Chapter 21 , Controlled Rectifiers

Example 21.1 , Page Number 508

In [1]:
import math

#Variables

RL = 100.0                                   #Resistance (in ohm)
Vm = 300.0                                   #Maximum voltage (in volts) 
P1 = 25.0                                    #Load power1 (in watt)
P2 = 80.0                                    #Load power2 (in watt)

#Calculation 

Vdc = Vm / (2 * math.pi)                     #dc voltage (in volts)
#When power is 25 watt
cosinealpha = (P1 * RL / Vdc**2)**0.5 -1     #cos of alpha       
alpha = math.acos(cosinealpha)               #Value of alpha (in radians)

#When power is 80 watt
cosinealpha1 = (P2 * RL / Vdc**2)**0.5 -1    #cos of alpha       
alpha1 = math.acos(cosinealpha1)             #Value of alpha (in radians)
#Result

print "Angular firing control when load power P = 25 W is ",round(alpha * 180.0 / math.pi,2)," degree.\nAngular firing control when load power P = 80 W is ",round(alpha1 * 180.0 / math.pi,2)," degree."

#Calculation difference in value of cosinealpha from book due to higher precision.
Angular firing control when load power P = 25 W is  87.29  degree.
Angular firing control when load power P = 80 W is  29.16  degree.

Example 21.2 , Page Number 509

In [2]:
import math

#Variables

Vm = 200.0                      #maximum voltage (in volts)
RL = 1.0                        #Resistance (in kilo-ohm)

#Calculation

#When alpha = 0 degree
Vdc = 0.318 * Vm                #dc voltage (in volts)
Idc = Vdc / RL                  #dc Current (in milli-Ampere)
P = Vdc * Idc                   #Power (in milli-watt) 

#When alpha = 45 degree         
Vdc1 = 0.27 * Vm                #dc voltage1 (in volts)
Idc1 = Vdc1 / RL                #dc current1 (in milli-Ampere)
P1 = Vdc1 * Idc1                #Power1 (in milli-watt)

#When alpha = 90 degree
Vdc2 = 0.159 * Vm               #dc voltage2 (in volts)
Idc2 = Vdc2 / RL                #dc current2 (in milli-Ampere)
P2 = Vdc2 * Idc2                #Power2 (in milli-watt)

#When alpha = 135 degree
Vdc3 = 0.0466 * Vm              #dc voltage3 (in volts)
Idc3 = Vdc3 / RL                #dc current3 (in milli-Ampere)
P3 = Vdc3 * Idc3                #Power3 (in milli-watt)

#Result

print "Power delivered when alpha = 0 degree is ",round(P)," mW.\nPower delivered when alpha = 45 degree is ",P1," mW.\nPower delivered when alpha = 90 degree is ",round(P2)," mW.\nPower delivered when alpha = 135 degree is ",round(P3,2)," mW." 
Power delivered when alpha = 0 degree is  4045.0  mW.
Power delivered when alpha = 45 degree is  2916.0  mW.
Power delivered when alpha = 90 degree is  1011.0  mW.
Power delivered when alpha = 135 degree is  86.86  mW.

Example 21.3 , Page Number 513

In [3]:
import math

#Variables

Vrms = 220.0                                 #rms voltage (in volts)
alpha = 60.0                                 #Firing angle (in degree)     

#Calculation

alpharad = alpha * math.pi/180.0             #Firing angle (in radians) 
Vm = 2**0.5 * Vrms                           #Maximum or Peak voltage (in volts)
Vdc = Vm /(2 * math.pi)*(1 + math.cos(alpharad))  #dc output voltage (in volts)                 

#Result

print "The d.c. output voltage is ",round(Vdc,2)," V."

#Slight variation in answer due to higher precision
The d.c. output voltage is  74.28  V.

Example 21.4 , Page Number 513

In [4]:
import math

#Variables

Idc = 0.5                                    #dc current (in Ampere)
Vrms = 100.0                                 #Rms voltage (in volts)
alpha = 45.0                                 #Firing angle (in degree)       
Idc = 0.5                                    #dc current (in Ampere)

#Calculation

alpharad = alpha * math.pi / 180.0           #Firing angle (in radians)
Vm = 2**0.5 * Vrms                           #Peak voltage (in volts) 
Vdc = Vm / (2 * math.pi)*(1 + math.cos(alpharad)) #Average voltage (in volts)
RL = Vdc / Idc                               #Load resistance (in ohm)  

#Result

print "The value of resistance to limit the average current to 0.5 A is ",round(RL,2)," ohm."

#Slight variation in answer due to higher precision
The value of resistance to limit the average current to 0.5 A is  76.85  ohm.

Example 21.5 , Page Number 520

In [5]:
import math

#Variables

TON = 30.0            #Chopper ON time (in milli-second)
TOFF = 10.0           #Chopper OFF time (in milli-second)   

#Calculation

T = TON + TOFF        #Total time (in milli-second)
cdc = TON / T         #Chopper duty cycle
f = 1 / T             #Chopping frequency (in Hertz)

#Result

print "Chopper duty cycle is ",cdc,".\nChopping frequency is ",f * 10**3," Hz."

#Correction to be done in book , the units mentioned are milli-Ampere but in the calculation it used micro-Ampere.
Chopper duty cycle is  0.75 .
Chopping frequency is  25.0  Hz.

Example 21.6 , Page Number 520

In [6]:
import math

#Variables

TON = 30.0                 #Chopper ON time (in milli-second)
TOFF = 10.0                #Chopper OFF time (in milli-second)   
Vdc = 200.0                #dc voltage (in volts)

#Calculation

T = TON + TOFF             #Total time (in milli-second)
cdc = TON / T              #Chopper duty cycle
VL = Vdc * cdc             #dc output voltage (in volts)  

#Result

print "Average valuye of dc voltage is ",VL," V."
Average valuye of dc voltage is  150.0  V.