# Chapter Four : Display Devices¶

## Example 4.1, Page Number 131¶

In [1]:
from math import exp

k=0.025 #in Electron Volts
E=0.4 #in Electron Volts (The difference between Ec & Ed)
Q=(10**8) # A given Constant in per second

j=E/k
O=Q*exp(-j) #Where O is the required probability
O=round(O,2)
Q1=1/O
Q1=round(Q1,2)
print "The probability of Escape per second of the trapped Electron "+str(O)+" /s"
print "The Luminescence Lifetime is "+str(Q1)+" seconds"

The probability of Escape per second of the trapped Electron 11.25 /s
The Luminescence Lifetime is 0.09 seconds


## Example 4.2, Page Number 152¶

In [2]:
from math import degrees
from math import asin

n1=3.6  #For a Given GaAs/Air Interface
n2=1    #For Air

#Using Equation 4.14
n3=n1-n2
n4=n1+n2
n6=(n3/n4)**2
n5=(n2/n1)**2

F=0.25*(n5)*(1-n6) #Where F is the Fractional Transmission for Isotropic Radiation Originating
F=round(F,3)

theta=degrees(asin(1/n1))  #Critical Angle in Degrees
theta=round(theta,0)

print "The Fractional Tranmission for Isotropic Radiation originating inside GaAs is "+str(F)
print "The Critical Angle which might explain the Low efficiency for the interface is "+str(theta)+" Degrees"

The Fractional Tranmission for Isotropic Radiation originating inside GaAs is 0.013
The Critical Angle which might explain the Low efficiency for the interface is 16.0 Degrees


## Example 4.3, Page Number 158¶

In [3]:
%matplotlib inline
import math
from matplotlib.pyplot import plot,suptitle,xlabel,ylabel

d=0.2*(10**-3) #Chip Diameter in meter
d1=1 #Distance in Meter
l=550*(10**-9 ) #Wavelength in Meter
q=0.001 #External Quantam Efficiency
i=50*(10**-3) #Operational Current
h=6.6*(10**-34)#Plancks Constant
c=3*(10**8)#Speed of Light
e=1.6*(10**-19)

theta=(d/2) #Whence theta is the angle emitting area subtends and is less than 1
print "Emitting Area subtends an angle Theta ="+str(theta)
print "Since theta is less than one, it acts as a Point Source"

W=((h*c)/l)*q*(i/e) #Where W is the total Radiant Power in terms of W
W=round(W,6)

print "The Total Radiant Power is "+str(W)+" W"

#From the above graph
l1=600 #Average Luminousity
print "Observing from the below graph at 550 nm"
lf=W*l1 #Where lf is the lumnious flux from the source
lf=round(lf,3)

print "The Luminous Flux from the source is"+str(lf)+" lm"

li=lf/(2*3.14)#Where li is the luminous intensity at normal incidence since flux is distributed over angle 2PI
li=round(li,4)

print "The Luminous Intensity is "+str(li)+" Candela"

X = [400,500,555,600,650,700]
V = [0.0,0.3,1.0,0.7,0.3,0.0]
plot(X,V);
xlabel("Wavelength in nm")
ylabel("V")
suptitle("Fig 1.24")

Emitting Area subtends an angle Theta =0.0001
Since theta is less than one, it acts as a Point Source
The Total Radiant Power is 0.000112 W
Observing from the below graph at 550 nm
The Luminous Flux from the source is0.067 lm
The Luminous Intensity is 0.0107 Candela

Out[3]:
<matplotlib.text.Text at 0x3d68198>