Chapter 2 Diode Applications

Exa 2.1 page No 112

In [5]:
from math import pi,sqrt
from __future__ import division
# Given data
N2 = 4
N1 = 1
R_L = 1*10**3# ohm
Vm = 40# V
V_Lav = Vm/pi# V
# The average load voltage 
V_LDC = V_Lav# V
print "The average load voltage = %.2f V"%V_LDC
Im = Vm/R_L# A
I_DC = Im/pi# A
I_DC= I_DC*10**3# mA
print "The average load current = %.2f mA"%I_DC
I_DC= I_DC*10**-3# A
# The RMS voltage
V_Lrms = Vm/2# V
print "The RMS voltage = %.2f V"%V_Lrms
# The RMS current
Irms = V_Lrms/R_L# A
Irms= Irms*10**3# mA
print "The RMS current = %.2f mA"%Irms
Irms= Irms*10**-3# A
#Eta = (P_ODC/P_iAC)*100
I_Loc = I_DC# A
P_ODC = (I_Loc**2)*R_L# W
P_iAC = (Irms**2)*R_L# W
# The efficiency of rectification 
Eta = (P_ODC/P_iAC)*100# %
print "The efficiency of rectification = %.2f %%"%Eta
V2rms = Vm/sqrt(2)
I2rms = Irms# A
# The value of TUF 
TUF = ((P_ODC)/(V2rms*I2rms))*100# %
print "The value of TUF = %.2f %%"%TUF
# The ripple factor 
Gamma = (sqrt((V_Lrms**2)-(V_LDC**2)))/V_LDC
Gamma = round(Gamma * 100)# % done by own
print "The ripple factor = %.2f %%"%Gamma

The average load voltage = 12.73 V
The average load current = 12.73 mA
The RMS voltage = 20.00 V
The RMS current = 20.00 mA
The efficiency of rectification = 40.53 %
The value of TUF = 28.66 %
The ripple factor = 121.00 %

Exa 2.2 page No 113

In [6]:
from math import pi,sqrt
from __future__ import division
# Given data
Rf = 10# ohm
R_L = 1# k ohm
R_L = R_L * 10**3# ohm
Vi = 230# V
Vm = Vi*sqrt(2)
#I_DC = Im/pi
I_DC = Vm/((R_L+Rf)*pi)# A
Irms = Vm/((R_L+Rf)*2)# A
# The input ac power 
P_iAC = (Irms**2)*(Rf+R_L)# W 
print "The input ac power = %.2f W"%P_iAC
# The output ac power 
P_ODC = (I_DC**2)*R_L# W
print "The output ac power = %.2f W"%P_ODC
# The efficiency 
Eta = (P_ODC/P_iAC)*100# %
print "The efficiency = %.2f %%"%Eta
# The percentage regulation 
R = (Rf/R_L)*100# %
print "The percentage regulation = %.2f %%"%R

# Note: The calculated value of input a.c. power in the book is wrong.

The input ac power = 26.19 W
The output ac power = 10.51 W
The efficiency = 40.13 %
The percentage regulation = 1.00 %

Exa 2.3 page No 115

In [7]:
from math import pi,sqrt
from __future__ import division
# Given data
V1 = 230# V
N2= 1
N1 = 4
R_L = 1# k ohm
R_L = R_L * 10**3# ohm
Vd = 0.7# V
# V_LDC = (Vm-Vd)/pi# V
V2 = V1*(N2/N1)# V
# Vm = sqrt(2)*Vrms
Vm = sqrt(2)*V2# V
# The output dc voltage 
V_LDC = (Vm-Vd)/pi# V
print "The output dc voltage = %.2f V"%V_LDC
# The current for a load resistance 
I_LDC = V_LDC/R_L# A
I_LDC = I_LDC * 10**3# mA
print "The current for a load resistance = %.2f mA"%I_LDC

The output dc voltage = 25.66 V
The current for a load resistance = 25.66 mA

Exa 2.7 page No 120

In [8]:
from math import pi,sqrt
from __future__ import division
# Given data
R1 = 2.2# k ohm
R2 = 2.2#  kohm
R3 = 4.7# k ohm
R = (R2*R3)/(R2+R3)# k ohm
Vin = 200
# Vo = (R/(R1+R))*Vin
# The PIV rating for first diode 
Vomax= round(R/(R1+R)*Vin)# V
print "The PIV rating for first diode = %.2f V"%Vomax
Rdas = (R1*R3)/(R1+R3)# k ohm
# Vo = (Rdas/(R1+Rdas))*(-Vin)
# The PIV rating for second diode 
Vomax=round(Rdas/(R1+Rdas)*Vin)# V
print "The PIV rating for second diode = %.2f V"%Vomax

The PIV rating for first diode = 81.00 V
The PIV rating for second diode = 81.00 V

Exa 2.9 page No 124

In [9]:
from math import pi,sqrt
from __future__ import division

# Given data
# Vi = 15*sind(314*t)
Vm = 15# V
R_L = 1# k ohm
R_L = R_L * 10**3# ohm
Im = Vm/R_L# A
Im = Im * 10**3# mA
Idc = Im/pi# mA
# The average current 
I_Dav = Idc# mA
print "The average current = %.2f mA"%I_Dav
# The RMS current 
I_Drms = Im/2# mA
print "The RMS current = %.2f mA"%I_Drms
# The peak diode current 
I_Dpeak = Im# mA
print "The peak diode current = %.2f mA"%I_Dpeak
# The PIV of diode 
PIV = 2*Vm# V
print "The PIV of diode = %.2f V"%PIV

The average current = 4.77 mA
The RMS current = 7.50 mA
The peak diode current = 15.00 mA
The PIV of diode = 30.00 V

Exa 2.10 page No 125

In [10]:
from math import sqrt,pi
from __future__ import division
# Given data
R_L = 1# k ohm
R_L = R_L * 10**3# ohm
Rf = 1# ohm
R2 = 2# ohm
N1 = 4
N2 = 1
V1rms = 240# V
V2rms = (N2/N1)*V1rms# V
Vm = sqrt(2)*V2rms# V
# The average load current 
I_LDC = (2*Vm)/(pi*(R2+Rf+R_L))# A
I_LDC= I_LDC *10**3# mA
print "The average load current = %.2f mA"%I_LDC
I_LDC= I_LDC *10**-3# A
# The average load voltage at no load 
V_NL = (2*Vm)/pi# V
print "The average load voltage at no load = %.2f V"%V_NL
# The average load voltage at full load 
V_LDC = I_LDC*R_L# V
print "The average load voltage at full load = %.2f V"%V_LDC
# The percentage load regulation 
Per_loadReg= (V_NL-V_LDC)/V_LDC*100# %
print "The percentage load regulation = %.2f %%"%Per_loadReg
Im = Vm/(R_L+R2+Rf)# A
Irms = Im/2# A
P_iAC = (Vm**2)/(2*(R2+Rf+R_L))# W 
P_ODC = (I_LDC**2)*R_L# W
# The rectification efficiency 
Eta = (P_ODC/P_iAC)*100# %
print "The rectification efficiency = %.2f %%"%Eta

The average load current = 53.86 mA
The average load voltage at no load = 54.02 V
The average load voltage at full load = 53.86 V
The percentage load regulation = 0.30 %
The rectification efficiency = 80.81 %

Exa 2.11 page No 127

In [11]:
from __future__ import division
# Given data
R1 = 2# k ohm
R2 = 2# k ohm
V_AB = 20# V
Vo = V_AB*(R1/(R1+R2))# V
# The required PIV 
V_AC = Vo# V
print "The required PIV = %.2f V"%V_AC

The required PIV = 10.00 V

Exa 2.12 page No 127

In [12]:
from math import sqrt,pi
from __future__ import division
# Given data
Vrms = 10# V
r_f = 0.3# ohm
R_L = 2# ohm
Vm = sqrt(2)*Vrms# V
Im = Vm/(R_L+r_f)# A
# The value of Idc 
Idc = Im/pi# A
print "The value of Idc = %.2f A"%Idc
# The RMS value of output current 
Irms = Im/2# A
print "The RMS value of output current = %.2f A"%Irms

The value of Idc = 1.96 A
The RMS value of output current = 3.07 A

Exa 2.13 page No 128

In [15]:
from __future__ import division
# Given data
Vdc = 50# V
Vrms = 5# V
# The ripple factor,
Gamma = Vrms/Vdc
print "The ripple factor = %.1f"%Gamma

The ripple factor = 0.1

Exa 2.14 page No 128

In [16]:
from math import sqrt,pi
from __future__ import division
# Given data
Vrms = 50# V
r_f = 20# ohm
R_L = 980# ohm
Vm = sqrt(2)*Vrms# V
Im = (Vm)/(R_L+r_f)
# The mean load current 
Idc = (2*Im)/pi# A
Idc = round(Idc * 10**3)# mA
print "The mean load current = %.2f mA"%Idc
# The RMS load current 
Irms = Im/sqrt(2)# A
Irms = Irms*10**3# mA
print "The RMS load current = %.2f mA"%Irms
a = 0.812## assumed
# The output efficiency 
Eta = a/(1+(r_f/R_L))
Eta = Eta * 100# %
print "The output efficiency = %.2f %%"%Eta

The mean load current = 45.00 mA
The RMS load current = 50.00 mA
The output efficiency = 79.58 %

Exa 2.15 page No 129

In [17]:
from __future__ import division
# Given data
V = 10# V
V1 = 2.5# V
R = 1# Mohm
R = R * 10**6# ohm
i = (V-V1)/R# A
i = i * 10**6# µA
# The output voltage for a complete cycle 
Vo1 = (i*10**-6*R)+V1# V
print "The output voltage for a complete cycle = %.2f V"%Vo1
# The output voltage for half neagtive cycle 
Vo2 = V1# V
print "The output voltage for half negative cycle = %.2f V"%Vo2

The output voltage for a complete cycle = 10.00 V
The output voltage for half negative cycle = 2.50 V

Exa 2.17 page No 130

In [18]:
from __future__ import division
from math import sqrt,pi
# Given data
V1 = 230## iin V
N2 = 1
N1 = 5
R_L = 100# ohm
V2 = V1*N2/N1# V
Vrms = V2# V
Vs = V2/2# V
Vm = sqrt(2)*Vs# V
# The dc output voltage 
Vdc = (2*Vm)/pi# V
print "The dc output voltage = %.2f V"%Vdc
# The PIV value 
PIV = round(2*Vm)# V
print "The PIV value = %.2f V"%PIV
# The rectification efficiency 
Eta = 0.812
Eta = Eta*100# %
print "The rectification efficiency = %.2f %%"%Eta

The dc output voltage = 20.71 V
The PIV value = 65.00 V
The rectification efficiency = 81.20 %

Exa 2.18 page No 131

In [19]:
from math import sqrt,pi
from __future__ import division
# Given data
V1 = 220# V
N2 = 1
N1 = 10
R_L = 250# ohm
V2 = V1 * (N2/N1)# V
Vm = sqrt(2)*V2# V
Im =Vm/R_L# A
Iav = (2*Im)/pi# A
Idc = Iav# A
# The dc output volatge 
Vdc = Idc* R_L# V
print "The dc output volatge = %.2f V"%Vdc
Pdc = (Idc**2)*R_L# W
Irms = (Im)/sqrt(2)# A
Pac = (Irms**2)*R_L# W
# The rectification efficiency 
Eta = (Pdc/Pac)*100# %
print "The rectification efficiency = %.2f %%"%Eta
# The peak inverse volatge 
PIV = Vm# V
print "The peak inverse volatge = %.2f V"%PIV
f_in = 50# Hz
# The output frequency 
f_out = 2*f_in# Hz
print "The output frequency = %.2f Hz"%f_out

# Note: The answer of rectification efficiency in the book is not accurate.

The dc output volatge = 19.81 V
The rectification efficiency = 81.06 %
The peak inverse volatge = 31.11 V
The output frequency = 100.00 Hz

Exa 2.21 page No 136

In [21]:
from math import sqrt,pi
from __future__ import division
# Given data
R_F = 0.01
Vdc = 30# V
R_L = 1#  k ohm
R_L = R_L * 10**3# ohm
Idc = Vdc/R_L# A
Idc = Idc * 10**3# mA
# Vdc = Vm-( (5000*Idc)/C )
Gamma = 0.01## ripple factor
#Gamma = 2900/(C*R_L)
C = 2900/(Gamma*R_L)# F
Vm = Vdc + ((5000*Idc*10**-3)/C)# V
# The input voltage required 
V2 = (2*Vm)/sqrt(2)# V 
print "The input voltage required = %.2f V"%V2

#Note: The value of Vm in the book is not accurate, So the answer in the book is wrong.

The input voltage required = 43.16 V

Exa 2.22 page No 137

In [22]:
from math import sqrt,pi
from __future__ import division
# Given data
V_L = 25# V
I_L = 200# mA
I_L = I_L * 10**-3# A
R_L = V_L/I_L# ohm
Gamma = 3/100
#Gamma = 1/(6*sqrt(2)*(omega**2)*L*C)
f = 50# Hz
omega = 2*pi*f# rad/sec
#LC = 1/( 6*sqrt(2)*(omega**2)*Gamma )
L = R_L/(3*omega)# H
print "The value of L = %.2f H"%L
C = 1/( 6*sqrt(2)*(omega**2)*Gamma*L )# F
C = C * 10**6# µF
print "The value of C = %.2f µF"%C

The value of L = 0.13 H
The value of C = 300.11 µF

Exa 2.23 page No 138

In [24]:
from math import sqrt,pi
from __future__ import division
# Given data
Vm  = 15# V
# The output voltage 
Vdc = (2*sqrt(2)*Vm)/pi# V
print "The output voltage = %.2f V"%Vdc
R_L = 5# ohm
Idc = Vdc/R_L# A
print "The current = %.2f A"%Idc
L = 50# mH
L = L * 10**-3# H
C = 1000# µF
C = C * 10**-6# F
f = 50# Hz
omega = 2*pi*f# rad/sec
# The ripple factor 
Gamma = 1/( 6*sqrt(2)*(omega**2)*L*C )
print "The ripple factor = %.2f"%Gamma
# Im =Vm/X_L = (Vm*sqrt(2))/(2*pi*f*L)
Im = (Vm*sqrt(2))/(2*pi*f*L)# A
I_Lmin = Im# A
# The maximum value of R_L 
R_Lmax = Vdc/I_Lmin# ohm
print "The maximum value of R_L = %.2f ohm"%R_Lmax

The output voltage = 13.50 V
The current = 2.70 A
The ripple factor = 0.02
The maximum value of R_L = 10.00 ohm

Exa 2.24 page No 139

In [25]:
from math import sqrt,pi
from __future__ import division
# Given data
f = 50# Hz
Vrms = 280# V
Vm = sqrt(2)*Vrms# V
V_Lmax = Vm# V
Idc = 100# mA
Idc =Idc * 10**-3# A
C2 = 10# µF
C2 = C2 * 10**-6# F
C1 = C2 # F
R_L =V_Lmax/Idc# ohm
L = 104# H
omega = 2*pi*f# rad/sec
# The percentage ripple 
Gamma = sqrt(2)/( 8*omega**3*C1*C2*L*R_L)*100# %
print "The percentage ripple = %.2f %%"%Gamma

# Note: There is calculation error to find the value of gamma, So the answer in the book is wrong.

The percentage ripple = 0.01 %

Exa 2.25 page No 140

In [27]:
from __future__ import division
# Given data
Vin = 15# V
# Peak output voltage,
Vout = Vin# V
print "Peak output voltage = %.2f V"%Vout

Peak output voltage = 15.00 V

Exa 2.26 page No 140

In [28]:
from __future__ import division
# Given data
R_L = 2# k ohm
R_L = R_L * 10**3# ohm
R = 2# k ohm
R = R * 10**3# ohm
Vin = 5# V
# The peak output voltage 
Vout = (R_L/(R+R_L))*Vin# V
print "The peak output voltage = %.2f V"%Vout

The peak output voltage = 2.50 V

Exa 2.27 page No 141

In [29]:
from __future__ import division
# Given data
Vi = 10# V
V1 = 6# V
R = 10# k ohm
R = R * 10**3# ohm
# The value of i_Rmax
i_Rmax = (Vi-V1)/R# A
i_Rmax = i_Rmax * 10**3# mA
Vi = -10# V
V1 = 8# V
# The value of i_R 
i_R =(Vi+V1)/R# A
i_R  =i_R * 10**3# mA
print "The value of i_Rmax = %.2f mA"%i_Rmax
print "The value of i_R = %.2f mA"%i_R

The value of i_Rmax = 0.40 mA
The value of i_R = -0.20 mA

Exa 2.28 page No 142

In [30]:
from __future__ import division
# Given data
Vout = -0.7# V
V = -12# V
# The output voltage 
V_R =V-Vout# V
print "The output voltage = %.2f V"%V_R

The output voltage = -11.30 V

Exa 2.29 page No 143

In [32]:
from __future__ import division
# Given data
Vin = 10# V
V1 = 2# V
# Vin -V_C - V1 = 0
V_C =Vin-V1# V
# During positive half cycle the output voltage 
Vout = Vin-V_C# V
print "During positive half cycle the output voltage = %.2f V"%Vout
Vin = -10# V
V1 = 8# V
# Vin-V1-Vout = 0
# During negative half cycle the output voltage 
Vout = Vin-V1# V
print "During negative half cycle the output voltage = %.2f V"%Vout

During positive half cycle the output voltage = 2.00 V
During negative half cycle the output voltage = -18.00 V

Exa 2.30 page No 145

In [33]:
from __future__ import division
# Given data
Vin = 10# V
V1 = 2# V
# Vin-V_C+V1 = 0
V_C = Vin+V1# V
#During positive half cycle the output voltage 
Vout = Vin-V_C# V
print "During positive half cycle the output voltage = %.2f V"%Vout
Vin = -10## iin V
V1 = 12# V
# Vin-V1-Vout = 0
#During negative half cycle the output voltage 
Vout = Vin-V1# V
print "During negative half cycle the output voltage = %.2f V"%Vout

During positive half cycle the output voltage = -2.00 V
During negative half cycle the output voltage = -22.00 V

Exa 2.31 page No 146

In [34]:
from __future__ import division
# Given data
Vi = 20# V
V1 = 5# V
Vc = Vi-V1# V
Vo = -5# V
# The value of Vo,
Vo = Vi+Vc# V
print "The value of Vo = %.2f V"%Vo

The value of Vo = 35.00 V