Ch-6 Optical Receiver Systems

Example 6.31 Page no 489

In [24]:
from __future__ import division
from math import sqrt, pi


# Calculation of a)peak photocurrent , b)shot noise and c)mean square shot noise current

#Given data
n=0.7#                               # Efficiency
lamda=0.9*10**-6#                    # Wavelength
R=5*10**3#                           # Load resistance
I=2*10**-9#                          # Dark current
P=300*10**-6#                        # Incident power
B=15*10**6#                          # Bandwidth
T=298#                             # Room temperature
h=6.62*10**-34#
c=3*10**8#
e=1.602*10**-19#                    # Charge of an electron
k=1.381*10**-23#                    # Boltzman constant

# a)Peak photocurrent
I=(n*P*e*lamda)/(h*c)#
I=I*10**6#

#b) Shot noise and mean square shot noise current
s=2*e*B*(2+I)#
s=s*10**11#

#c) mean square shot noise current
t=(4*k*T*B)/R#
t=t*10**17#
            
print"Peak photocurrent = %0.3f nA "%(I)#
print"Shot noise = %0.1f 10**-20 A "%(s)#
print"Mean square shot noise current = %0.2f*10**-17 A  "%(t)#

# The answers vary due to round off error
Peak photocurrent = 152.456 nA 
Shot noise = 74.2 10**-20 A 
Mean square shot noise current = 4.94*10**-17 A  

Example 6.32 Page no 495

In [25]:
from math import log10
# Calculation of signal to noise ratio

#Given data
I=152.3*10**-9#                     # Peak photocurrent
s=74.15*10**-20#                    # Shot noise 
t=4.94*10**-17#                    # Mean square shot noise current
F=10*log10(3)#                    # Noise figure
B=15*10**6#                        # Bandwidth
T=298#                            # Room temperature
k=1.381*10**-23#                   # Boltzman constant
R=5*10**3#                         # Load resistance
e=1.602*10**-19#                  # Charge of an electron


# Signal to noise ratio
S=(I**2)/((2*e*B*(2+I))+(4*k*T*B*F)/R)#
S=S*10**3#
            
print"Signal to noise ratio = %0.2f  "%(S)#


# The answers vary due to round off error
Signal to noise ratio = 2.41  

Example 6.33 Page no 495

In [26]:
# Calculation of a)load resistance and b)bandwidth

#Given data
Cd=5*10**-12                         # Capacitance of pin photodiode
B=10*10**6#                         # Bandwidth
Ca=10*10**-12#                      # Input capacitance


# a)Load resistance
R=1/(2*pi*B*Cd)#
R=R*10**-3#
# b)Bandwidth
Bm=1/(2*pi*(Cd+Ca)*R)#
Bm=Bm*10**-9#
            
print"Wavelength of photodiode = %0.2f kohm "%(R)#
print"Bandwidth = %0.3f MHz "%(Bm)#

# The answers vary due to round off error
Wavelength of photodiode = 3.18 kohm 
Bandwidth = 3.333 MHz 

Example 6.34 Page no 491

In [27]:
# Calculation of signal to noise ratio.

#Given data

h=6.62*10**-34#                       # Planck constant
c=5*10**-12#                          # capacitor
lamda=1.55*10**-6#                  # Wavelength
B=50*10**6#                         # Speed of communication
s=2*10**-9#
I=10**-7#
k=1.381*10**-23#
T=291#
x=0.3#
e=1.602*10**-19#

# Maximum load resistance is
R=1/(2*pi*c*B)#

S=I**2/((2*e*B*I)+(4*k*T*B/R))#
M=((4*k*T)/(e*x*R*I))**(0.435)#
S1=((((M**2)*(I**2))/(2*e*B*I*M**2.3))+((4*k*T*B)/R))#
S1=10*log10(S1)#
            
print"Load resistor = %0.1f ohm "%(R)#
print"S/N = %0.2f dB "%(S)#
print"M  = %0.2f  "%(M)#
print"S/N = %0.2f dB "%(S1)#

# The answers vary due to round off error
Load resistor = 636.6 ohm 
S/N = 7.91 dB 
M  = 41.53  
S/N = 33.10 dB 

Example 6.35 Page no 493

In [28]:
# Calculation of a) responsivity b)incident optical power

#Given data
n=0.6#                         # Quantum efficiency
e=1.602*10**-19#                # Charge of electron
lamda=0.9*10**-6#              # Wavelength 
h=6.626*10**-34#                # Planck constant 
c=3*10**8#                      # Velocity of light
I=2*10**-6#                     # Photocurrent

# a)Responsivity
R= (n*e*lamda)/(h*c)#

# b)Incident power
P=I/R#
P=P*10**6#

print"Responsivity = %0.3f A/W "%(R)#
print"Incident power = %0.3f microwatt "%(P)#

# The answers vary due to round off error
Responsivity = 0.435 A/W 
Incident power = 4.596 microwatt 

Example 6.36 Page no 493

In [29]:
# Calculation of a) responsivity b)Multiplication factor

#Given data
n=0.8#                            # Quantum efficiency
e=1.602*10**-19#                   # Charge of an electron
lamda=0.9*10**-6#                 # Wavelength 
h=6.626*10**-34#                   # Planck constant 
c=3*10**8#                        # Velocity of light
I=15*10**-6#                      # Photocurrent
P=0.6*10**-6#

# a)Responsivity
R= (n*e*lamda)/(h*c)#
# b)Multiplication factor
Ip=P*R#
M=I/Ip#
            
print"Responsivity = %0.3f A/W "%(R)#
print"Multiplication factor = %0.2f  "%(M)#

# The answers vary due to round off error
Responsivity = 0.580 A/W 
Multiplication factor = 43.08  

Example 6.37 Page no 494

In [30]:
# Calculation of a) quantum efficiency b) responsivity 

#Given data
e5=500#                            # No of incident photons
e8=800#                            # No of incident electrons
e=1.602*10**-19#                    # Charge of an electron
lamda=1.3*10**-6#                  # Wavelength 
h=6.626*10**-34#                    # Planck constant 
c=3*10**8#                         # Velocity of light
I=15*10**-6#                       # Photocurrent
P=0.6*10**-6#

# a)Quantum efficiency
n=e5/e8#
# b)Responsivity
R=(n*e*lamda)/(h*c)#

print"Quantum efficiency = %0.1f %% "%(n*100)#
print"Responsivity = %0.3f A/W "%(R)#

# The answers vary due to round off error
Quantum efficiency = 62.5 % 
Responsivity = 0.655 A/W 

Example 6.38 Page no 494

In [31]:
# Calculation of a) quantum efficiency b) responsivity

#Given data
e5=1.2*10**11#                            # No of electrons collected
e8=3.6*10**11#                            # No of incident photon
e=1.602*10**-19#                         # Charge of an electron
lamda=0.85*10**-6#                      # Wavelength 
h=6.626*10**-34#                        # Planck constant 
c=3*10**8#                              # Velocity of light
I=15*10**-6#                            # Photocurrent
P=0.6*10**-6# 

# a)Quantum efficiency
n=e5/e8#
# b)Responsivity
R=(n*e*lamda)/(h*c)#
            
print"Quantum efficiency = %0.2f  "%(n)#
print"Responsivity = %0.3f A/W"%(R)#

# The answers vary due to round off error
Quantum efficiency = 0.33  
Responsivity = 0.228 A/W

Example 6.39 Page no 495

In [32]:
# Calculation of a) operating wavelength b) incidence optical power

#Given data
n=0.60                            # Quantum efficiency
E=1.5*10**-19#                    # Photons of energy
e=1.602*10**-19#                  # Charge of an electron
h=6.626*10**-34#                  # Planck constant 
c=3*10**8#                        # Velocity of light
I=2*10**-6#                      # Photocurrent


# a)Operating wavelength
lamda=(h*c)/E#
lamda=lamda*10**6#

# b)Incident optical power
R=(n*e)/E#
P=I/R#
P=P*10**6#
            
print"Wavelength of photodiode = %0.2f micrometer "%(lamda)#
print"Incident optical power = %0.2f microWatt "%(P)#

# The answers vary due to round off error
Wavelength of photodiode = 1.33 micrometer 
Incident optical power = 3.12 microWatt 

Example 6.40 Page no 495

In [33]:
# Calculation of load resistance and bandwidth



#Given data
Cd=6*10**-12                         # Capacitance of pin photodiode
B=20*10**6#                         # Bandwidth
Ca=6*10**-12#                      # Input capacitance


# a)Load resistance
R=1/(2*pi*B*Cd)#
R=R*10**-3#
# b)Bandwidth
Bm=1/(2*pi*(Ca+Ca)*R)#
Bm=Bm*10**-9#

            
print"Load resistance = %0.2f kohm "%(R)#
print"Bandwidth = %0.0f MHz "%(Bm)#

# The answers vary due to round off error
Load resistance = 1.33 kohm 
Bandwidth = 10 MHz 

Example 6.41 Page no 496

In [34]:
# Calculation of maximum bandwidth

#Given data
t=5*10**-12                         # Electron transit time
G=70#                              # Gain of the device

# Maximum bandwidth
Bm=1/(2*pi*t*G)#
Bm=Bm*10**-6#
            
print"Bandwidth = %0.1f MHz "%(Bm)#

# The answers vary due to round off error
Bandwidth = 454.7 MHz