This document discusses diodes and their applications. It begins by defining a diode as an electronic component that allows current to flow in only one direction. The main application of diodes is rectification, where an alternating current is converted to direct current. The document describes different types of diodes and how they work in forward and reverse bias. Examples are provided to calculate current flow through an ideal diode under forward and reverse bias conditions.

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What is a Diode?
An electronic component that allows the current flow in one
direction only. It has with two electrodes called the anode and the
cathode. Most diodes are made with semiconductor materials such
as silicon, germanium.
 Application of diode
• Rectification :-
A rectifier changes AC (Alternating Current) to DC (Direct Current(
This is the most important application of diodes.
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4. Types of Diodes
• Normal Diodes
• Light-Emitting Diodes (LEDs)
• Schottky Diodes
• Zener Diodes
4
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How a diode works
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Forward Operation
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In forward operation the anode connected to the positive
electrode of the source and the cathode connected to the
negative electrode of the source. In this situation the current
flows in the circuit.
The current flows in the
Circuit.

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Reverse Operation
In reverse operation the anode connected to the
negative electrode of the source and the cathode
connected to the positive electrode of the source.
No current ( very small ) flows in
in the circuit.
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curve for a real diode.

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Forward and Reverse Operation
• Practical Exercise :-

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Partner Work

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Partner Work
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In short sentence:-
Q1 :- Define a diode ?
 A diode is a electronic component
that allows the current flow in one direction
only.
Q2 :- Draw a symbol of diode ?
Q3 :- what is the difference between
forward biased and reverse biased ?
In forward biased the current flows in the
circuit.
In reverse biased no current flows in the
circuit.

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Example ( 1 ) :-Example ( 1 ) :-
Assume the diode in the circuit below is ideal.
Determine the value of ID if :
a) VA = 5 volts (Forward Bias)
b) VA = -5 volts (Reverse Bias)

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Solution:-
a) With VA > 0 the diode is in
forward bias and is acting like a
perfect conductor so:
ID = VA / RS = 5 V / 50 ΩΩ =
0.1 A
b) With VA < 0 the diode is in
reverse bias and is acting like a
perfect insulator, therefore no
current can flow and ID = 0.

15. Group Work
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Group Work
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Assume the diode in the circuit below is ideal. Determine the value
of ID if :
a) VA = 10 v (Forward Bias)
b) VA = -10 v (Reverse Bias)

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Group Work
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Solution:-
a) With VA > 0 the diode is in forward
bias and is acting like a perfect
conductor so:
ID = VA / RS = 10 V / 80 =
0.125 A
b) With VA < 0 the diode is in reverse
bias and is acting like a perfect
insulator, therefore no current can flow
and ID = 0.

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