01-pn junction and diodes
TRANSCRIPT
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EE
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112 Basic Electronics Dr. Faraz Akram
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112 Basic Electronics Dr. Faraz Akram
• Electronic Devices: Conventional Current
Version, by Thomas L. Floyd
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112 Basic Electronics Dr. Faraz Akram
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Insulators:
• Material that does not conduct electrical current under normal condition•
Valence electrons are tightly bound to the atoms; therefore very few free electrons
•
e.g. rubber, plastic, glass, wood
Conductors:
• Material
that
easily conducts
electrical
current
• Valence electrons are
loosely
bound to the
atoms
• Copper, silver,
g
old, aluminum•
One valance electron, very loosely bound to atom
Semiconductors:
•
Material
that is between conductors and
insulators in
its ability to conduct electricalcurrent
•
Neither
a good conductor nor a good
insulator in its pure state
• Silicon, germanium•
Four valance electrons
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112 Basic Electronics Dr. Faraz Akram
Insulator Semiconductor Conductor
(a) (b) (c)
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Bohr diagram of the silicon (Semiconductor) and copper(Conductor) atoms
• The valence electron in the silicon atom “feels” an attractive force of +4
• The valence electron in the copper atom “feels” an attractive force of +1
Core has a net charge of
+4 (14p
-
10e)
net charge +
1
(29p
-
28e)
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•
Intrinsic Semiconductor:• Pure semiconductor containing no impurities
• Extrinsic Conductors
– Impure semiconductors
» P-type
» N-type
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• Both silicon and germanium have the characteristic four valence electrons.
• The valence electrons in germanium are in the fourth shell while those in silicon
are in the third shell, closer to the nucleus.
• This means that the germanium valence electrons require a smaller amount of
additional energy to escape from the atom.
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Energy band diagram for an unexcited (no external energy such as heat) atom in a
pure silicon crystal. This condition occurs only at a temperature of absolute 0 Kelvin.
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An intrinsic (pure) silicon crystal at room temperature has sufficient
heat (thermal) energy for some valence electrons to jump the gap from
the valence band into the conduction band, becoming free electrons.
Free electrons are also called conduction electrons .
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Dopping
Two categories of impurities
• n-type Adding Group V dopant such as arsenic (As), phosphorus
(P), bismuth (Bi), and antimony (Sb)
•p-type Adding Group III dopant such as boron(B), Indium (In),
gallium (Ga)
*Doping: Adding impurities to intrinsic semiconductors
Doping increases the number of current carriers (Electronsor holes)
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N-Type Material
N-Type Material:
An n-type material is produced when extra
valence electrons are introduced into a
material (silicon) by putting impurities or
dopants (Group V elements) into thesilicon.
The diagram shows the extra electron that
will be present when a Group V dopant is
introduced to a material such as silicon.This extra electron is very mobile.
+4+4
+5
+4
+4+4+4
+4+4
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P-Type Material
P-Type Material:
P-type material is produced when the dopant
that is introduced is from Group III. Group
III elements have only 3 valence electrons
and therefore there is an electron missing.
This creates a hole (h+), or a positive charge
that can move around in the material.
The diagram shows the hole that will be
present when a Group III dopant is
introduced to a material such as silicon.This hole is quite mobile in the same way the
extra electron is mobile in a n-type material.
+4+4
+3
+4
+4+4+4
+4+4
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p-n junction is the arrangement obtained when a p-type semiconductor is
joined to a n-type semiconductor
P-N Junction
Note:
P has a surplus of holes.
N has a surplus of electrons.
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As soon as p-n junction is formed electrons from n type material and
holes from p-type material diffuse into p-type and n-type material
respectively.
P-N Junction
Electrons and holes combine at junction.
Each recombination eliminates as electron and a hole.
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As a result
• n-region loses free electrons, this creates a layer of positive charge
near the junction.
• p-region loses holes as the electrons and hole combine. This
creates a layer of negative charge near the junction.
Depletion region:
The region of the p-n junction containing the uncompensated
receptor and donor ions is known as depletion region
The Width of
depletion region isExaggerated for
illustration
Note
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P N
+
+
+-
-
-
-
-
- +
+
+
Diode
Diodes are electronic components functions as a one-way valve
it means it allow current to flow in one direction.
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DiodeA diode is made from a small piece of semiconductor material,
usually silicon, in which half is doped as a p-region and half is
doped as an n-region with a PN junction and depletion region inbetween.
The p region is called the anode and is connected to a
conductive terminal. The n region is called the cathode and is
connected to a second conductive terminal
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Forward BiasPositive terminal of battery to P region
Negative terminal of the battery to N region
+
+
+-
-
-
-
-
- +
+
+
+
+
+
+
+
+
+
+
+ -
-
-
-
-
-
-
-
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N-typeP-type
+ -
+ -
Vapplied
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Forward Bias
Effect of forward bias on depletion region
+
+
+-
-
-
-
-
- +
+
+
+
+
+
+
+
+
+
+
+ -
-
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-
-
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N-typeP-type
+ -
+ -
Positive terminal
repels holes
Negative terminal
repels electrons
Vapplied
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Forward Bias
Due to this repulsion, the depletion region narrows down
+
+
+-
-
-+
+
+
+
+
+
+
+
+ -
-
-
-
-
-
-
-
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N-typeP-type
+ -
+ -
+
+
+ -
-
-
Vapplied
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Forward BiasIf the voltage is above a specified range, electrons in the N-region
drifts through the junction and migrates to the P region. And
holes from p region migrate to N- region.
+
+
+-
-
-+
+
+
+
+
+
+
+
+ -
-
-
-
-
-
-
-
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N-typeP-type
+ -
+ -
+
+
+ -
-
-
Now the current flows across the circuit
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Reverse Bias
Positive terminal of battery to N region
Negative terminal of the battery to P region
+
+
+-
-
-
-
-
- +
+
+
+
+
+
+
+
+
+
+
+ -
-
-
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-
-
-
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N-typeP-type
- +
- +
Vapplied
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Reverse Bias
+
+
+-
-
-
-
-
- +
+
+
+
+
+
+
+
+
+
+
+ -
-
-
-
-
-
-
-
-
N-typeP-type
- +
- +
Holes get attracted to
Negative terminal of the
battery
Electrons get attracted to
the positive terminal of
the battery
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Reverse Bias
As a result, the depletion region increases
+
+
+-
-
-
-
-
- +
+
+
+
+
+
+
+
+
+
+
+ -
-
-
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-
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N-typeP-type
- +
- +-
-
-
Now the PN junction acts as an insulator and will not allow
any current to flow in the circuit
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Forward Bias vs Reverse Bias
Reverse BiasForward Bias
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• What is an Ideal Diode?
An ideal diode is a diode that acts like a perfect conductor when
voltage is applied forward biased and like a perfect insulator
when voltage is applied reverse biased.
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Voltage Current (v-i) Characteristics
of a Diode
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1. Ideal Diode Model2. Practical Diode Model
3. Complete Diode Model
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1
Ideal Diode Model
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1-Ideal Diode Model
The Ideal Diode
Model• Least accurate approximation and can be
represented by a simple switch.
• In many situations, using the ideal diodeapproximation is acceptable.
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Example: Ideal Diode Model
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Example: Ideal Diode Model
Example: Assume the diode in the circuit below is ideal. Determine the
value of ID ifa) VA = 5 volts (forward bias) and
b) VA = -5 volts (reverse bias)
+
_ VA
ID
RS = 50
a) With VA > 0 the diode is in forward bias
and is acting like a perfect conductor so:
ID = VA /RS = 5 V / 50 = 100 mA
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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2
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Practical Diode Model
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2-Practical Diode Model
The Ideal Diode with
Barrier PotentialThis model is more accurate than the simple
ideal diode model because it includes the
approximate barrier potential voltage.Remember the barrier potential voltage is the
voltage at which appreciable current starts to
flow.
+
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Example: Practical Diode Model
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Example: Practical Diode Model
Example: To be more accurate than just using the ideal diode model
include the barrier potential. Assume VF = 0.3 volts (typical for agermanium diode) Determine the value of ID if VA = 5 volts (forward bias).
With VA > 0 the diode is in forward bias and
is acting like a perfect conductor so write a
KVL equation to find ID:
0 = VA – IDRS - VF
=
−
=
.7
= 94
+
_ VA
ID
RS = 50
VF
+
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3
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Complete
Diode Model
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3-Complete Diode Model
• This model is the most accurate of the three. It Includes
• Barrier potential
• Small forward dynamic resistance ( )
• Large internal reverse resistance ( )
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Example:
Complete
Diode Model
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Example: Complete Diode Model
Example: Assume the diode with a forward resistance value of 5 ohms.
The barrier potential voltage is still: VF
= 0.3 volts. Determine the value
of ID if VA = 5 volts.
+
_ VA
ID
RS = 50
VF
+
r d
Once again, write a KVL equation
for the circuit:
0 = VA – IDRS - VF - IDr d
ID = VA - VF = 5 – 0.3 = 85.5 mA
RS + r d 50 + 5
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Diode Circuit Models
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Diode Circuit Models
Values of ID for the Three Different Diode Circuit Models
Ideal Diode
Model
Practical Diode
Model
Complete Diode
Model
ID 100 mA 94 mA 85.5 mA
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Example
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Example
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