Chapter 27 : Diodes and Diodes Applications

Example No. 27_1 Page No. 864

In [1]:
# For the diode curve, calculate the dc resistance, RF, at points A and B.

# Given data

Vf1 = 0.65#         # Forward votage 1=0.65 Volts
If1 = 11*10**-3      # Forward current 1=11 mAmps
Vf2 = 0.7#          # Forward votage 2=0.7 Volts
If2 = 22.5*10**-3    # Forward current 2=22.5 mAmps

Rf1 = Vf1/If1#
print 'The Forward Resistance at Point A = %0.2f Ohms'%Rf1
print 'Approx 59.1 Ohms'

Rf2 = Vf2/If2#
print 'The Forward Resistance at Point B = %0.2f Ohms'%Rf2

The Forward Resistance at Point A = 59.09 Ohms
Approx 59.1 Ohms
The Forward Resistance at Point B = 31.11 Ohms

Example No. 27_2 Page No. 865

In [2]:
# A silicon diode has a forward voltage drop of 1.1 V for a forward diode current, If, of 1 A. Calculate the bulk resistance, Rb.

# Given data

Vf1 = 1.1#  # Forward votage 1=1.1 Volts
If1 = 1.     # Forward current 1=1 Amps
Vf2 = 0.7#  # Fwd. vltg. 2=0.7 Volts (min working vltg of diode is 0.7 V)
If2 = 0     # Forward current=0 Amps

delV = Vf1-Vf2#     # diff. between max. min. Voltages
delI = If1-If2#     # diff. between max. min. Currents

Rb = delV/delI#
print 'The Bulk Resistance = %0.2f Ohms'%Rb

The Bulk Resistance = 0.40 Ohms

Example No. 27_3 Page No. 868

In [3]:
# Solve for the load voltage and current using the first, second, and third diode approximations.

# Given data

Rl = 100.#       # Load resistance=100 Ohms
Rb = 2.5#       # Resistance=2.5 Ohms
Vin = 10.#       # Input voltage=10 Volts
Vb = 0.7#       # Voltage=0.7 Volts


# Using first approximation

Vl1 = Vin
print 'The Load Voltage of First Approximation = %0.2f Volts(dc)'%Vl1

Il1 = Vl1/Rl#
print 'The Load Current of First Approximation = %0.2f Amps'%Il1
print 'i.e 100 mAmps'

# Using second approximation

Vl2 = Vin-Vb
print 'The Load Voltage of Second Approximation = %0.2f Volts'%Vl2

Il2 = Vl2/Rl#
print 'The Load Current of Second Approximation = %0.2f Amps'%Il2
print 'i.e 93 mAmps'

# Using third approximation

Il3 = (Vin-Vb)/(Rl+Rb)#
print 'The Load Current of Third Approximation = %0.2f Amps'%Il3
print 'i.e 90.73 mAmps'

Vl3 = Il3*Rl#
print 'The Load Voltage of Third Approximation %0.2f Volts'%Vl3

The Load Voltage of First Approximation = 10.00 Volts(dc)
The Load Current of First Approximation = 0.10 Amps
i.e 100 mAmps
The Load Voltage of Second Approximation = 9.30 Volts
The Load Current of Second Approximation = 0.09 Amps
i.e 93 mAmps
The Load Current of Third Approximation = 0.09 Amps
i.e 90.73 mAmps
The Load Voltage of Third Approximation 9.07 Volts

Example No. 27_4 Page No. 875

In [4]:
# If the turns ratio Np:Ns is 3:1, calculate the following: Vs, Vdc, Il, Idiode, PIV for D1, and fout.

# Given data

Vp = 120.#       # Primary voltage=120 Vac
A = 3/1.#        # Turns ratio Np:Ns=3:1
B = 1/3.#        # Turns ratio Ns:Np=1:3
Rl = 100.#       # Load resistance=100 Ohms
fi = 60.#        # Input frequency=60

Vs = B*Vp#
print 'The Secondary Voltage = %0.2f Volts(ac)'%Vs

Vspk = (Vs*1.414)#

C = Vspk-0.7#

Vdc = 0.318*C#
print 'The DC Voltage = %0.2f Volts'%Vdc

Il = Vdc/Rl#
print 'The Load Current = %0.2f Amps'%Il

Idiode = Il#
print 'The DC Diode Current = %0.2f Amps'%Idiode

PIV = Vspk#
print 'The PIV for Diode-1 = %0.2f Volts'%PIV

fo =fi#
print 'The Output Frequency = %0.2f Hertz'%fo

The Secondary Voltage = 40.00 Volts(ac)
The DC Voltage = 17.76 Volts
The Load Current = 0.18 Amps
The DC Diode Current = 0.18 Amps
The PIV for Diode-1 = 56.56 Volts
The Output Frequency = 60.00 Hertz

Example No. 27_5 age No. 878

In [5]:
# If the turns ratio Np:Ns is 3:1, calculate the following: Vdc, Il, Idiode, PIV for D1, and fout.

# Given data

Vp = 120.       # Primary voltage=120 Vac
A = 3/1.        # Turns ratio Np:Ns = 3:1
B = 1/3.        # Turns ratio Ns:Np = 1:3
Rl = 100.       # Load resistance=100 Ohms

Vs = B*Vp#
Vspk = 1.414*(Vs/2)#
Vopk = Vspk-0.7#

Vdc = 0.636*Vopk#
print 'The DC Voltage = %0.2f Volts'%Vdc

Il = Vdc/Rl#
print 'The Load Current = %0.2f Amps'%Il
print 'i.e 175.4 mAmps'

Idiode = Il/2#
print 'The DC Diode Current = %0.2f Amps'%Idiode
print 'i.e 87.7 mAmps'

C = (Vspk*2)-0.7#

PIV = C#
print 'The PIV for Diode-1 = %0.2f Volts'%PIV

f =120#
print 'The Output Frequency = %0.2f Hertz'%f

The DC Voltage = 17.54 Volts
The Load Current = 0.18 Amps
i.e 175.4 mAmps
The DC Diode Current = 0.09 Amps
i.e 87.7 mAmps
The PIV for Diode-1 = 55.86 Volts
The Output Frequency = 120.00 Hertz

Example No. 27_6 Page No. 880

In [7]:
# If the turns ratio Np:Ns is 3:1, calculate the following: Vdc, Il, Idiode, PIV for each diode, and fout.

# Given data

Vp = 120.#       # Primary voltage=120 Vac
A = 3./1#        # Turns ratio Np:Ns = 3:1
B = 1./3#        # Turns ratio Ns:Np = 1:3
Rl = 100.#       # Load resistance=100 Ohms

Vs = B*Vp#
Vspk = 1.414*(Vs)#
Vopk = Vspk-1.4#

Vdc = 0.636*Vopk#
print 'The DC Voltage = %0.2f Volts'%Vdc

Il = Vdc/Rl#
print 'The Load Current = %0.2f Amps'%Il
print 'i.e 350.8 mAmps'

Idiode = Il/2#
print 'The DC Diode Current = %0.4f Amps'%Idiode
print 'i.e 175.4 mAmps'

C = Vspk-0.7#

PIV = C#
print 'The PIV for each Diode = %0.2f Volts'%PIV

f =120#
print 'The Output Frequency = %0.2f Hertz'%f

The DC Voltage = 35.08 Volts
The Load Current = 0.35 Amps
i.e 350.8 mAmps
The DC Diode Current = 0.1754 Amps
i.e 175.4 mAmps
The PIV for each Diode = 55.86 Volts
The Output Frequency = 120.00 Hertz

Example No. 27_7 Page No. 883

In [11]:
from math import exp
# Assume the transformer turns ratio Np:Ns = 4:1 in Fig. 27–21 a and 2:1 in Fig. 27–22a. Compare Vripple and Vdc if C = 500 uF and Rl = 250.
    
# Given data

A1 = 4./1#           # Turns ratio Np:Ns=4:1
B1 = 1./4#           # Turns ratio Ns:Np=1:4
A2 = 2./1#           # Turns ratio Np:Ns=2:1
B2 = 1./2#           # Turns ratio Ns:Np=1:2
Vp = 120.#           # Primary voltage=120 Vac
Vb = 0.7#           # 
t1 = 16.67*10**-3#   # Charging Time of Capacitor of Turns ratio Np:Ns=4:1=16.67 mSec
t2 = 8.33*10**-3#    # Charging Time of Capacitor of Turns ratio Np:Ns=4:1=8.33 mSec
Rl = 250.#           # Load resistance=250 Ohms
C = 500.*10**-6#      # Capacitor=500 uFarad

# Calculations for Turns Ratio Np:Ns=4:1

Vs1 = B1*Vp#
Vspk1 = Vs1*1.414#
Vopk1 = Vspk1 - Vb#
D = -t1/(Rl*C)#

Vrp1 = Vopk1*(1-exp(D))#
print 'The Ripple Voltage for Turns Ratio Np:Ns=4:1 = %0.2f Volts(p-p)'%Vrp1
print 'Approx 5.21 Volts(p-p)'

Vdc1 = Vopk1-(Vrp1/2)#
print 'The DC Voltage for Turns Ratio Np:Ns=4:1 = %0.2f Volts'%Vdc1
print 'Approx 39.12 Volts'

# Calculations for Turns Ratio Np:Ns = 2:1

Vs2 = B2*Vp#
V2 = Vs2/2#
V2pk2 = V2*1.414
Vopk2 = V2pk2 - Vb#
E = -t2/(Rl*C)#

Vrp2 = Vopk2*(1-(exp(E)))
print 'The Ripple Voltage for Turns Ratio Np:Ns=2:1 = %0.2f Volts(p-p)'%Vrp2
print 'Approx 2.69 Volts(p-p)'

Vdc2 = Vopk2-(Vrp2/2)#
print 'The DC Voltage for Turns Ratio Np:Ns=2:1 = %0.2f Volts'%Vdc2
print 'Approx 40.38 Volts'

The Ripple Voltage for Turns Ratio Np:Ns=4:1 = 5.21 Volts(p-p)
Approx 5.21 Volts(p-p)
The DC Voltage for Turns Ratio Np:Ns=4:1 = 39.12 Volts
Approx 39.12 Volts
The Ripple Voltage for Turns Ratio Np:Ns=2:1 = 2.69 Volts(p-p)
Approx 2.69 Volts(p-p)
The DC Voltage for Turns Ratio Np:Ns=2:1 = 40.38 Volts
Approx 40.38 Volts

Example No. 27_8 Page No. 885

In [12]:
#  Calculate the LED current.

# Given data

Vin = 24.#       # Input voltage=24 Volts
Vled = 2.#       # Voltage drop at LED=2 Volts
Rs = 2.2*10**3#  # Source Resistance=2.2 kOhms

Iled = (Vin-Vled)/Rs#
print 'The LED Current = %0.2f Amps'%Iled
print 'i.e 10 mAmps'

The LED Current = 0.01 Amps
i.e 10 mAmps

Example No. 27_9 Page No. 888

In [13]:
# Calculate the resistance Rs, required to provide an LED current of 25 mA.

# Given data

Vin = 24.#           # Input voltage=24 Volts
Vled = 2.#           # Voltage drop at LED=2 Volts
Iled = 25.*10**-3#    # LED Current=25 mAmps

Rs = (Vin-Vled)/Iled#
print 'The Resistance Rs, Required to Provide an LED Current of 25 mA = %0.2f Ohms'%Rs

The Resistance Rs, Required to Provide an LED Current of 25 mA = 880.00 Ohms

Example No. 27_10 Page No. 889

In [14]:
# Calculate the maximum rated zener current for a 1 W, 10 V zener.

# Given data

Pzm = 1.#        # Power rating of zener= 1 Watts
Vz = 10.#        # Voltage rating of zener= 10 Volts

Izm = Pzm/Vz#
print 'The Maximum Rated Current of Zener = %0.2f Amps'%Izm
print 'i.e 100 mAmps'

The Maximum Rated Current of Zener = 0.10 Amps
i.e 100 mAmps

Example No. 27_11 Page No. 890

In [15]:
# If Vz=10 V, calculate Iz.

# Given data

Vin = 25.#       # Input voltage=25 Volts
Vz = 10.#        # Zener voltage=10 Volts
Rs = 1.*10**3#    # Source Resistance=1 kOhms

Iz = (Vin-Vz)/Rs#
print 'The Zener Current = %0.2f Amps'%Iz
print 'i.e 15 mAmps'

The Zener Current = 0.01 Amps
i.e 15 mAmps

Example No. 27_12 Page No. 891

In [20]:
# If R L increases to 250 Ohms, calculate the following: Is, Il, Iz, and Pz.

# Given data

Vin = 25#       # Input voltage=25 Volts
Vz = 7.5#        # Zener voltage=7.5 Volts
Rl = 250#    # Load Resistance=250 Ohms
Is = 75*10**-3#  # Source current=75 mAmps

print 'The Source Current = %0.3f Amps'%Is
print 'i.e 75 mAmps'

Il = Vz/Rl#
print 'The Load Current = %0.2f Amps'%Il
print 'i.e 30 mAmps'

Iz = Is-Il#
print 'The Zener Current = %0.3f Amps'%Iz
print 'i.e 45 mAmps'

Pz = Vz*Iz#
print 'The Power Dissipation of Zener = %0.4f Watts'%Pz
print 'i.e 337.5 mWatts'

The Source Current = 0.075 Amps
i.e 75 mAmps
The Load Current = 0.03 Amps
i.e 30 mAmps
The Zener Current = 0.045 Amps
i.e 45 mAmps
The Power Dissipation of Zener = 0.3375 Watts
i.e 337.5 mWatts

Example No. 27_13 Page No. 892

In [21]:
# Calculate Is, Il and Iz for (a)Rl=200 ohms# (b)Rl=500 ohms.

# Given data

Vin = 16.#       # Vin=16 Volts given
Vz = 10.#        # Vz=10 Volts given
Rs = 100.#       # Source Resistance = 100 ohms given
Rla = 200.#      # Load Resistance A = 200 ohms given
Rlb = 500.#      #Load Resistance B = 500 ohms given

# For Rl 200 ohms

Is = (Vin-Vz)/Rs#
print 'The Source Current = %0.2f Amps.'%Is
print 'i.e 60 mAmps'

Ila = Vz/Rla#
print 'The Load Current for 200 ohms Load = %0.2f Amps.'%Ila
print 'i.e 50 mAmps'

Iza = Is-Ila
print 'The Zener Current for 200 ohms Load = %0.2f Amps.'%Iza
print 'i.e 10 mAmps'

# For Rl 500 ohms

Ilb = Vz/Rlb#
print 'The Load Current for 500 ohms Load = %0.2f Amps.'%Ilb
print 'i.e 20 mAmps'

Izb = Is-Ilb
print 'The Zener Current for 500 ohms load = %0.2f Amps.'%Izb
print 'i.e 40 mAmps'

The Source Current = 0.06 Amps.
i.e 60 mAmps
The Load Current for 200 ohms Load = 0.05 Amps.
i.e 50 mAmps
The Zener Current for 200 ohms Load = 0.01 Amps.
i.e 10 mAmps
The Load Current for 500 ohms Load = 0.02 Amps.
i.e 20 mAmps
The Zener Current for 500 ohms load = 0.04 Amps.
i.e 40 mAmps