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


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1 R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
2 18.5 RMS values of rectifier waveforms Doublymodulated 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
3 RMS transistor current RMS transistor current is i Q I Qrms = 1 Tac T ac i Q 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 Q dt Tac Ts n =1 nt s (n1)t s Quantity in parentheses is the value of i Q, averaged over the n th switching period. Fundamentals of Power Electronics 78 Chapter 18: PWM Rectifiers
4 Approximation of RMS expression T ac /T s n =1 I Qrms = 1 T 1 s i Q dt Tac Ts nt s (n1)t s When T s << T ac, then the summation can be approximated by an integral, which leads to the doubleaverage: I Qrms 1 Tac T ac /T s nt s lim Ts T 1 s i Q (τ)dτ Ts n=1 (n1)t s = 1 1 i Q (τ)dτdt Tac Ts T ac t t+t s = i Q Ts T ac Fundamentals of Power Electronics 79 Chapter 18: PWM Rectifiers
5 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 Q is therefore If the input voltage is i Q t+t s t = 1 i T s T Q dt s = di ac v ac = sin ωt then the input current will be given by i ac = Re and the duty cycle will ideally be sin ωt v ac = 1 1d (this neglects converter dynamics) Fundamentals of Power Electronics 8 Chapter 18: PWM Rectifiers
6 Boost rectifier example Duty cycle is therefore d=1 Evaluate the first integral: i Q = M T s sin ωt Now plug this into the RMS formula: I Qrms = 1 Tac i Q 1 sin ωt sin ωt T ac T s dt = 1 Tac T ac 1 sin ωt sin ωt dt I Qrms = Tac sin ωt sin 3 T ac / ωt dt Fundamentals of Power Electronics 81 Chapter 18: PWM Rectifiers
7 Integration of powers of sin θ over complete halfcycle n 1 π π sin n (θ)dθ 1 π π sin n (θ)dθ = 4 6 (n 1) π n (n 1) 4 6 n if n is odd if n is even 1 π π π Fundamentals of Power Electronics 8 Chapter 18: PWM Rectifiers
8 Boost example: transistor RMS current I Qrms = 1 8 3π = 1 8 3π Transistor RMS current is minimized by choosing as small as possible: =. This leads to I Qrms =.39 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
9 Table of rectifier current stresses for various topologies Table Summary of rectifier current stresses for several converter topologies rms Average Peak CCM boost Transistor 1 8 3π π 1 π 8 Diode I dc 16 3π I dc I dc M Inductor π CCM flyback, with n:1 isolation transformer and input filter Transistor, xfmr primary L π n π π 1+ n C 1 I 8 ac rms 3π n max 1, n Diode, xfmr secondary I dc π n I dc I dc 1+ n Fundamentals of Power Electronics 84 Chapter 18: PWM Rectifiers
10 Table of rectifier current stresses continued CCM SEPIC, nonisolated Transistor L 1 C π I ac rms π 1+ π 8 3π max 1, L Diode 3 I dc π I dc I dc 1+ CCM SEPIC, with n:1 isolation transformer transistor L 1 C 1, xfmr primary Diode, xfmr secondary π I 8 ac rms 3π n n π 1+ n π I dc π n I dc with, in all cases, = I dc, ac input voltage = sin(ωt) dc output voltage = Fundamentals of Power Electronics 85 Chapter 18: PWM Rectifiers I dc max 1, n 1+ n
11 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 Buckboost, 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
12 Comparison of rectifier topologies 1kW, 4rms example. Output voltage: 38dc. Input current: 4.Arms Converter Transistor rms current Transistor voltage Diode rms current Transistor rms current, 1 Diode rms current, 1 Boost A A 6.6 A 5.1 A Nonisolated SEPIC Isolated SEPIC 5.5 A A 9.8 A 6.1 A 5.5 A A 11.4 A 4.5 A Isolated SEPIC example has 4:1 turns ratio, with 4 3.8A dc load Fundamentals of Power Electronics 87 Chapter 18: PWM Rectifiers
13 18.6 Modeling losses and efficiency in CCM highquality 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 steadystate waveforms and average power. Boost example i L D g R L 1 i + i g R L D D' : 1 F + i + v g + Q 1 C R v v g + R v Dcdc boost converter circuit Averaged dc model Fundamentals of Power Electronics 88 Chapter 18: PWM Rectifiers
14 Modeling the acdc 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 d' : 1 F + i d i = I + v g + C R (large) v = Fundamentals of Power Electronics 89 Chapter 18: PWM Rectifiers
15 Boost rectifier waveforms v g 3 i g 1 Typical waveforms 1 v g i g (low frequency components) i g = v g d i d i = I ωt Fundamentals of Power Electronics 9 Chapter 18: PWM Rectifiers
16 Example: boost rectifier with MOSFET onresistance i g d d' : 1 i d i = I + v g + C R (large) 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
17 Expression for controller duty cycle d Solve input side of model: i g d = v g d'v with i g = v g v g = sin ωt v g i g d d' : 1 i d i = I + C R (large) + v = eliminate i g : v g d = v g d'v solve for d: d= v v g v v g Again, these expressions neglect converter dynamics, and assume that the converter always operates in CCM. Fundamentals of Power Electronics 9 Chapter 18: PWM Rectifiers
18 18.6. 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 v g i g d d' : 1 i d i = I + C R (large) + v = Butd is: d'= v g 1 v v g hence i d can be expressed as i d = v g 1 v v g 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
19 Dc load current I Now substitute v g = sin ωt, and integrate to find i d T ac : I = i d T ac = T ac T ac / 1 This can be written in the normalized form v sin ωt sin ωt dt I = T ac 1 R sin ωt on 1asin ωt T ac / dt with a = Fundamentals of Power Electronics 94 Chapter 18: PWM Rectifiers
20 Integration By waveform symmetry, we need only integrate from to T ac /4. Also, make the substitution θ = ωt: I = M 1 R π/ on sin θ π 1asin θ dθ This integral is obtained not only in the boost rectifier, but also in the buckboost and other rectifier topologies. The solution is 4 π π/ sin θ 1asin θ dθ = F(a)= a π 4 sin1 a π + a + cos 1 1a a Result is in closed form a is a measure of the loss resistance relative to a is typically much smaller than unity Fundamentals of Power Electronics 95 Chapter 18: PWM Rectifiers
21 The integral F(a) 4 π π/ sin θ 1asin θ dθ Approximation via polynomial: F(a) a +.78a For a.15, this approximate expression is within.1% of the exact value. If the a term is omitted, then the accuracy drops to ±% for a.15. The accuracy of F(a) coincides with the accuracy of the rectifier efficiency η. = F(a)= a π F(a) Fundamentals of Power Electronics 96 Chapter 18: PWM Rectifiers sin1 a π + a + cos 1 a 1a a
22 Solution for converter efficiency η Converter average input power is P in = p in Tac = Average load power is P out = I = 1 F(a) with a = So the efficiency is η = P out P in = 1 F(a) Polynomial approximation: η Fundamentals of Power Electronics 97 Chapter 18: PWM Rectifiers
23 Boost rectifier efficiency η 1.95 / =.5 η = P out P in = 1 F(a) / =.1 / =.15 / = / To obtain high efficiency, choose slightly larger than Efficiencies in the range 9% to 95% can then be obtained, even with as high as. Losses other than MOSFET onresistance are not included here Fundamentals of Power Electronics 98 Chapter 18: PWM Rectifiers
24 Design example Let us design for a given efficiency. Consider the following specifications: Output voltage 39 Output power 5 W rms input voltage 1 Efficiency 95% Assume that losses other than the MOSFET conduction loss are negligible. Average input power is P in = P out η = 5 W.95 = 56 W Then the emulated resistance is = g, rms = P in (1 ) 56 W = 7.4 Ω Fundamentals of Power Electronics 99 Chapter 18: PWM Rectifiers
25 Design example η Also, 1 = 1 39 = % efficiency with / =.435 occurs with / / =.5 / =.1 / =.15 / =. So we require a MOSFET with on resistance of (.75) = (.75) (7.4 Ω)=Ω / Fundamentals of Power Electronics 1 Chapter 18: PWM Rectifiers
R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
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