R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

17.1 The single-phase full-wave rectifier i g i L L D 4 D 1 v g Z i C v R D 3 D 2 Full-wave rectifier with dc-side L-C filter Two common reasons for including the dc-side L-C filter: Obtain good dc output voltage (large C) and acceptable ac line current waveform (large L) Filter conducted EMI generated by dc load (small L and C) Fundamentals of Power Electronics 2 Chapter 17: Line-commutated rectifiers

17.1.1 Continuous conduction mode Large L v g THD = 29% Typical ac line waveforms for CCM : As L, ac line current approaches a square wave i g 10 ms 20 ms 30 ms 40 ms CCM results, for L : distortion factor = I 1, rms I rms = 4 π 2 = 90.0% t THD = 1 distortion factor 2 1 = 48.3% Fundamentals of Power Electronics 3 Chapter 17: Line-commutated rectifiers

17.1.2 Discontinuous conduction mode Small L Typical ac line waveforms for DCM : As L 0, ac line current approaches impulse functions (peak detection) v g i g THD = 145% 10 ms 20 ms 30 ms 40 ms As the inductance is reduced, the THD rapidly increases, and the distortion factor decreases. Typical distortion factor of a full-wave rectifier with no inductor is in the range 55% to 65%, and is governed by ac system inductance. t Fundamentals of Power Electronics 4 Chapter 17: Line-commutated rectifiers

17.1.3 Behavior when C is large Solution of the full-wave rectifier circuit for infinite C: Define cos (ϕ 1 θ 1 ), PF, M 1.0 0.9 0.8 β cos (ϕ 1 θ 1 ) PF THD 200% 150% β 180 135 K L = 2L RT L 0.7 M 100% 90 M = V V m 0.6 0.5 DCM CCM THD 50% 45 0.4 0 0 0.0001 0.001 0.01 0.1 1 10 Fundamentals of Power Electronics 5 Chapter 17: Line-commutated rectifiers K L

17.1.4 Minimizing THD when C is small Sometimes the L-C filter is present only to remove high-frequency conducted EMI generated by the dc load, and is not intended to modify the ac line current waveform. If L and C are both zero, then the load resistor is connected directly to the output of the diode bridge, and the ac line current waveform is purely sinusoidal. An approximate argument: the L-C filter has negligible effect on the ac line current waveform provided that the filter input impedance Z i has zero phase shift at the second harmonic of the ac line frequency, 2 f L. i g i L L D 4 D 1 v g Z i C v R D 3 D 2 Fundamentals of Power Electronics 6 Chapter 17: Line-commutated rectifiers

Approximate THD f 0 = 1 2π LC R 0 = L C 50 10 THD=30% THD=10% THD=3% Q = R R 0 f p = 1 2πRC = f 0 Q Q THD=0.5% THD=1% 1 0.1 1 10 100 f 0 / f L Fundamentals of Power Electronics 7 Chapter 17: Line-commutated rectifiers

Example v g THD = 3.6% i g 10 ms 20 ms 30 ms 40 ms t Typical ac line current and voltage waveforms, near the boundary between continuous and discontinuous modes and with small dc filter capacitor. f 0 /f L = 10, Q = 1 Fundamentals of Power Electronics 8 Chapter 17: Line-commutated rectifiers

Chapter 18 Pulse-Width Modulated Rectifiers 18.1 Properties of the ideal rectifier 18.2 Realization of a near-ideal rectifier 18.3 Control of the current waveform 18.4 Single-phase converter systems employing ideal rectifiers 18.5 RMS values of rectifier waveforms 18.6 Modeling losses and efficiency in CCM high-quality rectifiers 18.7 Ideal three-phase rectifiers Fundamentals of Power Electronics 1 Chapter 18: PWM Rectifiers

18.1 Properties of the ideal rectifier It is desired that the rectifier present a resistive load to the ac power system. This leads to unity power factor ac line current has same waveshape as voltage i ac = v ac R e i ac R e is called the emulated resistance v ac R e Fundamentals of Power Electronics 2 Chapter 18: PWM Rectifiers

Control of power throughput P av = V 2 ac,rms R e (v control ) i ac Power apparently consumed by R e is actually transferred to rectifier dc output port. To control the amount of output power, it must be possible to adjust the value of R e. v ac R e (v control ) v control Fundamentals of Power Electronics 3 Chapter 18: PWM Rectifiers

Output port model The ideal rectifier is lossless and contains no internal energy storage. Hence, the instantaneous input power equals the instantaneous output power. Since the instantaneous power is independent of the dc load characteristics, the output port obeys a power source characteristic. v ac i ac ac input p= R e (v control ) v control v 2 ac R e (v control ) Ideal rectifier (LFR) p = v ac 2 /R e i v dc output vi=p= v ac 2 R e Fundamentals of Power Electronics 4 Chapter 18: PWM Rectifiers

The dependent power source i i i vi = p p v v p v power source power sink i-v characteristic Fundamentals of Power Electronics 5 Chapter 18: PWM Rectifiers

Equations of the ideal rectifier / LFR Defining equations of the ideal rectifier: i ac = vi=p v ac R e (v control ) When connected to a resistive load of value R, the input and output rms voltages and currents are related as follows: V rms V ac,rms = R Re p= v 2 ac R e (v control ) I ac,rms I rms = R Re Fundamentals of Power Electronics 6 Chapter 18: PWM Rectifiers

18.2 Realization of a near-ideal rectifier Control the duty cycle of a dc-dc converter, such that the input current is proportional to the input voltage: i g i ac dcdc converter 1 : M(d) i v ac v g v C R d i g Controller v g Fundamentals of Power Electronics 7 Chapter 18: PWM Rectifiers

Waveforms v ac V M i g t v V i ac V M /R e t M M min v g V M v ac =V M sin (ωt) v g =V M sin (ωt) M(d) = v v g = M min = V V M V M V sin (ωt) Fundamentals of Power Electronics 8 Chapter 18: PWM Rectifiers

Output-side current Averaged over switching period i= v gi g V 2 = v g 2 VR e i= V M sin VR 2 (ωt) e 2 = V M 2VR e 1 cos (2ωt) Averaged over ac line period i g i ac dcdc converter I = i TL = V 2 M 2VR e P = V 2 M 2R e 1 : M(d) i v ac v g v C R d i g Controller Fundamentals of Power Electronics 9 Chapter 18: PWM Rectifiers v g

Choice of converter M(d) = v v g = V M V sin (ωt) M M min To avoid distortion near line voltage zero crossings, converter should be capable of producing M(d) approaching infinity Above expression neglects converter dynamics Boost, buck-boost, Cuk, SEPIC, and other converters with similar conversion ratios are suitable We will see that the boost converter exhibits lowest transistor stresses. For this reason, it is most often chosen Fundamentals of Power Electronics 10 Chapter 18: PWM Rectifiers

18.2.1 CCM Boost converter with controller to cause input current to follow input voltage i g Boost converter i i ac L D 1 v ac v g Q 1 C v R v g i g d Controller DC output voltage peak AC input voltage Controller varies duty cycle as necessary to make i g proportional to v g Fundamentals of Power Electronics 11 Chapter 18: PWM Rectifiers

Variation of duty cycle in boost rectifier M(d) = v v g = V M V sin (ωt) Since M 1 in the boost converter, it is required that V V M If the converter operates in CCM, then M(d) = 1 1d The duty ratio should therefore follow d=1 v g V in CCM Fundamentals of Power Electronics 12 Chapter 18: PWM Rectifiers

CCM/DCM boundary, boost rectifier Inductor current ripple is i g = v gdt s 2L Low-frequency (average) component of inductor current waveform is i g Ts = v g R e The converter operates in CCM when i g Ts > i g d< 2L R e T s Substitute CCM expression for d: R e < T s 2L 1 v g V for CCM Fundamentals of Power Electronics 13 Chapter 18: PWM Rectifiers

CCM/DCM boundary R e < T s 2L 1 v g V for CCM Note that v g varies with time, between 0 and V M. Hence, this equation may be satisfied at some points on the ac line cycle, and not at others. The converter always operates in CCM provided that R e < 2L T s The converter always operates in DCM provided that R e > T s 2L 1 V M V For R e between these limits, the converter operates in DCM when v g is near zero, and in CCM when v g approaches V M. Fundamentals of Power Electronics 14 Chapter 18: PWM Rectifiers

Static input characteristics of the boost converter A plot of input current i g vs input voltage v g, for various duty cycles d. In CCM, the boost converter equilibrium equation is v g V =1d The input characteristic in DCM is found by solution of the averaged DCM model (Fig. 11.12(b)): i g p V v g v g 2L p d 2 T s Beware! This DCM R e (d) from Chapter 11 is not the same as the rectifier emulated resistance R e = v g /i g V Solve for input current: i g = v g p 2L V v g d 2 T s with p= v g 2 2L d 2 T s Fundamentals of Power Electronics 15 Chapter 18: PWM Rectifiers

Static input characteristics of the boost converter Now simplify DCM current expression, to obtain 2L i VT g 1 v g s V = d 2 v g V CCM/DCM mode boundary, in terms of v g and i g : 2L i VT g > v g s V 1 v g V Fundamentals of Power Electronics 16 Chapter 18: PWM Rectifiers

Boost input characteristics with superimposed resistive characteristic 1 0.75 d = 1 d = 0.8 d = 0.6 d = 0.4 d = 0.2 d = 0 CCM: v g V =1d j g = VT 2L i g s 0.5 0.25 i g =v g /R e DCM CCM DCM: 2L i VT g 1 v g s V CCM when 2L i VT g > v g s V = d 2 v g V 1 v g V 0 0 0.25 0.5 0.75 1 m g = v g V Fundamentals of Power Electronics 17 Chapter 18: PWM Rectifiers

Open-loop DCM approach We found in Chapter 11 that the buck-boost, SEPIC, and Cuk converters, when operated open-loop in DCM, inherently behave as loss-free resistors. This suggests that they could also be used as near-ideal rectifiers, without need for a multiplying controller. Advantage: simple control Disadvantages: higher peak currents, larger input current EMI Like other DCM applications, this approach is usually restricted to low power (< 200W). The boost converter can also be operated in DCM as a low harmonic rectifier. Input characteristic is i g Ts = v g v g2 R e vv g R e Input current contains harmonics. If v is sufficiently greater than v g, then harmonics are small. Fundamentals of Power Electronics 18 Chapter 18: PWM Rectifiers

Other similar approaches Use of other converters (in CCM) that are capable of increasing the voltage: SEPIC, Cuk, buck-boost Flyback, isolated versions of boost, SEPIC, Cuk, etc. Boundary or critical conduction mode: operation of boost or other converter at the boundary between CCM and DCM Buck converter: distortion occurs but stresses are low Resonant converter such as parallel resonant converter or some quasi-resonant converters Converters that combine the functions of rectification, energy storage, and dc-dc conversion Fundamentals of Power Electronics 19 Chapter 18: PWM Rectifiers

18.2.2 DCM flyback converter v ac i ac EMI filter i g v g Flyback converter n : 1 L D 1 i v C R Q 1 D Operation in DCM: we found in Chapter 11 that the converter input port obeys Ohm s law with effective resistance R e = 2n 2 L/D 2 T s. Hence, simply connect input port to AC line. Fundamentals of Power Electronics 20 Chapter 18: PWM Rectifiers

Averaged large-signal model EMI filter i ac i g T Averaged model i T v ac v g 2n 2 L D 2 T s p T v T C R D Under steady-state conditions, operate with constant D Adjust D to control average power drawn from AC line Fundamentals of Power Electronics 21 Chapter 18: PWM Rectifiers

Converter design Select L small enough that DCM operation occurs throughout AC line cycle. DCM occurs provided that d 3 > 0, or d 2 <1D But d 2 =D v g nv Substitute and solve for D: D < 1 1 v g nv Converter operates in DCM in every switching period where above inequality is satisfied. i 1 Area q 1 i pk i 1 Ts t d 1 T s d 2 T s d 3 T s T s To obtain DCM at all points on input AC sinusoid: worst case is at maximum v g = V M : D < 1 1 V M nv Fundamentals of Power Electronics 22 Chapter 18: PWM Rectifiers

Choice of L to obtain DCM everywhere along AC sinusoid We have: D < 1 1 V M nv with V rms V ac,rms = R Re Substitute expression for R e to obtain Solve for L: D = 2nV V M L < L crit = L RTs RT s 4 1 nv V M 2 Worst-case design For variations in load resistance and ac input voltage, the worst case occurs at maximum load power and minimum ac input voltage. The inductance should be chosen as follows: L < L crit-min = R min T s 4 1 nv V M-min 2 Fundamentals of Power Electronics 23 Chapter 18: PWM Rectifiers

18.3 Control of the Current Waveform 18.3.1 Average current control Feedforward 18.3.2 Current programmed control 18.3.3 Critical conduction mode and hysteretic control 18.3.4 Nonlinear carrier control Fundamentals of Power Electronics 24 Chapter 18: PWM Rectifiers

18.3.1 Average current control i g Boost example Low frequency (average) component of input current is controlled to follow input voltage v g L Gate driver v Pulse width modulator Current reference v r v a R s i g Ts G c (s) Compensator Fundamentals of Power Electronics 25 Chapter 18: PWM Rectifiers

Block diagram Current reference derived from input voltage waveform v ac Multiplier allows control of emulated resistance value Compensation of current loop i ac v control Multiplier X i g v g v g R s i g v a Controller Boost converter L v r = k x v g v control v err Q 1 D 1 PWM G c (s) Compensator i C v R Fundamentals of Power Electronics 26 Chapter 18: PWM Rectifiers

The emulated resistance Current sensor has gain R s : i g i ac Boost converter L D 1 i v a =R s i g Ts v ac v g Q 1 C v R If loop is well designed, then: v a v r Multiplier: v r =k x v g v control Hence the emulated resistance is: R e = v g i g = v r k x v control v control Multiplier X v g R s i g v a v r = k x v g v control Controller v err PWM G c (s) Compensator which can be simplified to v a R s R e v control = R s k x v control Fundamentals of Power Electronics 27 Chapter 18: PWM Rectifiers

System model using LFR Average current control i g Ts Ideal rectifier (LFR) i Ts i ac p Ts v ac v g Ts R e C v Ts R R e R e = R s k x v control v control Fundamentals of Power Electronics 28 Chapter 18: PWM Rectifiers

Use of multiplier to control average power As discussed in Chapter 17, an output voltage feedback loop adjusts the emulated resistance R e such that the rectifier power equals the dc load power: P av = V 2 g,rms R e = P load v g v g i g Gate driver Pulse width modulator C v An analog multiplier introduces the dependence of R e on v. x y Multiplier k x xy v ref1 v control v a v err G c (s) Compensator G cv (s) v v ref2 Voltage reference Fundamentals of Power Electronics 29 Chapter 18: PWM Rectifiers

Feedforward Feedforward is sometimes used to cancel out disturbances in the input voltage v g. v g i g v To maintain a given power throughput P av, the reference voltage v ref1 should be Gate driver v ref 1 = P avv g R s 2 V g,rms v g Pulse width modulator Peak detector VM x z y multiplier k v xy z 2 v a v ref1 v control G c (s) Compensator G cv (s) v ref2 Voltage reference Fundamentals of Power Electronics 30 Chapter 18: PWM Rectifiers

Feedforward, continued Controller with feedforward produces the following reference: i g v ref 1 = k vv control v g V M 2 The average power is then given by P av = k vv control 2R s v g v g Gate driver Pulse width modulator v Peak detector VM x z y multiplier k v xy z 2 v ref1 v control v a G c (s) Compensator G cv (s) v ref2 Voltage reference Fundamentals of Power Electronics 31 Chapter 18: PWM Rectifiers

Modeling the inner wide-bandwidth average current controller Averaged (but not linearized) boost converter model: L i Ts i 1 Ts v g Ts v 1 Ts i 2 Ts v 2 Ts C R v Ts Averaged switch network In Chapter 7, we perturbed and linearized using the assumptions v g Ts = V g v g d=d d d'=d' d i Ts = i 1 Ts = I i v Ts = v 2 Ts = V v v 1 Ts = V 1 v 1 i 2 Ts = I 2 i 2 Problem: variations in v g, i 1, and d are not small. So we are faced with the design of a control system that exhibits significant nonlinear time-varying behavior. Fundamentals of Power Electronics 32 Chapter 18: PWM Rectifiers

Linearizing the equations of the boost rectifier When the rectifier operates near steady-state, it is true that v Ts = V v with v << V In the special case of the boost rectifier, this is sufficient to linearize the equations of the average current controller. The boost converter average inductor voltage is L substitute: L di g Ts dt di g Ts dt = v g Ts d'v d'v = v g Ts d'v d'v Fundamentals of Power Electronics 33 Chapter 18: PWM Rectifiers

Linearized boost rectifier model L di g Ts dt = v g Ts d'v d'v The nonlinear term is much smaller than the linear ac term. Hence, it can be discarded to obtain L Equivalent circuit: di g Ts dt = v g Ts d'v L i g Ts i g (s) d(s) = V sl v g Ts d'v Fundamentals of Power Electronics 34 Chapter 18: PWM Rectifiers

The quasi-static approximation The above approach is not sufficient to linearize the equations needed to design the rectifier averaged current controllers of buck-boost, Cuk, SEPIC, and other converter topologies. These are truly nonlinear timevarying systems. An approximate approach that is sometimes used in these cases: the quasi-static approximation Assume that the ac line variations are much slower than the converter dynamics, so that the rectifier always operates near equilibrium. The quiescent operating point changes slowly along the input sinusoid, and we can find the slowly-varying equilibrium duty ratio as in Section 18.2.1. The converter small-signal transfer functions derived in Chapters 7 and 8 are evaluated, using the time-varying operating point. The poles, zeroes, and gains vary slowly as the operating point varies. An average current controller is designed, that has a positive phase margin at each operating point. Fundamentals of Power Electronics 35 Chapter 18: PWM Rectifiers

Quasi-static approximation: discussion In the literature, several authors have reported success using this method Should be valid provided that the converter dynamics are suffieiently fast, such that the converter always operates near the assumed operating points No good condition on system parameters, which can justify the approximation, is presently known for the basic converter topologies It is well-understood in the field of control systems that, when the converter dynamics are not sufficiently fast, then the quasi-static approximation yields neither necessary nor sufficient conditions for stability. Such behavior can be observed in rectifier systems. Worstcase analysis to prove stability should employ simulations. Fundamentals of Power Electronics 36 Chapter 18: PWM Rectifiers

18.3.2 Current programmed control Current programmed control is a natural approach to obtain input resistor emulation: Peak transistor current is programmed to follow input voltage. Peak transistor current differs from average inductor current, because of inductor current ripple and artificial ramp. This leads to significant input current waveform distortion. v control i g v g v g Multiplier X i s i a Boost converter L D 1 Q 1 m a Clock 0 i c Comparator = k x v g v control Current-programmed controller i 2 T s C S Q R Latch v R Fundamentals of Power Electronics 37 Chapter 18: PWM Rectifiers

CPM boost converter: Static input characteristics i g Ts = v g Li 2 c f s V v g v g m a L i c 1 v g V m a v g L T s in DCM in CCM 1 Static input characteristics of CPM boost, with minimum slope compensation: Mode boundary: CCM occurs when or, i g Ts > T sv 2L i c > T sv L It is desired that v g V m a L V 1 v g V v g V Minimum slope compensation: 1 v g V i c = v 0.2 g R e 0 m a = V 2L Fundamentals of Power Electronics 38 Chapter 18: PWM Rectifiers R j g = i g base Ts V 0.8 0.6 0.4 R e = 0.1R base R e = 0.2R base R e = 0.33R base DCM R e = 0.5R base CCM m a = V 2L R base = 2L T s R e = R base R e = 2R base R e = 4R base R e = 0.0 0.2 0.4 0.6 0.8 1.0 v g V 10R base

Input current waveforms with current mode control i g Peak i g 1.0 0.8 0.6 0.4 0.2 0.0 Sinusoid R e = 0.1R base Re = 0.33Rbase R e = 2R base m a = V 2L R base = 2L T s ωt Substantial distortion can occur Can meet harmonic limits if the range of operating points is not too large Difficult to meet harmonic limits in a universal input supply Fundamentals of Power Electronics 39 Chapter 18: PWM Rectifiers

18.3.3 Critical conduction mode and hysteretic control Variable switching frequency schemes Hysteretic control Critical conduction mode (boundary between CCM and DCM) i g i g Hysteretic control ωt Critical conduction mode t on ωt Fundamentals of Power Electronics 40 Chapter 18: PWM Rectifiers

An implementation of critical conduction control EMI filter i g Boost converter i i ac L D 1 v ac v g Q 1 C v R v control v g Multiplier X R s va Zero current i g detector S Q v r = k x v g v control Comparator Controller R Latch Fundamentals of Power Electronics 41 Chapter 18: PWM Rectifiers

Pros and cons of critical conduction control Simple, low-cost controller ICs Low-frequency harmonics are very small, with constant transistor on-time (for boost converter) Small inductor Increased peak current Increased conduction loss, reduced switching loss Requires larger input filter Variable switching frequency smears out the current EMI spectrum Cannot synchronize converter switching frequencies Fundamentals of Power Electronics 42 Chapter 18: PWM Rectifiers

Analysis i g Transistor is on for fixed time t on Transistor off-time ends when inductor current reaches zero Ratio of v g to i g is R e = 2L t on t on ωt On time, as a function of load power and line voltage: t on = 4LP V M 2 Inductor volt-second balance: v g t on v g V t off =0 Solve for t off : t off = t on v g V v g Fundamentals of Power Electronics 43 Chapter 18: PWM Rectifiers

Switching frequency variations Solve for how the controller varies the switching frequency over the ac line period: 1 T s = t off t on T s = 4LP V M 2 1 v g V For sinusoidal line voltage variations, the switching frequency will therefore vary as follows: f s = 1 = V 2 M T s 4LP 1V M V sin (ωt) Minimum and maximum limits on switching frequency: max f s = V 2 M 4LP min f s = V 2 M 4LP 1V M V These equations can be used to select the value of the inductance L. Fundamentals of Power Electronics 44 Chapter 18: PWM Rectifiers

18.3.4 Nonlinear carrier control Can attain simple control of input current waveform without sensing the ac input voltage, and with operation in continuous conduction mode The integral of the sensed switch current (charge) is compared to a nonlinear carrier waveform (i.e., a nonlinear ramp), on a cycle-bycycle basis Carrier waveform depends on converter topology Very low harmonics in CCM. Waveform distortion occurs in DCM. Peak current mode control is also possible, with a different carrier Fundamentals of Power Electronics 45 Chapter 18: PWM Rectifiers

Controller block diagram Nonlinear carrier charge control of boost converter Boost converter i g L D 1 i s v g n : 1 i s Q 1 C v R C i i s /n v i Comparator Latch R Q v c v i v c Nonlinear carrier generator 0 T s S Q 0 v i dt s T s v control Clock Nonlinear-carrier charge controller Fundamentals of Power Electronics 46 Chapter 18: PWM Rectifiers

Derivation of NLC approach The average switch current is i s Ts = 1 T s t t T s i s (τ)dτ We could make the controller regulate the average switch current by Integrating the monitored switch current Resetting the integrator to zero at the beginning of each switching period Turning off the transistor when the integrator reaches a reference value In the controller diagram, the integrator follows this equation: i.e., v i = 1 C i 0 dt s v i (dt s )= i s T s nc i f s i s (τ) n dτ for 0 < t < dt s for interval 0 < t < T s Fundamentals of Power Electronics 47 Chapter 18: PWM Rectifiers

How to control the average switch current Input resistor emulation: i g Ts = v g Ts R e (v control ) Relate average switch current to input current (assuming CCM): i s Ts = d i g Ts Relate input voltage to output voltage (assuming CCM): v g Ts = d v Ts Substitute above equations to find how average switch current should be controlled: i s Ts = d 1d v Ts R e (v control ) Fundamentals of Power Electronics 48 Chapter 18: PWM Rectifiers

Implementation using nonlinear carrier Desired control, from previous slide: i s Ts = d 1d v Ts R e (v control ) Generate carrier waveform as follows (replace d by t/t s ): v c =v control t Ts 1 t T s for 0 t T s v c (t T s )=v c The controller switches the transistor off when the integrator voltage equals the carrier waveform. This leads to: v i (dt s )=v c (dt s )=v control d 1d i s Ts nc i f s = v control d 1d R e (v control )=d 1d v Ts i s Ts = v Ts nc i f s v control Fundamentals of Power Electronics 49 Chapter 18: PWM Rectifiers

Generating the parabolic carrier v control Removal of dc component v c Integrator with reset Integrator with reset Clock (one approach, suitable for discrete circuitry) Note that no separate multiplier circuit is needed Fundamentals of Power Electronics 50 Chapter 18: PWM Rectifiers

18.4 Single-phase converter systems containing ideal rectifiers It is usually desired that the output voltage v 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: p load =vi=vi The instantaneous input power of a single-phase ideal rectifier is not constant: with p ac =v g i g v g =V M sin (ωt) i g = v g R e so p ac = V 2 M sin R 2 ωt = V 2 M e 2R e 1 cos 2ωt 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: p C = de C dt = d 1 2 Cv 2 C dt = p ac p load 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 p ac P load v c = d 1 2 Cv 2 C dt = p ac p load Fundamentals of Power Electronics 53 Chapter 18: PWM Rectifiers t

Single-phase system with internal energy storage i g Ideal rectifier (LFR) i 2 p load = VI = P load v ac i ac v g R e p ac Ts C v C Dcdc converter v i load Energy storage capacitor Energy storage capacitor voltage v C must be independent of input and output voltage waveforms, so that it can vary according to = d 1 2 Cv 2 C dt = p ac p load This system is capable of Wide-bandwidth control of output voltage Wide-bandwidth control of input current waveform Internal independent energy storage Fundamentals of Power Electronics 54 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 remains regulated after v ac 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 < v g. 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 P load = VI. To the extent that dc-dc converter losses can be neglected, then dc-dc converter input power is P load, regardless of capacitor voltage v c. Dc-dc converter input port behaves as a power sink. A low frequency converter model is i 2 p load = VI = P load i C v C P load V v load Energy storage capacitor Dc-dc converter Fundamentals of Power Electronics 59 Chapter 18: PWM Rectifiers

Low-frequency energy storage process, 1ø system A complete low-frequency system model: i g i 2 i ac p ac Ts p load = VI = P load i v ac v g R e C v C P load V v load Ideal rectifier (LFR) Energy storage capacitor 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 R e 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 i g i 2 p load = VI = P load i ac p ac Ts i v ac v g R e C v C P load V v load Ideal rectifier (LFR) Energy storage capacitor 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 v ac i g i ac v g Boost converter L D 1 Q 1 i 2 v C C DCDC Converter i Load v v control Multiplier X v g R s i g v a PWM d v ref1 = k x v g v control v err G c (s) Compensator Wide-bandwidth input current controller v Compensator and modulator v ref3 Wide-bandwidth output voltage controller v C Compensator v ref2 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 v c to be equal to some reference value Average rectifier power is controlled by variation of R e. R e must not vary too quickly; otherwise, ac line current harmonics are generated Extreme limit: loop has infinite bandwidth, and v c 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 = p ac v ac = p load v ac = P load V M sin ωt Fundamentals of Power Electronics 63 Chapter 18: PWM Rectifiers

Input current waveform, extreme limit i ac = p ac v ac = p load v ac = P load V M sin ωt THD Power factor 0 v ac i ac t So bandwidth of capacitor voltage loop must be limited, and THD increases rapidly with increasing bandwidth Fundamentals of Power Electronics 64 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 R e 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 T s Ideal rectifier (LFR) i g Ts p Ts i 2 Ts v g Ts R e (v control ) C v Ts Load ac input dc output v control 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 Ideal rectifier (LFR) i g Ts p Ts i 2 Ts v g = 2v g,rms sin ωt v g Ts R e (v control ) C v Ts Load Then the instantaneous power is: ac input v control dc output p Ts = 2 v g Ts R e (v control ) = 2 v g,rms R e (v control ) 1 cos 2ωt which contains a constant term plus a secondharmonic term Fundamentals of Power Electronics 67 Chapter 18: PWM Rectifiers

Separation of power source into its constant and time-varying components i 2 Ts V 2 g,rms cos R 2 2ωt e 2 V g,rms R e C v 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 v Ts v T2L t T 2L = 1 2 2π ω = π ω Fundamentals of Power Electronics 69 Chapter 18: PWM Rectifiers

Resulting averaged model i 2 T2L 2 V g,rms R e C v 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 T2L = p T 2L v T2L = 2 v g,rms R e (v control ) v T2L = f v g,rms, v T2L, v control ) Let with v T2L = V v i 2 T2L = I 2 i 2 v g,rms = V g,rms v g,rms v control =V control v control V >> v I 2 >> i 2 V g,rms >> v g,rms V control >> v control Fundamentals of Power Electronics 71 Chapter 18: PWM Rectifiers

Linearized result where I 2 i 2 =g 2 v g,rms j 2 v v control r 2 g 2 = df v g,rms, V, V control ) dv g,rms v g,rms = V g,rms = 2 R e (V control ) V g,rms V 1 r 2 = df V g,rms, v T2L, V control ) = I 2 dv V T2L v = V T2L j 2 = df V g,rms, V, v control ) dv control v control = V control = 2 V g,rms VR 2 e (V control ) dr e (v control ) dv control v control = V control Fundamentals of Power Electronics 72 Chapter 18: PWM Rectifiers

Small-signal equivalent circuit i 2 g 2 v g,rms j 2 v control C v R r 2 Rectifier output port Predicted transfer functions Control-to-output v(s) v control (s) = j 2 R r 2 1 1sC R r 2 Line-to-output v(s) v g,rms (s) = g 2 R r 2 1 1sC 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 g 2 j 2 r 2 Average current control with feedforward, Fig. 18.14 Current-programmed control, Fig. 18.16 Nonlinear-carrier charge control of boost rectifier, Fig. 18.21 Boost with critical conduction mode control, Fig. 18.20 0 2P av VV g,rms 2P av VV g,rms 2P av VV g,rms P av VV control V 2 P av P av VV control V 2 P av P av VV control V 2 2P av P av VV control V 2 P av DCM buck-boost, flyback, SEPIC, or Cuk converters 2P av VV g,rms 2P av VD V 2 P av Fundamentals of Power Electronics 74 Chapter 18: PWM Rectifiers

Constant power load i g i 2 p load = VI = P load i ac p ac Ts i v ac v g R e C v C P load V v load Ideal rectifier (LFR) Energy storage capacitor Dc-dc converter Rectifier and dc-dc converter operate with same average power Incremental resistance R of constant power load is negative, and is R = V 2 P av which is equal in magnitude and opposite in polarity to rectifier incremental output resistance r 2 for all controllers except NLC Fundamentals of Power Electronics 75 Chapter 18: PWM Rectifiers

Transfer functions with constant power load When r 2 = R, the parallel combination r 2 R becomes equal to zero. The small-signal transfer functions then reduce to v(s) v control (s) = j 2 sc v(s) v g,rms (s) = g 2 sc Fundamentals of Power Electronics 76 Chapter 18: PWM Rectifiers

18.5 RMS values of rectifier waveforms Doubly-modulated transistor current waveform, boost rectifier: i Q 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 t Fundamentals of Power Electronics 77 Chapter 18: PWM Rectifiers

RMS transistor current RMS transistor current is i Q I Qrms = 1 Tac 0 T ac i Q 2 dt Express as sum of integrals over all switching periods contained in one ac line period: t T ac /T s I Qrms = 1 T 1 s i 2 Q dt Tac Ts n =1 nt s (n-1)t s Quantity in parentheses is the value of i Q2, averaged over the n th switching period. Fundamentals of Power Electronics 78 Chapter 18: PWM Rectifiers

Approximation of RMS expression T ac /T s n =1 I Qrms = 1 T 1 s i 2 Q dt Tac Ts nt s (n-1)t s When T s << T ac, then the summation can be approximated by an integral, which leads to the double-average: I Qrms 1 Tac T ac /T s nt s lim Ts T 1 0 s i 2 Q (τ)dτ Ts n=1 (n-1)t s = 1 1 i 2 Q (τ)dτdt Tac Ts 0 T ac t tt s = i Q 2 Ts T ac Fundamentals of Power Electronics 79 Chapter 18: PWM Rectifiers

18.5.1 Boost rectifier example For the boost converter, the transistor current i Q 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 i Q2 is therefore If the input voltage is i Q 2 tt s t = 1 i T s T 2 Q dt s = di 2 ac v ac =V M sin ωt then the input current will be given by i ac = V M Re and the duty cycle will ideally be sin ωt V v ac = 1 1d (this neglects converter dynamics) Fundamentals of Power Electronics 80 Chapter 18: PWM Rectifiers

Boost rectifier example Duty cycle is therefore d=1 V M V Evaluate the first integral: i Q 2 = V 2 M T 2 s R e sin ωt Now plug this into the RMS formula: I Qrms = 1 Tac i Q 2 1 V M V sin ωt sin 2 ωt 0 T ac T s dt = 1 Tac 0 T ac 2 V M 2 R e 1 V M V sin ωt sin 2 ωt dt I Qrms = 2 2 V M Tac 2 sin 2 ωt V M sin 3 R e V 0 T ac /2 ωt dt Fundamentals of Power Electronics 81 Chapter 18: PWM Rectifiers

Integration of powers of sin θ over complete half-cycle n 1 π 0 π sin n (θ)dθ 1 π 0 π sin n (θ)dθ = 2 2 4 6 (n 1) π 1 3 5 n 1 3 5 (n 1) 2 4 6 n if n is odd if n is even 1 2 π 2 1 2 3 4 3π 4 3 8 5 6 16 15π 15 48 Fundamentals of Power Electronics 82 Chapter 18: PWM Rectifiers

Boost example: transistor RMS current I Qrms = V M 2R e 1 8 3π V M V = I ac rms 1 8 3π V M V Transistor RMS current is minimized by choosing V as small as possible: V = V M. 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 Table 18. 3 Summary of rectifier current stresses for several converter topologies rms Average Peak CCM boost Transistor I ac rms 1 8 3π V M V I ac rms 2 2 π 1 π 8 V M I ac rms 2 V Diode I dc 16 3π V V M I dc 2 I dc V VM Inductor I ac rms I ac rms 2 2 π I ac rms 2 CCM flyback, with n:1 isolation transformer and input filter Transistor, xfmr primary L 1 I ac rms 1 8 3π V M nv I ac rms I ac rms 2 2 π I ac rms 2 2 π I ac rms 2 1 V n I ac rms 2 C 1 I 8 V M ac rms 3π nv 0 I ac rms 2 max 1, V M nv Diode, xfmr secondary I dc 3 2 16 3π nv V M I dc 2I dc 1 nv V M Fundamentals of Power Electronics 84 Chapter 18: PWM Rectifiers

Table of rectifier current stresses continued CCM SEPIC, nonisolated Transistor L 1 C 1 I ac rms 1 8 3π I ac rms V M I 2 2 ac rms V π I ac rms 2 1 V M V I ac rms 2 2 π I ac rms 8 3π V M V 0 I ac rms 2 I ac rms max 1, V M V L 2 Diode I ac rms V M V 3 2 I ac rms 2 I dc 3 2 16 3π V V M I dc V M V I ac rms V M V 2 2I dc 1 V V M CCM SEPIC, with n:1 isolation transformer transistor L 1 C 1, xfmr primary Diode, xfmr secondary I ac rms 1 8 3π I ac rms I 8 V M ac rms 3π nv V M nv I ac rms 2 2 π I ac rms 2 1 V M nv I ac rms 2 2 π I dc 3 2 16 3π nv V M I dc with, in all cases, I ac rms = I dc 2 V V M, ac input voltage = V M sin(ωt) dc output voltage = V I ac rms 2 Fundamentals of Power Electronics 85 Chapter 18: PWM Rectifiers 0 I ac rms 2I dc 2 max 1, n 1 nv V M

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 2 A 380 V 3.6 A 6.6 A 5.1 A Nonisolated SEPIC Isolated SEPIC 5.5 A 719 V 4.85 A 9.8 A 6.1 A 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 i L D g R L 1 i i g R L DR on D' : 1 V F i v g Q 1 C R v v g R v Dc-dc boost converter circuit Averaged dc model Fundamentals of Power Electronics 88 Chapter 18: PWM Rectifiers

Modeling the ac-dc boost rectifier Boost rectifier circuit v ac i g i ac v g R L L Q 1 D 1 i d C i v R controller Averaged model i g R L d R on d' : 1 V F i d i = I v g C R (large) v = V Fundamentals of Power Electronics 89 Chapter 18: PWM Rectifiers

Boost rectifier waveforms v g 300 i g 10 Typical waveforms 200 100 v g i g 8 6 4 2 (low frequency components) i g = v g R e 0 0 30 60 90 120 150 180 0 d 1 6 0.8 5 i d 0.6 0.4 4 3 2 i = I 0.2 1 0 0 30 60 90 120 150 180 0 0 30 60 90 120 150 180 ωt Fundamentals of Power Electronics 90 Chapter 18: PWM Rectifiers

Example: boost rectifier with MOSFET on-resistance i g d R on d' : 1 i d i = I v g C R (large) v = 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 Solve input side of model: i g dr on = v g d'v with i g = v g R e v g =V M sin ωt v g i g d R on d' : 1 i d i = I C R (large) v = V eliminate i g : v g R e dr on = v g d'v solve for d: d= v v g v v g R on R e 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 = i d T ac i d =d'i g =d' v g R e v g i g d R on d' : 1 i d i = I C R (large) v = V Butd is: d'= v g 1 R on R e v v g R on R e hence i d can be expressed as i d = v g 2 R e 1 R on R e v v g R on R e Next, average i d 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 v g = V M sin ωt, and integrate to find i d T ac : I = i d T ac = 2 T ac 0 T ac /2 1 2 V M R e This can be written in the normalized form v V M R on R e R on R e sin 2 ωt sin ωt dt I = 2 T ac 2 V M VR e 1 R sin 2 ωt on R e 1asin ωt 0 T ac /2 dt with a = V M V R on R e Fundamentals of Power Electronics 94 Chapter 18: PWM Rectifiers

Integration By waveform symmetry, we need only integrate from 0 to T ac /4. Also, make the substitution θ = ωt: I = V 2 M VR e 1 R π/2 on 2 sin 2 θ R e π 1asin θ dθ 0 This integral is obtained not only in the boost rectifier, but also in the buck-boost and other rectifier topologies. The solution is 4 π 0 π/2 sin 2 θ 1asin θ dθ = F(a)= 2 a 2 π 4 sin1 2a π a 2 cos 1 1a 2 a Result is in closed form a is a measure of the loss resistance relative to R e a is typically much smaller than unity Fundamentals of Power Electronics 95 Chapter 18: PWM Rectifiers

The integral F(a) 4 π 0 π/2 sin 2 θ 1asin θ dθ Approximation via polynomial: 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 a 2 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. = F(a)= 2 a 2 π F(a) 0.85 0.15 0.10 0.05 0.00 0.05 0.10 0.15 Fundamentals of Power Electronics 96 Chapter 18: PWM Rectifiers 1.15 1.1 1.05 1 0.95 0.9 4 sin1 2a π a 2 cos 1 a 1a 2 a

18.6.3 Solution for converter efficiency η Converter average input power is P in = p in Tac = V M 2R e 2 Average load power is P out = VI = V V M 2 VR e 1 R on R e F(a) 2 with a = V M V R on R e So the efficiency is η = P out P in = 1 R on R e F(a) Polynomial approximation: η 1 R on R e 1 0.862 V M V R on R e 0.78 V M V R on R e 2 Fundamentals of Power Electronics 97 Chapter 18: PWM Rectifiers

Boost rectifier efficiency η 1 0.95 R on /R e = 0.05 η = P out P in = 1 R on R e F(a) 0.9 0.85 0.8 0.75 R on /R e = 0.1 R on /R e = 0.15 R on /R e = 0.2 0.0 0.2 0.4 0.6 0.8 1.0 V M /V To obtain high efficiency, choose V slightly larger than V M Efficiencies in the range 90% to 95% can then be obtained, even with R on as high as 0.2R e Losses other than MOSFET on-resistance are not included here Fundamentals of Power Electronics 98 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 P in = P out η = 500 W 0.95 = 526 W Then the emulated resistance is 2 R e = V g, rms = P in (120 V)2 526 W = 27.4 Ω Fundamentals of Power Electronics 99 Chapter 18: PWM Rectifiers

Design example η Also, 1 V M V = 120 2 V 390 V = 0.435 95% efficiency with V M /V = 0.435 occurs with R on /R e 0.075. 0.95 0.9 0.85 0.8 R on /R e = 0.05 R on /R e = 0.1 R on /R e = 0.15 R on /R e = 0.2 So we require a MOSFET with on resistance of R on (0.075) R e = (0.075) (27.4 Ω)=2Ω 0.75 0.0 0.2 0.4 0.6 0.8 1.0 V M /V Fundamentals of Power Electronics 100 Chapter 18: PWM Rectifiers