Chapter 10 : Diodes

Pg: 445 Ex: 10.1

In [1]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import plot,title,xlabel,ylabel,show
V_ss=2#
R=1*10**3#
V_D=arange(0,2+0.001,0.001)
I_D=[]
for x in V_D:
    I_D.append(10**3*(V_ss-x)/R)
plot(V_D,I_D)    
title('load line plot')
xlabel('voltage in volts')
ylabel('current in milli-amperes')      #milli-10**-3
show()
#we use the equation V_ss=R*i_D+V_D
#at point B
i_D=V_ss/R#      #as V_D=0
#at point A
V_D=V_ss#      #as i_D=0
#now we see intersection of load line with characteristic and we get following at operating point
V_DQ=0.7#      #voltage
I_DQ=1.3*10**-3#      #current
#diode characteristic cannot be plotted
print 'diode voltage at operating point = %0.2f volts'%V_DQ
print I_DQ*10**3,'current at opeating point = %0.2f milli-amperes'%(I_DQ*10**3)      #milli-10**-3

diode voltage at operating point = 0.70 volts
1.3 current at opeating point = 1.30 milli-amperes

Pg: 446 Ex: 10.2

In [2]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import plot,title,xlabel,ylabel,show
V_ss=10#
R=10*10**3#
V_D=arange(0,2+0.001,0.001)
I_D=[]
for x in V_D:
    I_D.append(10**3*(V_ss-x)/R)
plot(V_D,I_D)    
title('load line plot')
xlabel('voltage in volts')
ylabel('current in milli-amperes')      #milli-10**-3
show()

#we use the equation V_ss=R*i_D+V_D
#at point C
i_D=V_ss/R#      #as V_D=0
#now if we take i_D=0, we get V_D=10 which plots at a point far off the page
#so we take the value on the right-hand edge of V-axis i.e.,V_D=2
#at point D
V_D=2#
i_D=(V_ss-V_D)/R#
#from the intersection of load line with characteristic
V_DQ=0.68#
I_DQ=0.93*10**-3#
#diode characteristic cannot be plotted
print 'diode voltage at operating point = %0.2f volts'%V_DQ
print 'current at opeating point = %0.2f milli-amperes'%(I_DQ*10**3)      #milli-10**-3

diode voltage at operating point = 0.68 volts
current at opeating point = 0.93 milli-amperes

Pg: 448 Ex: 10.3

In [3]:
from numpy import arange
R=1*10**3#
#diode characteristic cannot be plotted
#case a)V_ss=15
V_ss=15#
V_D=arange(-15,0+0.001,0.001)
#from the intersection of load line and diode characteristic
V_o=10#
print 'output voltage for Vss=15 = %0.2f volts'%V_o
#case b)V_ss=20
V_ss=20#
V_D=arange(-20,0+0.001,0.001)
#from the intersection of load line and diode characteristic
V_o=10.5#
print 'output voltage for Vss=20 = %0.2f volts'%V_o

output voltage for Vss=15 = 10.00 volts
output voltage for Vss=20 = 10.50 volts

Pg: 449 Ex: 10.4

In [4]:
V_ss=24#
R=1.2*10**3#
R_L=6*10**3#
#by grouping linear elements together on left side of diode
V_T=V_ss*R_L/(R+R_L)#      #thevenin voltage
#zeroing sources 
R_T=1/((1/R)+(1/R_L))#      #thevenin resistance
#load-line equation is V_T+R_T*i_D+V_D=0
#locating the operating point
V_D=-10#
V_L=-V_D#      #load voltage
I_s=(V_ss-V_L)/R#      #source current
#diode characteristic cannot be plotted
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'load voltage = %0.2f volts'%V_L
print 'source current = %0.2f amperes'%I_s      #milli-10**-3

 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
load voltage = 10.00 volts
source current = 0.01 amperes

Pg: 451 Ex: 10.5

In [5]:
V_1=10#
V_2=3#
R_1=4*10**3#
R_2=6*10**3#
#1)analysis by assuming D1 off and D2 on
I_D_2=V_2/R_2#      #ohm's law
#applying KVL
V_D_1=7#      #contradiction to 'D1 is off'
#this assumption is not correct

#2)analysis by assuming D1 on and D2 off
I_D_1=V_1/R_1#      #ohm's law
#applying KVL
V_D_2=-V_1+V_2+I_D_1*R_1#
#we get V_D_2 which is consistent
print 'correct assumption is D2 off and D1 on'

correct assumption is D2 off and D1 on

Pg: 453 Ex: 10.7

In [6]:
V_1=3#
R_1=20#
#As given voltage source results in forward bias, we assume operating point is on line segment A
#replacing diode with the equivalent circuit
V_2=0.6#
R_2=10#
i_D=(V_1-V_2)/(R_1+R_2)#      #KVL around the circuit
print 'current in the circuit = %0.2f milli-amperes'%(i_D*10**3)      #milli-10**-3

current in the circuit = 80.00 milli-amperes