# Chapter 2 : Special Purpose Diodes¶

## Example 2.1, Page No. 68¶

In [6]:
# maximum current

import math
#Variable declaration
Pmax=364.0               #dissipation in milliwatt
Vz=9.1                   #in V

#Calculations
Izmax=Pmax/Vz            #in mA

#Result
print("maximum current the diode can handle is ,(mA)= %.f"%Izmax)

maximum current the diode can handle is ,(mA)= 40


## Example 2.2, Page No. 68¶

In [5]:
# resistance

import math
#Vaariable declaration
mip=15.0                  #in volt
op=6.8                    #output potential in volt
pd=mip-op                 #potential difference across series resistor
nmip=20                   #new maximum input voltage in volt
pd1=nmip-op               #new potential difference across series resistor
Il1=20                    #new load current in mA

#Calculations
R=((pd1-pd)/((Il1-Il)*10**-3))
print("value of series resistance is,(ohm)= %.1f"%R)

value of series resistance is,(ohm)= 333.3


## Example 2.3, Page No.69¶

In [8]:
# Current

import math
#VAriable declaration
V=120.0              #in V
Vz=50.0              #in V
R=5.0                #in ohm
Rl=10.0              #in k-ohm

#Calculations
vd5=V-Vz             #voltage drop across 5 ohm resistor
I5=vd5/R             #current through 5 ohm resistor

current through zener diode is ,(A)= 13.995