Chapter 3 Zener Diode and LED

Exa 3.1 page No 164

In [1]:
from __future__ import division
# Given data
I1 = 30# mA
I2 = 20# mA
delI_Z = I1-I2# mA
delI_Z = delI_Z * 10**-3# A
V1 = 5.75# V
V2 = 5.6# V
delV_Z = V1-V2# V
# The resistance of the device 
r_Z = delV_Z/delI_Z# ohm
print "The resistance of the device = %.2f ohm"%r_Z

The resistance of the device = 15.00 ohm

Exa 3.2 page No 164

In [2]:
from __future__ import division
# Given data
V_Z = 4.7# V
r_Z = 15# ohm
I_Z = 20# mA
I_Z = I_Z * 10**-3# A
# The terminal voltage 
VdasZ = V_Z + (I_Z*r_Z)# V
print "The terminal voltage = %.2f V"%VdasZ

The terminal voltage = 5.00 V

Exa 3.3 page No 165

In [3]:
from __future__ import division
# Given data
P_ZM = 500# mW
P_ZM = P_ZM * 10**-3# W
V_Z = 6.8# V
# The value of I_ZM 
I_ZM = P_ZM/V_Z# A
I_ZM = I_ZM * 10**3# mA
print "The value of I_ZM = %.2f mA"%I_ZM

The value of I_ZM = 73.53 mA

Exa 3.4 page No 165

In [1]:
from __future__ import division
# Given data
P_ZM = 1500# mW
Deratingfactor = 3.33# mW
T1 = 85# degree C
T2 = 60# degree C
total = Deratingfactor*(T1-T2)## total derating factor = %.2f mW
# The maximum power dissipation 
P_ZM = P_ZM - total# mW
print "The maximum power dissipation = %.2f mW"%P_ZM

The maximum power dissipation = 1416.75 mW

Exa 3.5 page No 166

In [8]:
from __future__ import division
# Given data
R_L = 1.2# k ohm
R_L = R_L * 10**3# ohm
Vi = 16# V
R = 1# k ohm
R = R * 10**3# ohm
# The value of V_L 
V_L = (R_L*Vi)/(R+R_L)# V
print "The value of V_L = %.2f V"%V_L
V_Z = 10# V
# The value of V_R 
V_R = Vi-V_L# V
print "The value of V_R = %.2f V"%V_R
# The value of I_Z 
I_Z = 0# A
print "The value of I_Z = %.2f A"%I_Z
# The value of P_Z 
P_Z =V_Z*I_Z# W
print "The value of P_Z = %.2f W"%P_Z

The value of V_L = 8.73 V
The value of V_R = 7.27 V
The value of I_Z = 0.00 A
The value of P_Z = 0.00 W

Exa 3.6 page No 167

In [10]:
from __future__ import division
# Given data
R_L = 1.2# k ohm
R_L = R_L * 10**3# ohm
R = 220# ohm
V_Z = 20# V
# The minimum value of Vi 
Vimin = ((R_L+R)/R_L)*V_Z# V
print "The minimum value of Vi = %.2f V"%Vimin
I_L = V_Z/R_L# A
I_ZM = 60# mA
I_ZM = I_ZM * 10**-3# A
# I_ZM = I_R-I_L
I_Rmax = I_ZM + I_L
# The maximum value of Vi 
Vimax = (I_Rmax*R)+V_Z# V
print "The maximum value of Vi = %.2f V"%Vimax

The minimum value of Vi = 23.67 V
The maximum value of Vi = 36.87 V

Exa 3.7 page No 168

In [11]:
from __future__ import division
# Given data
R_L = 1# k ohm
R_L = R_L * 10**3# ohm
R = 270# ohm 
V = 18# V
V_Z= 10# V
V_L = (R_L/(R_L+R))*V# V
if V_L > V_Z:
    print "As the value of V_L (%.2f"%(V_L),"V) is greater than the value of V_Z (%.2f"%(V_Z)," V), So"
    print "The zener diode is operating in the breakdown region."

As the value of V_L (14.17 V) is greater than the value of V_Z (10.00  V), So
The zener diode is operating in the breakdown region.

Exa 3.8 page No 169

In [12]:
from __future__ import division
# Given data
V1 = 18# V
V2 = 10# V
R = 250# ohm
I_R = (V1-V2)/R# A
I_R = I_R * 10**3# mA
R_L = 1# k ohm
R_L = R_L * 10**3# ohm
I_L = V2/R_L# A
I_L =I_L * 10**3# mA
# I_R = I_L+I_Z
# So, the value of zener current 
I_Z = I_R - I_L# mA
print "The value of zener current = %.2f mA"%I_Z

The value of zener current = 22.00 mA

Exa 3.9 page No 169

In [13]:
from __future__ import division
# Given data
R = 10# k ohm
R = R * 10**3# ohm
Vi = 20# V
V_Z = 10# V
I_L = 1# mA
I_L = I_L * 10**-3# A
R_L = 10# k ohm
R_L = R_L * 10**3# ohm
V_L = (R_L/(R_L+R))*Vi# V
# The current flowing through the resistance 
I = (Vi-V_Z)/R# A
I= I*10**3# mA
print "The current flowing through the resistance = %.2f mA"%I

The current flowing through the resistance = 1.00 mA

Exa 3.10 page No 170

In [14]:
from __future__ import division
# Given data
R = 200#  ohm
Vi = 20# V
V_Z = 10# V
R_L = 300# ohm
P_Zmax = 400# mW
# The value of V_L 
V_L = (R_L/(R_L+R))*Vi# V
print "The value of V_L = %.2f V"%V_L
V_L = V_Z# V
# The value of I_L 
I_L = V_L/R_L# A
I_L = I_L * 10**3# mA
print "The value of I_L = %.2f mA"%I_L
# The value of I_R 
I_R = (Vi-V_Z)/R# A
I_R = I_R * 10**3# mA
print "The value of I_R = %.2f mA"%I_R
# The value of I_Z 
I_Z = I_R-I_L# mA
print "The value of I_Z = %.2f mA"%I_Z
# V_L >= V_Z and V_L= R_L*Vi/(R_L+R)
# So, the minimum value of R_L 
R_L= R*V_Z/(Vi-V_Z)# ohm
print "The minimum value of R_L = %.2f ohm"%R_L

The value of V_L = 12.00 V
The value of I_L = 33.33 mA
The value of I_R = 50.00 mA
The value of I_Z = 16.67 mA
The minimum value of R_L = 200.00 ohm

Exa 3.11 page No 171

In [15]:
from __future__ import division
# Given data
I_Z = 0.2# A
R = 10# ohm
V_Z = 8 + (R*I_Z)# V
I_Lmin = V_Z/R# A
I_Zmax = 0.2##in A
Vimax = 15# V
# The minimum value of R
Rimin = (Vimax-V_Z)/(I_Zmax+I_Lmin)# ohm
print "The minimum value of R = %.2f ohm"%Rimin

# Note: The calculation in the book is not accurate, So the answer in the book is not accurate.

The minimum value of R = 4.17 ohm

Exa 3.12 page No 172

In [17]:
from __future__ import division
# Given data
R_L = 10# k ohm
R_S = 5# k ohm
V_S = 12# V
V_Z = 8# V
V_L = (R_L/(R_S+R_L))*V_S# V
#The output voltage 
Vo = V_L# V
print "The output voltage = %.2f V"%Vo
# The voltage drop across R_S 
R_S = V_S-Vo# V
print "The voltage drop across R_S = %.2f V"%R_S
# The current through the zener diode 
I_Z = 0# A
print "The current through the zener diode = %.2f A"%I_Z

The output voltage = 8.00 V
The voltage drop across R_S = 4.00 V
The current through the zener diode = 0.00 A

Exa 3.13 page No 173

In [26]:
from __future__ import division
# Given data
R_S = 1# k ohm
R_L = 1.2# k ohm
V_Z = 10# V
V_S = 16# V
V_L = (R_L/(R_L+R_S))*V_S# V
#The value of I_Z 
I_Z = 0# A
print "The value of I_Z = %.2f A"%I_Z
# The value of P_Z 
P_Z = 0
print "The value of P_Z = %.2f W"%P_Z
#The value of Vo
Vo = V_L# V
print "The value of Vo = %.2f V"%Vo

The value of I_Z = 0.00 A
The value of P_Z = 0.00 W
The value of Vo = 8.73 V

Exa 3.14 page No 173

In [21]:
from __future__ import division
# Given data
R_L = 200# ohm
Vin = 20# V
V_Z = 10# V
P_Zmaz = 400# mW
R_S = 220# ohm
#The value of V_L with 200 ohm 
V_L =(R_L/(R_S+R_L))*Vin# V
print "The value of V_L with 200 ohm = %.2f V"%V_L
# The value of I_Z with 200 ohm 
I_Z = 0# A
print "The value of I_Z with 200 ohm = %.2f A"%I_Z
# The value of I_L with 200 ohm 
I_L = Vin/(R_S+R_L)*10**3# mA
print "The value of I_L with 200 ohm = %.2f mA"%I_L
#The value of I_R with 200 ohm 
I_R = I_L# mA
print "The value of I_R with 200 ohm = %.2f mA"%I_R
R_L = 50# ohm
V_L = (R_L/(R_S+R_L))*Vin# V
#The value of I_L with 50 ohm 
I_L = Vin/(R_S+R_L)*10**3# mA
print "The value of I_L with 50 ohm = %.2f mA"%I_L
# The value of I_R with 50 ohm 
I_R = I_L# mA
print "The value of I_R with 50 ohm = %.2f mA"%I_R
# The value of I_Z with 50 ohm 
I_Z = 0# A
print "The value of I_Z with 50 ohm = %.2f A"%I_Z
print "The value of V_L with 50 ohm = %.2f V"%V_L

The value of V_L with 200 ohm = 9.52 V
The value of I_Z with 200 ohm = 0.00 A
The value of I_L with 200 ohm = 47.62 mA
The value of I_R with 200 ohm = 47.62 mA
The value of I_L with 50 ohm = 74.07 mA
The value of I_R with 50 ohm = 74.07 mA
The value of I_Z with 50 ohm = 0.00 A
The value of V_L with 50 ohm = 3.70 V

Exa 3.15 page No 175

In [22]:
from __future__ import division
# Given data
V_Z = 15# V
Vin = 24# V
R = 27# ohm
I = (Vin-V_Z)/R# A
# The minimum value of R_L 
R_Lmin = V_Z/I# ohm
print "The minimum value of R_L = %.2f ohm"%R_Lmin

The minimum value of R_L = 45.00 ohm

Exa 3.16 page No 175

In [23]:
from __future__ import division
# Given data
R = 50# ohm
Vin = 10# V
V_Z = 6# V
I = (Vin-V_Z)/R# A
I = I * 10**3# mA
I_Zmin = 5# mA
# I = I_Z+I_L
I_Rmax = I-I_Zmin# mA
# The minimum value of R 
R = V_Z/(I_Rmax*10**-3)# ohm
print "The minimum value of R = %.2f ohm"%R

The minimum value of R = 80.00 ohm

Exa 3.17 page No 176

In [27]:
from __future__ import division
# Given data
R = 150*10**3# ohm
R_L = 75# k ohm
R_L = R_L * 10**3# ohm
V_Z = 15# V
Vin = 50# V
R_Z = 0
Rth = (R*R_L)/(R+R_L)# ohm
Vth = Vin * ( R_L/(R_L+R) )# V
I_Z = Vth/Rth# A
# The power dissipation = %.2f the zener diode 
P_Z = V_Z*I_Z# W
P_Z= P_Z*10**3##in mW
print "The power dissipation in the zener diode = %.2f mW"%P_Z

# Note: The calculation in the last line is wrong as 15*0.333 = 5 mW not 0.5mW, So the answer in the book is wrong. 

The power dissipation in the zener diode = 5.00 mW

Exa 3.18 page No 176

In [28]:
from __future__ import division
# Given data
R = 222# ohm
Vin = 20# V
V_Z = 10# V
P = 400# mW
P= P*10**-3# W
I_Zmax = P/V_Z# A
#I = I_Z+I_L
I = (Vin-V_Z)/R# A
I_Lmin = I - I_Zmax# mA
# The value of R_L 
R_L =V_Z/I_Lmin# ohm
R_L= R_L*10**-3# k ohm
print "The value of R_L = %.2f k ohm"%R_L

The value of R_L = 1.98 k ohm

Exa 3.19 page No 177

In [29]:
from __future__ import division
# Given data
R = 12# k ohm
Vin = 6.3# V
V_Z = 4.8# V
I = (Vin-V_Z)/R*10**3# mA
I_Z = 5# mA
# Maximum value of load current,
I_Lmax = I - I_Z# mA
I_Z = 100# mA
# Minimum value of load current,
I_Lmin = I - I_Z# mA
print "The range of possible load current is : %.2f"%(I_Lmin),"mA <= I_L <= %.2f"%(I_Lmax),"mA"
# Minimum value of load resistance,
R_Lmin= I_Lmin*10**-3*V_Z# ohm
# Maximum value of load resistance,
R_Lmax= I_Lmax*10**-3*V_Z# ohm
print "The range of possible load resistance is : %.2f"%(R_Lmin),"ohm <= R_L <= %.2f"%(R_Lmax)," ohm"
# The power rating required for load resistance 
P_Zmax = I_Lmax*10**-3*V_Z# W
P_Zmax= P_Zmax*10**3# mW
print "The power rating required for load resistance = %.2f mW"%P_Zmax

# Note: The calculated value of P_Zmax is wrong as 120*10**-3*4.8= 576 mW not 5.76 mW

The range of possible load current is : 25.00 mA <= I_L <= 120.00 mA
The range of possible load resistance is : 0.12 ohm <= R_L <= 0.58  ohm
The power rating required for load resistance = 576.00 mW