R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 18.4 Single-phase converter systems containing ideal rectifiers • It is usually desired that the output voltage v(t) be regulated with high accuracy, using a wide-bandwidth feedback loop • For a given constant load characteristic, the instantaneous load current and power are then also constant: pload (t) = v(t)i(t) = VI • The instantaneous input power of a single-phase ideal rectifier is not constant: pac(t) = vg(t)i g(t) with so vg(t) = VM sin (ωt) vg(t) i g(t) = Re 2 V 2M V pac(t) = sin 2 ωt = M 1 – cos 2ωt Re 2Re Fundamentals of Power Electronics 51 Chapter 18: PWM Rectifiers Power flow in single-phase ideal rectifier system • Ideal rectifier is lossless, and contains no internal energy storage. • Hence instantaneous input and output powers must be equal • An energy storage element must be added • Capacitor energy storage: instantaneous power flowing into capacitor is equal to difference between input and output powers: d EC(t) pC(t) = = dt d 1 2 Cv 2C(t) dt = pac(t) – pload(t) Energy storage capacitor voltage must be allowed to vary, in accordance with this equation Fundamentals of Power Electronics 52 Chapter 18: PWM Rectifiers Capacitor energy storage in 1ø system pac(t) Pload vc(t) = d 1 2 Cv 2C(t) dt = pac(t) – pload(t) t Fundamentals of Power Electronics 53 Chapter 18: PWM Rectifiers Single-phase system with internal energy storage + iac(t) vac(t) pload(t) = VI = Pload Ideal rectifier (LFR) i (t) 2 ig(t) vg(t) 〈 pac(t)〉T + + s Re C – vC(t) Dc–dc converter i(t) v(t) load – – Energy storage capacitor Energy storage capacitor voltage vC(t) must be independent of input and output voltage waveforms, so that it can vary according to = d 1 2 Cv 2C(t) dt This system is capable of = pac(t) – pload(t) Fundamentals of Power Electronics 54 • Wide-bandwidth control of output voltage • Wide-bandwidth control of input current waveform • Internal independent energy storage Chapter 18: PWM Rectifiers Hold up time Internal energy storage allows the system to function in other situations where the instantaneous input and output powers differ. A common example: continue to supply load power in spite of failure of ac line for short periods of time. Hold up time: the duration which the dc output voltage v(t) remains regulated after vac(t) has become zero A typical hold-up time requirement: supply load for one complete missing ac line cycle, or 20 msec in a 50 Hz system During the hold-up time, the load power is supplied entirely by the energy storage capacitor Fundamentals of Power Electronics 55 Chapter 18: PWM Rectifiers Energy storage element Instead of a capacitor, and inductor or higher-order LC network could store the necessary energy. But, inductors are not good energy-storage elements Example 100 V 100 µF capacitor 100 A 100 µH inductor each store 1 Joule of energy But the capacitor is considerably smaller, lighter, and less expensive So a single big capacitor is the best solution Fundamentals of Power Electronics 56 Chapter 18: PWM Rectifiers Inrush current A problem caused by the large energy storage capacitor: the large inrush current observed during system startup, necessary to charge the capacitor to its equilibrium value. Boost converter is not capable of controlling this inrush current. Even with d = 0, a large current flows through the boost converter diode to the capacitor, as long as v(t) < vg(t). Additional circuitry is needed to limit the magnitude of this inrush current. Converters having buck-boost characteristics are capable of controlling the inrush current. Unfortunately, these converters exhibit higher transistor stresses. Fundamentals of Power Electronics 57 Chapter 18: PWM Rectifiers Universal input The capability to operate from the ac line voltages and frequencies found everywhere in the world: 50Hz and 60Hz Nominal rms line voltages of 100V to 260V: 100V, 110V, 115V, 120V, 132V, 200V, 220V, 230V, 240V, 260V Regardless of the input voltage and frequency, the near-ideal rectifier produces a constant nominal dc output voltage. With a boost converter, this voltage is 380 or 400V. Fundamentals of Power Electronics 58 Chapter 18: PWM Rectifiers Low-frequency model of dc-dc converter Dc-dc converter produces well-regulated dc load voltage V. Load therefore draws constant current I. Load power is therefore the constant value Pload = VI. To the extent that dc-dc converter losses can be neglected, then dc-dc converter input power is Pload , regardless of capacitor voltage vc(t). Dc-dc converter input port behaves as a power sink. A low frequency converter model is p (t) = VI = P i (t) load 2 + C vC(t) + Pload V + – – Energy storage capacitor Fundamentals of Power Electronics 59 load i(t) v(t) load – Dc-dc converter Chapter 18: PWM Rectifiers Low-frequency energy storage process, 1ø system A complete low-frequency system model: iac(t) vac(t) pload(t) = VI = Pload i2(t) ig(t) + vg(t) + 〈 pac(t)〉Ts Re C vC(t) + Pload V + – – – Ideal rectifier (LFR) Energy storage capacitor i(t) v(t) load – Dc-dc converter • Difference between rectifier output power and dc-dc converter input power flows into capacitor • In equilibrium, average rectifier and load powers must be equal • But the system contains no mechanism to accomplish this • An additional feeback loop is necessary, to adjust Re such that the rectifier average power is equal to the load power Fundamentals of Power Electronics 60 Chapter 18: PWM Rectifiers Obtaining average power balance iac(t) vac(t) pload(t) = VI = Pload i2(t) ig(t) + vg(t) + 〈 pac(t)〉Ts Re C vC(t) + Pload V + – v(t) load – – Ideal rectifier (LFR) Energy storage capacitor i(t) – Dc-dc converter If the load power exceeds the average rectifier power, then there is a net discharge in capacitor energy and voltage over one ac line cycle. There is a net increase in capacitor charge when the reverse is true. This suggests that rectifier and load powers can be balanced by regulating the energy storage capacitor voltage. Fundamentals of Power Electronics 61 Chapter 18: PWM Rectifiers A complete 1ø system containing three feedback loops Boost converter i2(t) + ig(t) + iac(t) vac(t) L D1 vg(t) Q1 vC(t) – vcontrol(t) vg(t) Multiplier X i(t) DC–DC Converter C Load v(t) – – ig(t) Rs d(t) PWM v(t) va(t) vref1(t) = kxvg(t)vcontrol(t) + v (t) +– err Gc(s) Compensator and modulator –+ vref3 Compensator Wide-bandwidth output voltage controller Wide-bandwidth input current controller vC(t) Compensator –+ vref2 Low-bandwidth energy-storage capacitor voltage controller Fundamentals of Power Electronics 62 Chapter 18: PWM Rectifiers Bandwidth of capacitor voltage loop • The energy-storage-capacitor voltage feedback loop causes the dc component of vc(t) to be equal to some reference value • Average rectifier power is controlled by variation of Re. • Re must not vary too quickly; otherwise, ac line current harmonics are generated • Extreme limit: loop has infinite bandwidth, and vc(t) is perfectly regulated to be equal to a constant reference value • Energy storage capacitor voltage then does not change, and this capacitor does not store or release energy • Instantaneous load and ac line powers are then equal • Input current becomes i ac(t) = Fundamentals of Power Electronics pac(t) p (t) Pload = load = vac(t) vac(t) VM sin ωt 63 Chapter 18: PWM Rectifiers Input current waveform, extreme limit i ac(t) = pac(t) p (t) Pload = load = vac(t) vac(t) VM sin ωt THD → ∞ Power factor → 0 vac(t) iac(t) t Fundamentals of Power Electronics 64 So bandwidth of capacitor voltage loop must be limited, and THD increases rapidly with increasing bandwidth Chapter 18: PWM Rectifiers 18.4.2 Modeling the outer low-bandwidth control system This loop maintains power balance, stabilizing the rectifier output voltage against variations in load power, ac line voltage, and component values The loop must be slow, to avoid introducing variations in Re at the harmonics of the ac line frequency Objective of our modeling efforts: low-frequency small-signal model that predicts transfer functions at frequencies below the ac line frequency Fundamentals of Power Electronics 65 Chapter 18: PWM Rectifiers Large signal model averaged over switching period Ts Ideal rectifier (LFR) 〈 ig(t)〉Ts s 〈 p(t)〉T 〈 vg(t)〉T s + – 〈 i2(t)〉T + s C Re (vcontrol ) 〈 v(t)〉T s Load – ac input dc output vcontrol Ideal rectifier model, assuming that inner wide-bandwidth loop operates ideally High-frequency switching harmonics are removed via averaging Ac line-frequency harmonics are included in model Nonlinear and time-varying Fundamentals of Power Electronics 66 Chapter 18: PWM Rectifiers Predictions of large-signal model If the input voltage is vg(t) = 2 vg,rms sin ωt Ideal rectifier (LFR) 〈 ig(t)〉Ts 〈 p(t)〉T 〈 vg(t)〉T s + – Re (vcontrol ) 〈 i2(t)〉T s + s C 〈 v(t)〉T s – Then the instantaneous power is: vg(t) ac input dc output vcontrol 2 v 2g,rms p(t) T = = 1 – cos 2ωt s Re(vcontrol(t)) Re(vcontrol(t)) Ts which contains a constant term plus a secondharmonic term Fundamentals of Power Electronics 67 Chapter 18: PWM Rectifiers Load Separation of power source into its constant and time-varying components 〈 i2(t)〉T s + V 2g,rms – cos 2 2ωt Re V 2g,rms Re C 〈 v(t)〉Ts Load – Rectifier output port The second-harmonic variation in power leads to second-harmonic variations in the output voltage and current Fundamentals of Power Electronics 68 Chapter 18: PWM Rectifiers Removal of even harmonics via averaging v(t) 〈 v(t)〉Ts 〈 v(t)〉T 2L t T2L = Fundamentals of Power Electronics 1 2 69 2π = π ω ω Chapter 18: PWM Rectifiers Resulting averaged model 〈 i2(t)〉T2L + V 2g,rms Re C 〈 v(t)〉T2L Load – Rectifier output port Time invariant model Power source is nonlinear Fundamentals of Power Electronics 70 Chapter 18: PWM Rectifiers Perturbation and linearization The averaged model predicts that the rectifier output current is i 2(t) T 2L = p(t) v(t) T 2L T 2L = v 2g,rms(t) Re(vcontrol(t)) v(t) = f vg,rms(t), v(t) Let T 2L , vcontrol(t)) T 2L with v(t) i 2(t) T 2L T 2L = V + v(t) V >> v(t) = I 2 + i 2(t) I 2 >> i 2(t) vg,rms = Vg,rms + vg,rms(t) vcontrol(t) = Vcontrol + vcontrol(t) Fundamentals of Power Electronics Vg,rms >> vg,rms(t) Vcontrol >> vcontrol(t) 71 Chapter 18: PWM Rectifiers Linearized result vcontrol(t) I 2 + i 2(t) = g 2vg,rms(t) + j2v(t) – r2 where g2 = df vg,rms, V, Vcontrol) dvg,rms – 1 = r2 j2 = Vg,rms 2 = Re(Vcontrol) V v g,rms = V g,rms df Vg,rms, v d v , Vcontrol) T 2L T 2L I2 =– V v T =V 2L df Vg,rms, V, vcontrol) dvcontrol Fundamentals of Power Electronics v control = V control 72 V 2g,rms dRe(vcontrol) =– VR 2e (Vcontrol) dvcontrol v control = V control Chapter 18: PWM Rectifiers Small-signal equivalent circuit i2 + r2 j2 vcontrol g 2 vg,rms C v R – Rectifier output port Predicted transfer functions Control-to-output v(s) 1 = j2 R||r 2 vcontrol(s) 1 + sC R||r 2 Line-to-output v(s) 1 = g 2 R||r 2 vg,rms(s) 1 + sC R||r 2 Fundamentals of Power Electronics 73 Chapter 18: PWM Rectifiers Model parameters Table 18.1 Small-signal model parameters for several types of rectifier control schemes Controller type g2 j2 r2 Average current control with feedforward, Fig. 18.14 0 Pav VVcontrol V2 Pav Current-programmed control, Fig. 18.16 2Pav VVg,rms Pav VVcontrol V2 Pav Nonlinear-carrier charge control of boost rectifier, Fig. 18.21 2Pav VVg,rms Pav VVcontrol V2 2Pav Boost with critical conduction mode control, Fig. 18.20 2Pav VVg,rms Pav VVcontrol V2 Pav DCM buck-boost, flyback, SEPIC, or Cuk converters 2Pav VVg,rms 2Pav VD V2 Pav Fundamentals of Power Electronics 74 Chapter 18: PWM Rectifiers Constant power load ig(t) + iac(t) vg(t) vac(t) pload(t) = VI = Pload i2(t) Re C – + + 〈 pac(t)〉Ts vC(t) Pload V + – v(t) load – – Ideal rectifier (LFR) Energy storage capacitor i(t) Dc-dc converter Rectifier and dc-dc converter operate with same average power Incremental resistance R of constant power load is negative, and is 2 V R=– Pav which is equal in magnitude and opposite in polarity to rectifier incremental output resistance r2 for all controllers except NLC Fundamentals of Power Electronics 75 Chapter 18: PWM Rectifiers Transfer functions with constant power load When r2 = – R, the parallel combination r2 || R becomes equal to zero. The small-signal transfer functions then reduce to j2 v(s) = vcontrol(s) sC g2 v(s) = vg,rms(s) sC Fundamentals of Power Electronics 76 Chapter 18: PWM Rectifiers 18.5 RMS values of rectifier waveforms Doubly-modulated transistor current waveform, boost rectifier: iQ(t) t Computation of rms value of this waveform is complex and tedious Approximate here using double integral Generate tables of component rms and average currents for various rectifier converter topologies, and compare Fundamentals of Power Electronics 77 Chapter 18: PWM Rectifiers RMS transistor current RMS transistor current is I Qrms = 1 Tac T ac iQ(t) i 2Q(t)dt 0 Express as sum of integrals over all switching periods contained in one ac line period: I Qrms = 1 T Tac s T ac/T s ∑ n=1 1 Ts t nT s i 2Q(t)dt (n-1)T s Quantity in parentheses is the value of iQ2, averaged over the nth switching period. Fundamentals of Power Electronics 78 Chapter 18: PWM Rectifiers Approximation of RMS expression I Qrms = T ac/T s 1 T Tac s 1 Ts ∑ n=1 nT s i 2Q(t)dt (n-1)T s When Ts << Tac, then the summation can be approximated by an integral, which leads to the double-average: I Qrms ≈ = = Fundamentals of Power Electronics 1 lim T Tac T s→0 s 1 Tac T ac 0 i 2Q(t) 79 Ts 1 Ts T ac/T s ∑ n=1 t+T s 1 Ts nT s i 2Q(τ)dτ (n-1)T s i 2Q(τ)dτ dt t T ac Chapter 18: PWM Rectifiers 18.5.1 Boost rectifier example For the boost converter, the transistor current iQ(t) is equal to the input current when the transistor conducts, and is zero when the transistor is off. The average over one switching period of iQ2(t) is therefore i 2 Q T s t+T s = 1 i 2Q(t)dt Ts t = d(t)i 2ac(t) If the input voltage is vac(t) = VM sin ωt then the input current will be given by VM i ac(t) = sin ωt Re and the duty cycle will ideally be V = 1 vac(t) 1 – d(t) Fundamentals of Power Electronics 80 (this neglects converter dynamics) Chapter 18: PWM Rectifiers Boost rectifier example Duty cycle is therefore VM d(t) = 1 – sin ωt V Evaluate the first integral: i 2 Q T s V 2M VM = 2 1– sin ωt V Re sin 2 ωt Now plug this into the RMS formula: I Qrms = = I Qrms = Fundamentals of Power Electronics 1 Tac T ac i 2Q 0 T ac 1 Tac 0 2 M 2 e 2 V Tac R 81 Ts dt V 2M VM 1– sin ωt 2 V Re T ac/2 sin 2 ωt – 0 sin 2 ωt dt VM sin 3 ωt dt V Chapter 18: PWM Rectifiers Integration of powers of sin θ over complete half-cycle n 1 π π n sin (θ)dθ = 0 2 2⋅4⋅6 (n – 1) if n is odd π 1⋅3⋅5 n 1⋅3⋅5 (n – 1) if n is even 2⋅4⋅6 n Fundamentals of Power Electronics 82 1 π π sin n (θ)dθ 0 1 2 π 2 1 2 3 4 3π 4 3 8 5 16 15π 6 15 48 Chapter 18: PWM Rectifiers Boost example: transistor RMS current VM I Qrms = 2 Re VM 8 1– 3π V = I ac rms VM 8 1– 3π V Transistor RMS current is minimized by choosing V as small as possible: V = VM. This leads to I Qrms = 0.39I ac rms When the dc output voltage is not too much greater than the peak ac input voltage, the boost rectifier exhibits very low transistor current. Efficiency of the boost rectifier is then quite high, and 95% is typical in a 1kW application. Fundamentals of Power Electronics 83 Chapter 18: PWM Rectifiers Table of rectifier current stresses for various topologies Tabl e 18. 3 Summary of rectifier current stresses for several converter topologies rms Average Peak CCM boost Transistor I ac rms Diode I dc VM 1 – 8 3π V V I ac rms 2 π2 1 – π M 8 V 16 V 3π V M I dc I ac rms 2 π2 I ac rms Inductor I ac rms 2 2 I dc V VM I ac rms 2 CCM flyback, with n:1 isolation transformer and input filter Transistor, xfmr primary I ac rms L1 C1 Diode, xfmr secondary Fundamentals of Power Electronics I ac rms I dc V 1+ 8 M 3π nV I ac rms 2 π2 I ac rms I ac rms 2 π2 8 VM 3π nV 0 3 + 16 nV 2 3π V M I dc 84 I ac rms 2 1 + V n I ac rms 2 I ac rms 2 max 1, VM nV 2I dc 1 + nV VM Chapter 18: PWM Rectifiers Table of rectifier current stresses continued CCM SEPIC, nonisolated Transistor V 1+ 8 M 3π V I ac rms I ac rms L1 C1 8 VM 3π V I ac rms L2 I dc I ac rms 2 1 + I ac rms 2 π2 I ac rms 2 0 VM 3 V 2 I ac rms V M 2 V 3 + 16 V 2 3π V M I dc I ac rms Diode I ac rms 2 π2 I ac rms max 1, I ac rms VM V VM V VM 2 V 2I dc 1 + V VM CCM SEPIC, with n:1 isolation transformer transistor I ac rms V 1+ 8 M 3π nV I ac rms L1 C1, xfmr primary Diode, xfmr secondary I ac rms I dc 8 VM 3π nV 3 + 16 nV 2 3π V M I ac rms 2 π2 I ac rms 2 1 + I ac rms 2 π2 I ac rms 2 0 I dc VM nV I ac rms 2 max 1, n 2I dc 1 + nV VM I ac rms = 2 V , ac input voltage = V M sin(ω t) VM I dc dc output voltage = V with, in all cases, Fundamentals of Power Electronics 85 Chapter 18: PWM Rectifiers Comparison of rectifier topologies Boost converter • Lowest transistor rms current, highest efficiency • Isolated topologies are possible, with higher transistor stress • No limiting of inrush current • Output voltage must be greater than peak input voltage Buck-boost, SEPIC, and Cuk converters • Higher transistor rms current, lower efficiency • Isolated topologies are possible, without increased transistor stress • Inrush current limiting is possible • Output voltage can be greater than or less than peak input voltage Fundamentals of Power Electronics 86 Chapter 18: PWM Rectifiers Comparison of rectifier topologies 1kW, 240Vrms example. Output voltage: 380Vdc. Input current: 4.2Arms Converter Transistor rms current Transistor voltage Diode rms current Transistor rms current, 120V Diode rms current, 120V Boost 2A 380 V 3.6 A 6.6 A 5.1 A Nonisolated SEPIC 5.5 A 719 V 4.85 A 9.8 A 6.1 A Isolated SEPIC 5.5 A 719 V 36.4 A 11.4 A 42.5 A Isolated SEPIC example has 4:1 turns ratio, with 42V 23.8A dc load Fundamentals of Power Electronics 87 Chapter 18: PWM Rectifiers 18.6 Modeling losses and efficiency in CCM high-quality rectifiers Objective: extend procedure of Chapter 3, to predict the output voltage, duty cycle variations, and efficiency, of PWM CCM low harmonic rectifiers. Approach: Use the models developed in Chapter 3. Integrate over one ac line cycle to determine steady-state waveforms and average power. Boost example L D1 ig(t) i(t) + vg(t) + – Q1 C R v(t) vg(t) RL DRon D' : 1 VF i(t) + – ig(t) RL + – + R – – Dc-dc boost converter circuit Fundamentals of Power Electronics Averaged dc model 88 v(t) Chapter 18: PWM Rectifiers Modeling the ac-dc boost rectifier Boost rectifier circuit ig(t) + iac(t) id(t) RL L + D1 vg(t) vac(t) i(t) Q1 C v(t) R – – controller Averaged model vg(t) RL d(t) Ron + – d'(t) : 1 VF + – ig(t) id(t) i(t) = I + C (large) R v(t) = V – Fundamentals of Power Electronics 89 Chapter 18: PWM Rectifiers Boost rectifier waveforms vg(t) ig(t) 300 10 vg(t) Typical waveforms 8 200 ig(t) (low frequency components) 6 vg(t) ig(t) = Re 4 100 2 0 0 0° d(t) 30° 60° 90° 120° 150° 180° 1 6 0.8 5 id(t) 4 0.6 i(t) = I 3 0.4 2 0.2 1 0 0 0° 30° 60° 90° 120° 150° 180° 0° 30° 60° 90° 120° 150° 180° ωt Fundamentals of Power Electronics 90 Chapter 18: PWM Rectifiers Example: boost rectifier with MOSFET on-resistance ig(t) + id(t) d(t) Ron vg(t) i(t) = I d'(t) : 1 + – C (large) R v(t) = V – Averaged model Inductor dynamics are neglected, a good approximation when the ac line variations are slow compared to the converter natural frequencies Fundamentals of Power Electronics 91 Chapter 18: PWM Rectifiers 18.6.1 Expression for controller duty cycle d(t) ig(t) Solve input side of model: d(t) Ron i g(t)d(t)Ron = vg(t) – d'(t)v with vg(t) ig(t) = Re vg(t) i(t) = I d'(t) : 1 + – + id(t) C (large) R v(t) = V – vg(t) = VM sin ωt eliminate ig(t): solve for d(t): v – vg(t) d(t) = Ron v – vg(t) Re vg(t) d(t)Ron = vg(t) – d'(t)v Re Again, these expressions neglect converter dynamics, and assume that the converter always operates in CCM. Fundamentals of Power Electronics 92 Chapter 18: PWM Rectifiers 18.6.2 Expression for the dc load current Solve output side of model, using charge balance on capacitor C: I = id T ac ig(t) d(t) Ron vg(t) + – i(t) = I d'(t) : 1 + id(t) C (large) R vg(t) i d (t) = d'(t)i g(t) = d'(t) Re v(t) = V – Butd’(t) is: hence id(t) can be expressed as Ron vg(t) 1 – Re d'(t) = Ron v – vg(t) Re Ron 1– 2 Re v g(t) i d (t) = Re Ron v – vg(t) Re Next, average id(t) over an ac line period, to find the dc load current I. Fundamentals of Power Electronics 93 Chapter 18: PWM Rectifiers Dc load current I Now substitute vg (t) = VM sin ωt, and integrate to find 〈id(t)〉Tac: T ac/2 I = id V 2M Re 2 = T ac Tac Ron 1– sin 2 ωt Re v– 0 VM Ron sin ωt Re dt This can be written in the normalized form T ac/2 2 M Ron V 2 I= 1– Tac VRe Re with a= Fundamentals of Power Electronics VM V sin 2 ωt 1 – a sin ωt 0 dt Ron Re 94 Chapter 18: PWM Rectifiers Integration By waveform symmetry, we need only integrate from 0 to Tac/4. Also, make the substitution θ = ωt: I= π/2 2 M V R 2 1 – on π VRe Re 0 sin 2 θ 1 – a sin θ dθ This integral is obtained not only in the boost rectifier, but also in the buck-boost and other rectifier topologies. The solution is π/2 4 π 0 sin 2 θ dθ = F(a) = 22 aπ 1 – a sin θ 1 – a2 • a is typically much smaller than unity • Result is in closed form • a is a measure of the loss resistance relative to Re Fundamentals of Power Electronics – 2a – π + 4 sin – 1 a + 2 cos – 1 a 95 Chapter 18: PWM Rectifiers The integral F(a) π/2 4 π 0 sin 2 θ dθ = F(a) = 22 aπ 1 – a sin θ Approximation via polynomial: Fundamentals of Power Electronics 1 – a2 1.15 1.1 F(a) ≈ 1 + 0.862a + 0.78a 2 For | a | ≤ 0.15, this approximate expression is within 0.1% of the exact value. If the a2 term is omitted, then the accuracy drops to ±2% for | a | ≤ 0.15. The accuracy of F(a) coincides with the accuracy of the rectifier efficiency η. – 2a – π + 4 sin – 1 a + 2 cos – 1 a 1.05 F(a) 1 0.95 0.9 0.85 –0.15 –0.10 –0.05 0.00 0.05 0.10 0.15 a 96 Chapter 18: PWM Rectifiers 18.6.3 Solution for converter efficiency η Converter average input power is V 2M Pin = pin(t) T = ac 2Re Average load power is Pout = VI = V V 2M Ron F(a) 1– VRe Re 2 VM a= V with Ron Re So the efficiency is Pout Ron η= = 1– F(a) Pin Re Polynomial approximation: η≈ 1– Ron Re Fundamentals of Power Electronics 1 + 0.862 VM Ron V R + 0.78 M on V Re V Re 97 2 Chapter 18: PWM Rectifiers Boost rectifier efficiency 1 Pout Ron η= = 1– F(a) Pin Re η .05 R on /R e = 0 0.95 = R on/R e 0.9 0.1 • To obtain high efficiency, choose V slightly larger than VM 0.15 = R R on/ e 0.2 = /R e R on 0.85 0.8 • Efficiencies in the range 90% to 95% can then be obtained, even with Ron as high as 0.2Re 0.75 0.0 0.2 0.4 0.6 0.8 VM /V Fundamentals of Power Electronics 98 1.0 • Losses other than MOSFET on-resistance are not included here Chapter 18: PWM Rectifiers 18.6.4 Design example Let us design for a given efficiency. Consider the following specifications: Output voltage 390 V Output power 500 W rms input voltage 120 V Efficiency 95% Assume that losses other than the MOSFET conduction loss are negligible. Average input power is Pout 500 W Pin = η = = 526 W 0.95 Then the emulated resistance is V 2g, rms (120 V) 2 Re = = = 27.4 Ω Pin 526 W Fundamentals of Power Electronics 99 Chapter 18: PWM Rectifiers Design example Also, VM 120 2 V = = 0.435 V 390 V 95% efficiency with VM/V = 0.435 occurs with Ron/Re ≈ 0.075. 1 η .05 R on /R e = 0 0.95 = R on/R e 0.9 So we require a MOSFET with on resistance of 0.1 0.15 = R R on/ e 0.2 = /R e R on 0.85 0.8 Ron ≤ (0.075) Re = (0.075) (27.4 Ω) = 2 Ω 0.75 0.0 0.2 0.4 0.6 0.8 1.0 VM /V Fundamentals of Power Electronics 100 Chapter 18: PWM Rectifiers

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