import math

#(a)
#initialisation of variables

E=10      #E in V
R=1     #R in Kohm


#Calculations
 
Id=E/R    #Eq.(2.2)
Vd=E
print "The current Ic is= %fmA "%(Id),";Vd=0V"
print "The diode voltage is= %fV"%(Vd),";Id=0A"
print "The resulting load line appears in Fig. 2.4. The intersection between the load line and the characteristic curve defines the Q-point as"
print "The level of VD is certainly an estimate, and the accuracy of ID is limited by the chosenscale. A higher degree of accuracy would require a plot that would be much large and perhaps unwieldy"


#(B)
print "(B)"
Ir=9.25 #Ir in mA
Vdq=0.78 #Vdq in v
Vr=Ir*R
print "Vr = Ir*R = Idq*R  %d="%(Vr),"or"
Vr = E-Vdq
print "Vr = E-Vdq = %f" %(Vr)
print "The difference in results is due to the accuracy with which the graph can be read. Ideally,the results obtained either way should be the same."

#Graph solution to example 2.1

import numpy as np
import matplotlib.pyplot as plt

Vd = np.linspace(0.0,10.0)
Id = np.linspace(0.0,10.0)
Id= -Vd + 10
plt.plot(Vd, Id)
Vd = [0,0,0.1,0.1,0.2,0.2,0.3,0.3,0.3,0.3,0.4,0.5,0.6,0.7]
Id = [0,0,0,0,0,0,0,0,0.1,0.1,0.3,0.7,2.0,10.0]

plt.plot(Vd, Id,'yo-')

plt.xlabel('Voltage (v)')
plt.ylabel('current (mA)')
plt.title('About as simple as it gets, folks')
plt.grid(True)
plt.savefig("test.png")

plt.show()

print "example 2.2:"
print "repeat the example 2.1 for R =2"