Chapter-2 Diode Applications

Example-2.1(a) Page Number-62

In [12]:
#from the figure given in the question
E=10         #in volts (applied voltage)
r=0.5       #in kohm

Id=E/r
V=E         #V=E at Id=0
#sketching the load line and with the intersection of load line and the characterstics curve,we gqt Q-point
Vdq=0.78     #in volts
Idq=18.5     #in mA
print "Vdq=",Vdq,'V'
print "Idq=",Idq,'mA'

Vdq= 0.78 V
Idq= 18.5 mA

Example-2.1(b) Page Number-62

In [13]:
#from the results of example 2.1(a), we get
Ir=Idq=18.5      #in mA
r=1             #in kohm
Vr=Ir*r          #in volts
print "Vr=",Vr,'V'

Vr= 18.5 V

Example-2.2 Page Number-63

In [14]:
#repeating the Example-2.1 with approximate equialent model for silicon diode, we get
Vdq=0.7           #in volts
Idq=18.5          #in mA
print "Vdq=",Vdq,'V'
print "Idq=",Idq,'mA'

Vdq= 0.7 V
Idq= 18.5 mA

Example-2.3 Page Number-64

In [15]:
#repeating the Example-2.1 with ideal model for silicon diode, we get
Vdq=0           #in volts
Idq=20          #in mA
print "Vdq=",Vdq,'V'
print "Idq=",Idq,'mA'

Vdq= 0 V
Idq= 20 mA

Example-2.4 Page Number-67

In [16]:
#the applied voltage makes the diode forward biased, hence
Vd=0.7       #in volts (silicon diode)
E=8          #in volts (applied voltage)
r=2.2        #in kohm
Vr=E-Vd      #volatge across resistance r
Ir=Vr/r      #current through resistance r
Id=Ir        #current through diode
print "Vd=",Vd,'V'
print "Vr=",Vr,'V'
print "Id=",round(Id,2),'mA'

Vd= 0.7 V
Vr= 7.3 V
Id= 3.32 mA

Example-2.5 Page Number-68

In [17]:
#the applied voltage makes the diode reverse biased,hence, the circuit is open
E=8          #in volts (applied voltage)
r=2.2        #in kohm
Ir=0         #in mA(since no current flow across resistor in open circuit)
Vr=0         #in volts(since no voltage drop occur across resistor as Ir=0)

Vd=E-Vr       #volatge across diode
Id=Ir         #current through diode
print "Vd=",Vd,'V'
print "Vr=",Vr,'V'
print "Id=",round(Id,2),'mA'

Vd= 8 V
Vr= 0 V
Id= 0.0 mA

Example-2.6 Page Number-68

In [18]:
#since the applied voltage is less than than the threshold voltage of diode,hence the diode is in off state.
#considering the diode as open circuit
E=0.5        #in volts (applied voltage)
r=1.2        #in kohm
Ir=0         #in mA(since no current flow across resistor in open circuit)
Id=Ir         #in mA(series configuration)
Vr=Ir*r       #volatge across resistance r
Vd=E-Vr       #volatge across diode
print "Vd=",Vd,'V'
print "Vr=",Vr,'V'
print "Id=",round(Id,2),'mA'

Vd= 0.5 V
Vr= 0.0 V
Id= 0.0 mA

Example-2.7 Page Number-69

In [19]:
#the applied voltage makes both the diode forward biased, hence
Vd=0.7         #in volts (silicon diode)
Vdred=1.8      #in volts (red diode)
E=12          #in volts (applied voltage)
r=680         #in ohm

Vo=E-Vd-Vdred
Ir=Vo/r       #in A
Ir=Ir*1000    #in mA
Id=Ir         #in mA(series configuration)
print "Vo=",Vo,'V'
print "Id=",round(Id,2),'mA'

Vo= 9.5 V
Id= 13.97 mA

Example-2.8 Page Number-69

In [20]:
#he applied volatge puts the silicon diode forward biased and germanium diode in reverse bias,hnce an open circuit occur.
E=20           #in volts (applied voltage)
r=5.6          #in kohm
Id=0           #in mA(current flowing in an open circuit is 0)
Vsi=0          #in volts(since current throgh it is 0)
Ir=Id          #in mA(series configuration)
Vo=Ir*r
#applying Kirchoff's Voltage Law:- E-Vsi-Vge-Vo=0,Therefore
Vge=E-Vsi-Vo    #voltage across germanium diode
print "Id=",Id,'mA'
print "Vd2=",Vge,'V'
print "Vo=",Vo,'V'

Id= 0 mA
Vd2= 20.0 V
Vo= 0.0 V

Example-2.9 Page Number-70

In [21]:
#the applied voltage sets the diode in forward bias,therefore
Vd=0.7         #in volts (silicon diode)
E1=10          #in volts (applied voltage)
E2=-5          #in volts (applied voltage)
r1=4.7          #in kohm
r2=2.2          #in kohm

#applying Kirchoff's Voltage Law to the input section of the circuit:-
Ic=(E1-E2-Vd)/(r1+r2)
Vr1=Ic*r1       #voltage across r1
Vr2=Ic*r2       #voltage across r2
#applying Kirchoff's Voltage Law to the output section of the circuit:-
Vo=Vr2+E2
print "I=",round(Ic,2),'mA'
print "V1=",round(Vr1,2),'V'
print "V2=",round(Vr2,2),'V'
print "Vo=",round(Vo,2),'V'

I= 2.07 mA
V1= 9.74 V
V2= 4.56 V
Vo= -0.44 V

Example-2.10 Page Number-71

In [22]:
#the applied volatge sets both the parallel diode in forward bias,hence
E=10           #in volts
r=0.33         #in kohm
V1=0.7         #in volts (silicon diode)
V2=0.7         #in volts (silicon diode)
#applying Kirchoff's Voltage Law to the first loop of the circuit:-
I=(E-V1)/r      #current in the circuit
#assuming diodes of similar characterstics, we have
Id1=I/2
Id2=I/2
Vo=V2           #Vo is the voltage across the diodes in parallel
print "Vo=",round(Vo,2),'V'
print "I1=",round(I,2),'mA'
print "Id1=",round(Id1,2),'mA'
print "Id2=",round(Id2,2),'mA'

Vo= 0.7 V
I1= 28.18 mA
Id1= 14.09 mA
Id2= 14.09 mA

Example-2.11 Page Number-72

In [5]:
#the applied volatge sets the green diode in forward bias and red diode in negative bias,hence
E=8          #in volts
Vled=2       #in volts(turn-on voltage)
I=20         #in mA
#applying Ohm's law:
R=((E-Vled)*1000)/I
print "Required Resistance=",R,'ohm'

Required Resistance= 300 ohm

Example-2.12 Page Number-73

In [6]:
#the applied volatge sets the green and red diode forward bias ,hence
Vg=0.7           #in volts
Vr=0.7           #in volts
E=12             #in volts
R=2.2            #in kohm
#applying Kirchoff's voltage law,
Vo=E-Vr
print "Voltage Vo=",Vo,'V'

Voltage Vo= 11.3 V

Example-2.13 Page Number-74

In [8]:
#the applied volatge sets both the diodes  forward bias ,hence
Vk1=0.7           #in volts (silicon diode)
Vk2=0.7           #in volts (silicon diode)
E=20              #in volts
R1=3.3            #in kohm
R2=5.6            #in kohm
#applying Kirchoff's Voltage Law to the 2nd loop section of the circuit:-
I1=Vk1/R1          
#applying Kirchoff's Voltage Law to the 1st loop section of the circuit:-
V2=E-Vk1-Vk2
I2=V2/R2
#applying Kirchoff's current Law at the junction node of two diodes:-
Id2=I2-I1
print "I1=",round(I1,3),'mA'
print "I2=",round(I2,2),'mA'
print "Id2=",round(Id2,2),'mA'

I1= 0.212 mA
I2= 3.32 mA
Id2= 3.11 mA

Example-2.14 Page Number-74

In [10]:
#from the circuit given in the question
E1=10             #in volts (applied voltage)
E2=0              #in volts (applied voltage)
r=1               #in kohm
Vd1=0.7           #in volts (silicon diode)
Vd2=0.7           #in volts (silicon diode)
Vo=E1-Vd1
I=(E1-Vd1)/r       #current in the circuit
print "Vo=",round(Vo,2),'V'
print "I=",round(I,2),'mA'

Vo= 9.3 V
I= 9.3 mA

Example-2.15 Page Number-75

In [9]:
#from the circuit given in the question,
E1=10             #in volts (applied voltage)
E2=0              #in volts (applied voltage)
E3=10             #in volts (applied voltage)
r=1               #in kohm
Vd1=0.7           #in volts (silicon diode)
Vd2=0.7           #in volts (silicon diode)
Vo=E2+Vd2
I=(E3-Vo)/r        #current in the circuit
#Diode1 is in off state,because Vanode=Vo=0.7V and Vcathode=10V,hence Reverse biased
print "Vo=",round(Vo,2),'V'
print "I=",round(I,2),'mA'

Vo= 0.7 V
I= 9.3 mA

Example-2.24 Page Number-93

In [7]:
#from the circuit given in the question,
E=40              #in volts (applied voltage)
Vk=0.7            #in volts (white led )
Vled=4            #in volts (silicon diode)
R=1.3             #in kohm
Vz1=6             #in volts (zener diode)
Vz2=3.3           #in volts (zener diode)
Vo1=Vz2+Vk       
Vo2=Vo1+Vz1
print "Reference voltage Vo1=",round(Vo1,2),'V'
print "Reference voltage Vo2=",round(Vo2,2),'V'
Ir=(E-Vo2-Vled)/R    #current across led
Ps=E*Ir              #power delivered by supply
Pled=Vled*Ir         # power absorbed by led
Pz=Vz1*Ir           #power absorbed by diode
print "power delivered by supply=",round(Ps,0),'mW'
print "power absorbed by LED=",round(Pled,1),'mW'
print "power absorbed by zener diode=",round(Pz,1),'mW'

Reference voltage Vo1= 4.0 V
Reference voltage Vo2= 10.0 V
power delivered by supply= 800.0 mW
power absorbed by LED= 80.0 mW
power absorbed by zener diode= 120.0 mW

Example-2.25 Page Number-94

In [23]:
#Case-1:positive voltage supplied>20V. therefore zener diode is  forward biased
Vo=20      #in volts(volatge across parallel zener diode is 20volts)

#Case-2:negative voltage supplied. therefore zener diode is  reverse biased
Vo=0      #in volts(volatge across parallel zener diode is 0volts because of open circuit)

Example-2.26(a) Page Number-96

In [25]:
Vi=16           #in volts(applied voltage)
R=1             #in kohms
Rl=1.2          #in kohms
Vz=10           #in volts(zener diode)
#removing the diode and calclating voltage across the open circuit
V=(Rl*Vi)/(R+Rl)
#since V=8.73V<Vz,therefore diode is in off state
Vl=V             #voltage across resistor Rl
Vr=Vi-Vl         #voltage across resistor R
Iz=0             #in A(current through diode=0A due to open circuit)
Pz=Iz*Vz         #in W(power dissipiated by zener diode)

print "Vl",round(Vl,2),'V'
print "Vr",round(Vr,2),'V'
print "Iz",Iz,'A'
print "Pz",Pz,'W'

Vl 8.73 V
Vr 7.27 V
Iz 0 A
Pz 0 W

Example-2.26(b) Page Number-96

In [28]:
Vi=16           #in volts(applied voltage)
R=1             #in kohms
Rl=3.0          #in kohms
Vz=10           #in volts(zener diode)
#removing the diode and calclating voltage across the open circuit
V=(Rl*Vi)/(R+Rl)
#since V=12V>Vz,therefore diode is in on state
Vl=Vz             #voltage across resistor Rl
Vr=Vi-Vl         #voltage across resistor R
Il=Vl/Rl         #current through resistor Rl
Ir=Vr/R          #current through resistor R
Iz=Ir-Il         #in mA(current through diode=0A due to open circuit)
Pz=Iz*Vz         #in mW(power dissipiated by zener diode)

print "Vl",round(Vl,2),'V'
print "Vr",round(Vr,2),'V'
print "Iz",round(Iz,2),'mA'
print "Pz",round(Pz,2),'mW'

Vl 10.0 V
Vr 6.0 V
Iz 2.67 mA
Pz 26.67 mW

Example-2.27(a) Page Number-98

In [37]:
Vi=50           #in volts(applied voltage)
R=1             #in kohms
Vz=10           #in volts(zener diode)
Izm=32          #in mA(maximum current through zener diode)
Rlmin=(R*1000*Vz)/(Vi-Vz)   #in ohm (minimum value of Rl)
Vr=Vi-Vz               #voltage across resistor R
Ir=Vr/R                #current through resistor R
Ilmin=Ir-Izm
Rlmax=Vz*1000/Ilmin         #in ohm (maximum value of Rl)
print "Range of Rl is",Rlmin,"ohm to ",(Rlmax),"ohm"
print "Il=",Ilmin,"mA"

Range of Rl is 250 ohm to  1250 ohm
Il= 8 mA

Example-2.27(b) Page Number-98

In [32]:
Vz=10           #in volts(zener diode)
Izm=32          #in mA(maximum current through zener diode)
Pmax=Vz*Izm
print "maximum wattage rating of the diode=",Pmax,"mW"

maximum wattage rating of the diode= 320 mW

Example-2.28 Page Number-99

In [48]:
R=220           #in ohms
Rl=1.2          #in kohm
Vz=20           #in volts(zener diode)
Vl=Vz           #in volts(parallel connection)
Izm=60          #in mA(maximum current through zener diode)
Vimin=(((Rl*1000)+R)*Vz)/(Rl*1000)   #in volts(minimum value of  Volatge Vi)
Il=Vl/Rl            #in mA(current through resistor Rl)
Irmax=Izm+Il        #maximum value of current through resistor R
Vimax=((Irmax/1000)*R)+Vz    #maximum value of volatge Vi
print "Range of voltage Vi is",round(Vimin,2),"V to ",round(Vimax,2),"V"

Range of voltage Vi is 23.67 V to  36.87 V