# Chapter 4 Diode Applications¶

## Example 4.1 Page No 83¶

In :
# given data
V2rms= 40.0##  V
R_L= 20.0##  Ω
V2peak= V2rms/0.707##  V
Vout_peak= V2peak##  V
# The dc voltage across the load resistor
Vdc=0.318*Vout_peak##  V
#The peak inverse voltage across the diode
PIV= V2peak##  V
Idc= Vdc/R_L##  A
# The dc current through the diode
I_diode= Idc##  A
print "The dc voltage across the load resistor = %.2f volts"%Vdc
print "The peak inverse voltage across the diode = %.2f volts"%PIV
print "The dc current through the diode = %.2f A"%I_diode

The dc voltage across the load resistor = 17.99 volts
The peak inverse voltage across the diode = 56.58 volts
The dc current through the diode = 0.90 A


## Example 4.2 Page No 86¶

In :
# given data
Vrms= 40.0##  V
R_L= 20.0##  Ω
V2peak= Vrms/0.707##  V
Vout_peak= V2peak/2##  V
# The dc load voltage
Vdc=0.636*Vout_peak##  V
# The peak inverse voltage across each diode
PIV= V2peak##  V
Idc= Vdc/R_L##  A
# The dc current through each diode
I_diode= Idc/2##  A
print "The dc load voltage = %.2f volts"%Vdc
print "The peak inverse voltage across each diode = %.2f volts"%PIV
print "The dc current through each diode = %.2f A"%I_diode

The dc load voltage = 17.99 volts
The peak inverse voltage across each diode = 56.58 volts
The dc current through each diode = 0.45 A


## Example 4.3 Page No 88¶

In :
# given data
Vrms= 40.0##  V
R_L= 20.0##  Ω
V2peak= Vrms/0.707##  V
Vout_peak= V2peak##  V
# The value of Vdc
Vdc=0.636*Vout_peak##  V
# The value of PIV
PIV= V2peak##  V
Idc= Vdc/R_L##  A
#The value of I_diode
I_diode= Idc/2##  A
print "The dc load voltage = %.2f volts"%Vdc
print "The peak inverse voltage across each diode = %.2f volts"%PIV
print "The dc current through each diode = %.2f A"%I_diode

The dc load voltage = 35.98 volts
The peak inverse voltage across each diode = 56.58 volts
The dc current through each diode = 0.90 A


## Example 4.6 Page No 94¶

In :
# given data
Vdc= 56.6##  V
R_L= 100.0##  Ω
f=120.0##  Hz
C= 1000.0##  µF
C= C*10**-6##  F
V2peak= Vdc##  V
Idc= Vdc/R_L##  A
# The peak-to-peak ripple
Vrip= Idc/(f*C)##  V
# The dc load voltage
Vdc= V2peak-Vrip/2##  V
print "The peak-to-peak ripple = %.2f volts"%Vrip
print "The dc load voltage = %.2f volts"%Vdc

The peak-to-peak ripple = 4.72 volts
The dc load voltage = 54.24 volts


## Example 4.8 Page No 98¶

In :
# given data
V2rms= 12.6##  V
V_Z= 6.8##  V
V2peak= V2rms/0.707##  V
Vin= V2peak##  V
Vout= V_Z##  V
R_L= 1.2##  kΩ
R_L= R_L*10**3##in Ω
Rs= 1##  kΩ
Rs= Rs*10**3##  Ω
Is= (Vin-Vout)/Rs##  A
I_L= Vout/R_L##  A
# The zener current
Iz= Is-I_L##  A
Iz= Iz*10**3##  mA
print "The zener current = %.2f mA"%Iz
# Note: The calculation = %.2f the book is not accurate.

The zener current = 5.36 mA


## Example 4.9 Page No 98¶

In :
# given data
C= 100.0##in µF
C= C*10**-6##  F
Rz= 5.0##in Ω
Rs= 1.0*10**3##in Ω
Idc= 11.0*10**-3##in A
f=120.0##in Hz
Vin_rip= Idc/(f*C)##  V
# The ripple across the load resistance
Vout_rip= Rz*Vin_rip/(Rs+Rz)##in A
Vout_rip= Vout_rip*10**3##  mV
print "The ripple across the load resistance = %.2f mV"%Vout_rip

The ripple across the load resistance = 4.56 mV