VIEWS: 89 PAGES: 42 CATEGORY: Technology POSTED ON: 5/30/2012
DC to DC CONVERTER (CHOPPER) www.sayedsaad.com www.tkne.net www.sayedsaad.com www.tkne.net Chapter 3 DC to DC CONVERTER (CHOPPER) • General • Buck converter • Boost converter • Buck-Boost converter • Switched-mode power supply • Bridge converter • Notes on electromagnetic compatibility (EMC) and solutions. Power Electronics and 1 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net DC-DC Converter (Chopper) • DEFINITION: Converting the unregulated DC input to a controlled DC output with a desired voltage level. • General block diagram: DC supply (from rectifier- filter, battery, DC output LOAD fuel cell etc.) Vcontrol (derived from feedback circuit) • APPLICATIONS: – Switched-mode power supply (SMPS), DC motor control, battery chargers www.sayedsaad.com www.tkne.net Linear regulator • Transistor is operated in linear (active) mode. IL + VCE − + • Output voltage Vs RL Vo − Vo = I L RT MODEL OF LINEAR • The transistor can REGULATOR be conveniently modelled by an + VCE − IL equivalent RT + variable resistor, Vs RL Vo as shown. − EQUIVALENT • Power loss is high CIRCUIT at high current due to: Po = I L 2 RT Power Electronics and 3 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Switching Regulator • Power loss is zero (for ideal switch): + VCE − IL – when switch is RL open, no current Vs + flow in it, Vo − – when switch is MODEL OF LINEAR closed no voltage REGULATOR drop across it. IL – Since power is a SWITCH + product of voltage Vs RL Vo − and current, no losses occurs in the switch. EQUIVALENT CIRCUIT – Power is 100% Vo transferred from source to load. (ON) (OFF) (ON) closed open closed DT T • Switching regulator OUTPUT VOLTAGE is the basis of all DC-DC converters Power Electronics and 4 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Buck (step-down) converter S L + Vd D C RL Vo − CIRCUIT OF BUCK CONVERTER iL S + vL − + Vd D RL Vo − CIRCUIT WHEN SWITCH IS CLOSED S iL + vL − + Vd RL Vo D − CIRCUIT WHEN SWITCH IS OPENED Power Electronics and 5 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Circuit operation when switch is turned on (closed) • Diode is reversed + vL - biased. Switch S iL + conducts inductor + C Vo Vd VD RL current − − • This results in vL positive inductor Vd−Vo voltage, i.e: closed opened closed opened vL = Vd − Vo t • It causes linear −Vo increase in the iL inductor current iLmax diL vL = L IL iLmin dt 1 ⇒ iL = ∫ vL dt DT T t L Power Electronics and 6 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Operation when switch turned off (opened) + vL - S • Because of iL + C Vo inductive energy Vd D RL − storage, iL continues to flow. vL Vd−Vo • Diode is forward opened opened biased closed closed t • Current now flows through the diode −Vo iL and iLmax vL = −Vo IL iLmin (1-D)T t DT T Power Electronics and 7 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Analysis for switch closed The inductor voltage, vL = Vd − Vo vL diL =L Vd− Vo dt closed t diL Vd − Vo ⇒ = dt L iL Note : since the deri - vative of iL is a posi - iL max tive constant. I ∆iL Therefore iL must L increase linearly. iL min t From Figure DT T diL ∆iL ∆iL Vd − Vo = = = dt ∆t DT L V −V (∆iL )closed = d o ⋅ DT L Power Electronics and 8 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Analysis for switch opened For switch opened, vL vL = −Vo opened Vd− Vo diL =L dt t di −V ⇒ L= o dt L iL Note : since the deri - vative of iL is a nega - iL max tive constant, iL must IL ∆iL decrease linearly. iL min From Figure t DT T diL ∆iL ∆iL − Vo = = = (1− D)T dt ∆t (1 − D)T L −V (∆iL )opened = o ⋅ (1 − D)T L Power Electronics and 9 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Steady-state operation iL Unstable current t iL Decaying current t iL Steady-state current t Steady - state operation requires that i L at the end of switching cycle is the same at the begining of the next cycle. That is the change of i L over one period is zero, i.e : (∆iL )closed + (∆iL )opened = 0 Vd − Vo − Vo ⋅ DTs − ⋅ (1 − D)Ts = 0 L L ⇒ Vo = DVd Power Electronics and 10 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Average, Maximum and Minimum inductor current iL Imax IL ∆iL Imin t Average inductor current = Average current in R L V ⇒ IL = IR = o R Maximum current : ∆i V 1 V I max = I L + L = o + o (1 − D)T 2 R 2 L 1 (1 − D) = Vo + R 2 Lf Minimum current : ∆i 1 (1 − D ) I min = I L − L = Vo − 2 R 2 Lf Power Electronics and 11 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Continuous current operation iL Imax Imin t 0 From previous analysis, ∆i 1 (1 − D) I min = I L − L = Vo − 2 R 2 Lf For continuous operation, I min ≥ 0, 1 (1 − D) ⇒ Vo − ≥0 R 2 Lf (1 − D) ⇒ L ≥ Lmin = ⋅R 2f This is the minimum inductor current to ensure continous mode of operation. Normally L is chosen be be >> Lmin Power Electronics and 12 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Output voltage ripple iL imax iL iR iL=IR L Vo / R iC + imin Vo 0 − 0 ic = iL + iR ∆Q Q = CVo ⇒ ∆Q = C∆V ⇒ o ∆Vo = C From figure, use triangle area formula : 1 T ∆iL T∆iL ∆Q = = 2 2 2 8 T∆iL (1 − D ) ∴ ∆Vo = = 8C 8 LCf 2 So, the ripple factor, ∆Vo (1 − D) r= = Vo 8 LCf 2 Power Electronics and 13 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Design procedures for Buck SWITCH L Lmin= ? RL L = 10Lmin Vd Po = ? f=? D (input C Io = ? D=? ripple ? spec.) TYPE ? • Calculate D to obtain required output voltage. • Select a particular switching frequency: – preferably >20KHz for negligible acoustic noise – higher fs results in smaller L, but higher device losses. Thus lowering efficiency and larger heat sink. Also C is reduced. – Possible devices: MOSFET, IGBT and BJT. Low power MOSFET can reach MHz range. Power Electronics and 14 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Design procedures for Buck • Determine Lmin. Increase Lmin by about 10 times to ensure full continuos mode. • Calculate C for ripple factor requirement. • Capacitor ratings: – must withstand peak output voltage – must carry required RMS current. Note RMS current for triangular w/f is Ip/3, where Ip is the peak capacitor current given by ∆iL/2 • Wire size consideration: – Normally rated in RMS. But iL is known as peak. RMS value for iL is given as: 2 2 ∆iL 2 I L , RMS = I L + 3 Power Electronics and 15 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Examples of Buck converter • A buck converter is supplied from a 50V battery source. Given L=400uH, C=100uF, R=20 Ohm, f=20KHz and D=0.4. Calculate: (a) output voltage (b) maximum and minimum inductor current, (c) output voltage ripple. • A buck converter has an input voltage of 50V and output of 25V. The switching frequency is 10KHz. The power output is 125W. (a) Determine the duty cycle, (b) value of L to limit the peak inductor current to 6.25A, (c) value of capacitance to limit the output voltage ripple factor to 0.5%. • Design a buck converter such that the output voltage is 28V when the input is 48V. The load is 8Ohm. Design the converter such that it will be in continuous current mode. The output voltage ripple must not be more than 0.5%. Specify the frequency and the values of each component. Suggest the power switch also. Power Electronics and 16 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Boost (step-up) converter L D Vd C + S RL Vo − CIRCUIT OF BOOST CONVERTER iL L D + vL − Vd + S C RL Vo − CIRCUIT WHEN SWITCH IS CLOSED L D + vL - + Vd C RL S Vo − CIRCUIT WHEN SWITCH IS OPENED Power Electronics and 17 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Boost analysis:switch closed iL L D + vL − + Vd C vo S − vL = Vd vL diL Vd =L dt CLOSED diL Vd ⇒ = t dt L Vd − V o diL ∆iL ∆iL iL = = ∆iL dt ∆t DT diL Vd ⇒ = dt L t DT T Vd DT (∆iL )closed = L Power Electronics and 18 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Switch opened iL D + vL - + Vd C vo S - vL = Vd − Vo vL di =L L Vd dt V − Vo OPENED di ⇒ L= d t dt L Vd − V o diL ∆iL = iL dt ∆t ∆iL ∆iL = (1 − D)T ( 1-D )T DT T t diL Vd − Vo ⇒ = dt L (V − Vo )(1 − DT ) ⇒ (∆iL )opened = d L Power Electronics and 19 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Steady-state operation (∆iL )closed + (∆iL )opened = 0 Vd DT (Vd − Vo )(1 − D)T − =0 L L Vd ⇒ Vo = 1− D • Boost converter produces output voltage that is greater or equal to the input voltage. • Alternative explanation: – when switch is closed, diode is reversed. Thus output is isolated. The input supplies energy to inductor. – When switch is opened, the output stage receives energy from the input as well as from the inductor. Hence output is large. – Output voltage is maintained constant by virtue of large C. Power Electronics and 20 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Average, Maximum, Minimum inductor current Input power = Output power Vo 2 Vd I d = R 2 Vd (1 − D ) Vd 2 Vd I L = = R (1 − D) 2 R Average inductor current Vd IL = (1 − D) 2 R Max, min inductor current ∆iL Vd Vd DT I max = I L + = 2 + 2 (1 − D) R 2L ∆iL Vd Vd DT I min = IL − = 2 − 2 (1 − D) R 2L Power Electronics and 21 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net L and C values For continous operation, Vd I min ≥ 0 vL Vd Vd DT 2 − ≥0 (1 − D) R 2L Vd−Vo D(1 − D )2 TR Lmin = imax 2 iL D(1 − D ) R 2 = imin 2f imax Ripple factor iD imin V ∆Q = o DT = C∆Vo Io=Vo / R R Vo DT Vo D ∆Vo = = RCf RCf ic ∆V D r= o = ∆Q Vo RCf DT T Power Electronics and 22 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Examples • The boost converter has the following parameters: Vd=20V, D=0.6, R=12.5ohm, L=65uH, C=200uF, fs=40KHz. Determine (a) output voltage, (b) average, maximum and minimum inductor current, (c) output voltage ripple. • Design a boost converter to provide an output voltage of 36V from a 24V source. The load is 50W. The voltage ripple factor must be less than 0.5%. Specify the duty cycle ratio, switching frequency, inductor and capacitor size, and power device. Power Electronics and 23 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Buck-Boost converter S D + Vd C L RL Vo − CIRCUIT OF BUCK-BOOST CONVERTER S D + + Vd iL vL Vo − − CIRCUIT WHEN SWITCH IS CLOSED S D + + Vd iL vL Vo − − CIRCUIT WHEN SWITCH IS OPENED Power Electronics and 24 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Buck-boost analysis Switch closed Vd di vL = Vd = L L vL dt di V ⇒ L= d dt L Vd−Vo ∆iL ∆iL Vd imax = = iL ∆t DT L V DT imin (∆iL ) closed = d L imax iD Switch opened imin di vL = Vo = L L Io=Vo / R dt di V ⇒ L= o dt L ic ∆iL ∆iL V = = o ∆Q ∆t (1 − D)T L DT T Vo (1 − D)T (∆iL ) opened = L Power Electronics and 25 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Output voltage Steady state operation : Vd DT Vo (1 − D )T + =0 L L D ⇒ Vo = −Vs 1− D • NOTE: Output of a buck-boost converter either be higher or lower than the source voltage. – If D>0.5, output is higher – If D<0.5, output is lower • Output voltage is always negative • Note that output is never directly • connected to load. Energy is stored in inductor when switch is closed and transferred to load when switch is opened. Power Electronics and 26 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Average inductor current Assuming no power loss in the converter, power absorbed by the load must equal power supplied the by source, i.e. Po = Ps Vo2 = Vd I s R But average source current is related to average inductor current as : Is = ILD Vo2 ⇒ = Vd I L D R Substituting for Vo , Vo2 Po Vd D ⇒ IL = = = Vd RD Vd D R (1 − D ) 2 Power Electronics and 27 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net L and C values Max and min inductor current, ∆iL Vd D Vd DT I max = I L + = 2 + 2 R (1 − D ) 2L ∆iL Vd D Vd DT I min = I L − = 2 − 2 R (1 − D) 2L For continuous current, Vd D Vd DT 2 + =0 R (1 − D ) 2L (1 − D ) 2 R ⇒ Lmin = 2f Output voltage ripple, Vo ∆Q = DT = C∆Vo R Vo DT Vo D ∆Vo = = RC RCf ∆Vo D r= = Vo RCf Power Electronics and 28 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Control of DC-DC converter using pulse width modulation- PWM Vo (desired) + Vcontrol Switch control signal Vo (actual) Comparator - Sawtooth Waveform Sawtooth Waveform Vcontrol 1 Vcontrol 2 Switch control ton 2 signal ton 1 T Power Electronics and 29 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Switch-mode power supply (SMPS) • Advantages over linear power -Efficient (70-95%) -Weight and size reduction • Disadvantages -Complex design -EMI problems • However above certain ratings, SMPS is the only feasible choice • Types of SMPS -Flyback -forward -Push-pull -Bridge (half and full) Power Electronics and 30 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Linear and switched mode power supplies block diagram Basic Block diagram of linear power supply C E Vce=Vd-Vo +Vo Rectifier B + + Base/gate RL Vd Drive Vo Line Input - 1φ / 3φ 50/60 Hz - Error Vo Isolation Amp. Transformer Vref Basic Block diagram of SMPS DC-DC CONVERSITION + ISOLATION DC High Regulated RECTIFIER Vo EMI Frequency AND FILTER rectifier FILTER and DC filter Unregulated Vref Base/ PWM error gate Controller Amp drive Power Electronics and 31 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net High frequency transformer Basic function : i) Input - output electrical isolation ii) step up/down time - varying voltage Basic input - output relationship v1 N1 i1 N 2 = ; = v2 N 2 i2 N1 Models : i1 N1 N2 i2 + + V1 V2 Ideal model − − i1 N1 N2 i2 + + Lm Model used for V1 V2 − most PE application − Power Electronics and 32 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Flyback Converter + LM C R Vo Vs − Flyback converter circuit iD iS N1 N2 i1 - +v - + + D iLM v2 iR Vo v1 iC Vs + − - + vSW − i2 Model with magnetising inductance 0 N1 N2 is=iLM + + iLM v1 Vo Vs - − v1=Vs 0 Switch closed iD N1 N2 + − + iLM v1 v2= -VS Vo Vs − + − + vSW − N v1 − Vo 1 N Voltage and current 2 conditions when switch N Vsw = Vs + Vo 1 N opened 2 Power Electronics and 33 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Flyback waveforms v1 Vs DT T -V(N1/N2) t is ∆iLM DT T t iD DT T t iLm t iC DT T DT T Vo/ R t Power Electronics and 34 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Analysis: switched closed diLm v1 = Vd = Lm dt diLm ∆iLm ∆iLm Vd = = = dt dt DT Lm ( ⇒ ∆iLm )closed = VdLDT m On the load side of the transformer, N2 N2 v2 = v1 = Vd N1 N1 N vD = −Vo − Vd 2 < 0 N1 Therefore, i1 = 0 i2 = 0 Power Electronics and 35 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Analysis: switch opened N v1 = −V0 1 ; N v2 = −V0 2 N N ⇒ v1 = v2 1 = −V0 1 N N 2 2 diL m N Lm = v1 = −V0 1 N dt 2 diL m ∆iL m ∆iL m − V0 N1 = = = dt dt (1 − D )T Lm N 2 V (1 − D)T N1 ⇒ (∆iL m )open = − 0 N Lm 2 For steady - state operation, (∆iL )closed + (∆iL )opened = 0 m m V DT V0 (1 − D )T N1 ⇒ d + N =0 Lm Lm 2 D N1 ⇒ V0 = Vd (1 − D) N 2 Power Electronics and 36 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Output voltage • Input output relationship is similar to buck- boost converter. • Output can be greater of less than input,depending upon D. • Additional term, i.e. transformer ratio is present. Power Electronics and 37 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Average inductor current Ps = P0 V0 2 Vd I s = R I s is related to I Lm as : I Lm DT Is = T ( ) = I Lm D Substitute and solving for I Lm V02 ( ) Vd I Lm D = R V0 2 ⇒ I Lm = Vd DR The average inductor current is also written as : 2 Vd D N 2 V0 N 2 I Lm = = 2 (1 − D) R N1 (1 − D) R N1 Power Electronics and 38 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Max, Min inductor current, Lmin, C values 2 ∆iLm N 2 V d DT Vd D I Lm = I Lm + = 2 + ,max 2 (1 − D ) R N1 2 Lm 2 ∆iLm N 2 Vd DT Vd D I Lm ,min = I Lm − = 2 − 2 (1 − D ) R N1 2 Lm For continuos operation, I Lm , min = 0 2 Vd D N 2 Vd DT Vd D 2 = = (1 − D ) R N1 2 Lm 2 Lm f 2 2 Vd (1 − D) R N1 (Lm )min = 2f N2 The ripple calculation is similar to boost converter, ∆V0 D r= = V0 RCf Power Electronics and 39 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Full-bridge converter SW1 SW3 Lx + + + NS vx Vo C R − − vp VS NS − SW4 SW2 SW1,SW2 DT T SW3,SW4 T T + DT VP 2 2 VS -VS Vx N VS S N P DT T T T + DT 2 2 Power Electronics and 40 Drives (Version 2): Dr. Zainal Salam www.sayedsaad.com www.tkne.net Full bridge: basic operation • Switch “pair”: [S1 & S2];[S3 & S4]. • Each switch pair turn on at a time as shown. The other pair is off. • “AC voltage” is developed across the primary. Then transferred to secondary via high frequency transformers. • On secondary side, diode pair is “high frequency full wave rectification”. • The choke (L) and © acts like the “buck converter” circuit. Ns • Output Voltage Vo = 2Vs N ⋅ D p Power Electronics and 41 Drives (Version 2): Dr. Zainal Salam