Chapter No.1: Special Diodes

Example 1.1, Page No. 9

In [14]:
#Reverse Saturation current of diode
import math

#variable declaration

I=40*10**-3           #forword bias current in A
V=0.25                #forword bias voltage in Volt
T=20                  #junction temperature in degree C
T=T+273               #junction temperature in degree K
ETA=1                 #For Ge
e=1.6*10**-19         #in Coulamb(electronic charge)
k=1.38*10**-23        #in J/K(Boltzman Constant)

#Calculation
#Formula : I=Io*(exp(e*V/(ETA*k*T))-1)
x=math.ceil((e*V/(ETA*k*T)))
Io=I/(math.exp(x)-1)
Io=math.ceil(Io*10**8)/10**8

#Result
print("Reverse saturation current in micro Ampere : %.2f "%(Io*10**6))

Reverse saturation current in micro Ampere : 1.82

example 1.2, Page No. 9

In [21]:
#Value of forword voltage
import math
#variable declaration
Io=10*10**-6                   # reverse saturation currrent in A
I=1                            # forword current in Ampere
ETA=2                          # For Si
T=27                           # room temperature in degree C
T=T+273                        # room temperature in degree K
e=1.6*10**-19                  # in Coulamb(electronic charge)
k=1.38*10**-23                 # in J/K(Boltzman Constant)

#Calculation
#Formula : I=Io*(exp(%e*V/(ETA*k*T))-1)
V=(ETA*k*T/e)*math.log(I/(Io)+1)
V=math.floor(V*100)/100
#result
print("Forward Voltage across the diode in Volt :%.2f"%V)

Forward Voltage across the diode in Volt :0.59

example 1.3 , Page No. 23

In [31]:
#DC Current, DC Voltage, Ripple Factor
import math
#Variable Declaration
RL=1                #load resistance in kOhm
#rf<<RL
Vrms=200            #in Volt

#Part (i)
Vo=Vrms*math.sqrt(2)        #in volt
Idc=Vo/(RL*10**3*math.pi)   #in Ampere
print("(i)\nDC current in load in mA :%.0f"%(math.floor((Idc*10**3))))

#Part (ii)
Vdc=RL*10**3*Idc            #in Volt
print("\n(ii)\nDC voltage across load in volt :%.0f"%(math.floor(Vdc)))

#Part (iii)
#Gamma=sqrt((Irms/Idc)^2-1)=sqrt((Io/2)/(Io/%pi)-1)=sqrt((%pi/2)^2-1)
Gamma=math.sqrt((math.pi/2)**2-1)    #unitless
print("\n(iii)\nRipple factor :%.2f "%(math.floor(Gamma*100)/100))

#Part (iv)
PIV=Vrms*math.sqrt(2)       #in volt
print("\n(iv)\nPeak Inverse Voltage in volt :%.0f"%(math.floor(PIV)))

(i)
DC current in load in mA :90

(ii)
DC voltage across load in volt :90

(iii)
Ripple factor :1.21 

(iv)
Peak Inverse Voltage in volt :282

example 1.4 , Page No. 23

In [38]:
#Average dc current rms current rectifcation efcieny PIV
import math
#variable declaration
rf=20.0                      #in ohm
RL=980.0                     #in Ohm
Vrms=50.0                    #in Volt
Vo=Vrms*math.sqrt(2)         #in Volt
Io=Vo/(RL+rf)                #in Ampere

#Part (i)
Idc=2*Io/math.pi             #in Ampere
print("(i)\nAverage DC current in mA :%.0f"%(math.floor(Idc*10**3)))

#Part (ii)
Irms=Io/math.sqrt(2)          #in Ampere
print("\n(ii)\nrms value of load current in mA :%.0f"%(math.ceil(Irms*1000)))

#Part (iii)
Vdc=RL*Idc                    #in Volt
print("\n(iii)\nDC output voltage in volt :%.1f"%(math.floor(Vdc*10)/10))

#Part (iv)
ETA=(Idc**2*RL/(Irms**2*(RL+rf)))*100          #Rectification Efficiency in %
print("\n(iv)\nRectification Efficiency is %.1f%%"%(math.ceil(ETA*10)/10))

#Part (v)
PIV=2*Vo                       #in volt
print("\n(v)\nPeak Inverse Voltage in volt :%.1f"%PIV)

(i)
Average DC current in mA :45

(ii)
rms value of load current in mA :50

(iii)
DC output voltage in volt :44.1

(iv)
Rectification Efficiency is 79.5%

(v)
Peak Inverse Voltage in volt :141.4

example 1.5, Page No.26

In [40]:
#Minimum value of resistance
import math
#Variable declaration

Vin=40                       #in volt
VZ=10                        #in volt
Vo=10                        #in volt
IZmax=50                     #in mA
IL=0                         #in mA

#calculation
#Formula : I=IZ+IL=IZmax+0
I=IZmax+0                    #in mA
#Formula : VZ=Vin-R*I
Rmin=(Vin-VZ)/(I*10**-3)     #in Ohm
#Result
print("Minimum value of resistance in Ohm :%.0f "%Rmin)

Minimum value of resistance in Ohm :600

example 1.6, Page No.26

In [51]:
#Value of series resistor
import math
#variable declaration
Vmin=15                    #Minimum input voltage in volt
VZ=6.8                     #Voltage across zener in volt
Vo=VZ                      #output voltage in volt
Vsr1=Vmin-Vo               #Voltage aross series resistance in volt
print("If R is the series resistance, Total current in series resistance in Ampere : I=Vsr/R=8.2/R ")
ILmin=5                    #in mA
print("current in zener diode in Ampere :IZ=I-IL=(8.2/R-IL*10-3).............eqn(1)\n");
Vmax=20                    #mximum output voltage
Vo=VZ                      #output voltage in volt
Vsr2=Vmax-Vo               #Voltage aross series resistance in volt
print("Current in series resistance circuit in Ampere : I=Vsr/R\n")
ILmax=15                   #in mA
print("current in zener diode in Ampere :IZ=I-IL=(Rs/R-IL*10-3)..............eqn(2)\n")
print("For Zener diode to work as voltage regulator,(1) and (2) must be same.");
print("(8.2/R-IL*10-3)=(13.2/R-IL*10-3)")
R=(Vsr2-Vsr1)/(ILmax*10**-3-ILmin*10**-3)            #in Ohm
print("\nRequired value of Series Resistor in ohm : %.0f"%R)

If R is the series resistance, Total current in series resistance in Ampere : I=Vsr/R=8.2/R 
current in zener diode in Ampere :IZ=I-IL=(8.2/R-IL*10-3).............eqn(1)

Current in series resistance circuit in Ampere : I=Vsr/R

current in zener diode in Ampere :IZ=I-IL=(Rs/R-IL*10-3)..............eqn(2)

For Zener diode to work as voltage regulator,(1) and (2) must be same.
(8.2/R-IL*10-3)=(13.2/R-IL*10-3)

Required value of Series Resistor in ohm : 500

example 1.7 Page No.27

In [45]:
#Current limiting resistance and dissipated power
import math
#variable declaration

Vin=18                     #in volt
IZ=20                      #in mA
ILav=(5+35)/2              #in mA
VZ=12                      #in volt
Vo=12                      #in volt

#Calculation
I=IZ+ILav                  #in mA
R=(Vin-Vo)/(I*10**-3)      #in Ohm
P=(I*10**-3)**2*R          #in Watts

#Result
print("Current limiting resistance in Ohm : %.0f"%R);
print("Power disspation in resistance in Watt :%.2f "%P);

Current limiting resistance in Ohm : 150
Power disspation in resistance in Watt :0.24

example 1.8, Page No. 28

In [53]:
#Maximum and minimum input supply voltage
import math
#Variable declaration
R=1                            #in kOhm
RL=5                           #in kOhm
VZ=10                          #in volt
Vo=10                          #in volt
P=250                          #in mW

#Calculation
IL=Vo/RL                       #in mA
IZmin=0                        #in mA
IZmax=P/VZ                     #in mA
Imin=IZmin+IL                  #in mA
Imax=IZmax+IL                  #in mA
Vin_min=VZ+Imin*10**-3*R*10**3   #in volt
Vin_max=VZ+Imax*10**-3*R*10**3   #in volt

#Result
print("The input voltage ranges from %.0fV to %.0fV"%(Vin_min,Vin_max));

The input voltage ranges from 12V to 37V

example 1.9, Page No.28

In [61]:
#Output voltage voltage drop and current in zener diode
import math
#Variable Declaration
R=5                             #in kOhm
R=R*1000                        #in Ohm
RL=10.0                         #in kOhm
RL=RL*1000                      #in Ohm
Vin=120.0                       #in Volt
VZ=50.0                         #in Volt

#Part (i)

#calculation
Vo=VZ                           #in Volt
#Result
print("\n(i)\nOutput voltage in volt :%.0f"%Vo)

#Part (ii)

#calculation
VR=Vin-VZ                       #in Volt
#Result
print("\n(ii)\nVoltage drop across series resistance in volt :%.0f"%VR);



#Part (iii) 

#Calculation
IL=Vo/RL                        #in Ampere
I=VR/R                          #in Ampere
IZ=I-IL                         #in Ampere
#Result
print("\n(iii)\nLoad Current in mA :%.0f"%(IL*1000))
print("Current through resistance R in mA :%.0f"%(I*1000));
print("Load Current in mA :%.0f"%(IZ*1000))

(i)
Output voltage in volt :50

(ii)
Voltage drop across series resistance in volt :70

(iii)
Load Current in mA :5
Current through resistance R in mA :14
Load Current in mA :9

example 1.10, Page No. 30

In [65]:
#Maximum and Minimum LED current
import math

#Variable declaration
VDmin=1.5              #in Volt
VDmax=2.3              #in Volt
VS=5.0                 #in Volt
RS=270.0               #in Ohm

#Calculation
Imin=(VS-VDmax)/RS     #in Ampere
Imax=(VS-VDmin)/RS     #in Ampere

#Result
print("Minimum value of LED current in mA : %.0f"%(Imin*1000))
print("Maximum value of LED current in mA : %.0f"%(math.ceil(Imax*1000)))

Minimum value of LED current in mA : 10
Maximum value of LED current in mA : 13

example 1.11, Page No. 33

In [68]:
#Frequency range of tuning circuit
import math

#VAriable declarion
C1min=10                 #in pF
C2max=50                 #in pF
L=5                      #in mH
L=L*10**-3               #in H

#Calculation

#Formula : CT=C1*C2/(C1+C2)
#Minimum
C1=10                    #in pF
C2=10                    #in pF
CTmin=C1*C2/(C1+C2)      #in pF
CTmin=CTmin*10**-12      #in F
#Maximum
C1=50                    #in pF
C2=50                    #in pF
CTmax=C1*C2/(C1+C2)      #in pF
CTmax=CTmax*10**-12      #in F

#Formula : f=1/(2*%pi*sqrt(L*C))
#maximum :
fmax=1/(2*math.pi*math.sqrt(L*CTmin));
#minimum :
fmin=1/(2*math.pi*math.sqrt(L*CTmax));

#Result
print("The frequency of tuning circuit ranges from %.3fMHz to %.3fMHz."%(fmin/10**6,fmax/10**6));
#Note : Answer in the book is wrong.

The frequency of tuning circuit ranges from 0.450MHz to 1.007MHz.

example 1.12, page No.33

In [72]:
#Diode Capacitance
import math

#Variable Declaration
C1=21.0                 # in pF
V1=4.0                  # in volt
V2=9.0                  # in volt

#Calculations
print("C is proportional to 1/sqrt(V)")
print("So, C2/C1=sqrt(V1/V2)")
C2=math.sqrt(V1/V2)*C1     #in pF

#Result
print("At reverse bias 9V, Diode capacitance in pF : %.0f"%math

C is proportional to 1/sqrt(V)
So, C2/C1=sqrt(V1/V2)
At reverse bias 9V, Diode capacitance in pF : 14

example 1.13, Page No.39

In [74]:
#Photocurrent
import math
#Variable Declaration
R=0.90               #in A/W
Pop=1.0              #in mW

#Part (i)
#calculation
IP=R*Pop             #in mA
#Result
print("(i)\nPower of incident light 1mW, Photocurrent in mA is :%.2f"%IP);

#Part (ii)
#Result
print("\n(ii)\nHere IP is not proportional to Pop(for Pop>1.5mW)")
print("Hence Photourrent can not be calculated.")

(i)
Power of incident light 1mW, Photocurrent in mA is :0.90

(ii)
Here IP is not proportional to Pop(for Pop>1.5mW)
Hence Photourrent can not be calculated.

example 1.14, Page No. 39

In [78]:
#Responsivity of InGaAs photodiode
import math
#variable declaration
ETA=70.0                #in %
Eg=0.75                 #in eV
Eg=Eg*1.6*10**-19       #in Joule
h=6.63*10**-34          #Planks constant in J-s
c=3*10**8               #speed of light in m/s
e=1.6*10**-19           #in coulamb

#Calcualtions
lambda1=h*c/Eg           #in meter
R=(ETA/100)*e*lambda1/(h*c)     #in A/W

#Result
print("Wavelength in nm :%.1f"%(lambda1*10**9))
print("Responsivity of InGaAs photodiode in A/W :%.3f"%R)

Wavelength in nm :1657.5
Responsivity of InGaAs photodiode in A/W :0.933

example 1.15, Page No. 41

In [80]:
#Equilibrium contact potential
import math
#variable declaration
W1=2.5                    #in eV
W2=1.9                    #in eV

#Calculation
ContactPotential=W1-W2    #in Volt

#Result
print("Contact potential in Volts :%.1f "%ContactPotential)

Contact potential in Volts :0.6