Project 1
Question 1 - buck converter ccm vin = 40-50v
Vmin = 40; % input supply 40- 50 volt
Vmax = 50;
in this simulation ckt use maximum input voltage
P = 300; % 300w
Vout = 30; % out put voltage
dVo = 0.02; % Vout ripple voltage 0.020000000000000
fs = 50000; %assumed switching frequncy
Dmin = Vout/Vmax;
Dmax = Vout/Vmin;
= 30/40 = 0.750000000000000
D = Dmin; % 0.600000000000000
Io = P/Vout;
= 300/30 - 10 amp %Io out put current 10amp
R = Vout*Vout/P ;
=30 *30/300 = 9 ohm % load resistance
dIl = 0.2*Io; % ripplle is 20 % output current 2amp
L = Vout*(1-D)/(fs*dIl);
= 30*(1-0.75)/(50000*2) % filtter inductor 1.200000000000000e-04
=1.200000000000000e-04 henry
C = (1-D)/(8*L*fs*fs*(dVo)); % filtter Capacitor 8.333333333333334e-06
= (1-0.75)/(8*1.200000000000000e-04 *1.200000000000000e-04 *0.02)
= 8.333333333333334e-06 f
The full schematics of your converter including gate drivers, PWM chips, RC circuits, current sensors, voltage sensors, etc…
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1- A Design report to show the reasoning behind your choices of inductance and capacitance levels in your design.
* 8.333333333333334e-06 f of capacitor value is sufficient for decreases lower than 2%of out put voltage
* 1.200000000000000e-04 henry of inductor is sufficient for decreases lower than 30% of out put voltage
input voltage
output voltage
pwm
mosfet current
mosfet voltge
inductor current and voltage
the needed components in your circuit.
L=1.200000000000000e-04
C = 8.333333333333334e-06 f
R = 3 ohm
2nd questuion
boost ccm converter 20-30/40 volt 600watt
% boost vin - 25 , vout = 40 , p = 600w
taken avarage value of input value
Vin = 25;
Vout = 40 ;
fs = 50000;
p = 600;
dv = 0.02 % 2% of out put voltage is capacitor voltage ripple 0.020000000000000
R = Vout*Vout/p=2.666666666666667
D = 1-(Vin/Vout) =0.375000000000000
Il = Vin/(((1-D)^2)*R)=24.000000000000004
dIl = 0.4*Il; = 9.600000000000001 % ripple 40 % of out put current 9.600000000000001
Lmin = Vin*D/(fs*dIl)=1.953125000000000e-05
L = 1.3*Lmin =2.539062500000000e-05
C = D/(R*fs*dv); =1.406250000000000e-04
components
L =2.539062500000000e-05h
C =1.406250000000000e-04f
R = 2.666666666666667ohm
vin
Vout
inductor current
pwm
mosfet current , voltage
Doiode currernt voltage
3rd question inverting buck boost ccm30-60/-45 volt 300 watt
Vmin = 30 ;
Vmax = 60 ;
Vin = Vmax;
Vout = -45;
dVo = 0.02;
fs = 50000;
P = 300 ;
D = Vout/(Vout-Vin) = 0.428571428571429
R = Vout*Vout/P = 6.750000000000000
Io = -Vout/R = 6.666666666666667
Lmin = (1-D)*(1-D)*R/(2*fs) = 2.204081632653061e-05
L = 1.2*Lmin = 2.644897959183673e-05
C = D/(R*fs*dVo) = 6.349206349206349e-05
out put waveform
Vout
pwm
inductor current
mosfet current voltage
diode current and voltage
all component value selecte in all model acording to desied output