Chapter 10 Switched Mode Power Supplies¶

Example 10.1 , Page no:296¶

import math
from __future__ import division

#initialisation of variables
V2=12 #load
D=0.8 #duty cycle
P0=20 #average power

#CALCULATIONS
V1=V2/D #V
I1=P0/V1

#RESULTS
print"The value of I1=",round(I1,3),"A";

The value of I1= 1.333 A

Example 10.2 , Page no:296¶

import math
from __future__ import division

#initialisation of variables
Rl=7 #Load Ω

#CALCULATIONS
fs=30*10**3 #kHz.
Lc=50*10**-6 #Inductor H
D=1-((2*fs*Lc)/Rl)

#RESULTS
print"The value of D=",round(D,3);

The value of D= 0.571

Example 10.4 , Page no:296¶

import math
from __future__ import division

#initialisation of variables
V2=5 #V
V1=12 #V
Rl=5 #Ω

#CALCULATIONS
D=V2/V1
p0=V2**2/Rl

#RESULTS
print"The value of D=",round(D,3);
print"The value of p0=",round(p0,3);

The value of D= 0.417
The value of p0= 5.0

Example 10.7, Page no:297¶

import math
from __future__ import division

#initialisation of variables
D=0.6 #Duty cycle
V1=24 #V
Rl=7

#CALCULATIONS
fs=30*10**3
L=50*10**-6
V2=D*V1
Imax=V2/Rl+((V1-V2)*D)/(2*fs*L) #maximum  values of the inductor current
Imin=V2/Rl-((V1-V2)*D)/(2*fs*L) #minimum values of the inductor current

#RESULTS
print"The value of Imax=",round(Imax,3),"A";
print"The value of Imin=",round(Imin,3),"A";

The value of Imax= 3.977 A
The value of Imin= 0.137 A

Example 10.9, Page no:298¶

import math
from __future__ import division

#initialisation of variables
V1=12
D=0.6
P0=60 #w Supplying power 

#CALCULATIONS
V2=V1/(1-D) #output voltage
Rl=V2**2/P0 #load resistance
I2=V2/Rl #load current

#RESULTS
print"The value of V2=",round(V2,3),"V";
print"The value of Rl=",round(Rl,3),"ohm";
print"The value of I2=",round(I2,3),"A";

The value of V2= 30.0 V
The value of Rl= 15.0 ohm
The value of I2= 2.0 A