3semicond Diode.1

8/2/2019 3semicond Diode.1

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PHYSICAL

ELECTRONICSECX 5239

PRESENTATION 01

Name : A.T.U.N Senevirathna.

Reg, No : 20661910

Center : Kandy


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Introduction

What is a semiconductor ?

Introduction about Current density, Resistivity, Conductivity,Drift velocity, Mobility.

How to solve my problems according to above equations.


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What Is a Semiconductor?

Many materials, such as most metals, allow electrical current to

flow through them

These are known as conductors

Materials that do not allow electrical current to flow throughthem are called insulators

A material whose properties are such that it is not quite a

conductor, not quite an insulator

That material called as semiconductor.


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Semiconductors

Some common semiconductors

Si - Silicon (most common)

GeGermanium

Silicon is the best and most widely used semiconductor.

The main characteristic of a

semiconductor element is thatit has four electrons in its

outer or valence orbit.


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Doping

To make the semiconductor conduct electricity, otheratoms called impurities must be added.

Impurities are different elements.

This process is called doping.

Some impurities are As, P, B


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Doping with Boron

Boron has 3 electrons are in its

outer shell. We remove a siliconatom from the

crystal lattice.

Then we replace it with a boron

atom.

Notice we have a hole in a bondthis hole is thus free for conduction

This type of silicon is called p-type

p-type

will be

shown

like

this.


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Semiconductors can be Conductors

An impurity, or element likearsenic, has 5 valenceelectrons.

we remove a silicon atomfrom the crystal lattice andreplace it with a arsenicatom.

We now have an electronthat is not bondedit isthus free for conduction.

This type of silicon is

called n-type .

n-type

will be

shown

like this.


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Carrier Drift

When apply an electric field tosemiconductor , charged particles

move according to electric field.

This process is called drift.

Charged particles move with an

average velocity. This velocity

proportional to the electric field.

The proportionality constant is the

carrier mobility.

Hole velocity

Electron velocity

Notation:

mp hole mobility (cm2/Vs)mn electron mobility (cm2/Vs)

Hole velocity

Electron velocity


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Drift current is proportional to the

carrier velocity and carrierconcentration:

Drift Current

(current density) J= =


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Drift Current Equations


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Electrical Resistance

1

2

Using 1,2 we can get eq. 3 3


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Problems

Question No 08 :

A current density of 10 A/m2 flows through an n-typegermanium which has resistivity 0.05 ohm-m. Calculate the

time taken for electrons in the material to travel 50 m.

According to question we can get

Current density = 10 A/m2

Resistivity = 0.05 ohm-m

Distance = 50mCharge of electron = 1 .6 x 10-19 c

Electron mobility = 0.39 m2 / vs because ntype germanium.

Where is Drift velocity

J

e

em

nevJv

1

D


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en

em

2Density of electrons

2005.0

11

3Conductivity

Using 2,3 we can get eq. 4

20

20102051.3

1024.6

20

en

em

4

Using 1,4 we can get eq. 5

5.19106.1102051.3

10

1920

3

v 5

v

DTime 6

Using 5,6 can get,

ss

ms

mTime m5.2105641.2

5.19

1050 61

6


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Question No 09 :

Intrinsic silicon has a resistivity of 2000 ohm-m at R.T. and the density of

conduction electrons is 1 .4 x 1016calculate resistivity's of samples

containing acceptor concentrations of 1021 and 1023 m -3 Assume that hremains as for intrinsic silicon and thath = 0.25e .

According to question we can get

Density of electrons 316104.1 mn

Resistivity 2000

4105

2000

11

Conductivity of Intrinsic silicon

Charge of electron ce19

106.1

Using above data we can get ,

Electron mobility 2232.0106.1104.1

105

1916

4

nee

m 4

3


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Hole mobility of intrinsiceh

mm 25.0 B

Using 4,B we can get 0558.02232.025.0 h

m 5

Now we can get eq. for conductivity of

sample he pne mm 1

We use another eq. for number of electronsand number of holes

ad NNn

da NNp

Where Na is acceptor concentration and

Nd Is donor atom concentration/impurity concentration

Let 2110aN And assume this is p-type semiconductor

Hence 2110 aNp 2 Because room temperature

2,3,4,5,6 apply to 1 we can get 0558.0102232.0104.1106.1211619


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9284.8 Hence resistivity of sample () = 1/ 8.9284= 0.112 ohm.m

Similarly , when 2310a

N We can get () = 0.00112 ohm.m

Question No 10 :A rod of p-type germanium 6mm long , 1mm wide and 0.5mm thick has and

electrical resistance of 120 ohm. What is the impurity concentration ? Assume

e= 0.39,h = 0.19 m2 /v.s and ni = 2.5 10

19 m -3 what proportion of the

conductivity is due to electrons in the conduction band ?

According to question we can get ,e= 0.39,h = 0.19 m2 /v.s, ni = 2.5 10

19 m -3

Length of rod (L) = 6mm, Resistance of rod (R) =120, Area of rod (A) =

0.5 m2 Then we can get conductivity of rod

100

105.0120

106

6

3

RA

L 1

LetaNp

a

i

N

nn

2

As p-type germanium2

he epen mm 3


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Using 1,2,3 we can gethae

a

i eNN

ne mm

2

4

Solving eq. 4 Using previous data we can get21

10289.3 aN or17

109.3 aN

Solving eq. 2 we can get 17109.1 n Hence let21

10289.3 aN

Because p>n as p-type

Finally we can get impurity concentration

32121171029.31028.31097.1

mNnN ad


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Reference

Course material of physical electronics.

END

Electronic materials and Devices.

(By S.O.Kasap)

Internet resources.

Electronic materials and Devices.

John Allis0n

3semicond Diode.1

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