DC-DC开关模式变换器

1、Chapter 7： DC-DC Switch-Mode Converters:7-1 Introduction7-2 Control of dc-dc Converters7-3 Step-Down (Buck) Converter 7-4 Step-Up (Boost) Converter 7-5 Buck-Boost Converter7-6 Cuk dc-dc Converter7-7 Full Bridge dc-dc Converter7-8 dc-dc Converter Comparison7-3 Step-Down (Buck) Converter7-3 BUCK CONVE

2、RTER Practical Buck Converter7-3 BUCK CONVERTER Simplifying Assumptions1.Transistors, diodes and other passive components are all ideal.2. The input is a pure dc voltage Vd.3. The output voltage is purely dc without any ripple, that is vo(t) Vo, 4. Continuous conduction mode, CCMvoiVdOtontt toffoffs

3、SfT1=(b)+Vd-C+ vL -R+vo=Vo-+voi-(a)LidiLioVo+Vd-C+ vL -R+vo=Vo-+voi-(a)LidiLiott0ABTstont0t0iD(b)(c)(d)(e)iLidvL(Vd-Vo)(-Vo)(IL=Io)toffOperation principle 7-3 BUCK CONVERTER 1) Turning on the switch increase the inductor current.2) When the transistor is turned off, the inductor current “freewheels”

4、 through the diode. )37()()0=DTtVVtTVtVVVsondoonsoonodL（In the dc steady state: 7-3 BUCK CONVERTER Vo can be controlled by varying the duty ratio D.In the dc steady state: RVIIIooLC= 0)()(,titirippleLC)47(1:=DIIandIVIVThereforePPdoooddod7-3 BUCK CONVERTER Neglecting power losses associate with all t

5、he circuit elements.In the dc steady state: RVIIIooLC= 0)()(,titirippleLC7-3 BUCK CONVERTER )()(,tiItirippleLLL=）（227)1 (1)(1=sosodLTDVLDTVVLi)237(8)1 (2=CLDVTVosoExample 7-1 In the Buck dc-dc converter of Fig. 3-5a, L =24H . It is operating in dc steady state under the following conditions: Vin = 2

6、0V , D = 0.6 , Po = 12W , and fs = 200kHz . Assuming ideal components, calculate and draw the waveforms shown earlier in Fig. 3-5d.7-3 BUCK CONVERTER Solution With fs = 200kHz , Ts = 5 s and Ton = DTs = 3s . Vo=DVin =12V .The inductor voltage vL fluctuates between (Vin Vo ) = 8V and (Vo) =12V , as s

7、hown inFig. 3-6.ADTVVLisoinL1245.608)(1=AVPIIoooL11212= Therefore, iL=IL+iL,ripple , as shown in Fig. 3-6. When the transistor is on, iin=iL, otherwise zero. The average input currents is Iin=DIo =0.6A .7-3 BUCK CONVERTER 7-4 Step-Up (Boost) Converter7-4 BOOST CONVERTER Step-up converter is mainly a

8、pplied in regulated dc power supplies and regenerative braking of dc motors.+-LVLiDLDi+oV-RC+-dVsiOperation principle1) Turning on the switch applies the voltage Vd across the inductor, vL=Vd, and iL linearly ramps up, increasing the energy in the inductor.7-4 BOOST CONVERTER tt0(Vd-Vo)Tstont0t0iD(b

9、)(c)(d)(e)vLiLis(Vd)ILtoff+-LVLiDLiD+oV-RC+-dVsi(a)2) turning off the switch forces the inductor current to flow through the diode, some of the stored energy is transferred to the output stage.In the dc steady state: )267(110)(=DtTVVtVVtVoffsdooffoinond Assuming a lossless circuit,)277()D1:=（doooddo

10、dIIandIVIVThereforePP7-4 BOOST CONVERTER tt0(Vd-Vo)Tstont0t0iD(b)(c)(d)(e)vLiLis(Vd)ILtoff+-LVLiDLiD+oV-RC+-dVsi(a)odioderipplediodeCIititi=)()(,7-4 BOOST CONVERTER sdosdLTDVVLDTVLi)1)(1)(1=RVDIIodL=11tt0(Vd-Vo)Tstont0t0iD(b)(c)(d)(e)vLiLis(Vd)ILtoff+-LVLiDLiD+oV-RC+-dVsi(a)297()1 (2)287()1 (2222,=D

11、DLVTIDDLVTLtViIosoBosondpeakLLBExample 7-2 In a Boost converter of Fig. 3-8a, the inductor current has iL= 2 A. It is operating in dc steady state under the following conditions: Vin =5V , Vo =12V ,Po=10W , and fs=200kHz . Assuming ideal components, calculate L and draw the waveforms as shown in Fig

12、. 3-8c.7-4 BOOST CONVERTER Solution The duty-ratio D=0.583 . With fs=200kHz , Ts= 5s and Ton=DTs =2.917s . vL fluctuates between Vin =5V and (Vo Vin )=7V . HDTiVLsLin29. 7917. 225=7-4 BOOST CONVERTER ripplLLLinoinLiIAVPII,i2510=When the transistor is on, the idiode = 0; otherwise idiode =iL. AIDIIin

13、odiode834. 02)583. 01 ()1 (=AIIitioodiodeC834. 0)(=The capacitor current jumps to a value of 2.167 A and drops to （1 0.833 ）= 0.167 A.7-4 BOOST CONVERTER 7-5 BUCK-BOOST CONVERTER7-5 BUCK-BOOST CONVERTER Buck-Boost converter is mainly used in regulated dc power supplies, where a negative-polarity out

14、put voltage may be desired with respect to the common terminal of the input voltage, and the output voltage can be either higher or lower than the input voltage.L+vL-+Vd-C-Vo+RidDiLiDioOperation principle1) Turning on the switch applies the voltage Vd across the inductor, vL=Vd, and iL linearly ramp

15、s up, increasing the enery in the inductor.7-5 BUCK-BOOST CONVERTER tt0VdTstont0t0iD(b)(c)(d)(e)vLiLidIL=(Id+Io)toff(-Vo)L+vL-+Vd-C-Vo+RidDiLiDio2) turning off the switch results in the inductor current flowing through the diode, some of the stored energy is transferred to the output stage.In the dc

16、 steady state: )427(10)1)(=DDVVTDVDTVdososd7-5 BUCK-BOOST CONVERTER )437(1:min=DDIIPPgassudoodtt0VdTstont0t0iD(b)(c)(d)(e)vLiLidIL=(Id+Io)toff(-Vo)L+vL-+Vd-C-Vo+RidDiLiDioIn the dc steady state: )()(,titiripplediodeC7-5 BUCK-BOOST CONVERTER RVDIIIoodL=11sosdLTDVLDTVLi)1 (1)(1=tt0VdTstont0t0iD(b)(c)(

17、d)(e)vLiLidIL=(Id+Io)toff(-Vo)L+vL-+Vd-C-Vo+RidDiLiDio)477()1 (2)1 ()467()1 (2222,=DLVTIDIDLVTDLVTiIosLBoBosdspeakLLBExample 7-3 A Buck-Boost converter of 3-11b is operating in dc steady state under the following conditions: Vin=14V , Vo=2V , Po=21W , iL =2A and fs = 200kHz . Assuming ideal componen

18、ts, calculate L and draw the waveforms as shown in Fig. 3-12c.7-5 BUCK-BOOST CONVERTER Solution D = 0.75 .Ts = 1/fs=5s and Ton = DTs =3.75s as shown in Fig. 3-13. The inductor voltage vL fluctuates between Vin=14V and Vo=42V . 7-5 BUCK-BOOST CONVERTER sDTTsfTVVVDsonssinoo75. 35175. 0144242=HDTiVLsLin25.2675. 3214=When the transistor is on, the diode current is zero; otherwise i diode=iL. AIIAVPIAVPPIoinoooinoinin2I Therefore,