Pulse Width Modulation

ulseDuty-Cycle Estimates

Estimate the power-switch duty cycle over the range of input voltages using:

D=Vo+VdVi-VSat

D=24+0.637.2-0.5

=0.67

The duty cycle for Vi = 24v is 0.67.

Inductor

Choose the inductance value to maintain continuous-mode operation

∆IO=2 x 0.1 x IoMax=2 x 0.1 x 200=0.04 A peak to peak

Inductance is given by:

LI = Vi-VSat-VoDt∆IO= 37.2-0.5-240.675x10-60.04=1mH

To fulfill CCM requirement, LI has to be greater or equal to 1mH.

Capacitor

Assuming all the inductor ripple current flows through the filter capacitor and the ESR is zero (??)Capacitor needed to limit the ripple to 50 mV peak-to-peak is:

C= ∆IO8 x fs x ∆VO= 0.048 x 200 x 103 x 0.05=0.5μF

Assuming the capacitance is very large, the ESR needed to limit the ripple to 50mV peak-to-peak is:

ESR=∆Vo∆IO=0.050.04=1.25Ω

RMS current is shown as:

∆IO(rms)=0.6 x 0.289=0.1734 Arms

Catch-Rectifier Snubber Network

R10= 50 x 10-9C8= 50 x 10-9820 x 10-12=60.98Ω

Step 1

flc≤0.1 x fRHP

flc= 1-D2π1 x 220=112Hz fRHP= rload x (1-D)22π x L = 3.12 KHz

Step 2

fc=0.33 x fRHP cross over point= 0.33 x 3.12 KHz=1.03 KHz

Step 3

Open loop gain

GOL(fc)= 20logVout1-DVComp+40logflcfc=20log241-0.670.7+40log(1121.03k)

=40.33-38.54=1.785dB

Step 4+5

fcompz=0.7 x flc=78.4

Gcompz= 20log(fcompzfc)-GOLfc=20log⁡(78.41.03k) – 1.785 = -24.16

Step 6+7

R1 = 100k (R5)

RBIAS= VREF x R1Vout- VREF= 1 x 100k24-1=4.35 kΩ : choose 3.9k Ω

Step 8

Gcompz=20log(RZ2R1)

RZ2R1=10GCompz20

RZ2= R1 x 10(-24.1620)=6.2K Ω (R4) : choose 6.8k Ω

Step 9

fcompz= 12π x RZ2 x CZ2 x 78.4=0..29μF : (C2) choose 0.33 μF

Step 10

fcompz= 12π x R1 x CZ3

CZ3= 12π x R1 x fcompz= 12π x 100k x 78.4=20.3nF :choose 0.022μF

Step 11

fcompz=fRHP= 12π x Rz2 x CP1

CP1=12π x 6.8k x 3.12k=7.5nF

Step 12

fcompz= fRHP= 12π x Rz3 x Cz3

RZ3= 12π x 20.3nF x 3.12k=2.5kΩ

Aea(s)= -1SR5C2+C1[S(R5+R7(3+1)(SR4(2+1)SR73+1[SR4C11(2+1)

= -[2.08ms+12.03ms+10.0335s50.75μs+150μs+1]...