Chapter 4:Semiconductor Diode

Example 4.1 Page No.85

In [1]:
Vrms=220             #Volts, power supply
n2=1                     #Assumption
n1=12*n2            #Turns Ratio

#Calculation
import math
Vp=math.sqrt(2)*Vrms     #Maximum(Peak) Primary Voltage
Vm=n2*Vp/n1                  #Maximum Secondary Voltage
Vdc=Vm/math.pi                   #DC load Voltage 
# Results 
print "The DC load Voltage is = ",round(Vdc,2),"V"
print "The Peak Inverse Voltage(PIV) is = ",round(Vm,1),"V"

The DC load Voltage is =  8.25 V
The Peak Inverse Voltage(PIV) is =  25.9 V

Example 4.2 Page No.90

In [2]:
Vrms=220                   #Volts, power supply  rms voltage
n2=1                           #Assumption
n1=12*n2                 #Turns Ratio

#Calculation
import math
Vp=math.sqrt(2)*Vrms  #Maximum(Peak Primary Voltage
Vm=n2*Vp/n1                #Maximum Secondary Voltage
Vdc=2*Vm/math.pi              #DC load Voltage 

# Results 
print "The DC load Voltage is = ",round(Vdc,1),"V"
print "The Peak Inverse Voltage(PIV of Bridge Rectifier is =  ",round(Vm,1),"V"
print "The Peak Inverse Voltage(PIV of Centre-tap Rectifier is = ",round(2*Vm,1),"v"

The DC load Voltage is =  16.5 V
The Peak Inverse Voltage(PIV of Bridge Rectifier is =   25.9 V
The Peak Inverse Voltage(PIV of Centre-tap Rectifier is =  51.9 v

Example 4.3 Page No.95

In [3]:
import math
Rl=1000.0                     #Ohms, load resistance
rd=10.0                          #Ohms   forward base dynamic resistance
Vm=220.0                     #Volts(Peak Value of Voltage)
#Calculation
Im=Vm/(rd+Rl)                       #Peak Value of Current

# Result
print "The Peak Value of Current is =  ",round(Im*1000,1),"mA"

#(b) dc or av value of current

Idc=2*Im/math.pi                        #DC Value of Current
# Results 
print "The DC or Average Value of Current is ",round(Idc*1000,2),"mA"

#(c)
Irms=Im/math.sqrt(2)              #RMS Value of Current
# Results 
print "The RMS Value of Current is = ",round(Irms*1000,1),"mA"

#(d)
r=math.sqrt((Irms/Idc)**2-1)     #Ripple Factor
# Results i
print " The Ripple Factor r = ",round(r,3)

#(e)
Pdc=Idc**2*Rl
Pac=Irms**2*(rd+Rl)
n=Pdc/Pac                                  #Rectification Efficiency
# Results
print "The Rectification EFficiency n(eeta) =  percent.",round(n*100,2)

The Peak Value of Current is =   217.8 mA
The DC or Average Value of Current is  138.67 mA
The RMS Value of Current is =  154.0 mA
 The Ripple Factor r =  0.483
The Rectification EFficiency n(eeta) =  percent. 80.25

Example 4.4 Page No.103

In [4]:
Vz=9.1                #Volts
P=0.364              #Watts
#Calculation
Iz=P/Vz
#Result
print " The Maximum permissible Current is ",Iz*1000,"mA"

 The Maximum permissible Current is  40.0 mA

Example 4.5 Page No.105

In [5]:
Ci=18*10**(-12)            #i.e. 18 pF,  capacitance of diode
Vi=4                                 #volt, initial  voltage
Vf=8                               #v, final voltage

#Calculation
import math
Vf=8                             
K=Ci*math.sqrt(Vi)
Cf=K/math.sqrt(Vf)
#Result
print " The Final Value of Capacitance is C = ",round(Cf/10**(-12),3),"pF"

 The Final Value of Capacitance is C =  12.728 pF