Chapter 1: Special Diodes

Example 1.1, Page No. 6

In [2]:
# resistance

import math
#Variable declaration
Iz=10*10**-3              # reverse current in ampere
Vz=0.05                   # zener voltage in volts

#Calculations
Rz=Vz/Iz                  # resistance in ohm


#Result
print("resistance (ohm) = %.f "%Rz);

resistance (ohm) = 5

Example 1.2, Page No. 6

In [1]:
# terminal voltage 

import math
#Variable declaration
v=4.7                 # in volts
r=15                  # in ohm
i=20*10**-3           # in ampere

#Calculations
Vz=(v+(i*r))          # terminal voltage in volts

#Result
print("terminal voltage  =  %.f V"%Vz)

terminal voltage  =  5 V

Example 1.3, Page No.14

In [6]:
# tuning range of the circuit

import math
#Variable declaration
C1=5.0*10**-12              # minimum capacitance in  farad
C2=50.0*10**-12             # maximum capacitance in  farad
L=10.0*10**-3               # in henry

#Calculations
CTmin= (C1/2)             # minimum total capacitance of varactor diode
p= (math.sqrt(L*CTmin))   # calculating square root
q= (2*math.pi*p)
fomax= (1/q)              # maximum resonant frequency
CTmax= ((C2*C2)/(C2+C2))  # maximum total capacitance of varactor diode
r= (math.sqrt(L*CTmax))   # calculating square root
s= (2*math.pi*r)
fomin= (1/s)              # minimum resonant frequency

#Result
print("maximum resonant frequency = %.f MHz"%(fomax/10**6))
print("minimum resonant frequency = %.f kHz"%(fomin/1000))

maximum resonant frequency = 1 MHz
minimum resonant frequency = 318 kHz

Example 1.4, Page No.24

In [9]:
# standard resistor

import math
#Variable declaration
vf=1.8          # in volts
If=16*10**-3    # in ampere
vo=8            # in volts

#Calculations
rs=(vo-vf)/If   # resistor in ohm

#Result
print("standard resistor (ohm) = %.1f"%rs)

standard resistor (ohm) = 387.5

Example 1.5, Page No. 24

In [9]:
# min and max value of led current

import math
#Variable declaration
v1=1.5         # in volts
v2=2.3         # in volts
vs=10.0        # in volts
r1=470.0       # in ohm

#Calculations
I1=(vs-v1)/r1  # in ampere
I2=(vs-v2)/r1  # in ampere

#Result
print("maximum current = %.1f mA"%(I1*10**3))
print("minimum current = %.1f mA"%(I2*10**3))

maximum current = 18.1 mA
minimum current = 16.4 mA

Example 1.6, Page No. 24

In [13]:
# which supply voltage will keep brighness of diode constant

import math
#Variable declaration
v1=1.8             # in volts
v2=3.0             # in volts
vs=24.0            # in volts
rs=820.0           # in ohms
vs1=5.0            # in volts
rs1=120.0          # in ohms
r1=470.0           # in ohmI1=(vs-v1)/r1; // in ampere

#Calcualtions
# case1
Imin=((vs-v2)/rs)
Imax=((vs-v1)/rs)
# case2
Imin1=((vs1-v2)/rs1)
Imax1=((vs1-v1)/rs1)

#Result
print("maximum current in ampere in case1 = %.1f mA"%(Imax*10**3))
print("minimum current in ampere in case1 = %.1f mA"%(Imin*10**3))
print("maximum current in ampere in case2 = %.1f mA"%(Imax1*10**3))
print("minimum current in ampere in case2 = %.1f mA"%(Imin1*10**3))
print("\nBrightness in the first case will remain constant")
print("where as in second case it will be changing,")
print("Therefore, in order to get an approximately constant")
print("brighntness we use as large a supply voltage as possible.")

maximum current in ampere in case1 = 27.1 mA
minimum current in ampere in case1 = 25.6 mA
maximum current in ampere in case2 = 26.7 mA
minimum current in ampere in case2 = 16.7 mA

Brightness in the first case will remain constant
where as in second case it will be changing,
Therefore, in order to get an approximately constant
brighntness we use as large a supply voltage as possible.

Example 1.7.a, Page No. 33

In [14]:
# photocurrent

import math
#Variable declaration
r=0.85               # reponsivity of a photodiode in apmere per watt
p1=1.0               # incident light power in milli watt

#Calculations
Ip=r*p1

#Result
print("photocurrent  = %.2f mA"%Ip)

photocurrent  = 0.85 mA

Example 1.7.b, Page No. 33

In [20]:
# photocurrent

import math
#Variable declaration
r=0.85               # reponsivity of a photodiode in apmere per watt
p1=2.0               # incident light power in milli watt

#Result
print("Given input power saturation is 1.5mw so Ip is not proportional to Pop")
print("hence we cannot find the value of photocurrent")

Given input power saturation is 1.5mw so Ip is not proportional to Pop
hence we cannot find the value of photocurrent

Example 1.8, Page No. 34

In [15]:
# quantum efficiency

import math
#Variable declaration
EHP=5.4*10**6
photons=6*10**6 

#Calcualtions
n=EHP/photons

#Result
print("Quantum efficiency = %.1f"%n)

Quantum efficiency = 0.9

Example 1.9, Page No. 34

In [1]:
# Responsivity

import math
#Variable declaration
h=6.62*10**-34          # planc's constant
c=3*10**8               # speed of light in vaccum
e=0.70                  # efficiency 
Eg=0.75*1.6*10**-19     # Energy gap in volts

#Calculations
w=((h*c)/Eg)            # wavelength in meters (This answer is wrong in the book)
w = 1664*10**-9         # value used in book 
R=((e/1248.0)*w)        # in ampere per watt

#Result
print("Responsivity = %.3f * 10^-10 AW^-1"%(R*10**10))

Responsivity = 9.333 * 10^-10 AW^-1