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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"
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