CHAPTER 3 DIODE THEORY

Example 3-2, Page 63

In [1]:
Pr=5     #Power rating(W)
Vd=1.2   #diode voltage(V)
Id=1.75  #diode current(A)

PD=Vd*Id #Power dissipaion(W)

print 'Power Dissipation =',PD,'W'
print 'PD(',PD,'W) < ''Pr(',Pr,'W), So diode will not be destroyed.'

Power Dissipation = 2.1 W
PD( 2.1 W) < Pr( 5 W), So diode will not be destroyed.

Example 3-3, Page 65

In [3]:
Vs=10     #Source voltage(V)
RL=1      #Load resistance(KOhm)
VL=Vs     #LOad voltage(V)

IL=VL/RL  #Load current(mA)

print 'Vs will be appearing across RL'
print 'Load voltage VL =',Vs,'V'
print 'Load current IL =',IL,'mA'

Vs will be appearing across RL
Load voltage VL = 10 V
Load current IL = 10 mA

Example 3-4, Page 65

In [4]:
print 'As per figure 3-6b, Thevenize the circuit to the left of the diode'
R1=6   #Resistance(KOhm)
R2=3   #Resistance(KOhm)
RL=1   #Load Resistance(KOhm)
Vs=36  #Supply voltage(V)

Vth=R2*Vs/(R1+R2)   #ThCevenin voltage(V)
Rth=(R1*R2)/(R1+R2) #Thevenin resistance(KOhm)
Rt=Rth+RL           #total resistance(KOhm)
IL=Vth/Rt           #Load current(mA)
VL=IL*RL            #Load voltage(V)

print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'
print 'Visualize diode as closed switch,'
print 'IL =',IL,'mA'
print 'VL =',VL,'V'

As per figure 3-6b, Thevenize the circuit to the left of the diode
Vth = 12 V & Rth = 2 KOhm
Visualize diode as closed switch,
IL = 4 mA
VL = 4 V

Example 3-5, Page 67

In [5]:
print 'As per Second approximation in Fig.3-8,'
Vd=0.7    #diode voltgage(V)
Vs=10     #supply voltage(V)
RL=1      #Load resistance(KOhm)

VL=Vs-Vd  #Load voltage(v)
IL=VL/RL  #Load current(mA)
PD=Vd*IL  #diode power(mW)

print 'IL =',IL,'mA & VL =',VL,'V'
print 'Diode power PD =',PD,'mW'

As per Second approximation in Fig.3-8,
IL = 9.3 mA & VL = 9.3 V
Diode power PD = 6.51 mW

Example 3-6, Page 67

In [7]:
print 'As per Second approximation in Fig.3-9a,'
Vd=0.7  #diode voltgage(V)
Vs=36   #supply voltage(V)
R1=6    #Resistance(KOhm)
R2=3    #Resistance(KOhm)
RL=1    #Load resistance(KOhm)

Vth=R2*Vs/(R1+R2)#Thevenin Voltage(V)
Rth=(R1*R2)/(R1+R2)#Thevenin resistance(KOhm)
IL=(Vth-Vd)/R2#Load current(mA)
VL=IL*RL#Load voltage(V)
PD=Vd*IL#diode power(mW)

print 'Thevenize the circuit to the left of the diode'
print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'
print 'VL =',round(VL,2),'V & IL =',round(IL,2),'mA'
print 'Diode power PD =',round(PD,2),'mW'

As per Second approximation in Fig.3-9a,
Thevenize the circuit to the left of the diode
Vth = 12 V & Rth = 2 KOhm
VL = 3.77 V & IL = 3.77 mA
Diode power PD = 2.64 mW

Example 3-7, Page 68

In [2]:
print 'In Fig.3-11a,'
Vd=0.7    #diode voltgage(V)
Vs=10     #supply voltage(V)
RL=1000L  #Load resistance(Ohm)
Rb=0.23   #bulk resistance

print 'As per third approximation, we get fig.3-11b'
if Rb<(RL/100):
    print'If Rb < 0.01RL than ignore Rb & use second approximation.'
    VL=Vs-Vd          #Load voltage(V)
    IL=(VL/RL)*1000   #Load current(mA)
    PD=Vd*IL          #diode power(mW)

    print 'IL =',IL,'mA & VL =',VL,'V'
    print 'Diode power PD =',PD,'mW'

In Fig.3-11a,
As per third approximation, we get fig.3-11b
If Rb < 0.01RL than ignore Rb & use second approximation.
IL = 9.3 mA & VL = 9.3 V
Diode power PD = 6.51 mW

Example 3-8, Page 69

In [9]:
print 'In Fig.3-11a, take RL = 10 Ohm'
Vd=0.7    #diode voltgage(V)
Vs=10     #supply voltage(V)
RL=10     #Load resistance(Ohm)
Rb=0.23   #bulk resistance
RT=Rb+RL  #Total reistance(Ohm)
VT=Vs-Vd  #total voltage(V)

print 'RT =',RT,'Ohm & VT =',VT,'V'
IL=VT/RT      #Load current(mA)
VL=IL*RL      #Load voltage(V)
VD=Vd+(IL*Rb) 
PD=VD*IL      #diode power(W)

print 'IL =',round(IL,2),'mA & VL =',round(VL,2),'V'
print 'VD =',round(VD,2),'V'
print 'Diode power PD =',round(PD,2),'W'

In Fig.3-11a, take RL = 10 Ohm
RT = 10.23 Ohm & VT = 9.3 V
IL = 0.91 mA & VL = 9.09 V
VD = 0.91 V
Diode power PD = 0.83 W
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