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FIRST EDDITION

SEMICONDUCTORS

CONDUCTION IN SEMICONDUCTORS
Electronic devices are practically, almost based on the study of semiconductor
devices, due to their unique electrical properties which vary between that of
conductors and insulators
...
Also, are capable
of amplifying weak signals
...


Theory of Semiconductors
The Atomic Structure
Atoms are the basic structures of all fundamental building blocks
...

Each element is unique because that common structure of its atoms is also unique
...
A complete nucleus is surrounded by negatively charged particles called electrons,
these electrons are always orbiting or revolving around in region of space called orbitals
...
Since the positive
charge on each protons is equal in magnitude to the negative charge on each electron, the complete
atom is electrically neutral
...
The principal quantum shells are numbered according how far they are from the
nucleus
...


An atom showing energy levels or principle quantum shell, n
1

The total number or maximum number (Ne) of electrons that can occupy any principal quantum
number n is
Ne = 2n2

Quantum Sub-shells

The principal quantum shells, apart from the first, are split into sub-shells
...
The first energy level contains one sub-shell, the
second energy level contains two and so on
...
The energy of electrons in the sub-shells increases in the order s < p < d < f
...


Principal

Max Number

Quantum Shell

of Electrons

Sub-shell

Capacity

Name of subshell

K (n = 1)

2

s

2

1s

L (n = 2)

8

s

2

2s

p

6

2p

s

2

3s

p

6

3p

d

10

3d

s

2

4s

p

6

4p

d

10

4d

f

14

4f

M (n = 3)

N (n = 4)

18

32

Not every electron is constrained to forever occupy a certain shell or subshell of an atomic nucleus
...
g
...
These electrons that have escaped
from their shells are called free electrons
...

The outermost shell in an atom is called the valence shell, and the number of electrons in the
valence shell in the valence shell has a significant influence on the electrical properties of an
element
...
Moreover, valence electrons being further away from the nucleus than electron in the inner
shells, are the ones that experience the least force of attraction to the nucleus
...
g
...
When an electrical potential is applied across the ends of
a conductor, free electrons (free delocalized electrons) readily move from one end to the other,
creating a transfer of charge through the conductor i
...
an electrical current
...


Semiconductor Materials
Virtually all modern electronic devices are constructed from semiconductor material
...
g
...
g
...
The mechanism by which charge flows through
a semiconductor cannot be entirely by the process known to cause charge to flow through other
materials
...
The valence shell of a semiconductor atom is such that
it can just fill an incomplete shell by acquiring four more electrons
...
e
...
In turn the neighboring atom, shares each of
its own four valence electrons with its neighboring atoms, contributing to the filling of their
subshells
...

3

Covalent bonding in a silicon semiconductor crystal

Current in a Semiconductor
The source of electrical charge available to establish a charge is the large number of free electrons
in the material
...
However, in semiconductors great amount of energy is required,
because the electrons are held more tightly in covalent bonds
...

Looking at the liberation of an electron from the perceptive of the quantity of energy possessed by
electrons, the energies that we deal with for the liberation or emission of electrons are very small
such that a Joule is a very high amount of energy
...

1eV = 1,602 x 10-19 J
In accordance to modern quantum theory, an electron in an isolated atom must acquire a specific
amount of energy in order to be freed, depending on the kind of atom and subshell it occupies
...
Also,
they can lose energy and fall back into lower energy shell
...

4

Energy Bands
When electrons are in close proximity, as they are when interlocked to form a crystal, the adjoin
atoms make the energy levels less distinct, in this case, it is possible to visualise a continuous
energy band
...

Within any given material there are two distinct energy bands in which electrons may exist
...
Free electrons exist in the conduction band as charge carriers for conduction of
current
...
The energy band of interest is the highest band or valence
band, electrons in this band have less energy and are shown lower in the energy diagram
energy

energy

energy

Insulators have a high forbidden band, meaning electrons have to acquire a very great deal of
energy to reach the conduction band of which that is practically impossible
...

In semiconductors, the valance band and the conduction band are so close (1
...
72 eV for germanium) that electron can be lifted from the valence band to the conduction
band by imparting some energy to it
...
Higher temperatures mean more heat and therefore greater electron
energies
...
The inner orbit electrons are bound to nucleus
whereas, the valence electrons are bound by the forces of covalent bonds
...


But as the temperature is raised, more and more electrons acquire sufficient energy and covalent
bonds in semiconductors ruptures allowing electrons to cross the forbidden gap into the conduction
band
...

Conduction can then occur by electron movement and by hole-transfer with the increase in
temperature, the rate of generation of electron-hole pairs is increased
...
Thus with the increase in temperature, the concentration of charge carriers
increases
...


For conductors the energy gap is quite small, (about 0
...
The valance band and conduction band are partially filled at room temperature, electrons
can easily jump from valance band to the conduction band and due to this reason current can easily
pass through conductors
...
Consequently it becomes more difficult to establish a
uniform flow of charge at higher temperatures, resulting in a positive temperature coefficient of
resistance for conductors
...
In good conductors (e
...
copper) the flow of current is
due to electrons only
...
The direction of the flow of current is always opposite to that of electrons
in both conductors and semiconductors
...
Both electron and holes are referred to as charge
carriers
...
Since the atom has lost an electron it now has a net positive charge,
we can regard that hole as representing a unit positive charge
...
This transfer of a hole from one atom to another constitutes a flow
of (positive) charge and therefore represents a component of electric current
...

NB: Hole current occurs at the valence band level, because valence band electrons do not become free electrons when
they simply move from atom to atom
...
Also if an electron in the conduction band happen to fall into a hole (which it may) this does
not constitute current flow; in fact it is a cancellation of charge and we say a hole-electron pair has been annihilated
or recombination has occurred
...
g
...
The holes in a pure semiconductor are created by electrons that
have been freed their covalent bonds, the number free electrons must be equal to the number of
holes
...
It then follows that for an
intrinsic semiconductors, electron density (in electrons/cm3) is equal to hole density (in
holes/cm3)
...
5 × 1016 and 𝑛𝑖 = 𝑝𝑖 = 2
...

Mobility of charge carriers: Drift Current
When an electric potential is applied across a semiconductor, the electric field established in the
material causes electrons to drift in one direction and holes to drift in the other
...
The total current due to the electric field is called drift current
...
The measure of this ability to move
is called drift mobility (µ)
...
𝒔)

𝜇𝑛 = 𝟎
...
𝒔)

𝝁𝒑 = 𝟎
...
18𝒎𝟐 /(𝑽
...
Recall that the units of electric field
intensity are V/m, so µ measure carrier velocity (m/s) per unit field intensity:
̅ 𝝁𝒏
𝒗𝒏 = 𝑬

and

̅ 𝝁𝒑
𝒗𝒑 = 𝑬

𝑤ℎ𝑒𝑟𝑒 𝐸̅ is the electric field intensity in V/m and 𝒗𝒏 𝒂𝒏𝒅 𝒗𝒑 are the electron and hole velocities
in m/s
...

We can use carrier mobility to compute the total current density J in a semiconductor when the
electric field intensity is known
...

8

J = 𝐽𝑛 + 𝐽𝑝 = 𝑛𝑞𝑛 𝜇𝑛 𝐸̅ + 𝑝𝑞𝑝 𝜇𝑝 𝐸̅
= 𝑛𝑞𝑛 𝑣𝑛 + 𝑛𝑞𝑝 𝑣𝑝
𝑤ℎ𝑒𝑟𝑒

J = current density, A/𝑚2
𝑛, 𝑝 = electron and hole density, carrirs/𝑚3
𝑞𝑛 = 𝑞𝑝 = unit electron charge = 1
...
s)
𝐸̅ = electrivc field intensity, V/m
𝑣n , 𝑣p = electron and hole density, carrirs/𝑚3

The total density is the sum of the electron and hole components of current density, 𝐽𝑛 𝑎𝑛𝑑 𝐽𝑝
...
6 × 10−19 coulombs

Example 1:
A potential difference of 18 V is applied across the ends of an intrinsic silicon bar of length 0
...
Assuming that ni = 1
...
14 𝑚2 /(𝑉
...
05 𝑚2 /(𝑉
...


9

Solution
...

18 V
a) 𝐸̅ = 0
...
14m2
]
𝑣𝑛 = 𝐸̅ 𝜇𝑛 = (2 × 103 V/m) [
V
...
8 × 102 m/s
𝑣𝑝 = 𝐸̅ 𝜇𝑝 = (2 × 103 V/m) [

0
...
s

= 102 m/s
b) Since the material is intrinsic
1
...
5 × 1016 carriers/m3
𝑎𝑛𝑑
𝐽𝑛 = 𝑛𝑖 𝑞𝑛 𝑣𝑛 = (1
...
6 × 10−19 )(2
...
672 A/m2
𝐽𝑝 = 𝑝𝑖 𝑞𝑝 𝑣𝑝 = (1
...
6 × 10−19 )(102 )
= 0
...
672 + 0
...
912 A/m2
d) The cross-sectional area of the bar is 800 × 10−6 m2 = 8 × 10−4 m2
𝐼 = 𝐽𝐴 = (0
...
7296 mA

The resistance of anybody can be calculated using
𝑅=
where

𝜌𝑙
𝐴

R = resistance, ohms (Ω)
10

𝜌 = resistivity of the material,

Ω
...
This implies that conductivity, 𝜎, is the reciprocal of resistivity:
𝜎=

1
𝜌

Thus the units of conductivity are 1/(Ω
...


Recall that,

J = 𝐽𝑛 + 𝐽𝑝 = 𝑛𝑞𝑛 𝜇𝑛 𝐸̅ + 𝑝𝑞𝑝 𝜇𝑝 𝐸̅
= (𝑛𝑞𝑛 𝜇𝑛 + 𝑝𝑞𝑝 𝜇𝑝 )𝐸̅
= 𝜎𝐸̅

the expression (nqn μn + pqp μp )is equal to the conductivity, σ of the material
...

For an intrinsic semiconductor, 𝑛 = 𝑝 = 𝑛𝑖 , therefore,
𝐽 = 𝑛𝑖 (𝑞𝑛 + 𝑞𝑝 )𝑞𝐸̅
and conductivity of an intrinsic semiconductor is
𝜎𝑖 = 𝑛𝑖 𝑞(𝑞𝑛 + 𝑞𝑝 )
NB: In general the conductivity any semiconductor is computed as shown:
𝜎 = 𝑛𝑞𝑛 𝜇𝑛 + 𝑝𝑞𝑝 𝜇𝑝

Example 2
a) Compute the conductance and resistivity of the intrinsic silicon bar in Example 1
...


11

Solution
...
5 × 1016 carriers/m3
𝜎𝑖 = 𝑛𝑖 𝑞(𝑞𝑛 + 𝑞𝑝 )
= (1
...
6 × 10−19 )(0
...
05)
= 4
...
56 × 10−4

Then

= 2192
...
m

b) 𝑅 =

𝜌𝑙
𝐴

=

(2192
...
9×10−2 )
8×10−4

= 24
...
67 × 103
= 0
...
In other words, there is
a natural tendency for energetic carriers to disperse themselves to achieve a uniform concentration
...
This current is
called diffusion current
...


Extrinsic Semiconductor
Recall that an intrinsic semiconductor is one which is made of the semiconductor material in its
extremely pure form and has the same electron density as the hope density: 𝑛𝑖 = 𝑝𝑖
...
In the fabrication of semiconductor materials used in
practical applications (such as diodes, transistors, etc
...
Such materials are called extrinsic (or impure)
semiconductors
...
This process
adding impurity atoms to an intrinsic semiconductor so as to modify the number and type of charge
carriers is called doping
...


N-type Semiconductors
N-type semiconductor material is produced by doping a pure semiconductor with a pentavalent
atom (an atom having 5 valence electrons)
...
Typical examples of
pentavalent impurities are Arsenic, Antimony, Bismuth, Phosphorous etc
...
Then 4 out of 5 electrons can (and do) participate in the same covalent
bonding that holds all other atoms together
...
One atom will be
available in the in the semiconductor crystal for every atom of the pentavalent atom added
...
Such impurities which produce n-type semiconductor are known
as donor impurities because they donate or provide free electrons to the semiconductor crystal
...
So, n-type material electrons are called majority carriers
13

and holes are called minority carriers
...
Typical examples of trivalent impurities are gallium, indium,
boron etc
...
In other words a hole is created everywhere the impurity appears in the
crystal
...
The p-type material, like n-type material
is electrically neutral
...
Due to thermal energy, even though holes that are created due generation
of hole-electron pairs, the number of hole the number of holes is far much greater than that of
electron
...

NB: Even with a very small increase in the dopants, the electrical resistivity of an extrinsic
semiconductor decreases significantly because the number of charge carriers of the intrinsic
semiconductor is typically very small
...


Example 3:
A pure silicon semiconductor at 500K has an electron density of 1
...
5 × 1022 holes/m3
...
14 𝑚2 (𝑉
...
05 𝑚2 (𝑉
...


Solution:
𝑛𝑝 = 𝑛𝑖 2

a) Since
𝑛𝑖 2
=> 𝑛 =
𝑝
=

(1
...
5 × 1022

= 5 × 109
Given 𝑝 = 4
...
6 × 10−19 )(0
...
5 × 1022 )(1
...
05)
= 1,12 × 10−10 + 360
=>

σ ≈ 360 S/m

Minority Carrier Suppression
The conductivity of a semiconductor depends only on the majority carrier density as shown in
Example 3 above
...

Suppose that the majority carriers are electrons, and that there are substantially more electrons than
holes
...
Since there are very many
15

more electrons than holes, the resultant percent decrease in electrons is much less than the percent
decrease in holes
...


Fermi-level
Fermi level EF is present between the valence and conduction bands
...
The charge carriers in this state have their own quantum states
and generally do not interact with each other
...

Fermi-level of an intrinsic semiconductor
Recall that an intrinsic semiconductor acts as an insulator at absolute zero temperature because
are no free electrons available for conduction however as temperature increases electron hole
pairs are generated and hence the number of electrons will be equal to the of holes
...

So if EC is the lowest level of the conduction band and EV is the highest level of the valence
band the Fermi level EF lies in the middle of the two as shown by the graph below
...
The donor electron can easily be excited to the
conduction band, thus becoming one of the conduction electrons moving freely through the
semiconductor
...


Energy band diagram showing the Fermi level 𝐄𝐅
of a N-type semiconductor

17

Since the donor energy level ED , is just below the conduction band, in an N-type semiconductor,
as temperature T increases, the Fermi level E F moves towards the conduction band EC, due to large
numbers of electrons occupying the energy level towards the conduction band will be high, hence
the Fermi level EF shifts towards the conduction band
...
The valence electrons of the pure semiconductor can easily
be excited to this level, leaving behind a hole in the valence band
...


Energy band diagram showing the Fermi level 𝐄𝐅
of a P-type semiconductor

Now due to large numbers of holes the probability of holes occupying the energy level towards
the valence band will be high, hence the Fermi level 𝐄𝐅 get shifted towards the valence as shown
in the diagram above
...


When these two are joined together, (when a block of a p-type material is joined to a block of ntype material) a very important and useful structure is formed
...


Formation of a Depletion Region and Barrier Potential
Remember that diffusion current flows whenever there is a surplus of carries in one region and a
corresponding lack of carriers in the other
...
Simultaneously, the holes from the P-region diffuse into
the N-region and undergo hole-electron combination with the electrons available in the conduction
band
...
)
For each electron that leaves the N-region to cross into the P-region, there is left behind a positive
donor ions as they are robbed of the free electrons
...
The upshot of this process is that negatively charged accepter atoms begin to line the region
just inside the P-region and positively charged donor atoms accumulate just inside the N-region
...
The direction of the field (which by
convention is the direction of the force on a positive charge placed in a field) is from the positive
N-region to the negative P-region
...
In other words, the positive and negative charges whose location were caused by the
original diffusion current across the junction now inhibits the further flow of current across the
junction
...
Therefore, after the initial surge
of charge across the junction, the diffusion current dwindles to a negligible amount
...
This small drift of minority carriers opposes the direction of the diffusion current
...
Therefore, the net current
across the junction is zero
...
It is also called the barrier region because the
electric field therein acts as a barrier to further diffusion current
...


20

The electric field in the junction is a result of the potential difference that exists across the junction
due to oppositely charged side of the junction
...
It is also called junction potential or
diffusion potential
...


𝑉𝑜 =

𝑘𝑇
𝑁𝐴 𝑁𝐷
𝐼𝑛 ( 2 )
𝑞
𝑛𝑖

Where 𝑉𝑜 = 𝑏𝑎𝑟𝑟𝑖𝑒𝑟 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙, 𝑣𝑜𝑙𝑡𝑠
𝑘 = 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 1
...
6 × 10−19
𝑁𝐴 = 𝑎𝑐𝑐𝑒𝑝𝑡𝑜𝑟 𝑑𝑜𝑝𝑖𝑛𝑔 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑃 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑁𝐷 = 𝑑𝑜𝑛𝑜𝑟 𝑑𝑜𝑝𝑖𝑛𝑔 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑁 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑛𝑖 = 𝑖𝑛𝑡𝑟𝑖𝑛𝑠𝑖𝑐 𝑖𝑚𝑝𝑢𝑟𝑖𝑡𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

NB: The barrier potential is proportional to temperature
...
7V whereas for Germanium 𝑉𝑜 = 0
...
5 × 1021 donors/m3 and P material doped
with 1
...
Calculate the thermal voltage and the barrier voltage at 25℃
...
38 × 10−23 )(298)
=
𝑞
1
...
7 mV

𝑛𝑖 2 = (1
...
25 × 1032

Barrier voltage,

𝑉𝑜 = 𝑉𝑇 𝐼𝑛 (

𝑁𝐴 𝑁𝐷
)
𝑛𝑖 2

(1
...
5 × 1021 )
= 0
...
25 × 1032
= 0
...
The P-N junction can be biased by
connecting a dc voltage source across its P and N sides
...
In this circuit, the external
electric field 𝐸̅ opposes that of the junction, therefore the potential barrier is reduced at the
junction
...
3V for Ge and 0
...

In other words, the applied positive potential repels the holes in the ‘P’ region so that the holes
moves towards the junction and applied negative potential repels the electrons in the ‘N’ region
towards the junction, as a result the depletion region is lowered (reduced) in width and the barrier
potential is also reduced
...

Once the potential barrier is eliminated by a forward voltage, the junction establishes a low
resistance path for the entire circuit, thus the majority charge carriers flow across the junction,
resulting in the flow of a current (called forward current) in the circuit
...


Reverse Biasing:
In reverse bias the negative terminal is connected to the P-side and positive to the N-side of the
junction
...
The external and barrier voltage act in the same
direction
...

In other words, the applied reverse voltage establishes an electric field 𝐸̅ which acts in the same
direction of the potential barrier
...
This increase in the potential barrier prevents the flow of
charge carriers across the junction, results in a high resistance path, once depletion layer equals
the barrier potential, no current flows
...
Due to minority carriers, this results in build-up of reverse current
(known as reverse saturation current) and a reverse voltage to large values resulting in a reverse
breakdown

24

𝐸̅

I-V Characteristics of the P-N junction
A graph between current and voltage applied across the P-N junction is called characteristics of
the P-N junction
If the forward current is to be treated as positive (upwards), then the reverse current should be
below the horizontal axis, i
...
downwards or negative
...
e
...

Forward Characteristics:
When the external voltage is zero, i
...
, when the circuit is open, the potential barrier at the junction
does not allow the flow of current and, therefore, the circuit current is zero
...

In other words, with forward bias very little current, called the forward current flows until the
forward voltage exceeds the junction barrier potential
...
3V for Ge and 0
...


Example 5:
Why would an increase in the temperature, lower the threshold voltage
Solution
...
Therefore, the
junction resistance becomes very high and there is no possibility of a majority carriers flowing
across a reverse-biased junction
...
This results in a very small current which is known as reverse current
...
Leakage current is a
current that flows along the surface of a diode, obeys Ohm’s law relationship and is not accounted
for in the ideal diode equation so it deviates from that predicted by the diode equation if the reverse
voltage is allowed to approach a certain value called the reverse breakdown voltage, V BR
...
Furthermore, a very
small increase in the reverse bias voltage in vicinity of VBR results in a very small increase in the
reverse current
...
This
conveys that large increase in the reverse current result from a very small increase in reverse
voltage in the vicinity of VBR
...
As a result of collision, the liberated electrons in turn liberate more electrons and the current
becomes very large leading the breakdown of the crystal structure itself
...
The large number carriers freed in this way accounts for
the increase in the through the junction
...
The essentially vertical characteristics in the breakdown region means that the voltage
across the diode remains constant in that region, independent of the (reverse) current that flows
through it
...
Zener diodes are heavily doped than ordinary diodes,
they have narrower depletion regions and smaller breakdown voltage, less than about 5V
...
The very high electric field intensity
across the narrow depletion region directly forces carriers out of their bonds, i
...
will break the
covalent bonds of the semiconductor atoms leading to a large number of free minority carriers,
which suddenly increases the reverse current
...


Despite the name breakdown, nothing about the phenomenon is inherently damaging to a diode
...

Unless there is a sufficient current-limiting resistance connected in series with a diode, if the large
reverse current were allowed to approach breakdown could cause excessive heating
...

28

The ideal diode equation shows that both forward and reverse current magnitudes depend on
temperature
...
A commonly used
rule of thumbs is that 𝐼𝑠 doubles for every 10℃ rise in temperature
...
1 pA at 20℃
...
55 V
...

Solution:
At T = 20℃
𝑉𝑇 =

𝑘𝑇 (1
...
6 × 10−19
= 0
...
55/0
...
283 mA
At T = 100℃,
𝑉𝑇 =

1,38 × 10−23 (273 + 100)
1
...
0317 V
In going from 20℃ to 100℃, the temperature increases in 8 increments of 10℃ each: (100-20)
= 80; 80/10 = 8
...
e
...

𝐼 = 256 × 10−13 (𝑒 0
...
03217 − 1) = 0
...


29

Increasing the temperature causes the I-V
characteristic to shift left
Ratings:
Knee/Cut-in voltage – the forward voltage at which a conductor starts conducting
Maximum Forward Current – the maximum current that can be sustained or conducted by P-N
junction when in forward biased
...

Breakdown voltage – the reverse voltage at which the diode (p-n junction) breaks down with
sudden rise in reverse current
...

If the PIV is exceeded, the junction or diode breaks down due to excessive heat, it breaks down
only in the sense that it readily conducts current in reverse direction
...
C
...
Breakdown may
lead to permanent failure if the power dissipation rating of the diode is exceeded
Maximum Power Rating – the maximum power that a junction can dissipate without it being
damaged and is given by:
P = IV 𝑊𝑎𝑡𝑡𝑠

where 𝑽 is the across the device and 𝑰 is the current through it
...


ii)

The Zener breakdown occurs in heavily doped junctions (P-type semiconductor
moderately doped and N-type heavily doped), which produce Narrow depletion layers,
whereas Avalanche breakdown occurs in lightly doped junctions, which produce wide
depletion layers
...


iv)

The zener diodes have a negative temperature coefficient while Avalanche diodes have
a positive temperature coefficient
...
g
...
In both materials (the metal and the doped
N-type silicon), the electrons are the majority carriers and in the metal, the level of minority
carriers is insignificant
...
Just like a P-N
junction, the MS junction presents a low resistance to current flow when it is forward biased (metal
positive with respect to N-type silicon) and a high resistance when in reverse biased
...

Since when in forward biased electrons on the N-side gain enough energy to cross the junction and
enter the metal, they plunge into the metal with very large energy they are usually called the hot
carriers and the diode is called the hot carrier diode
...
In other
words it is a unipolar device
...
The current in a semiconductor device results from the motion of two different kinds of
charge carriers: electrons and holes
...
In a particular device, suppose the electron density is
2 × 1019 electrons/m3 and hole density is 5 × 1018 holes/m3
...
If the electrons are moving to the left at a velocity of 0
...
2 mm/s what are the:
a) The direction of the current in the semiconductor;
b) The magnitude of current in the device
2
...
Discuss the position of
Fermi level in each case
...
The current flowing in a certain PN junction at room temperature is 2 × 10−7 A, when a
large reverse bias voltage is applied
...
1V
is applied across the junction
...
Find the static and the dynamic resistance of a p-n junction Germanium diode, if the
temperature is 27oC and 𝐼𝑆 = 1𝜇𝐴 for an applied forward bias of 0
...

5
...
What is p-n-junction? Define depletion region and potential barrier
...

7
...

8
...
Is the net charge on P-type or N-type material zero? If so, why?
10
...

11
...

12
...
Discuss qualitatively the origin of the depletion region at a p-n junction and use this to
explain how a p-n junction can act as a rectifier
...
A bar of intrinsic silicon having a cross-sectional are of 3 × 10−4 𝑚2 has an electron
density of 1
...
If μn = 0
...
s), μp = 0
...
s), how

33

long should the bar be in order for the current in it to be 1
...
A silicon PN junction is from N material doped with 2
...
Find the thermal voltage and barrier voltage at
40oC
...
A silicon p-n junction has a saturation current of 1
...
Assuming 𝜂 = 1
find its current when it is forward bias voltage is 0
...
65V

References:
1
...
T
...
Boylestad
...
L (2006), Electronic Devices and Circuit Theory, Ninth
Edition, Prentice Hall, ISBN: 0131189050
...
Chapman
...
T (2005), Electric Machinery Fundamentals, Fourth Edition, 2005, McGrawHill, ISBN: 0072465239
4
...
Neamen (2006), An Introduction to Semiconductor Devices, 1st Ed
...
Mehta V K (1994), Principles of Electronics, S Chand & Company Ltd
6
...

7
...
S
...
India
...
Tokheim
...
L (2002), Digital Electronics: Principles and Applications, Student Text with
MultiSIM CD-ROM, Sixth Edition H, McGraw-Hill, ISBN: 0078309816

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