πα π π 2 1 VRMS = πα∫ [Vm sin (ωt )] dωt 1 α sin (2α ) = Vm − + 2 2π 4π The power absorbed by the R load is : VRMS 2 Po = R Power Electronics and 31 Drives (Version 2), Dr. Zainal Salam, 2002 Controlled, R-L load io iD1 is + vR + + vs _ _ vo + _ vL _ io α π β π+α 2π vo Discontinuous mode π+α io α π β 2π vo Continuous mode Power Electronics and 32 Drives (Version 2), Dr. Zainal Salam, 2002 Discontinuous mode Analysis similar to controlled half wave with R - L load : V Z [ i (ωt ) = m ⋅ sin(ωt − θ ) − sin(α − θ )e −(ωt −α ) ωτ ] for α ≤ ωt ≤ β Z = R 2 + (ωL) 2 −1 ωL L and θ = tan ; τ = R R For discontinous mode, need to ensure : β < (α + π ) Note that β is the extinction angle and must be solved numerically with condition : io ( β ) = 0 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 33 Drives (Version 2), Dr. Zainal Salam, 2002 Continuous mode i (π + α ) ≥ 0 sin(π + α − θ ) − sin(π + α − θ )e −(π +α −α ) ωτ ≥ 0 Using Trigonometry identity : sin(π + α − θ ) = sin(θ − α ), [ sin(θ − α ) 1 − e −(π ωτ ) ] ≥ 0, Solving for α −1 ωL α = tan R Thus for continuous current mode, −1 ωL α ≤ tan R Average (DC) output voltage is given as : 1 α +π 2Vm Vo = π α ∫ Vm sin (ωt )dωt = π cos α Power Electronics and 34 Drives (Version 2), Dr. Zainal Salam, 2002 Single-phase diode groups D1 io D3 vp + vs + _ vo D4 _ D2 vn 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 35 Drives (Version 2), Dr. Zainal Salam, 2002 Three-phase rectifiers D1 + van - io D3 + vbn - D5 n vpn + + vcn - vo D2 _ D6 vnn vo =vp −vn D4 van vbn vcn Vm vp Vm vn vo =vp - vn 0 π 2π 3π 4π Power Electronics and 36 Drives (Version 2), Dr. Zainal Salam, 2002 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, vn 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 37 Drives (Version 2), Dr. Zainal Salam, 2002 Three-phase, average voltage vo vo π/3 Vm, L-L 0 π/3 2π/3 Considers only one of the six segments. Obtain its average over 60 degrees or π 3 radians. Average voltage : 2π 3 1 Vo = π 3π3 ∫ Vm,L− L sin(ωt )dωt 3Vm, L − L = [cos(ωt )]ππ33 2 π 3Vm, L − L = = 0.955Vm, L − L π Note that the output DC voltage component of a three - phase rectifier is much higher than of a single - phase. Power Electronics and 38 Drives (Version 2), Dr. Zainal Salam, 2002 Controlled, three-phase D1 + van - io D3 + vbn - D5 vpn n + + vcn - vo D2 _ D6 vnn D4 α van vbn vcn Vm vo Power Electronics and 39 Drives (Version 2), Dr. Zainal Salam, 2002 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 : 2π 3+α 1 Vo = ∫ Vm, L− L sin(ωt )dωt π 3 π 3+α 3Vm, L − L = ⋅ cos α π • 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 40 Drives (Version 2), Dr. Zainal Salam, 2002

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