Chapter 6: Optical detectors

Example 6.1, Page number 201

In [2]:
'''Find the quantum efficiency'''

#Varaible declaration 
re = 4.2*10**6   #EHP's
rp = 6*10**6     #no. of photons
lamda = 1200     #wavelength(nm)

#Calculation
N = re/rp

#Result
print "Quantum efficiency =",N*100,"%"

Quantum efficiency = 70.0 %

Example 6.2, Page number 201

In [3]:
'''Find photocurrent'''

#Varaible declaration 
R = 0.85   #responsivity(A/W)
Po = 1     #incident power(mW)

#Calculation
Ip = R*Po

#Result
print "Photocurrent =",Ip,"mA"

Photocurrent = 0.85 mA

Example 6.3, Page number 201

In [4]:
'''Find responsivity of photo diode'''

#Varaible declaration 
E = 0.75*1.6*10**-19    #energy gap(V)
c = 3*10**8             #speed of light(m/s)
N = 60./100             #quantum efficiency
h = 6.62*10**-34        #Planck's constant

#Calculation
lamda = (h*c)/E
R = (N*lamda)/1248

#Result
print "Responsivity =",round((R/1E-9),1),"A/W"

Responsivity = 0.8 A/W

Example 6.4, Page number 202

In [6]:
'''Find responsivity and quantum efficiency'''

#Varaible declaration 
re = 1.5*10**12         #EHP's
rp = 3*10**12           #no. of photons
lamda = 0.65*10**-6     #wavelength(m)
h = 6.62*10**-34        #Palnck's constant
c = 3*10**8             #speed of light(m/s)
e = 1.6*10**-19         #charge of an electron(C)

#Calculation
N = re/rp

R = (N*e*lamda)/(h*c)

#Result
print "Quantum efficiency =",N*100,"%"
print "Responsivity =",round(R,3),"A/W"

Quantum efficiency = 50.0 %
Responsivity = 0.262 A/W

Example 6.5, Page number 202

In [7]:
'''Estimate -
a)operating walength
b)incident optical power'''

#Varaible declaration 
E = 1.5*10**-19         #photon energy(J)
c = 3*10**8             #speed of light(m/s)
N = 65./100             #quantum efficiency
Ip = 1.5*10**-6         #photocurrent(A)
h = 6.62*10**-34        #Planck's constant
c = 3*10**8             #speed of light(m/s)
e = 1.6*10**-19         #charge of an electron(C)

#Calculations
#Part a
lamda = (h*c)/E

#Part b
f = c/lamda
R = (N*e)/(h*f)
Po = Ip/R

#Results
print "a)Operating waelength =",lamda/1E-6,"um"
print "b)Optical power =",round((Po/1E-6),2),"uW"

a)Operating waelength = 1.324 um
b)Optical power = 2.16 uW

Example 6.6, Page number 203

In [9]:
'''Find wavelength'''

#Varaible declaration 
Eg = 1.43*1.6*10**-19   #energy gap(V)
c = 3*10**8             #speed of light(m/s)
h = 6.62*10**-34        #Planck's constant

#Calculations
lamda_c = (h*c)/Eg

#Result
print "Wavelength =",round((lamda_c/1E-6),2),"um"

Wavelength = 0.87 um

Example 6.7, Page number 203

In [11]:
'''Find -
a)responsivity
b)Optical power received
c)no. of received photons'''

#Varaible declaration 
N = 50./100    #quantum efficiency
lamda = 900    #wavelength(nm)
Ip = 10**-6    #photocurrent(A)
c = 3*10**8             #speed of light(m/s)
h = 6.62*10**-34        #Planck's constant

#Calculations
#Part a
R = (N*lamda)/1248

#Part b
Po = Ip/R

#Part c
n = (Po*lamda*10**-9)/(h*c)

#Results
print "a)Responsivity =",round(R,2),"A/W"
print "b)Optical power =",round((Po/1E-6),2),"*10^-6 W"
print "c)No. of photons =",round((n/1e+13),3),"*10^13(Calculation mistake in textbook)"

a)Responsivity = 0.36 A/W
b)Optical power = 2.77 *10^-6 W
c)No. of photons = 1.257 *10^13(Calculation mistake in textbook)

Example 6.8, Page number 204

In [12]:
'''Find the multiplication factor of photodiode'''

#Varaible declaration 
N = 80./100           #quantum efficiency
lamda = 0.9*10**-6   #wavelength(m)
I = 12*10**-6         #output current(A)
Po = 0.5*10**-6       #incident power(W)
c = 3*10**8           #speed of light(m/s)
h = 6.626*10**-34     #Planck's constant
e = 1.6*10**-19       #charge of an electron(C)

#Calculations
R = (N*e*lamda)/(h*c)  #responsivity(A/W)
Ip = Po*R              #photocurrent(A)
M = I/Ip

#Result
print "Multiplication factor =",M

Multiplication factor = 41.4125

Example 6.9, Page number 205

In [13]:
'''Find the responsivity and multiplication factor'''

#Varaible declaration 
N = 65./100           #quantum efficiency
lamda = 850*10**-9    #wavelength(m)
I = 10*10**-6         #output current(A)
Po = 0.5*10**-6       #incident power(W)
c = 3*10**8           #speed of light(m/s)
h = 6.626*10**-34     #Planck's constant
e = 1.6*10**-19       #charge of an electron(C)

#Calculations
R = (N*e*lamda)/(h*c)  #responsivity(A/W)

M = I/(R*Po)

#Result
print "Responsiviy =",round(R,3),"A/W"
print "Multiplication factor =",M

Responsiviy = 0.445 A/W
Multiplication factor = 44.9728506787

Example 6.10, Page number 205

In [14]:
'''Estimate -
a)noise equivalent power
b)specific detectivity'''

#Varaible declaration 
N = 55./100         #quantum efficiency
lamda = 1.3*10**-6  #wavelength(nm)
Id = 8*10**-9       #dark current(A)
A = 75*50*10**-12   #dimensions(m)
c = 3*10**8         #speed of light(m/s)
h = 6.62*10**-34    #Planck's constant

#Calculations
#Part a
NEP = (h*c*((2*e*Id)**0.5))/(N*e*lamda)

#Part b
D = A**0.5/NEP

#Results
print "a)Noise equivalent power =",round((NEP/1E-14),2),"*10^-14 W"
print "b)Specific detectivity =",round((D/1E+8),2),"*10^8"

a)Noise equivalent power = 8.78 *10^-14 W
b)Specific detectivity = 6.97 *10^8

Example 6.11, Page number 206

In [15]:
'''Find -
a)optical gain
b)CE current'''

#Varaible declaration 
N = 60./100           #quantum efficiency
lamda = 1.26*10**-6   #wavelength(m)
Ic = 15*10**-3        #output current(A)
Po = 125*10**-6       #incident power(W)
c = 3*10**8           #speed of light(m/s)
h = 6.626*10**-34     #Planck's constant
e = 1.6*10**-19       #charge of an electron(C)

#Calculations
#Part a
Go = (h*c*Ic)/(lamda*e*Po)

#Part b
Nfe = Go/N

#Results
print "a)Optical gain =",round(Go,1)
print "b)Common emiiter current =",round(Nfe,1),"A"

a)Optical gain = 118.3
b)Common emiiter current = 197.2 A

Example 6.12, Page number 207

In [16]:
'''Calculate the maximum 3dB bandwidth'''

import math

#Variable declaration
tf = 5*10**-12  #transit time(sec)
G = 70          #photoconductive gain

#Calculation
Bm = 1/(2*math.pi*tf*G)

#Result
print "The maximum 3dB bandwidth permitted by photoconductor is",round((Bm/1E+6),1),"MHz"

The maximum 3dB bandwidth permitted by photoconductor is 454.7 MHz

Example 6.13, Page number 207

In [17]:
'''Calculate the output photocurrent'''

#Variable declaration
rp = 10**11       #no. of photons/sec
hf = 1.28*10**19  #energy of photons(J)
e = 1.6*10**-19   #charge of an electron(C)

#Calculations
Po = rp/hf
N = 1        #efficiency for an ideal photodiode
Ip = (N*Po*e)/hf

#Result
print "Output photocurrent =",round((Ip/1E-47),2),"*10^-47 A(Calculation mistake in textbook)"

Output photocurrent = 9.77 *10^-47 A(Calculation mistake in textbook)

Example 6.14, Page number 207

In [5]:
'''Find the output photocurrent'''

#Variable declaration
R = 0.40       #responsivity(A/W)
phi = 100      #incident flux(uW/mm^2)
A = 2          #active area(mm^2)

#Calculations
Po = phi*A    #incident power(uW)
Ip = R*Po

#Result
print "Photocurrent =",Ip/1e+3,"mA"

Photocurrent = 0.08 mA

Example 6.15, Page number 208

In [19]:
'''Find the multiplication factor of photodiode'''

#Varaible declaration 
N = 50./100           #quantum efficiency
lamda = 1.3*10**-6    #wavelength(m)
I = 8*10**-6          #output current(A)
Po = 0.4*10**-6       #incident power(W)
c = 3*10**8           #speed of light(m/s)
h = 6.626*10**-34     #Planck's constant
e = 1.6*10**-19       #charge of an electron(C)

#Calculations
R = (N*e*lamda)/(h*c)  #responsivity(A/W)
Ip = Po*R              #photocurrent(A)
M = I/Ip

#Result
print "Multiplication factor =",round(M,2)

Multiplication factor = 38.23

Example 6.16, Page number 208

In [36]:
'''Calculate the maximum 3dB bandwidth'''

import math

#Variable declaration
tf = 4.5*10**-12  #transit time(sec)
G = 80            #photoconductive gain

#Calculation
Bm = 1/(2*math.pi*tf*G)

#Result
print "The maximum 3dB bandwidth permitted by photoconductor is",round((Bm/1E+10),4),"GHz(Calculation mistake in textbook)"

The maximum 3dB bandwidth permitted by photoconductor is 0.0442 GHz(Calculation mistake in textbook)

Example 6.17, Page number 209

In [20]:
'''Estimate responsivity and received optical power'''

#Varaible declaration 
N = 50./100         #quantum efficiency
lamda = 0.9*10**-6  #wavelength(m)
Ip = 10**-6         #photocurrent(A)
c = 3*10**8         #speed of light(m/s)
h = 6.62*10**-34    #Planck's constant
e = 1.6*10**-19     #charge of an electron(C)

#Calculations
R = (N*e*lamda)/(h*c)

Po = Ip/R


#Results
print "Responsivity =",round(R,2),"A/W"
print "Optical power =",round((Po/1E-6),2),"*10^-6 W"

Responsivity = 0.36 A/W
Optical power = 2.76 *10^-6 W

Example 6.18, Page number 209

In [21]:
'''Find the efficiency of a PIN silicon photodiode'''

#Varaible declaration 
R = 0.374            #responsivity(A/W)
lamda = 1300*10**-9  #wavelength(m)
c = 3*10**8          #speed of light(m/s)
h = 6.62*10**-34     #Planck's constant
e = 1.6*10**-19      #charge of an electron(C)

#Calculations
N = (R*h*c)/(e*lamda)

#Result
print "Efficiency =",round((N*100),1),"%"

Efficiency = 35.7 %

Example 6.20, Page number 209

In [22]:
'''Find the thickness of the intrinsic region'''

#Variable declaration
A = 1.5*10**-3    #area(mm^2)
R = 100           #load resistance(Ohms)
Eo = 1.04*10**-10 #permitivitty for Si(F/m)
vd = 10**7        #electron saturation velocity(m/s)

#Calculation
w = (R*Eo*A*vd)**0.5

#Result
print "The required thickness is",round(w/1E-6),"um(Calculation mistake in textbook)"

The required thickness is 12490.0 um(Calculation mistake in textbook)

Example 6.21, Page number 210

In [6]:
'''Find the dark current'''

import math

#Varaible declaration 
N = 64./100         #quantum efficiency
lamda = 0.85*10**-6 #wavelength(m)
B = 1               #bandwidth(Hz)
D = 7*10**10        #specific detectivity(/MHz-W)
A = 10*10**-6       #dimensions(m)
c = 3*10**8         #speed of light(m/s)
h = 6.62*10**-34    #Planck's constant
e = 1.602*10**-19   #charge of an electron(C)

#Calculations
Id = ((N*math.sqrt(e*A)*lamda)/(h*c*math.sqrt(2)*D))**0.5

#Result
print "Dark current =",round((Id/1E-3),2),"mA"
#Square root of Id has not been taken in the textbook.Hence, the difference in solution

Dark current = 5.92 mA