Chapter 3 Diode Applications

Ex 3_1 pg-3.14

In [3]:
from math import sqrt, pi
from __future__ import division
print "Refer to the circuit diagram shown in figure-3.7"
Rf=75 #  diode forward resistance
Rl=10e3 #  load resistance
Rs=10 #  transformer secondary resistance
Ep=230 #  rms value of primary voltage
N2byN1=1/3 #  turns ratio
Es=Ep*N2byN1 #  rms value of secondary voltage
Esm=sqrt(2)*Es #  peak value of secondary voltage
Im=Esm/(Rs+Rf+Rl)#
Im=Im*1e3#
print "\n Therefore peak value of current is %.2f mA \n"%(Im)
Idc=Im/pi#
print "\n Average current is %.3f mA \n"%(Idc)
Irms=Im/2 #  for half wave rectifier
print "\n rms current is %.3f mA \n"%(Irms)
Idc1=Idc*1e-3#
Edc=Idc1*Rl#
print "\n DC output voltage is %.2f V \n"%(Edc)
Pdc=Edc*Idc1 #  Dc output power
Irms1=Irms*1e-3#
Pac=Irms1**2*(Rs+Rf+Rl) #  AC output power
n=Pdc/Pac*100 #  efficiency
print "\n Efficiency is %.2f %% \n"%(n)
print "Ripple factor for half wave rectifier is 1.21"

Refer to the circuit diagram shown in figure-3.7

 Therefore peak value of current is 10.75 mA 


 Average current is 3.422 mA 


 rms current is 5.375 mA 


 DC output voltage is 34.22 V 


 Efficiency is 40.19 % 

Ripple factor for half wave rectifier is 1.21

Ex 3_2 PG-3.15

In [4]:
from math import sqrt, pi
from __future__ import division
print "Refer to the figure-3.8 shown"
Vin=0 #  cut-in voltage for an ideal diode is zero
Rf=0 #  forward resistance for an ideaal diode is zero
print "For an ideal diode"
Vm=15#
Vdc=-Vm/pi#
print "\n DC output voltage is %.2f V\n "%(Vdc)
print "-ve sign indicates that voltage is negative wrt ground"
print "For a silicon diode Vin=0.7V "
Vin=0.7#
Vdc=-(Vm-Vin)/pi#
print "\n DC output voltage is %.2f V\n "%(Vdc)
print "-ve sign indicates that voltage is negative wrt ground"

Refer to the figure-3.8 shown
For an ideal diode

 DC output voltage is -4.77 V
 
-ve sign indicates that voltage is negative wrt ground
For a silicon diode Vin=0.7V 

 DC output voltage is -4.55 V
 
-ve sign indicates that voltage is negative wrt ground

Ex 3_3 PG-3.16

In [5]:
from math import sqrt, pi
from __future__ import division
Rl=1 #  load resistance in kohm
Vm=10#
print "For an ideal diode Vin= 0V"
Vin=0 #  for ideal diode
Rf=0 #  for ideal diode
Edc=Vm/pi#
Idc=Edc/Rl#
print "\n Dc voltage is %.2f V and load current is %.2f mA \n"%(Edc,Idc)
print "For a silicon diode Vin=0.7 V"
Vin=0.7#
Edc=(Vm-Vin)/pi#
Idc=Edc/Rl
print "\n Dc voltage is %.2f V and load current is %.2f mA \n"%(Edc,Idc)

For an ideal diode Vin= 0V

 Dc voltage is 3.18 V and load current is 3.18 mA 

For a silicon diode Vin=0.7 V

 Dc voltage is 2.96 V and load current is 2.96 mA

Ex 3_4 Pg3.16

In [6]:
from math import sqrt, pi
from __future__ import division
Esm=300 #  peak rms voltage
Edc=Esm/pi #  average output voltage
print "\n average output voltage is %.3f V"%(Edc)

 average output voltage is 95.493 V

Ex 3_5 PG-3.28

In [7]:
from math import sqrt, pi
from __future__ import division
Es=30 #  rms voltage
Rf=2#
Rs=8#
Rl=1e3 #  in kohm
Esm=sqrt(2)*Es #  peak value of voltage
Im=Esm/(Rf+Rl+Rs) #  peak value of current
Idc=2*Im/pi #  average current for full wave rectifier
P=Idc**2*Rl#
print " Power delivered to the load is %.3f W \n"%(P)
Vdc_noload=2*Esm/pi#
print "\n Vnl=%f \n"%(Vdc_noload)
Vdc_fullload=Idc*Rl#
print "\n Vfl=%.2f \n"%(Vdc_fullload)
reg=(Vdc_noload-Vdc_fullload)/Vdc_fullload*100#
print "\n percentage regulation is %f %% \n"%(reg)
print "Efficiency of rectification =(DC output)/(AC output)"
x=(1+(Rf+Rs)/Rl)**(-1)#
n=8/pi**2*x*100#
print "\n Efficiency of rectification is %.1f %% \n"%(n)
Irms=Im/sqrt(2)#
AC_rating=Es*Irms#
TUF=P/AC_rating#
print "\n TUF of secondary is %.3f"%(TUF)
#The exact answer for percentage regulation is 1% not .97% as shown in the book ....
#because in the book it has rounded off the value 27.009 to 27 only

 Power delivered to the load is 0.715 W 


 Vnl=27.009489 


 Vfl=26.74 


 percentage regulation is 1.000000 % 

Efficiency of rectification =(DC output)/(AC output)

 Efficiency of rectification is 80.3 % 


 TUF of secondary is 0.803

Ex 3_6 PG-3.29

In [8]:
from math import sqrt, pi
from __future__ import division
print "Refer to the figure-3.16 shown"
Rl=100#
Idc=1 #  maximum value of DC current in each diode
print "We know Idc=Im/pi for each diode"
Vm=pi*Rl*Idc#
Vp=2*Vm#
print "\n maximum peak to peak voltage is %.2f V"%Vp

Refer to the figure-3.16 shown
We know Idc=Im/pi for each diode

 maximum peak to peak voltage is 628.32 V

Ex 3_7 PG-3.30

In [9]:
from math import sqrt, pi
from __future__ import division
Rf=1#
Es=10#
Rs=5#
Esm=sqrt(2)*Es#
Edc_nl=2*Esm/pi#
print "Therefore o load DC output voltage is %.4f V \n"%(Edc_nl)
Idc=100e-3#
print "We know that Idc=2Im/pi and Im=Esm/(Rf+rs+rl)"
Im=Idc*pi/2#
Rl=Esm/Im-Rf-Rs #  load resistance
Edc_ol=Idc*Rl #  DC output voltage at 100mA
print "\n DC output voltage at 100mA is %.4f V \n "%(Edc_ol)
reg=(Edc_nl-Edc_ol)/Edc_ol*100#
print "\n percentage regulation is %.2f %%"%(reg)

Therefore o load DC output voltage is 9.0032 V 

We know that Idc=2Im/pi and Im=Esm/(Rf+rs+rl)

 DC output voltage at 100mA is 8.4032 V 
 

 percentage regulation is 7.14 %

Ex 3_8 PG-3.31

In [10]:
from __future__ import division
Pdc=500 #  for half wave rectifier
n=40.6 #  maximum efficiency for half wave rectifier
print " We know that efficiency=Pdc/Pac "
print "for Half wave rectifier "
Pac=Pdc/n*100#
print "\n AC input power is %.3f W \n"%(Pac)
print "for Full wave rectifier "
n=81.2 #  maximum efficiency for full wave rectifier 
Pac=Pdc/n*100#
print "\n AC input power is %.3f W \n"%(Pac)

 We know that efficiency=Pdc/Pac 
for Half wave rectifier 

 AC input power is 1231.527 W 

for Full wave rectifier 

 AC input power is 615.764 W

Ex 3_9 PG-3.32

In [11]:
from math import sqrt, pi
from __future__ import division
Rf=10#
Rl=100#
Es=30 #  transformer rms voltage
Esm=sqrt(2)*Es #  peak value of the voltage
Im=Esm/(Rf+Rl)#
Idc=2*Im/pi#
Edc=Idc*Rl#
print "Average voltage is %.2f V \n"%(Edc)
Pdc=Idc**2*Rl#
Irms=Im/sqrt(2) #  rms value of the current
Pac=Irms**2*(Rf+Rl)#
n=Pdc/Pac*100#
print "\n Efficiency is %.2f %%"%n

Average voltage is 24.55 V 


 Efficiency is 73.69 %

Ex 3_10 PG-3.32

In [12]:
from math import sqrt, pi
from __future__ import division
Es=100 #  rms value of current
Rf=50 #  forward resistance
Rl=950#
Rs=0 #  resistance of the transformer secondary which is assumed to be 0 ohm
Esm=sqrt(2)*Es #  peak value of the input voltage
Im=Esm/(Rs+Rl+Rf)#
Irms=Im/sqrt(2)#
print "rms value of input current is %.4f A \n"%(Irms)
Idc=2*Im/pi#
print "\n load value of current is %.4f A"%(Idc)

rms value of input current is 0.1000 A 


 load value of current is 0.0900 A

Ex 3_11 PG-3.38

In [13]:
from math import sqrt, pi
from __future__ import division
Rf=0.1#
Idc=10#
Rs=0#
Es=30 #  rms value of input voltage
Esm=sqrt(2)*Es #  peak value of the input voltage
Im=Idc*pi/2 #  DC output current
print "We know that Im=Esm/(2Rf+Rs+Rl) for fullwave rectifier"
Rl=Esm/Im-2*Rf-0#
print "\n Therefore load resistance is %.1f ohm \n"%(Rl)
Pdc=Idc**2*Rl #  Dc output power rating
Irms=Im/sqrt(2) #  rms value of input current
Pac=Irms**2*(2*Rf+Rs+Rl) #  Ac input power
n=Pdc/Pac*100 #  efficiency
print "\n Therefore efficiency is %.2f %% \n"%(n)

We know that Im=Esm/(2Rf+Rs+Rl) for fullwave rectifier

 Therefore load resistance is 2.5 ohm 


 Therefore efficiency is 75.05 %

Ex 3_12 PG-3.39

In [14]:
from math import sqrt, pi
from __future__ import division
Rl=5e3#
N1toN2=2 #  transformer turns ratio
Ep=460 #  rms value of primary voltage
Es=Ep/N1toN2#
Esm=sqrt(2)*Es #  peak value of the secondary voltage
Im=Esm/Rl #  We neglect forward diode resistance
Idc=2*Im/pi#
print "\n  Therefore DC load current is %f A \n"%(Idc)
Edc=Idc*Rl#
print "\n  DC load voltage is %.3f V \n"%(Edc)
Rf=.482 #  ripple factor for full bridge rectifier
Vrip=Rf*Edc #  ripple voltage
print "\n  Therefore ripple voltage is %.1f V \n"%(Vrip)
print " Peak value of bridge rectifier=PIV rating of each diode"
PIV=Esm#
print "\n  Therefore PIV rating of each diode is %.2f V"%(PIV)

  Therefore DC load current is 0.041415 A 


  DC load voltage is 207.073 V 


  Therefore ripple voltage is 99.8 V 

 Peak value of bridge rectifier=PIV rating of each diode

  Therefore PIV rating of each diode is 325.27 V

Ex 3_13 PG-3.40

In [15]:
from math import sqrt, pi
from __future__ import division
Ep=230# #rms value of primary voltage
N2toN1=1/15 #  turns ratio
Rf=0 #  diode is ideal
Rs=0 #  transformer is ideal
Rl=50 #  load resistance
Es=Ep*N2toN1 #  rms vaue of primary voltage
Esm=sqrt(2)*Es #  peak value of input voltage
Im=Esm/(Rs+2*Rf+Rl)#
Idc=2*Im/pi#
Edc=Idc*Rl #  load voltage
print "\n Therefore load voltage is %.1f V\n"%(Edc)
Rf=.482 #  ripple factor for full wave rectifier
Vrip=Rf*Edc #  ripple voltage
print "\n ripple voltage is %.4f V"%(Vrip)
#in the book the ripple voltage has been rounded off to 
#6.6516V but the actual ans is 6.6539V

 Therefore load voltage is 13.8 V


 ripple voltage is 6.6539 V

Ex 3_14 PG-3.40

In [16]:
from math import sqrt, pi
from __future__ import division
Rf=0 #  diode forward resistance
Es=240 #  rms value of supply voltage
Rl=48 #  load resistance
Im=sqrt(2)*Es/(Rl+Rf)#
Idc=2*Im/pi#
print " Average load current is %.3f A \n"%(Idc)
Pdc=Idc**2*Rl#
Irms=Im/sqrt(2) #  rms value of input current
Pac=Irms**2*Rl#
n=Pdc/Pac*100 #  efficiency
print "\n Therefore efficiency is %.2f %% \n"%(n)

 Average load current is 4.502 A 


 Therefore efficiency is 81.06 %

Ex 3_15 PG-3.41

In [17]:
from math import sqrt, pi
from __future__ import division
Esm=100 #  peak value of supply voltage
Rf=25 #  diode forward voltage
Rl=950 #  load resistance
Rt=(2*Rf)+Rl #  total resistance
Im=Esm/Rt#
Idc=2*Im/pi#
Edc=Idc*Rl#
print "\n DC output voltage is %f V \n"%(Edc)
Irms=Im/sqrt(2) #  rms value of input current
x=(Irms/Idc)**2-1#
Rf=sqrt(x) #  ripple factor#
print "\n Therefore ripple factor is %.4f \n"%(Rf)
Pdc=Idc**2*Rl#
Pac=Irms**2*(2*Rf+Rl) #  Ac input power
n=Pdc/Pac*100 #  efficiency
print "\n Therefore efficiency is %.2f %% \n"%(n)
print "Peak value of bridge rectifier=PIV rating of each diode"
PIV=Esm#
print "\n Therefore PIV rating of each diode is %.2f V"%(PIV)
#In the book the answer for Edc=57.5985V which is wrong because they have taken 
#Rf=50 ohm instead of 25 ohm as given in the question similarly 
#the efficiency=73.3417% in the book is wrong 
#the correct answer for efficiency is 80.97%

 DC output voltage is 60.478878 V 


 Therefore ripple factor is 0.4834 


 Therefore efficiency is 80.97 % 

Peak value of bridge rectifier=PIV rating of each diode

 Therefore PIV rating of each diode is 100.00 V

Ex 3_16 PG-42

In [18]:
from math import sqrt, pi
from __future__ import division
Rl=100 #  load resistance
Es=230 #  rms value of input voltage
Rf=0 #  ideal diode resistance
Rs=0 #  neglecting transformer resistance
f=50 #  frequency of the supply
Esm=sqrt(2)*Es #  peak value of the input voltage
Edc=2*Esm/pi #  as Rf=0 ohm 
print "\n Average DC voltage is %.0f V \n"%(Edc)
Im=Esm/Rl#
Idc=2*Im/pi#
print "\n Therefore DC load current is %.2f A \n"%(Idc)
f=2*f #  frequency of output waveform
print "\n frequency of output waveform is %.0f Hz"%(f)

 Average DC voltage is 207 V 


 Therefore DC load current is 2.07 A 


 frequency of output waveform is 100 Hz

Ex 3_17 PG-3.56

In [19]:
from math import sqrt, pi
from __future__ import division
Edc=12 #  output DC voltage
f=50 #  frequency
Idc=50e-3#
C=100e-6 #  filter capacitor
Rl=2e3 #  load resistance
Vr=Edc/(2*f*C*Rl) #  rms value of ripple voltage
print "\n rms value of ripple voltage is %.1f V \n "%(Vr)
Rf=(4*sqrt(3)*f*C*Rl)**(-1)*100 #  ripple factor
print "\n ripple factor is %.2f %% \n"%(Rf)

 rms value of ripple voltage is 0.6 V 
 

 ripple factor is 1.44 %

Ex 3_18 PG-3.56

In [20]:
from math import sqrt, pi
from __future__ import division
Es=120 #  rms value of input voltage
f=50 #  frequency
Idc=50e-3#
C=100e-6 #  filter capacitor
Esm=sqrt(2)*Es#
Edc=Esm-Idc/(4*f*C)#
print "\n DC output voltage is %.4f V \n"%(Edc)
Vr=Idc/(4*sqrt(3)*f*C) #  rms value of ripple voltage
print "\n rms value of ripple voltage is %.4f V \n "%(Vr)
Rf=Vr/Edc#
print "\n ripple factor is %f \n"%(Rf)

 DC output voltage is 167.2056 V 


 rms value of ripple voltage is 1.4434 V 
 

 ripple factor is 0.008632

Ex 3_19 PG-3.56

In [21]:
from math import sqrt, pi
from __future__ import division
Ep=230 #  rms value of primary voltage
N1toN2=10 #  turns ratio
Rl=50 #  load resistance
f=50 #  frequency of the supply in Hz
Es=Ep/N1toN2 #  rms vaue of secondary voltage or the input voltage
Esm=sqrt(2)*Es #  peak value of input voltage
Im=Esm/(Rl)#
Idc=2*Im/pi#
Edc=Idc*Rl #  load voltage
print "\n Therefore load voltage is %.1f V \n"%(Edc)
Rf=.48 #  ripple factor for full wave rectifier without filter
Vrip=Rf*Edc #  ripple voltage
print "\n Ripple factor is 0.48"
print "\n ripple voltage is %.4f V \n"%(Vrip)
print "If the capacitor filter is used then"
C=470e-6 #  filter capacitor
Edc=Esm-Idc/(4*f*C)#
print "\n new DC load voltage is %.2f V \n"%(Edc)
Rf=((4*sqrt(3)*f*C*Rl))**(-1) #   new ripple factor
Vrip=Rf*Edc #  new ripple voltage
print "\n Ripple factor is %.4f"%(Rf)
print "\n rms value of ripple voltage is %.4f V \n "%(Vrip)
# in the book the new ripple factor is 3.256e-3 which is wrong 
#the actual answer is 0.1228 hence the new ripple voltage is 3.4544V 
# not 0.09156V as shown in the book 

 Therefore load voltage is 20.7 V 


 Ripple factor is 0.48

 ripple voltage is 9.9395 V 

If the capacitor filter is used then

 new DC load voltage is 28.12 V 


 Ripple factor is 0.1228

 rms value of ripple voltage is 3.4544 V

Ex 3_20 PG-3.57

In [22]:
from math import sqrt, pi
from __future__ import division
Rf=.01 #  ripple factor in percentage
Rl=2 #  load resistance in kohm
f=50 #  frequency
print "the %ripple factor=Rf=((4*sqrt(3)*f*C*Rl))**(-1)*100 "
C=((4*sqrt(3)*f*Rf*Rl))**(-1)*100 #  filter capacitor
print "\n the filter capacitor is %.3f mF"%(C)

the %ripple factor=Rf=((4*sqrt(3)*f*C*Rl))**(-1)*100 

 the filter capacitor is 14.434 mF

Ex 3_21 PG-3.58

In [23]:
from math import sqrt, pi
from __future__ import division
Idc=100e-3 #  average current
C=500e-6 #  filter capacitor
Esm=18 #  peak voltage
f=50 #  frequency of the supply in Hz
Edc=Esm-Idc/(4*f*C)#
Rl=Edc/Idc #  load resistance
Rf=(4*sqrt(3)*f*C*Rl)**(-1)*100 #  ripple factor
print "\n ripple factor is %.2f %% \n"%(Rf)

 ripple factor is 3.40 %

Ex 3_22 PG-3.58

In [24]:
from math import sqrt, pi
from __future__ import division
f=50 #  frequency
C=1000e-6 #  filter capacitor
Rl=500 #  load resistance
Vrms=120 #  rms value of voltage
T1=1e-3 #  conduction period of diode,T1=1ms
print "conduction period of diode,T1=1ms"
Esm=sqrt(2)*Vrms #  peak value of input voltage
print "Edc=Esm-Idc/(2*f*C) and Idc=Edc/Rl"
Edc=Esm/(1+(2*Rl*f*C)**(-1)) #  output dc voltage
Idc=Edc/Rl#
T=1/f#
#Idc*T=Ip*T1
#Ip is the surge current
Ip=Idc*T/T1#
print "\n hence the diode should be rated for a minimum surge\n current of %.2f A \n"%(Ip)

conduction period of diode,T1=1ms
Edc=Esm-Idc/(2*f*C) and Idc=Edc/Rl

 hence the diode should be rated for a minimum surge
 current of 6.66 A

Ex 3_23 PG-3.59

In [25]:
from math import sqrt, pi
from __future__ import division
Rf=0.1 #  riplle facto
Edc=10#
Vrip=Rf*Edc #  rms value of voltage
print "\n rms value of ripple voltage is %.0f V \n "%(Vrip)
Vp_p=2*sqrt(2)*Vrip#
print "\n peak to peak voltage is %.4f V "%(Vp_p)

 rms value of ripple voltage is 1 V 
 

 peak to peak voltage is 2.8284 V

Ex 3_24 PG-3.59

In [26]:
from math import sqrt, pi
from __future__ import division
Es=230 #  rms value of input voltage
f=50 #  frequency
Idc=50e-6#
Rl=100 #  load resistance
C=1000e-6 #  filter capacitor
Esm=sqrt(2)*Es#
Edc=2*Esm/pi#
print "\n Therefore DC output voltage is %.2f V \n"%(Edc)
Idc=Edc/Rl#
print "if the capacitor filter C=1000e-6 is use then "
Rf=(4*sqrt(3)*f*C*Rl)**(-1) #  ripple factor
print "\n ripple factor is %.4f \n"%(Rf)
Edc=Esm-Idc/(4*f*C)#
print "\n Therefore new DC load voltage is %.4f V \n"%(Edc)

 Therefore DC output voltage is 207.07 V 

if the capacitor filter C=1000e-6 is use then 

 ripple factor is 0.0289 


 Therefore new DC load voltage is 314.9155 V

Ex 3_25 Pg-3.60

In [27]:
from math import sqrt, pi
from __future__ import division
Es=230#
f=50#frequency
Rf=.005 #  ripple factor
Il=0.5 #  average load current
Esm=sqrt(2)*Es #  peak value of input voltage
print " For a half wave rectifier Ripple factor=(2*sqrt(3)*f*C*Rl)**(-1)"
Edc=Esm/pi#for half wave rectifier
Rl=Edc/Il#
C=(2*sqrt(3)*f*Rf*Rl)**(-1) #  for half wave rectifier
C=C*1e3#
print "\n  Therefore capacitance required is %.3f mF"%(C)

 For a half wave rectifier Ripple factor=(2*sqrt(3)*f*C*Rl)**(-1)

  Therefore capacitance required is 5.576 mF

Ex 3_26 PG-3.60

In [28]:
from math import sqrt, pi
from __future__ import division
Rl=1000#
C=500e-3
f=50#
Rf=(4*sqrt(3)*f*C*Rl)**(-1) #  ripple factor
Rf=Rf*1e6#
print "\n ripple factor is %.2f 10**(-6) \n"%(Rf)

 ripple factor is 5.77 10**(-6)

Ex 3_27 PG-3.60

In [29]:
from math import sqrt, pi
from __future__ import division
Il=12e-3 #  load current
Es=200 #  rms voltage
Rf=0.02 #  riplle factor
Esm=sqrt(2)*Es #  peak value of input voltage
Edc=2*Esm/pi#
Idc=Il#
Rl=Edc/Idc #  load resistance
f=50 #  frequency of the supply in Hz
print " For a half wave rectifier Ripple factor=1/(2*sqrt(3)*f*C*Rl)"
C=(4*sqrt(3)*f*Rf*Rl)**(-1) #  filter capacitor
print "\n  Therefore minimum value of capacitance required is %.3f microF"%(C*1e6)
#C=9.619 microF not 9.622 microF

 For a half wave rectifier Ripple factor=1/(2*sqrt(3)*f*C*Rl)

  Therefore minimum value of capacitance required is 9.619 microF

Ex 3_28 PG-3.61

In [30]:
from math import sqrt, pi
from __future__ import division
Es=230 #  rms voltage
f=50#
Il=10e-3 #  load current
Rf=.01 #  ripple factor
Esm=sqrt(2)*Es #  peak value of input voltage
Edc=2*Esm/pi #  for full wave
Rl=Edc/Il#
C=(4*sqrt(3)*f*Rf*Rl)**(-1) #  for full wave rectifier
C=C*1e6#
print "\n Therefore capacitance required is %.2f  microF"%(C)

 Therefore capacitance required is 13.94  microF

Ex 3_29 PG-3.61

In [31]:
from math import sqrt, pi
from __future__ import division
Rl=2e3 #  load resistance
Es=200 #  rms voltage
f=50#
Esm=sqrt(2)*Es #  peak value of input voltage
Rf=0 #  ideal diodes
Rs=0#
Ism=Esm/(Rf+Rs+Rl)#
Idc=2*Ism/pi#
print "\n Therefore Average DC load current is %.2f A \n"%(Idc)
Edc=Idc*Rl#
print "\n Therefore average DC voltage is %.0f V \n"%(Edc)
Rf=0.48 #  ripple factor
Vrip=Rf*Edc #   ripple voltage
print "\n rms value of ripple voltage is %.1f V \n "%(Vrip)
print "if a filter capacitor C=500 microF is used then"
C=500e-6 #  capacitor filter
Rf=(4*sqrt(3)*f*C*Rl)**(-1) #  for full wave rectifier
Vrip=Rf*Edc #  new ripple voltage
print "\n rms value of new ripple voltage is %.4f V \n "%(Vrip)

 Therefore Average DC load current is 0.09 A 


 Therefore average DC voltage is 180 V 


 rms value of ripple voltage is 86.4 V 
 
if a filter capacitor C=500 microF is used then

 rms value of new ripple voltage is 0.5198 V

Ex 3_30 PG-3.67

In [32]:
from __future__ import division
Vnl=10 #  no load output voltage
Vfl=9.8 #  full output voltage
LR=Vnl-Vfl #  load regulation
LR=(Vnl-Vfl)/Vfl*100 #  percentage load regulation
print "\n percentage load regulation is +%.2f %%"%(LR)

 percentage load regulation is +2.04 %

Ex 3_31 PG-3.67

In [33]:
from __future__ import division
LR=3e-3 #  load regulation
Vnl=15 #  no load voltage or maximum voltage
Vfl=Vnl-LR #  full load voltage
LR=(Vnl-Vfl)/Vfl*100 #  percentage load regulation
print "\n percentage load regulation is +%.2f %%"%(LR)

 percentage load regulation is +0.02 %

Ex 3_32 PG-3.67

In [34]:
from __future__ import division

Vhl=10+.3 #  high line voltage
Vll=10-.3 #  low line voltage
SR=Vhl-Vll #  source regulation
Vnom=10 #  nominal load voltage
SR=SR/Vnom*100 #  percentage source regulation
print "\n percentage source regulation is %.0f %%"%(SR)

 percentage source regulation is 6 %

Ex 3_33 PG-3.70

In [35]:
from __future__ import division
print "Refer to the figure-3.45 shown"
Vz=6.1 #  zener voltage
Iz_min=2.5e-3 #  minimum zener current
Iz_max=25e-3 #  maximum zener current
rZ=0 #  ideal zener diode
R=2.2e3#
Rl=1e3 #  loadd resistance
Il=Vz/Rl#
#For minimum input voltage(Vin_min)
Iz=Iz_min
I=Iz_min+Il#
Vin_min=Vz+I*R#
print "\n minimum input voltage(Vin_min) is %.2f V \n"%(Vin_min)
#For maximum input voltage(Vin_max)
I=Iz_max+Il#
Vin_max=Vz+I*R#
print "\n maximum input voltage(Vin_max) is %.2f V \n"%(Vin_max)
print "\n range of input voltage is from %.3f V to %.2f V \n"%(Vin_min,Vin_max)

Refer to the figure-3.45 shown

 minimum input voltage(Vin_min) is 25.02 V 


 maximum input voltage(Vin_max) is 74.52 V 


 range of input voltage is from 25.020 V to 74.52 V

Ex 3_34 PG-3.70

In [36]:
from __future__ import division
Vo=5 #  output voltage
Vin_min=12-3 #  min input voltage
Vin_max=12+3 #  max input voltage
Iz_min=10e-3 #  minimum zener current
Il=20e-3 #  load current
Pz=500e-3 #  Zener wattage
Vz=Vo #  zener voltage
print "Step1 : Maximum power dissipation correesponds to Iz_max"
Iz_max=Pz/Vz#
print "\n maximum current that should flow through the zener diode is %.1f A \n"%(Iz_max)
print "Step2 : We know that Il is constant"
#for Vin_max, Iz=Iz_max
I=Il+Iz_max#
Rmin=(Vin_max-Vz)/I#
print "\n minimum resistance required is %.2f ohm \n"%(Rmin)
print "Iz is maximum when R=Rminimum"
print "Step3 : for calculation of Rmax I must be minimum ie I=Iz_min "
I=Il+Iz_min
Rmax=(Vin_min-Vz)/I#
print "\n maximum resistance required is %.2f ohm \n"%(Rmax)
print "\n Thus R must be greater than %.2f ohm and less than \n %.2f ohm for proper regulation \n"%(Rmin,Rmax)

Step1 : Maximum power dissipation correesponds to Iz_max

 maximum current that should flow through the zener diode is 0.1 A 

Step2 : We know that Il is constant

 minimum resistance required is 83.33 ohm 

Iz is maximum when R=Rminimum
Step3 : for calculation of Rmax I must be minimum ie I=Iz_min 

 maximum resistance required is 133.33 ohm 


 Thus R must be greater than 83.33 ohm and less than 
 133.33 ohm for proper regulation

Ex 3_35 PG-3.72

In [37]:
from __future__ import division
print "refer to the figure-3.47 shown"
Vz=10 #  output voltage
Vin=20 #  input voltage
Iz_max=25e-3 #  maximum zener current
Iz_min=5e-3 #  minimum zener current
R=300#
Rz=0 #  zener resistance
I=(Vin-Vz)/R#
#for Il_min Iz=Iz_max
Il_min=I-Iz_max #  minimum load current
print "\n minimum load current is %.2f mA \n"%(Il_min*1e3)
#for Il_max, Iz=Iz_min
Il_max=I-Iz_min #  maximum load current
print "\n maximum load current is %.2f mA \n"%(Il_max*1e3)
Rl_min=Vz/Il_max #  minimum load resistance
print "\n minimum load resistance is %.3f ohm \n"%(Rl_min)
# in the book in the question it given that Iz_max=50mA
#but during the solution Iz_max is taken as 25mA I have taken Iz_max=25mA
# in this program 

refer to the figure-3.47 shown

 minimum load current is 8.33 mA 


 maximum load current is 28.33 mA 


 minimum load resistance is 352.941 ohm

Ex 3_38 Pg-3.75

In [38]:
from __future__ import division
Vo=5#
Il=20e-3#
Pz=500e-3#
Rl=Vo/Il#
Il_min=Il #  minimum load current
Il_max=Il #  maximum load current
Iz_max=Pz/Vo #  maximum zener current
Iz_min=5e-3 #  minimum zener current
V=12 #  input DC voltage
Vin_min=12-3 #  min input voltage
Vin_max=12+3 #  max input voltage
Rmax=(Vin_min-Vo)/(Il_max+Iz_min)#
print "\n maximum resistance required is %.0f ohm \n"%(Rmax)
Rmin=(Vin_max-Vo)/(Il_min+Iz_max)#
print "\n minimum resistance required is %.2f ohm \n"%(Rmin)
print "\n So series resistance must be selected between %.2f ohm to %.0f ohm \n"%(Rmin,Rmax)

 maximum resistance required is 160 ohm 


 minimum resistance required is 83.33 ohm 


 So series resistance must be selected between 83.33 ohm to 160 ohm

Ex 3_39 Pg-3.76

In [39]:
from __future__ import division
Vo=10#
Il_min=0 #  mainimum load current
Il_max=10e-3 #  maximum load current
Iz_max=50e-3 #  maximum zener current
Iz_min=2e-3 #  minimum zener current
Vin_min=20 #  min input voltage
Vin_max=30 #  max input voltage
Rl_min=Vo/Il_max#
Rmax=(Vin_min-Vo)/(Il_max+Iz_min)#
print "\n maximum resistance required is %.2f ohm \n"%(Rmax)
Rmin=(Vin_max-Vo)/(Il_min+Iz_max)#
print "\n minimum resistance required is %.0f ohm \n"%(Rmin)
print "\n So series resistance must be selected between %.0f ohm to %.2f ohm \n"%(Rmin,Rmax)

 maximum resistance required is 833.33 ohm 


 minimum resistance required is 400 ohm 


 So series resistance must be selected between 400 ohm to 833.33 ohm

Ex 3_40 Pg-3.76

In [40]:
from __future__ import division
Vo=24#
Il_min=0 #  minimum load current
Pz=600e-3#
Vin=32 #  input voltage
Iz_max=Pz/Vo#
Rmin=(Vin-Vo)/(Il_min+Iz_max)#
print "\n minimum resistance required is %.0f ohm \n"%(Rmin)
print "\n As Vin and Il are not changing R=Rmin=%.0f ohm\n is sufficient to work as a regulator\n"%(Rmin)
print "For Rl=1200 ohm"
Rl=1200#
Il=Vo/Rl#
print " \n load current is: %.2f A \n"%(Il)
R=Rmin
It=(Vin-Vo)/R#
Iz=It-Il#
print "\n zener current is :%.3f A \n"%(Iz)
print " As Iz=Iz_min=%.3f A, the circuit will work as a regulator"%(Iz)

 minimum resistance required is 320 ohm 


 As Vin and Il are not changing R=Rmin=320 ohm
 is sufficient to work as a regulator

For Rl=1200 ohm
 
 load current is: 0.02 A 


 zener current is :0.005 A 

 As Iz=Iz_min=0.005 A, the circuit will work as a regulator

Ex 3_41 PG-3.79

In [41]:
from __future__ import division
Pd=400e-3#
Vz=8.1 #  output voltage
Vo=Vz#
Zz=8#
Vin=15#
Izm=Pd/Vz#
Rmax=(Vin-Vz)/Izm#
print "\n maximum series resistance is %.3f ohm \n"%(Rmax)
Iz_min=5e-3 #  we select the minimum zener current
Il_max=Izm-Iz_min #  maximum load current
print "\n maximum load current is %.7f A \n"%(Il_max)
Rl=Vz/Il_max #  load resistance
deltaVin=.1*Vin #  change in input voltage is equal to 10% of the original input voltage
R=Rmax #  series resistance
x=(Rl*Zz)/(Rl+Zz)#
deltaVo=(deltaVin*x)/(R+x)#
Sv=deltaVo/deltaVin #  voltage stability factor 
print "\n voltage stability factor  is %.3f \n"%(Sv)
SR=deltaVo/Vo*100 #   line regulation for a 10% change in Vin
print "\n line regulation is %.4f %% \n"%(SR)
deltaIL=Il_max#
y=(R*Zz)/(R+Zz)
deltaVo=deltaIL*y#
LR=deltaVo/Vo*100 #  load regulation  
print "\n load regulation is %.4f %% \n"%(LR)
z=(Rl*Zz)/(Rl+Zz)
RR=z/(R+z) #  ripple rejection ratio
print "\n ripple rejection ratio is %.3f \n"%(RR)
Ro=y #  output resistance
print "\n output resistance is %.3f ohm \n"%(Ro)

 maximum series resistance is 139.725 ohm 


 maximum load current is 0.0443827 A 


 voltage stability factor  is 0.052 


 line regulation is 0.9629 % 


 load regulation is 4.1461 % 


 ripple rejection ratio is 0.052 


 output resistance is 7.567 ohm