1. SWITCH-MODE POWER SUPPLIES AND SYSTEMS Silesian University of Technology Faculty of Automatic Control, Electronics and Computer Sciences Ryszard Siurek Ph.D., El. Eng. Lecture No 4

2. Linear regulator versus switching regulator Basic functional diagrams Linear regulatorSwitching regulator - + Vr CC amp RLRLRLRL R1 R2  Uin Uout(Uo) Io Uout(Uo) RLRLRLRL fs fc Ku Io Uce Uo = Uin – Uce = const. Po = Uo. Io P Loss = Uce. Io  T  =  = = PoPin Uo. Io Uin. Io Uin - Uce Uin efficiency: Uce > Ucemin (~2,5V) 0,3 <  < 0,5 typical values fs = 1T U1 T  = U +  Usin(2i  ft) U1= U 1av +  U i sin(2i  f p t) i=1 Uin U1av = Uin =  Uin T  – duty cycle eliminated by low-pass filter T1 T1 Power loss in T1 - high Power losses in T1 = 0 Filter power losses (LC) = 0 U1av  = 1 (100%) Cross-over frequency

3. T L D C RLRL IoIo ILIL ICIC IDID ITIT U in U0U0 Assumptions: 1. Diode D and transitor T are perfect (ideal) switches 2. Series resistance of the choke L is negligible (r = 0) 2. Series resistance of the choke L is negligible (r L = 0) 3. Capacitance C is very large (  U << U) 3. Capacitance C is very large (  U c << U o ) EELEEL EECEEC D T EELEEL EECEEC D T U in I cycle II cycle T – closed, D – open T – open, D – closed IoIo IoIo STEP-DOWN (BUCK) SWITCHING REGULATOR U1U1 U0U0 UCUC ~ UCUC T 

4. t t t t t t U1U1 ITIT ILIL IDID ICIC UCUC ~ T I cycle - equivalent circuit 0 < t <  U 1 = U in r= 0 r L = 0 ~ ILIL L I I Lmin U0U0 Ro II0II0 Calculation of I – superposition method Calculation of I L – superposition method I I Lmax  <<1 inductor current swing I Lmin Basic waveforms in step-down switching regulator I Lmax ~ UCUC UCUC ~ << U 0 U(0) U c (0),,

5. t t t t t t U1U1 ITIT ILIL IDID ICIC UCUC ~ T I Lmin I Lmax r= 0 r L = 0 ~ ILIL L I I Lmax U0U0 Ro II0II0 ~ UCUC UCUC ~ << U 0 II cycle - equivalent circuit  t  T U0U0 „ Calculating I - superposition method Calculating I L - superposition method „ Similarly as before: inductor current swing in steady state: I Lmax I Lmin II0II0 U()Uc()U()Uc() Step-down regulator transfer function

6. Inductor & capacitor selection exemplary calculations for step-down regulator Design specification: U IN = 10 – 15V, Uo = 5V, Io = 10A, f = 100kHz Inductor current amplitude (general rule):  I L < 0,1 – 0,2 I 0max we assume:  I L < 1A we assume:  I L < 1A hence: hence: for  min for  max selected L = 3  H – 4  H The AC voltage component across the capacitor is approximately described by the equation: Selected C = 1000  F Przyjmijmy: