Chapter 18 - Optoelectronic Devices

Example E1 - Pg 901

In [1]:
#Ex18_1 Pg-901
#calculate Source resistance,Wattage rating
Vs=12. #supply voltage in V
Vd=2. #forward bias voltage in V
Id=20.*10.**(-3.) #forward bias current
Rs=(Vs-Vd)/Id #source resistor
print '%s %.0f' %("Source resistance = ohm",Rs)
P=Id**2.*Rs #power
print '%s %.1f' %("Wattage rating = mW",P*1e3)
print '%s' %("Therefore a standard size 0.25 watt = 250mW resistor is required")
Source resistance = ohm 500
Wattage rating = mW 200.0
Therefore a standard size 0.25 watt = 250mW resistor is required

Example E2 - Pg 945

In [2]:
#Ex18_2 Pg-945
#calculate R
import math
T=2000. #temperature in Kelvin
f=5.*10.**(14.) # frequency in Hz
h=6.6*10.**(-34.) #planck constant
k=1.38*10.**(-23.) #Boltzmann constant
R=math.exp((h*f)/(k*T)) #ratio of spontaneous and stimulated emisson
print '%s %.2f' %("R =",R*1e-5)
R = 1.56

Example E3 - Pg 946

In [3]:
#Ex18_3 Pg-946
#calculate Average wavelength of visible radiation
import math
print '%s' %("Average wavelength of visible radiation = 550 nm")
print '%s' %("      E1 - E2 = hc/lamda")
h=6.6*10.**(-34.) #planck constant
c=3.*10.**(8.) #speed of light in sec
lamda= 550.*10.**(-9.) #wavelength in m
E=h*c/lamda #difference in energy levels in Joules
e=1.6*10.**(-19.) #electron charge in eV
E_eV=E/e #difference in energy levels in electronVolt
print '%s %.1f' %("              = J",E*1e19)
print '%s %.2f' %("              = eV",E_eV)
T=300. #temperature in Kelvn
k=1.38*10.**(-23.) #Boltzmann constant
print '%s' %("Average room temperature=300K and g1=g2,we have")
print '%s %.2f' %("N2/N1 =",N*1e37)
#answer in the book is wrong
Average wavelength of visible radiation = 550 nm
      E1 - E2 = hc/lamda
              = J 3.6
              = eV 2.25
Average room temperature=300K and g1=g2,we have
N2/N1 = 0.17

Example E4 - Pg 946

In [5]:
#Ex18_4 Pg-946
#calculate Energy absorbed,recombination radiation
import math
w=0.3*10.**(-6.)*100. #width of silicon in cm
e=1.6*10.**(-19.) #electron charge in eV
print '%s' %("(1)Energy absorbed/sec is given by ")
E=phi*(1.-math.exp(alpha*w)) #energy absorbed(textbook answer is wrong)
print '%s %.1f' %("                = mW",abs(E)*1e3)
print '%s' %("(2)The portion of each photo energy that is converted into heat is obtained as hv-Eg/hv")
Heat=(3-1.12)/3.*100. #photon energy coverted to heat
print '%s %.0f' %("                = ",Heat)
E1=(62./100.)*0.0232 #energy dissipated/sec (textbook answer is wrong)
print '%s %.1f' %("\nObviously, the amount of energy dissipated/sec to lattice is = mW",E1*1e3)
print '%s' %("(3)Number of photons/sec from recombination is")
print '%s %.1f' %("                = photon/sec",num_photons*1e-19)
#textbook answer is wrong
print '%s' %("Therefore recombination radiation")
RR=abs(E)-E1 #recombination radiation (textbok answer is wrong)
print '%s %.1f' %("                = mW",RR*1e3)
(1)Energy absorbed/sec is given by 
                = mW 23.2
(2)The portion of each photo energy that is converted into heat is obtained as hv-Eg/hv
                =  63

Obviously, the amount of energy dissipated/sec to lattice is = mW 14.4
(3)Number of photons/sec from recombination is
                = photon/sec 1.3
Therefore recombination radiation
                = mW 8.8

Example E5 - Pg 947

In [6]:
#Ex18_5 Pg-947
#calculate Time taken to diffuse
d=5.*10.**(-6.) #thickness of silicon in m
Dc=3.4*10.**(-3.) #diffusion coefficient in m**2sec**(-1)
t=d**2./(2.*Dc) #time taken to diffuse
print '%s %.1f' %("Time taken to diffuse = sec",t*1e9)
Time taken to diffuse = sec 3.7

Example E6 - Pg 947

In [7]:
#Ex18_6 Pg-947
#calculate Diode capacitance
import math
A=10.**(-6.) #diode area in m
epsilon_r=11.7 #relative permitivity 
Nd=10.**(21.) #number of doping carriers
V=10. #bias potential in V
e=1.6*10.**(-19.) #electron charge in eV
epsilon_0=8.85*10.**(-12.) #permitivity of free space
print '%s %.f' %("Diode capacitance = pF",Cj*1e12)
#textbook answer is wrong
Diode capacitance = pF 29

Example E7 - Pg 947

In [8]:
#Ex18_7 Pg-947
#calculate Gain
import math
L=10.**(-6.) #length of cavity in m
r2=0.5 #relative coefficient of semiconductor
r1=1.5 #relative coefficient of semiconductor
print '%s' %("No internal loss means di=0; we have")
g=math.log10(1./(r1*r2))/(2.*L) #gain of the laser (textbook answer is wrong)
print '%s %.2f' %("Gain g = cm**(-1)",g*1e-3)
No internal loss means di=0; we have
Gain g = cm**(-1) 62.47

Example E8 - Pg 947

In [9]:
#Ex18_8 Pg-947
#calculate gain
L=100.*10.**(-6.) #length of semiconductor in m
A=10.**(-7.) #area of semiconductor in cm**2
V=10. #applied voltage in V
mew_n=1350. #mobility of electrons 
mew_p=480. #mobiltiy of protons
tp=10.**(-6.) #lifetime of protons in sec
tn=L/(mew_n*V) #lifetime of electrons in sec
Gain=tp/tn*(1+(mew_p/mew_n)) #gain of photoconductor
print '%s %.2f' %("Gain =",Gain*1e-2)
Gain = 1.83