# Chapter 2 Rectifier Diodes¶

## Example 2.1 Page No 29¶

In [2]:
# given data
Vin= 15.0##  V
R_L= 10.0##  kΩ
# The output voltage
Vout= Vin ##  V
# The current
I= Vout/R_L##  mA
print "The output voltage = %.2f volts"%Vout
print "The current = %.2f mA"%I

The output voltage = 15.00 volts
The current = 1.50 mA


## Example 2.2 Page No 30¶

In [6]:
# given data
Vin= 15.0##  V
I=0#
R_L= 10.0##  kΩ
R_L= R_L*10**3##  Ω
# The output voltage
Vout= I*R_L##  V
# The voltage across the diode
V_R= Vin-Vout##  V
print "The output voltage = %.f volts"%Vout
print "The voltage across the diode = %.2f volts"%V_R

The output voltage = 0 volts
The voltage across the diode = 15.00 volts


## Example 2.4 Page No 31¶

In [9]:
# given data
Vin= 15.0##  V
V_P= Vin##  V
R_L= 10##  kΩ
R_L= R_L*10**3##  Ω
Vout=0#
# The peak current through the diode
I_P= V_P/R_L##  A
# The maximum reverse voltage
V_R= Vin-Vout##  V
I_P= I_P*10**3##  mA
print "The peak current through the diode = %.2f mA"%I_P
print "The maximum reverse voltage = %.2f volts"%V_R

The peak current through the diode = 1.50 mA
The maximum reverse voltage = 15.00 volts


## Example 2.5 Page No 33¶

In [10]:
# given data
Vin= 15.0##  V
V_K= 0.7##  V
R_L= 10.0##  kΩ
R_L= R_L*10**3##  Ω
# The output voltage
Vout= Vin-V_K##  V
# The current
I= Vout/R_L##  A
# The power dissipation of the diode
P= V_K*I##  W
I=I*10**3##  mA
P= round(P*10**3)##  mW
print "The output voltage = %.2f volts"%Vout
print "The current = %.2f mA"%I
print "The power dissipation of the diode = %.2f mW"%P

The output voltage = 14.30 volts
The current = 1.43 mA
The power dissipation of the diode = 1.00 mW


## Example 2.6 Page No 34¶

In [11]:
# given data
Vin= 15.0##  V
V_K= 0.7##  V
Vout=0##  V
R_L= 10.0##  kΩ
R_L= R_L*10**3##  Ω
# The peak output voltage
V_P= Vin-V_K##  V
# The maximum forward current
I_P= V_P/R_L##  A
# The peak inverse voltage
PIV= Vin-Vout##  V
I_P= I_P*10**3##  mA
print "The peak output voltage = %.2f volts"%V_P
print "The maximum forward current = %.2f mA"%I_P
print "The peak inverse voltage = %.2f volts"%PIV

The peak output voltage = 14.30 volts
The maximum forward current = 1.43 mA
The peak inverse voltage = 15.00 volts


## Example 2.7 Page No 36¶

In [1]:
# given data
Vin= 10.0##  V
V_K= 0.7##  V
Vout=0##  V
R_L= 1000.0##  kΩ
r_B= 20.0##  Ω
# The peak forward current,
I_P= (Vin-V_K)/(R_L+r_B)##  A
# The peak voltage
V_P= I_P*R_L##  V
# The peak inverse voltage
PIV= Vin-Vout##  V
print "The peak voltage = %.2f volts"%V_P
print "The peak inverse voltage = %.2f volts"%PIV

The peak voltage = 9.12 volts
The peak inverse voltage = 10.00 volts