《Power_Electronics-13》由会员分享,可在线阅读,更多相关《Power_Electronics-13(49页珍藏版)》请在金锄头文库上搜索。
1、ECE 364 Power Electronics Session 13 Direct dc-dc Converters: Examples; Indirect dc-dc Converters,P. T. Krein Department of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, 2000 Board of Trustees of the University of Illinois,Power Electronics,Direct dc-dc converters R
2、educed switch circuits Analysis examplesIndirect dc-dc converters The buck-boost converter,Session #13 outline,Simplifications,In many applications, it is desirable to share a common input-output node (ground reference). This requires one switch always on and one always off.,Common-Ground Dc-Dc,Ex.:
3、 2x2 switch matrix, with common input-output ground,Common-Ground Dc-Dc,Common-Ground Dc-Dc,With two switches left, label them #1 and #2. One becomes and oneThis can be checked by testing current (on) polarity and voltage (off) polarity.,Switching Functions,With ideal, or near-ideal, current and vol
4、tage sources, KVL and KCL require q1 + q2 = 1. The buck converter:,Buck Converter,The voltage vout is the “switch matrix output.” The load voltage is since = 0.,Relationships,Average power: =,vout = q1 Vin,iin = q1 Iout, = D1 Vin, = D1 Iout,There is no loss.,Instantaneous power:,pin(t) = q1 Vin Iout
5、,= pout(t),=D1VinIout,Relationships,in,V,q,1,=,vout is the switching matrix output.,in,V,D,1,=,load voltage,out,v,=,load voltage,Relationships,Relationships,The RMS “output”,The voltage vout has an RMS value of,Is this relevant?,Notice that,A Design,A 24 V to 5 V converter, switching at 100 kHz. The
6、 nominal load is 25 W, and the ripple is to be less than 1% peak-to-peak. This could be met with a buck converter, since Vout 0.75 mH.,Boost Example,What about Vout? The capacitor current is Iin - Iload = 2.5 A - 1 A when switch #2 is on, and -1 A when switch #1 is on. We want + 1% of 5 V, or a peak
7、- to-peak change below 0.1 V.,Boost Example,#1 ON: iC= -1 A,#2 ON: iC= 1.5 A,Boost Example,With switch #2 on (duty ratio was found to be 0.4, so time is 5 us), iC = 1.5 A = C dv/dt = C Dv/Dt. (1.5 A)(5 us)/C = Dv 7.5 uF.,Boost Example,Practice: What if fs is changed to 40 kHz? average values are the
8、 same, ripple 2x,2 to 5 V, 80 kHz boost converter:,Comments,With a few practice examples, you should be able to design a common-ground buck or boost converter. Challenge: Think about effects of nonideal switching. It is not so difficult to include some basic nonideal effects, such as switching devic
9、e voltage drops and resistances. Consider an example with switch and diode voltage drop.,Nonideal boost,+,-,V,IN,I,IN,(Still) v,IN,L,i,OUT,+,-,V,OUT,C,R,0.5 V,Voltage drop,Nonideal boost,Nonideal boost,Switching function expressions still apply. Boost: vin = q1(0.5 V)+ q2(Vout+ 1 V). On average, = Vin = D1(0.5V) + (1-D1)(Vout + 1 V), and Vout = (Vin + 0.5D1 -1)/(1 - D1) For current, iout = q2 IL, = D2 IL. Since is the load current Iload, we have IL = Iload/D2 = Iload/(1 D1).,Nonideal boost,