Chapter 19 , Rectifiers and Filters

Example 19.1 , Page Number 430

In [1]:
import math

#Variables

V1 = 230.0                      #Primary voltage (in volts)
N2byN1 = 1.0/10                 #Turns ratio

#Calculation

V2 = V1 * N2byN1                #Secondary voltage (in ratio)
Vm = 2**0.5 * V2                #Maximum value of secondary voltage (in volts)
Vdc = 0.318 * Vm                #dc output voltage (in volts)
PIV = Vm                        #Peak inverse voltage (in volts)  

#Result

print "dc output voltage is ",round(Vdc,1)," V.\nPIV of the diode is ",round(PIV,1)," V." 

dc output voltage is  10.3  V.
PIV of the diode is  32.5  V.

Example 19.2 , Page Number 431

In [2]:
import math

#Variables

RL = 20                        #Load resistance (in kilo-ohm)
V2 = 24                        #Secondary voltage (in volts)

#Calculation

Vm = 2**0.5 * V2               #Maximum value of secondary voltage (in volts)
Im = Vm / RL                   #Maximumj value of load current (in milli-Ampere)
Idc = 0.318 * Im               #dc current (in milli-Ampere)

#Result

print "The value of dc load current is ",round(Idc,3)," mA."

The value of dc load current is  0.54  mA.

Example 19.3 , Page Number 431

In [3]:
import math

#Variables

V1 = 230                 #Primary voltage (in volts)
N2byN1 = 1.0/2.0         #Turns ratio
RL = 200                 #Resistance (in ohm)

#Calculation

V2 = V1 * N2byN1         #Secondary voltage (in volts)
Vm = 2**0.5 * V2         #Maximum value of secondary voltage (in volts)
Im = Vm / RL             #Maximum value of load current (in Ampere)
Pm = Im**2 * RL          #Maximum value of load power (in watt)
Vdc = 0.318 * Vm         #Average value of load power (in watt)
Idc = Vdc / RL           #Average value of load current (in Ampere)
Pdc = Idc**2 * RL        #Average value of load power (in watt)

#Result

print "Maximum value of load power is ",round(Pm,1)," W."
print "Average value of load power is ",round(Pdc,1)," W."

Maximum value of load power is  132.3  W.
Average value of load power is  13.4  W.

Example 19.4 , Page Number 432

In [2]:
import math

#Variables

Vdc = 30.0                    #Average value of voltage (in volts)
RL = 600.0                    #Load resistance (in ohm)
Rf = 25.0                     #forward resistance (in ohm)

#Calculation

Idc = Vdc / RL                #Average value of load current (in Ampere)
Im = round(math.pi * Idc,3)   #Maximum value of load current (in Ampere) 
Vinmax = Im * (Rf + RL)       #Maximum a.c. voltage required at the input (in volts)

#Result

print "Maximum a.c. voltage required at the input is ",Vinmax," V."

Maximum a.c. voltage required at the input is  98.125  V.

Example 19.5 , Page Number 436

In [5]:
import math

#Variables

V2 = 30.0                 #secondary voltage (in volts)
RL = 5.1                  #Load resistance (in kilo-ohm)

#Calculation

VS = V2 / 2               #Voltage between center - tap and either end of secondary winding (in volts)
Vm = 2**0.5 * VS          #maximum value of voltage (in volts)
Vdc = 0.636 * Vm          #dc output voltage (in volts)
Idc = Vdc / RL            #dc load current (in milli-Ampere)

#Result

print "The dc output voltage is ",round(Vdc,1)," V.\nThe dc output current is ",round(Idc,3)," mA." 

#Slight variation in value of Idc due to less precision used in book

The dc output voltage is  13.5  V.
The dc output current is  2.645  mA.

Example 19.6 , Page Number 440

In [6]:
import math

#Variables

V1 = 230.0                #primary voltage (in volts)
N2byN1 = 1.0/4.0          #Turns ratio
RL = 200                  #Load resistance (in ohm)
fin = 50                  #Frequency (in hertz)

#Calculation

V2 = V1 * N2byN1          #Secondary voltage (in volts)
Vm = 2**0.5 * V2          #Maximum value of voltage (in volts)
Vdc = 0.636 * Vm          #Average value of voltage (in volts)
PIV = Vm                  #peak inverse voltage (in volts)
fout = 2 * fin            #Output frequency (in volts)     

#Result

print "The dc output voltage is ",round(Vdc,1)," V.\nPeak inverse Voltage of a diode is ",round(PIV,1)," V.\nOutput frequency is ",fout," HZ."

The dc output voltage is  51.7  V.
Peak inverse Voltage of a diode is  81.3  V.
Output frequency is  100  HZ.

Example 19.7 , Page Number 444

In [7]:
import math

#Variables

V1 = 230.0                #primary voltage (in volts)
N2byN1 = 1.0/5.0          #Turns ratio
RL = 100                  #Load resistance (in ohm)

#Calculation

V2 = V1 * N2byN1          #Secondary voltage (in volts)
VS = V2 / 2               #Voltage between center - tap and either end of secondary winding (in volts)
Vm = 2**0.5 * VS          #Maximum value of voltage (in volts)
Vdc = 2/math.pi * Vm      #Averaage value of Voltage (in volts)
PIV = 2 * Vm              #Peak inverse voltage (in volts)
n = 0.812                 #Efficiency of full wave rectifier

#Result

print "The dc output voltage is ",round(Vdc,1)," V.\nPeak inverse voltage is ",round(PIV)," V.\nRectification efficiency is ",n,"."

The dc output voltage is  20.7  V.
Peak inverse voltage is  65.0  V.
Rectification efficiency is  0.812 .

Example 19.8 , Page Number 445

In [8]:
import math

#Variables

VS = 200                 #Voltage between center - tap and either end of secondary winding (in volts)
Imax = 700               #Maximum value of current (in milli-Ampere)
Iavg = 250               #Average value of current (in milli-Ampere)

#Calculation

Imax1 = 0.8 * Imax       #Maximum value of current for normal operation (in milli-Ampere)
Vm = 2**0.5 * VS         #Maximum value of voltage (in volts)
RL = Vm / Imax1          #Load resistance (in kilo-ohm)
Vdc = 2 * Vm / math.pi   #Average value of voltage (in volts)
Idc = Vdc / RL           #Average value of current (in milli-Ampere)
PIV = 2 * Vm             #peak Inverse voltage (in volts)

#Result

print "The value of load resistor is ",round(RL,3), "kilo-ohm.\nThe dc load voltage and current is ",round(Vdc)," V and ",Idc," mA.\nPeak inverse voltage is ",round(PIV,2)," V."  

The value of load resistor is  0.505 kilo-ohm.
The dc load voltage and current is  180.0  V and  356.507072526  mA.
Peak inverse voltage is  565.69  V.

Example 19.9 , Page Number 449

In [9]:
import math

#Variables

Vs = 150.0            #Voltage (in volts)
Idc = 2.0             #Average value of current (in Ampere)

#Calculation

Vdc = 2.34 * Vs       #Average calue of voltage (in volts)
Ipi = 1/0.955 * Idc   #Peak current per diode (in Ampere)
Iavg = 2.0/3.0        #Average current per diode (in AMpere)
Pdc = Vdc * Idc       #Average power delievered to the load (in watt)

#Result

print "The value of Vdc is ",Vdc," V.\nPeak current through each diode is ",round(Ipi,1)," A.\nAverage current through each diode is",round(Iavg,2)," A.\nAverage power delievered to the load is ",Pdc," W."

#Calculation error in calculating the value of Pdc in book. It's value is 702 but printed value is 701 watt.  

The value of Vdc is  351.0  V.
Peak current through each diode is  2.1  A.
Average current through each diode is 0.67  A.
Average power delievered to the load is  702.0  W.

Example 19.10 , Page Number 454

In [10]:
import math

#Case (a):

#Variables

f = 50.0                             #Frequency (in Hertz)
g = 0.05                             #Ripple factor
RL = 100.0                           #Resistance (in ohm)
w = 2 * math.pi * f                  #Angular frequency (in radians per second)       

#Calculation

L = RL / (3 * 2**0.5 * w * g)        #Inductance (in Henry) 

#Result

print "Value of inductance is ",round(L,1)," H."

#Case (b):

#Variables

f = 400.0                            #Frequency (in Hertz)
g = 0.05                             #Ripple factor
RL = 100.0                           #Resistance (in ohm)
w = 2 * math.pi * f                  #Angular frequency (in radians per second)       

#Calculation

L = RL / (3 * 2**0.5 * w * g)        #Inductance (in Henry) 

#Result

print "New Value of inductance is ",round(L,3)," H."

Value of inductance is  1.5  H.
New Value of inductance is  0.188  H.

Example 19.11 , Page Number 455

In [11]:
import math

#Variables

Vdc = 30.0                        #Average value of voltage (in volts)
RL = 1.0                          #Resistance (in kilo-ohm)
gamma = 0.01                      #Ripple factor
f = 50                            #Frequency (in Hertz)
#Calculation

C = 2890.0 / (gamma * RL)         #Capacitance (in nano Farad)

#Result

print "Value of capacitance is ",C * 10**-3," micro-Farad."

Value of capacitance is  289.0  micro-Farad.

Example 19.12 , Page Number 456

In [12]:
import math

#Variables

Vdc = 12.0                    #Average value of voltage (in volts)
Idc = 100.0                   #Average value of current (in milli-Ampere) 
gamma = 0.01                  #Ripple factor
L = 1.0                       #Inductance (in Henry) 

#Calculation

C = 1.195 / (gamma * L)       #Capacitance

#Result

print "Capacitance is ",C," micro-Farad."

Capacitance is  119.5  micro-Farad.

Example 19.13 , Page Number 457

In [13]:
import math

#Variables

Idc = 0.2                            #Average value of current (in Ampere)
Vdc = 30.0                           #Average value of voltage (in volts)
C1 = C2 = 100.0                      #Capacitance (in milli-Farad)
L = 5.0                              #Inductance (in Henry)
f = 50.0                             #Frequency (in Hertz) 

#Calculation

RL = Vdc / Idc                       #Load resistance (in ohm)
gamma = 5700.0 / (C1 * C2 * RL * L)  #Ripple factor
 
#Result

print "Ripple factor for 50 Hz supply is ",gamma,"."

#Correction to be done in the formula for ripple factor used in the question. 
#gamma = 5700 / (c1 * C2 * RL * L) --> right formula.

Ripple factor for 50 Hz supply is  0.00076 .