# Chapter 6: Diode Circuits¶

## Example 1, Page 114¶

In [2]:
import math

#variable declaration
Vrms=20;    #in volts
Vm=20*1.41; #in volts
Rf=50.;  #forward resistance in ohms

#Calculations&Results
print'Im=%.4f A'%Im

print 'Idc=%.2e A'%Idc

print 'rms ac load current is=%.3e A'%Irms1

Vdc=Idc*Rf;     #Dc voltage across the diode
print 'Dc voltage across the diode=%.4f V'%Vdc

Pdc=Idc*Idc*RL;     #Dc output power
print 'Dc output power=%.3f W'%Pdc

n=40.6/(1+(Rf/RL));     #conversion efficiency
print 'conversion efficiency=%.f %%'%n

s=Rf*100/RL;        #Pertcentage regulation
print 'Percentage regulation=%.2f %%'%s

Im=0.0226 A
Idc=7.18e-03 A
rms ac load current is=8.699e-03 A
Dc voltage across the diode=0.3591 V
Dc output power=0.062 W
conversion efficiency=39 %
Percentage regulation=4.17 %


## Example 2, Page 114¶

In [4]:
import math

#variable declaration
Rf=100; #forward resistance in ohms
n=10;   #Primary to secondary turns ratio
Vp=240; #Primary input V(rms)

#Calculations&Results
Vm=24*(2**(1./2))/2;   #secondary peak voltage from cenre tap
Vs=Vp/n;    #Secondary input voltage
Im=Vm/(Rf+Rl);  #peak current through the resistance in A
Idc=(2*Im)/math.pi;  #DC Load current in A
print 'DC load current Idc=%.2e A'%Idc
I=Idc/2;    #Direct current supplied by each diode in A
print 'Direct current supplied by each diode Idc=%.2e A'%I
Pdc=Idc*Idc*Rl; #DC power output
print 'Pdc=%.4f W'%Pdc
Irms=Im/(2**(1./2));
Vrp=math.sqrt((Irms*Irms)-(Idc*Idc))*Rl;   #Ripple voltage in V
print 'Ripple voltage Vrp=%.2f V'%Vrp
M=(Rf*100)/Rl;    #percentage regulation
print 'Percentage regulation=%.f %%'%M
n=81.2/(1+(Rf/Rl));  #Efficiency of rectification
print 'Efficiency of rectification=%.1f %%'%n

DC load current Idc=9.82e-03 A
Direct current supplied by each diode Idc=4.91e-03 A
Pdc=0.0965 W
Ripple voltage Vrp=4.75 V
Percentage regulation=10 %
Efficiency of rectification=73.8 %


## Example 3, Page 115¶

In [6]:
import math

#variable declaration
Rf=50.;  #forward resistance in ohms
Vp=30;   #Primary input V(rms)
Vm=30*math.sqrt(2);

#Calculations&Results
Im=Vm/(2*Rf+Rl);    #peak load current in A
Idc=2*Im/math.pi;
print 'Vdc=%.f V'%Vdc
Irms=Im/math.sqrt(2);
Vrp=Rl*math.sqrt(((Irms*Irms)-(Idc*Idc)));   #Ripple voltage in V
print 'Ripple voltage Vrp=%.1f V'%Vrp

M=(2*Rf/Rl)*100;    #Percentage regulation
print 'Percentage regulation=%.f %%'%M

Vdc=26 V
Ripple voltage Vrp=12.6 V
Percentage regulation=4 %


## Example 4, Page 115¶

In [9]:
import math

#variable declaration
Vdc=20;  #DC value in V
Vpp=1.;  #Peak to peak ripple voltage in V

#Calculations&Results
Vp=Vpp/2;   #Peak ripple voltage in V
Vrms=Vp/math.sqrt(2);    #Vrms voltage in V
S=Vrms/Vdc;     #Ripple Factor
print 'Ripple factor=%.4f'%S
T=S*100;
print 'Percentage Ripple=%.2f %%'%T

Ripple factor=0.0177
Percentage Ripple=1.77 %


## Example 5, Page 116¶

In [11]:
import math

#variable declaration
#For a full wave rectifier
#L-type LC filter
f=50#f=line frequency in Hz
w=2*math.pi*f
Vdc=10#Vdc=dc output voltage
y=0.02#y=allowable ripple factor

#Calculations
#y=sqrt(2)/(12*(w^2)*L*C)
#Let L*C=a...............(1)
a=math.sqrt(2)/(y*12*(w**2))
#Lc=critical inductance
#Lc=RL/(3*w)
#For line frequency of 50Hz,Lc=RL/(300*%pi)
#Lc=RL/950
Lc=RL/950
L=0.1#Assumed inductance in henry
C=a/L#C=capacitance calculated from equation (1)
L1=1#Assumed inductance in henry
C1=a/L1#C1=capacitance calculated from equation (1)
Rb=950*L1#Rb=bleeder resistance for good voltage regulation

#Results
print "The designed values of the components for a full wave rectifier with L-type LC filter are:"
print "The load resistance RL is =%.f ohm"%RL
print "The critical inductance Lc is = %.1f H"%Lc
print "The inductance L is=%.1f H"%L
print "The capacitance C is %.f µF"%(C/10**-6)#C is converted in terms of microfarad
#In textbook 957µF is approximately taken as 600µF
print "\nBut if the inductance L designed is of the value = %.f H"%L1
print "the capacitance C will be of the value = %.f µF"%(C1/10**-6)#C1 is converted in terms of microfarad
print "So,a standard value of 50µF can be used in practice"
print "The bleeder resistance Rb for good voltage regulation is=%.f ohm"%Rb
print "\nAs Rb is much greater than RL,little power is wasted in Rb.This reflects the advantage of selecting L>Lc"

The designed values of the components for a full wave rectifier with L-type LC filter are:
The load resistance RL is =100 ohm
The critical inductance Lc is = 0.1 H
The inductance L is=0.1 H
The capacitance C is 597 µF

But if the inductance L designed is of the value = 1 H
the capacitance C will be of the value = 60 µF
So,a standard value of 50µF can be used in practice
The bleeder resistance Rb for good voltage regulation is=950 ohm

As Rb is much greater than RL,little power is wasted in Rb.This reflects the advantage of selecting L>Lc