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Transcript:

Today we will discuss about semiconductor materials. Semiconductor materials are the

backbone of electronic devices and circuits. What is a semiconductor material? Probably

all of you know about conductors and insulators. Conductor is a material which when you apply

a voltage source creates a generous flow of charges like for example copper. Copper is

a metal and metals are conductors. Insulator is a material that offers a very low level

of conductivity when a voltage is applied; for example mica. Semiconductor material is

in between this conductor and insulator. It is a material that has conductivity level

somewhere between the extremes of an insulator and a conductor.

For example germanium and silicon are these two materials which are semiconductors showing

similar property. The resistivity of a material is inverse of conductivity. If we compare

the resistance levels of conductors, insulators and semiconductors we see that the conductor

has resistivity value which is the least among the three. That is, a copper which is a conductor

has a maximum conductivity so its resistivity value is around 10-6 ohm centimeter whereas

an insulator has a resistivity of 1012 ohm centimeter. This is much more than conductor

but semiconductor like germanium and silicon have resistivity values in between these two.

For example if we consider germanium semiconductor it has resistivity value of 50 ohm centimeter

and silicon has resistivity value of 50 x 103 ohm centimeter.

You can see that semiconductor has resistivity in between conductor and insulator. If we

consider these three, i.e insulators, conductors and semiconductors and see their energy band

diagrams, then we find that the energy gap between the valence band and the conduction

band in an insulator is very high.

It is around 6 electron volts as is seen here. This energy gap means it is a forbidden band

and the electrons will have to acquire this much of energy to become conductive and enter

the conduction band. But semiconductor does not have such a high forbidden energy gap

which is around 1 electron volt as is seen here. Electrons can acquire that much of energy

which is 1 electron volt only to jump from valence band to the conduction band in order

to become conductive and metals like copper which are conductors have overlapping conduction

band and valence band and they have zero electron volt. That is why they conduct because electrons

can very easily shift between valence band and conduction band and take part in conduction.

Semiconductor is a material that has conductivity level between the extremes of insulator and

conductor and why this is so?

That is because the semiconductor like germanium and silicon have some unique qualities which

are due to their position in the periodic table and their atomic structure. These are

group IV elements. The atoms of these materials like germanium and silicon have a very different

and definite pattern that is periodic in nature. That is it continuously repeats itself and

one complete pattern is called a crystal and the periodic arrangement of the atoms is a

lattice. Both germanium and silicon have 4 electrons in the outer most orbit which are

called valence electrons. In order to complete the structure of inert gas they will always

try to complete the 8 electrons in the outer most orbit.

In the pure germanium or silicon crystal these 4 valence electrons will be bonded to 4 neighboring

atoms as shown in this diagram. If I consider a germanium atom it has 1, 2, 3, 4 electrons

in the outer most orbit which are called valence electrons. Similarly silicon will also have

4 electrons in the outer most orbit because they are group IV materials.

They will try to complete the 8 electron structure. That is why they will share electrons with

the neighboring atoms as is seen here. If we consider silicon atom, these 4 electrons

in the outermost orbit are sharing electrons among themselves so that the total number

becomes 4 and 4, 8. These atoms which are nearby are sharing these covalent bonds.

These covalent bonds are formed between neighboring atoms of silicon and thereby they complete

the structure of the inert gas having 8 electrons in the outermost orbit so that they become

stable. These semiconductors which are in the purest form are known as intrinsic semiconductors.

In these intrinsic semiconductors the free electrons are due to only natural causes like

thermal energy or light energy which impart extra energy to the electrons to become free

and thereby they get additional kinetic energy from these natural causes and they break this

covalent bond. These covalent bonds will break in order to become free. 1 electron will become

free when it absorbs kinetic energy from either the surrounding or any light is falling on

them.

An increase in temperature is naturally the reason for more conductivity in semiconductors.

This increase in temperature of a semiconductor results in increase in the number of free

electrons in the material so they become more and more conductive. Even though they may

be not so conductive at certain temperature but if we increase the temperature they will

absorb more kinetic energy and will become more conductive. The semiconductors like germanium

and silicon they have negative temperature coefficient as the resistance will decrease

with increase in temperature. This is a peculiar phenomena that is seen in semiconductors which

is just opposite to your conductors like metal.

We have been talking about the intrinsic semiconductors. Those are the purest form of semiconductors

without any impurity in them. But there is another type of semiconductor which is extrinsic

semiconductor.

This extrinsic semiconductor has the characteristics such that their property can be significantly

altered by addition of certain impurity atoms into the pure semiconductor material. The

impurities added to the purest semiconductor to make an intrinsic into extrinsic semiconductor

are defined. We can add pentavalent impurity materials like antimony, arsenic and phosphorous

which are group V elements and have 5 electrons in the outer most orbit or we can add trivalent

materials like boron, gallium and indium which are group III materials and have 3 electrons

in their outer most orbit and the standard norm of adding is that they are added 1 part

in 10 millions. That is to 10 million atoms of intrinsic semiconductor you add one impurity

atom.

We will discuss how the addition of this impurity totally changes the electrical properties

of the intrinsic semiconducting material. When certain impurity atoms like antimony

or boron are added to a pure semiconductor material we get an extrinsic semiconductor

and this process of adding certain impurity atoms like antimony or boron into pure semiconductor

like silicon is called the doping process. As a result of this doping we get extrinsic

semiconductor either n-type or p-type.

That is you can get n-type extrinsic semiconductor if you add pentavalent impurity material like

antimony or phosphorous into a pure semiconductor like germanium or silicon or you can have

p-type extrinsic semiconductor when you add trivalent impurity material like boron or

indium into a pure germanium or silicon semiconductor.

What happens when you add an impurity atom to a pure semiconductor? Let us take for example

an n-type semiconductor. How it is formed? N-type semiconductor means we have to add

pentavalent impurity material. Let us take the example of adding antimony which is a

pentavalent impurity material which is a group V material having 5 electrons in the outer

most orbit to a silicon semiconductor. This silicon atom has 4 electrons in the outer

most orbit. These are silicon outermost orbit electrons and now as we have added 1 antimony

atom which has 5 electrons in the outer most orbit, it will share these electrons with

the silicon atoms and these covalent bonds will be completed. 4 and 4, 8 electrons are

there so these covalent bonds will be completed. But then 1 electron is in excess. That is

it is extra. This fifth electron is donated by this antimony atom. This fifth electron

is free. It can take part in conduction easily. That is why it is donating 1 electron. It

is excess of 1 electron which is negatively charged.

This is called n-type semiconductor. What happens when we add p-type semiconductor?

We have seen how n-type semiconductor is formed by adding pentavalent impurity material to

intrinsic semiconductor like germanium or silicon. Let us see how p-type semiconductor

is formed from pure semiconductor like germanium or silicon by adding trivalent impurity material

like boron or indium. Let us add boron impurity which is a group III material having 3 electrons

in the outer most orbit to a silicon semiconductor. This boron has 3 electrons in the outer most

orbit. It will complete its covalent bond with neighboring silicon atom having shared

these 2 electrons as is seen here. These 2 electrons will form a covalent bond. In this

process 3 covalent bonds are now completed. What about the other covalent bond? This covalent

bond is now devoid or it is short of 1 electron to complete its covalent bond. This shortage

of 1 electron means it is having less negative charge. This less negative charge or a positive

charge is there in this covalent bond. This is called a hole. Hole means it is basically

devoid of 1 electron. So there is a hole formed here in this covalent bond and this hole is

a positively charged particle.

So this p- type semiconductor basically is formed in this way which will have holes as

charge carriers.

If we consider again the n-type material we have seen that in n-type material which is

an extrinsic semiconductor it has donated 1 electron. As it has donated 1 electron,

the atom of this extrinsic semiconductor is now having 1 positive charge which is called

an ion. 1 atom of this extrinsic semiconductor when it donates one electron it becomes ion

having 1 positive charge which is shown here. The donor ions are having 1 positive charge

and there will be electrons which are in abundance in the n-type material. These are called majority

carriers. In an n-type material the majority carriers are electrons but there will also

be in less number the minority carriers which are holes. These holes will be from this intrinsic

material that is coming. If n-type material is considered altogether we will see majority

carriers as electrons, minority carriers as holes and immobile ions which are positively

charged or donor ions.

This is the structure of n-type material. Similarly if we consider a p-type material,

p-type material has acceptor ions. In the p-type material we have seen that there is

a hole and as it is devoid of an electron it will be always trying to get an electron

from the neighborhood atoms. The electron from the neighborhood atom will come and fill

up this hole and as a result this boron atom will be negatively charged. It is accepting

1 electron so it will be negatively charged.

That means in a p-type material the atom of the impurity will be an acceptor ion having

a negative charge and there will be holes which are positively charged as majority carriers

and also very less in numbers although there will be presence of minority carriers which

are electrons which are coming from the intrinsic semiconductor. This is p-type material where

all these carriers and ions you see here. We can summarize that in n-type semiconductor

the majority carriers are electrons and minority carriers are holes and in p-type semiconductor

majority carriers are holes but minority carriers are electrons.

These are the charge carriers available in n-type and p-type semiconductors. How does

the current flow in a semiconductor? Current flow is due to the movement of electrons and

conventional current is always against the direction of movement of electrons. Suppose

if electron moves in this way then conventional current direction will be in opposite way.

This is electron movement direction and this is conventional current direction. But we

have seen that another type of charge carrier is also there in semiconductors which are

holes and which are positively charged. So the current flow will be now due to these

two types of carriers; one is electron and other is hole and both are opposite in charge.

That is electron is negatively charged and hole is positively charged.

We are having a semiconductor. We have the silicon atom and this is an impurity atom

which is boron. It is creating a hole as is seen here. In order to fill up this hole electron

from a neighboring atom will come. This movement of electron will be in this direction. When

this electron comes and fills up this hole then this place will be now devoid of electron.

It will be now a hole. Now hole movement will be from this position to this position. It

is just opposite to the electron movement. Electron movement is in the direction of this

solid arrow and hole movement is in the direction of this dotted arrow. If we consider the conventional

current direction which is opposite to the direction of movement of electrons, then the

current direction will be in the same direction, because of both these components. If I consider

conventional current direction it will be like this because this electron movement is in this

direction and hole movement is in opposite direction. But the final direction of current

flow is opposite to the electron movement.

So it will be in opposition to this movement of electron but that is the same direction

of the hole movement. We will discuss a very fundamental and very important law which gives

the relationship between electron and hole concentration and that is known as mass action

law. What is that law? It states that under thermal equilibrium the product of the free

negative and positive concentrations is a constant independent of the amount of donor

and acceptor impurity doping. That is mathematically if you write np is equal to ni square where

n is the concentration of free electrons in thermal equilibrium, p is the concentration

of holes in thermal equilibrium and ni is the intrinsic carrier concentration.

This is a very fundamental law in electronics which is mass action law. Now let us consider

about the charge densities in a semiconductor. We will consider charge density later. Before

that you must also know what is the law of electrical neutrality that is satisfied in

the semiconductor? Let us consider a semiconductor material and consider a situation where it

is doped by both n-type and p-type materials and let ND be the concentration of donor atoms.

These donor atoms are all ionized and already we have seen that these will be positively

charged donor atoms. We have ND positive charges per cubic meter which are contributed by these

donor ions and let p be the hole concentration in the semiconductor so that the total positive

charge density in the semiconductor is ND plus p. That is total positive charge density

in the semiconductor which is contributed by donor atoms as well as holes. Similarly

if we consider that NA is the concentration of acceptor ions then these contribute NA

negative charges per cubic meter and let small n be the electron concentration in the semiconductor.

So the total negative charge density in the semiconductor will be now NA plus n and since

the semiconductor is electrically neutral we have this law to be satisfied that ND plus

p is equal to NA plus n. Let us number this equation as 2.

That means this electrical neutrality must be satisfied in a semiconductor given by this

equation. Consider solely n-type material. We can either dope by n-type or by p-type

impurity. So in an n-type material doping NA will be zero and then the equation 2 will

be now ND plus p is equal to n since NA is equal to zero or ND is equal to n minus p.

As it is an n-type material the concentration of electrons is much much more than concentration

of the holes. So n minus p we can roughly say that equal to n. Then from this equation

we will get that the electron concentration is almost equal to the concentration of the

donor atoms ND and if we use a subscript small n, the electron concentration in an n-type

material nn is equal to capital N subscript capital D. From this equation we can get now

hole concentration for this n-type material which is given by pn is equal to ni square

by ND. Here we are using equation number 1, mass action law.

Similarly if we consider a p-type material, the donor concentration is zero since it is

a p-type material. So the equation 2 will be now reduced to p is equal to NA plus small

n or NA is equal to p minus n and since it is a p-type material the hole concentration

is much, much more than the electron concentration, p minus n can be approximately written to

be equal to p. Finally in the equation 2 we get p almost equal to NA. So in a p-type material

hole concentration is approximately equal to the density of acceptor atoms. If I use

a subscript small p, p subscript p is almost equal to NA and similarly from equation 1,

mass action law if we use, we can find out the concentration of electron in a p-type

material which is equal to ni square by NA.

Here I am substituting in this equation np is equal to ni square. I am substituting this

p with NA. Then I get this n equal to ni square by NA. The fundamental difference between

a metal and a semiconductor is that metal is unipolar since current conduction is taking

place only by electrons which are having only one sign that is negative sign.

But in semiconductor there are two charge carriers having negative as well as positive

charges which are known as electrons and holes. Semiconductor is a bipolar material. There

are 2 types of charge carriers positive as well as negative. Holes are positively charged

and electrons are negatively charged. Whenever you apply an electric field, E then your holes

and electrons will move in opposite directions since they are of opposite sign but the current

due to both these two components will be in one direction.

If you want to find out the current density after the application of an electrical field

E, that is given by this expression. The current density J is equal to n times mu n plus p

times mu p into q into E which is known as sigma into E. What are those terms? n is magnitude

of free electron concentration, mu n is mobility of the electrons, p is magnitude of the hole

concentration and mu p is the mobility of holes, q is the charge of an electron and

sigma is known as conductivity which is given by this expression n mu n plus p mu p into

q. For intrinsic semiconductor as n is equal to p which is equal to ni that is intrinsic

concentration, we get for intrinsic semiconductor if you want to find out the current density

you will have to simply substitute ni both for n and p.

Let us take one example and see how you can find out the current density. Consider an

intrinsic silicon bar of cross section 5 centimeter square and length 0.5 centimeter at room temperature,

300 degree Kelvin. An average field of 20 volt per centimeter is applied across the

ends of the silicon bar. It is the silicon bar having a cross section of 5 centimeter

square and you are applying an average field of 20 volt per cm. Now you have to calculate

electron and hole component of current density, total current in the bar and resistivity of

the bar; these three you have to find out. You are given that the mobility of electrons

is 1400 centimeter square per voltage second and hole mobility is 450 centimeter square

by volt second and intrinsic carrier concentration of silicon at room temperature, 300 degree

Kelvin is generally taken as room temperature. Kelvin is small t degree centigrade plus 273

degree that gives you the Kelvin temperature. Here it is 300 degree Kelvin. You can find

out at what room temperature this example is carried out 300-273. At 27 degree centigrade

room temperature the sample is given and now at that temperature the intrinsic carrier

concentration of silicon is given as 1.5 into 10 to the power 10 per centimeter cube. You

have to find out electron and hole component of current density.

If we are to find out electron and hole component of current density we must use the equation

J equal to n mu n q E plus p mu p q E. Now you have to see the data given to you. We

know already n is equal to p is equal to ni in an intrinsic semiconductor, since this

is an intrinsic semiconductor and it’s value is given in the data as 1.5 into 10 to the

power 10 per centimeter cube. Electron mobility and hole mobility are also given. Electron

mobility mu n is 1400 centimeter square per volt second and hole mobility is 450 centimeter

square per volt second. But this charge of an electron is not given. It is supposed that

you know this and its value is standard 1.6 into 10 to the power -19 coulomb.

It is simple to find out this current density J given by this equation. In this equation

there are 2 components, one is due to the electrons and the other is due to the holes.

Electron component of current density we can find out from this part of this equation.

You substitute each of these values and by an intrinsic concentration mu n q E, E is

30 volt per cm and each of these are substituted by the given values then we get this equation

and finally the answer comes to 67.2 micro ampere per cm square. That is the electron

component of the current density.

This current density is due to the electrons. Similarly you can find out the hole component

of current density. That is p mu p q E. Here also you substitute all these given values

and you find 21.6 micro ampere per cm square, that comes to be the hole component of current

density. Another thing you have to find out is the total current. This is only current

density. In order to find out the total current you have to multiply it by the cross sectional

area which is given as 5 cm square. Current density mu A per cm square we have found out.

Each of these hole component and electron component you have to add up. Then you will

get the total component, total current density. 67.2 plus 21.6 is the total current density

multiplied by the cross sectional area 5 centimeter square. Finally we get 444 micro ampere which

is the current flowing in the silicon bar and finally the resistivity of the silicon

bar which is inverse of conductivity. 1 by conductivity gives you the resistivity. Conductivity

is sigma which is given by n mu n plus p mu p into q. So 1 by this whole term will give

you the resistivity and substituting each of these values we get finally the resistivity

of the silicon bar is 22.52 into 10 to the power 4 ohm centimeter.

In these examples we have used only one equation that is the current density equation. The

rest of the things are found from that.

In today’s class we have learnt about intrinsic semiconductor material as well as extrinsic

semiconductor material and their characteristics or properties we have studied and this semiconductor

is the main component in all electronic applications and devices which is made of intrinsic semiconductor

silicon or germanium and when it is doped with impurity atoms like pentavalent or trivalent

materials then you get n-type as well as p-type extrinsic semiconductors. These will be used

in all electronic devices. In today’s class we have learnt about intrinsic as well as

extrinsic semiconductors. How extrinsic semiconductors are formed from intrinsic semiconductors by

doping them with p-type or n-type materials and we have also seen the characteristics

or properties of these extrinsic semiconductors. This semiconductor is used as a backbone in

all semiconductor devices and circuits. Let us try one example.

Consider an intrinsic silicon bar of cross section 5 centimeter square and length 0.5

centimeter at room temperature 300 degree Kelvin. An average field of 20 volt per centimeter

is applied across the ends of the silicon bar. In part a, calculate number i, electron

and hole component of current density; number ii, total current in the bar and number iii,

resistivity of the bar. Part b of the problem is if now donor impurity to the extent of

1 part in 10 to the power 8 atoms of silicon is added find the density of minority carriers

and the resistivity. The data given to you are electron mobility is given as 1400 centimeter

square per volt second, hole mobility is 450 centimeter square per volt second, intrinsic

carrier concentration of silicon at room temperature 300 degree Kelvin is given as 1.5 into 10

to the power 10 per centimeter cube and number of silicon atoms which are doped per meter

cube is 4.99 into 10 to the power 28. This last data will be required for the part

b of the problem. Let us first of all do part a of the problem. It is given that it is an

intrinsic semiconductor. The hole concentration and electron concentration is same which is

the intrinsic concentration and it is given as 1.5 into 10 to the power 10 per centimeter

cube. Electron mobility is given as 1400 centimeter square per volt second. Hole mobility is given

450 centimeter square per volt second and charge of an electron is not given but you

should know this value which is 1.6 into 10 to the power -19 coulomb. With this data we

will now find out the current density first. As you know the current density J is given

by n mu n q into E plus p mu p q into E where E is the applied electric field and all the

other data are given to you. This current density has 2 components one due to the electron

and other due to the hole. Electron component of current density will be the first part;

this part will be the electron component of current density.

You just put down the values and you know that electron concentration is nothing but

intrinsic concentration. So it will be ni mu n q into E. Now substitute the values which

are known to you. Each one we will substitute. ni is 1.5 into 10 to the power 10 per cm cube

and electron mobility is 1400 centimeter square per volt second. q charge of an electron is

1.6 into 10 to the power -19 coulomb and the applied electrical field is 20 volt per cm.

All these are in cm that you must be careful because centimeter is the unit in all the

data. If you calculate this value then we are getting 67.2 micro ampere per centimeter

square. Part one of the problem a, is now 67.2 micro ampere per centimeter square.

Now in part 2, hole component of current density has to be found out and that is the second

part of the equation J which is p mu p q into E and that can be replaced by ni mu p q E

because ni is the intrinsic concentration that is equal to the hole concentration as

well as electron concentration in intrinsic semiconductor. Substituting each of these

values ni is 1.5 into 10 to the power 10. Then mu p is 450, q is 1.6 into 10 to the

power -19 and E is 20. We now calculate this and the result is 21.6 micro ampere per cm

square.

That is the hole component of current density. Third part of the problem says that you have

to find out the total current in the silicon bar. In order to find out the total current

you have to multiply the current density J by the cross sectional area. The total current

in the bar is equal to the current density into the cross sectional area of the bar.

Current density is the total current density due to the holes as well as electrons. We

have to sum up these two 67.2 plus 21.6 micro ampere per cm square and multiply it by the

5 cm square cross sectional area and that gives the result as 444 micro ampere which

is the total current and the next part of the problem is resistivity. This is part iii

and this is part ii.

Part iii is resistivity of the silicon bar. Resistivity is 1 upon conductivity and conductivity,

sigma is n mu n plus p mu p into q. You substitute all these values which we already have because

n and p is nothing but ni. ni can be taken out; ni into q then mu n plus mu p which will

give you the resistivity and that is equal to 22.52 into 10 to the power 4 ohm centimeter.

Now coming to the part b of the problem in which you have to find the resistivity of

the doped semiconductor as well as minority carrier concentration. If you look into the

problem it is doped and the doping impurity is an n-type of doping material. Donor impurity

is 1 part in 10 to the power 8 atoms of silicon. In part b of the problem you are doping silicon

with donor atom. The concentration of the donor atoms ND will have to be found out and

we have been told that the doping is 1 in 10 to the power 8 atoms of silicon. We also

know and it is given in the problem that number of silicon atoms per meter cube is 4.99 into

10 to the power 28. We are doping 1 in 10 to the power 8 silicon atoms. What will be

the donor concentration? Donor concentration will be 4.99 into 10 to the power 28 divided

by 10 to the power 8. This value is 4.99 into 10 to the power 20 atoms per meter cube. You

have to be careful. This is given in meter cube. This concentration is given in meter

cube. But other data in part of the problem were in centimeter cube. Be careful about

this and you have to divide wherever necessary to have conformity in the units. What will

be this majority carrier density nn which is almost equal to ND in n-type semiconductor?

We can write it as 4.99 into 10 to the power 20 atoms per meter cube.

Now we have to find out minority carrier density which is asked. Minority carrier here will

be holes. Minority carrier density that is hole density we have to find out and we know

from our previous discussion that pn, the minority carrier density in n-type of material

is ni square divided by ND. When you replace you have to see because ni in our example,

if you look into the problem is given as 1.5 into 10 to the power 10 per cm cube. But here

you see that ND is per meter cube. So we have to bring conformity by multiplying or dividing

accordingly. Now I will find out this hole carrier density that is minority carrier density.

That is 1.5 into 10 to the power 10 per cm cube is given divided by 4.99 into 10 to the

power 20. But this way it will be wrong if I do like this because this is in per meter

cube. So I have to bring this numerator to the same unit as in the denominator. I will

have to bring this centimeter cube to meter cube. So 10 to the power -2 we will have to

multiply, to get it into meter. So it will be 1.5 into 10 to the power 10 divided by

10 to the power -2 into 3; that becomes 10 to the power -6. Then it will be in per meter

cube. There is a square and in the denominator it is 4.99 into 10 to the power 20. So all

are now in per meter cube here in the denominator also.

If I calculate this, then I get this value as 4.51 into 10 to the power 11 holes per

meter cube because I have brought it into meter cube. The hole concentration now is

4.51 into 10 to the power 11 holes per meter cube.

Next part I have to find out the resistivity. Resistivity is 1 upon conductivity. You just

replace these values. Sigma is equal to nn mu n plus pn mu p into q. Here nn electron

concentration is 4.99 into 10 to the power 20 into mobility of electron is 1400. Electron

mobility was 1400 cm square per volt second. So I will convert everything to meter units.

I will have to multiply 10 to the power -4 with 1400 plus the hole concentration I have

found out 4.51 into 10 to the power 11 per meter cube into the mobility of holes is 450.

Again I will multiply by 10 to the power -4 to bring it into meter square unit. The whole

thing I’ll have to multiply by q which is 1.6 into 10 to the power -19 and this calculation

gives me the whole denominator will be 11.1777. So I will get 0.08946 ohm meter.

Now I can convert it to ohm centimeter just to see the difference how the resistivity

changes when you are doping an intrinsic semiconductor. If I convert it to centimeter multiplying

by 10 to the power 2 then I get it as 8.946 ohm centimeter. You compare what was the resistivity

when you had only intrinsic semiconductor and that intrinsic semiconductor had a resistivity

we have done in part a of the problem and that was 22.52 into 10 to the power 4 ohm

centimeter that was of the intrinsic material. Now when we doped it with a material then

that doping material is donor so that doping has brought down this resistivity to 8.946

ohm centimeter. If I find out how much ratio it has brought it down dividing this by 8.946

ohm centimeter, then I find that after doping 25,173.26 times the resistivity of the material

has been brought down.

This effect of doping an intrinsic semiconductor by an n-type or a doping agent is very important

because it has brought down the resistivity means it has increased or enhanced the conductivity

so much. This example illustrated the effect of doping an intrinsic semiconductor by an

external n-type impurity atom that we have learnt today.

backbone of electronic devices and circuits. What is a semiconductor material? Probably

all of you know about conductors and insulators. Conductor is a material which when you apply

a voltage source creates a generous flow of charges like for example copper. Copper is

a metal and metals are conductors. Insulator is a material that offers a very low level

of conductivity when a voltage is applied; for example mica. Semiconductor material is

in between this conductor and insulator. It is a material that has conductivity level

somewhere between the extremes of an insulator and a conductor.

For example germanium and silicon are these two materials which are semiconductors showing

similar property. The resistivity of a material is inverse of conductivity. If we compare

the resistance levels of conductors, insulators and semiconductors we see that the conductor

has resistivity value which is the least among the three. That is, a copper which is a conductor

has a maximum conductivity so its resistivity value is around 10-6 ohm centimeter whereas

an insulator has a resistivity of 1012 ohm centimeter. This is much more than conductor

but semiconductor like germanium and silicon have resistivity values in between these two.

For example if we consider germanium semiconductor it has resistivity value of 50 ohm centimeter

and silicon has resistivity value of 50 x 103 ohm centimeter.

You can see that semiconductor has resistivity in between conductor and insulator. If we

consider these three, i.e insulators, conductors and semiconductors and see their energy band

diagrams, then we find that the energy gap between the valence band and the conduction

band in an insulator is very high.

It is around 6 electron volts as is seen here. This energy gap means it is a forbidden band

and the electrons will have to acquire this much of energy to become conductive and enter

the conduction band. But semiconductor does not have such a high forbidden energy gap

which is around 1 electron volt as is seen here. Electrons can acquire that much of energy

which is 1 electron volt only to jump from valence band to the conduction band in order

to become conductive and metals like copper which are conductors have overlapping conduction

band and valence band and they have zero electron volt. That is why they conduct because electrons

can very easily shift between valence band and conduction band and take part in conduction.

Semiconductor is a material that has conductivity level between the extremes of insulator and

conductor and why this is so?

That is because the semiconductor like germanium and silicon have some unique qualities which

are due to their position in the periodic table and their atomic structure. These are

group IV elements. The atoms of these materials like germanium and silicon have a very different

and definite pattern that is periodic in nature. That is it continuously repeats itself and

one complete pattern is called a crystal and the periodic arrangement of the atoms is a

lattice. Both germanium and silicon have 4 electrons in the outer most orbit which are

called valence electrons. In order to complete the structure of inert gas they will always

try to complete the 8 electrons in the outer most orbit.

In the pure germanium or silicon crystal these 4 valence electrons will be bonded to 4 neighboring

atoms as shown in this diagram. If I consider a germanium atom it has 1, 2, 3, 4 electrons

in the outer most orbit which are called valence electrons. Similarly silicon will also have

4 electrons in the outer most orbit because they are group IV materials.

They will try to complete the 8 electron structure. That is why they will share electrons with

the neighboring atoms as is seen here. If we consider silicon atom, these 4 electrons

in the outermost orbit are sharing electrons among themselves so that the total number

becomes 4 and 4, 8. These atoms which are nearby are sharing these covalent bonds.

These covalent bonds are formed between neighboring atoms of silicon and thereby they complete

the structure of the inert gas having 8 electrons in the outermost orbit so that they become

stable. These semiconductors which are in the purest form are known as intrinsic semiconductors.

In these intrinsic semiconductors the free electrons are due to only natural causes like

thermal energy or light energy which impart extra energy to the electrons to become free

and thereby they get additional kinetic energy from these natural causes and they break this

covalent bond. These covalent bonds will break in order to become free. 1 electron will become

free when it absorbs kinetic energy from either the surrounding or any light is falling on

them.

An increase in temperature is naturally the reason for more conductivity in semiconductors.

This increase in temperature of a semiconductor results in increase in the number of free

electrons in the material so they become more and more conductive. Even though they may

be not so conductive at certain temperature but if we increase the temperature they will

absorb more kinetic energy and will become more conductive. The semiconductors like germanium

and silicon they have negative temperature coefficient as the resistance will decrease

with increase in temperature. This is a peculiar phenomena that is seen in semiconductors which

is just opposite to your conductors like metal.

We have been talking about the intrinsic semiconductors. Those are the purest form of semiconductors

without any impurity in them. But there is another type of semiconductor which is extrinsic

semiconductor.

This extrinsic semiconductor has the characteristics such that their property can be significantly

altered by addition of certain impurity atoms into the pure semiconductor material. The

impurities added to the purest semiconductor to make an intrinsic into extrinsic semiconductor

are defined. We can add pentavalent impurity materials like antimony, arsenic and phosphorous

which are group V elements and have 5 electrons in the outer most orbit or we can add trivalent

materials like boron, gallium and indium which are group III materials and have 3 electrons

in their outer most orbit and the standard norm of adding is that they are added 1 part

in 10 millions. That is to 10 million atoms of intrinsic semiconductor you add one impurity

atom.

We will discuss how the addition of this impurity totally changes the electrical properties

of the intrinsic semiconducting material. When certain impurity atoms like antimony

or boron are added to a pure semiconductor material we get an extrinsic semiconductor

and this process of adding certain impurity atoms like antimony or boron into pure semiconductor

like silicon is called the doping process. As a result of this doping we get extrinsic

semiconductor either n-type or p-type.

That is you can get n-type extrinsic semiconductor if you add pentavalent impurity material like

antimony or phosphorous into a pure semiconductor like germanium or silicon or you can have

p-type extrinsic semiconductor when you add trivalent impurity material like boron or

indium into a pure germanium or silicon semiconductor.

What happens when you add an impurity atom to a pure semiconductor? Let us take for example

an n-type semiconductor. How it is formed? N-type semiconductor means we have to add

pentavalent impurity material. Let us take the example of adding antimony which is a

pentavalent impurity material which is a group V material having 5 electrons in the outer

most orbit to a silicon semiconductor. This silicon atom has 4 electrons in the outer

most orbit. These are silicon outermost orbit electrons and now as we have added 1 antimony

atom which has 5 electrons in the outer most orbit, it will share these electrons with

the silicon atoms and these covalent bonds will be completed. 4 and 4, 8 electrons are

there so these covalent bonds will be completed. But then 1 electron is in excess. That is

it is extra. This fifth electron is donated by this antimony atom. This fifth electron

is free. It can take part in conduction easily. That is why it is donating 1 electron. It

is excess of 1 electron which is negatively charged.

This is called n-type semiconductor. What happens when we add p-type semiconductor?

We have seen how n-type semiconductor is formed by adding pentavalent impurity material to

intrinsic semiconductor like germanium or silicon. Let us see how p-type semiconductor

is formed from pure semiconductor like germanium or silicon by adding trivalent impurity material

like boron or indium. Let us add boron impurity which is a group III material having 3 electrons

in the outer most orbit to a silicon semiconductor. This boron has 3 electrons in the outer most

orbit. It will complete its covalent bond with neighboring silicon atom having shared

these 2 electrons as is seen here. These 2 electrons will form a covalent bond. In this

process 3 covalent bonds are now completed. What about the other covalent bond? This covalent

bond is now devoid or it is short of 1 electron to complete its covalent bond. This shortage

of 1 electron means it is having less negative charge. This less negative charge or a positive

charge is there in this covalent bond. This is called a hole. Hole means it is basically

devoid of 1 electron. So there is a hole formed here in this covalent bond and this hole is

a positively charged particle.

So this p- type semiconductor basically is formed in this way which will have holes as

charge carriers.

If we consider again the n-type material we have seen that in n-type material which is

an extrinsic semiconductor it has donated 1 electron. As it has donated 1 electron,

the atom of this extrinsic semiconductor is now having 1 positive charge which is called

an ion. 1 atom of this extrinsic semiconductor when it donates one electron it becomes ion

having 1 positive charge which is shown here. The donor ions are having 1 positive charge

and there will be electrons which are in abundance in the n-type material. These are called majority

carriers. In an n-type material the majority carriers are electrons but there will also

be in less number the minority carriers which are holes. These holes will be from this intrinsic

material that is coming. If n-type material is considered altogether we will see majority

carriers as electrons, minority carriers as holes and immobile ions which are positively

charged or donor ions.

This is the structure of n-type material. Similarly if we consider a p-type material,

p-type material has acceptor ions. In the p-type material we have seen that there is

a hole and as it is devoid of an electron it will be always trying to get an electron

from the neighborhood atoms. The electron from the neighborhood atom will come and fill

up this hole and as a result this boron atom will be negatively charged. It is accepting

1 electron so it will be negatively charged.

That means in a p-type material the atom of the impurity will be an acceptor ion having

a negative charge and there will be holes which are positively charged as majority carriers

and also very less in numbers although there will be presence of minority carriers which

are electrons which are coming from the intrinsic semiconductor. This is p-type material where

all these carriers and ions you see here. We can summarize that in n-type semiconductor

the majority carriers are electrons and minority carriers are holes and in p-type semiconductor

majority carriers are holes but minority carriers are electrons.

These are the charge carriers available in n-type and p-type semiconductors. How does

the current flow in a semiconductor? Current flow is due to the movement of electrons and

conventional current is always against the direction of movement of electrons. Suppose

if electron moves in this way then conventional current direction will be in opposite way.

This is electron movement direction and this is conventional current direction. But we

have seen that another type of charge carrier is also there in semiconductors which are

holes and which are positively charged. So the current flow will be now due to these

two types of carriers; one is electron and other is hole and both are opposite in charge.

That is electron is negatively charged and hole is positively charged.

We are having a semiconductor. We have the silicon atom and this is an impurity atom

which is boron. It is creating a hole as is seen here. In order to fill up this hole electron

from a neighboring atom will come. This movement of electron will be in this direction. When

this electron comes and fills up this hole then this place will be now devoid of electron.

It will be now a hole. Now hole movement will be from this position to this position. It

is just opposite to the electron movement. Electron movement is in the direction of this

solid arrow and hole movement is in the direction of this dotted arrow. If we consider the conventional

current direction which is opposite to the direction of movement of electrons, then the

current direction will be in the same direction, because of both these components. If I consider

conventional current direction it will be like this because this electron movement is in this

direction and hole movement is in opposite direction. But the final direction of current

flow is opposite to the electron movement.

So it will be in opposition to this movement of electron but that is the same direction

of the hole movement. We will discuss a very fundamental and very important law which gives

the relationship between electron and hole concentration and that is known as mass action

law. What is that law? It states that under thermal equilibrium the product of the free

negative and positive concentrations is a constant independent of the amount of donor

and acceptor impurity doping. That is mathematically if you write np is equal to ni square where

n is the concentration of free electrons in thermal equilibrium, p is the concentration

of holes in thermal equilibrium and ni is the intrinsic carrier concentration.

This is a very fundamental law in electronics which is mass action law. Now let us consider

about the charge densities in a semiconductor. We will consider charge density later. Before

that you must also know what is the law of electrical neutrality that is satisfied in

the semiconductor? Let us consider a semiconductor material and consider a situation where it

is doped by both n-type and p-type materials and let ND be the concentration of donor atoms.

These donor atoms are all ionized and already we have seen that these will be positively

charged donor atoms. We have ND positive charges per cubic meter which are contributed by these

donor ions and let p be the hole concentration in the semiconductor so that the total positive

charge density in the semiconductor is ND plus p. That is total positive charge density

in the semiconductor which is contributed by donor atoms as well as holes. Similarly

if we consider that NA is the concentration of acceptor ions then these contribute NA

negative charges per cubic meter and let small n be the electron concentration in the semiconductor.

So the total negative charge density in the semiconductor will be now NA plus n and since

the semiconductor is electrically neutral we have this law to be satisfied that ND plus

p is equal to NA plus n. Let us number this equation as 2.

That means this electrical neutrality must be satisfied in a semiconductor given by this

equation. Consider solely n-type material. We can either dope by n-type or by p-type

impurity. So in an n-type material doping NA will be zero and then the equation 2 will

be now ND plus p is equal to n since NA is equal to zero or ND is equal to n minus p.

As it is an n-type material the concentration of electrons is much much more than concentration

of the holes. So n minus p we can roughly say that equal to n. Then from this equation

we will get that the electron concentration is almost equal to the concentration of the

donor atoms ND and if we use a subscript small n, the electron concentration in an n-type

material nn is equal to capital N subscript capital D. From this equation we can get now

hole concentration for this n-type material which is given by pn is equal to ni square

by ND. Here we are using equation number 1, mass action law.

Similarly if we consider a p-type material, the donor concentration is zero since it is

a p-type material. So the equation 2 will be now reduced to p is equal to NA plus small

n or NA is equal to p minus n and since it is a p-type material the hole concentration

is much, much more than the electron concentration, p minus n can be approximately written to

be equal to p. Finally in the equation 2 we get p almost equal to NA. So in a p-type material

hole concentration is approximately equal to the density of acceptor atoms. If I use

a subscript small p, p subscript p is almost equal to NA and similarly from equation 1,

mass action law if we use, we can find out the concentration of electron in a p-type

material which is equal to ni square by NA.

Here I am substituting in this equation np is equal to ni square. I am substituting this

p with NA. Then I get this n equal to ni square by NA. The fundamental difference between

a metal and a semiconductor is that metal is unipolar since current conduction is taking

place only by electrons which are having only one sign that is negative sign.

But in semiconductor there are two charge carriers having negative as well as positive

charges which are known as electrons and holes. Semiconductor is a bipolar material. There

are 2 types of charge carriers positive as well as negative. Holes are positively charged

and electrons are negatively charged. Whenever you apply an electric field, E then your holes

and electrons will move in opposite directions since they are of opposite sign but the current

due to both these two components will be in one direction.

If you want to find out the current density after the application of an electrical field

E, that is given by this expression. The current density J is equal to n times mu n plus p

times mu p into q into E which is known as sigma into E. What are those terms? n is magnitude

of free electron concentration, mu n is mobility of the electrons, p is magnitude of the hole

concentration and mu p is the mobility of holes, q is the charge of an electron and

sigma is known as conductivity which is given by this expression n mu n plus p mu p into

q. For intrinsic semiconductor as n is equal to p which is equal to ni that is intrinsic

concentration, we get for intrinsic semiconductor if you want to find out the current density

you will have to simply substitute ni both for n and p.

Let us take one example and see how you can find out the current density. Consider an

intrinsic silicon bar of cross section 5 centimeter square and length 0.5 centimeter at room temperature,

300 degree Kelvin. An average field of 20 volt per centimeter is applied across the

ends of the silicon bar. It is the silicon bar having a cross section of 5 centimeter

square and you are applying an average field of 20 volt per cm. Now you have to calculate

electron and hole component of current density, total current in the bar and resistivity of

the bar; these three you have to find out. You are given that the mobility of electrons

is 1400 centimeter square per voltage second and hole mobility is 450 centimeter square

by volt second and intrinsic carrier concentration of silicon at room temperature, 300 degree

Kelvin is generally taken as room temperature. Kelvin is small t degree centigrade plus 273

degree that gives you the Kelvin temperature. Here it is 300 degree Kelvin. You can find

out at what room temperature this example is carried out 300-273. At 27 degree centigrade

room temperature the sample is given and now at that temperature the intrinsic carrier

concentration of silicon is given as 1.5 into 10 to the power 10 per centimeter cube. You

have to find out electron and hole component of current density.

If we are to find out electron and hole component of current density we must use the equation

J equal to n mu n q E plus p mu p q E. Now you have to see the data given to you. We

know already n is equal to p is equal to ni in an intrinsic semiconductor, since this

is an intrinsic semiconductor and it’s value is given in the data as 1.5 into 10 to the

power 10 per centimeter cube. Electron mobility and hole mobility are also given. Electron

mobility mu n is 1400 centimeter square per volt second and hole mobility is 450 centimeter

square per volt second. But this charge of an electron is not given. It is supposed that

you know this and its value is standard 1.6 into 10 to the power -19 coulomb.

It is simple to find out this current density J given by this equation. In this equation

there are 2 components, one is due to the electrons and the other is due to the holes.

Electron component of current density we can find out from this part of this equation.

You substitute each of these values and by an intrinsic concentration mu n q E, E is

30 volt per cm and each of these are substituted by the given values then we get this equation

and finally the answer comes to 67.2 micro ampere per cm square. That is the electron

component of the current density.

This current density is due to the electrons. Similarly you can find out the hole component

of current density. That is p mu p q E. Here also you substitute all these given values

and you find 21.6 micro ampere per cm square, that comes to be the hole component of current

density. Another thing you have to find out is the total current. This is only current

density. In order to find out the total current you have to multiply it by the cross sectional

area which is given as 5 cm square. Current density mu A per cm square we have found out.

Each of these hole component and electron component you have to add up. Then you will

get the total component, total current density. 67.2 plus 21.6 is the total current density

multiplied by the cross sectional area 5 centimeter square. Finally we get 444 micro ampere which

is the current flowing in the silicon bar and finally the resistivity of the silicon

bar which is inverse of conductivity. 1 by conductivity gives you the resistivity. Conductivity

is sigma which is given by n mu n plus p mu p into q. So 1 by this whole term will give

you the resistivity and substituting each of these values we get finally the resistivity

of the silicon bar is 22.52 into 10 to the power 4 ohm centimeter.

In these examples we have used only one equation that is the current density equation. The

rest of the things are found from that.

In today’s class we have learnt about intrinsic semiconductor material as well as extrinsic

semiconductor material and their characteristics or properties we have studied and this semiconductor

is the main component in all electronic applications and devices which is made of intrinsic semiconductor

silicon or germanium and when it is doped with impurity atoms like pentavalent or trivalent

materials then you get n-type as well as p-type extrinsic semiconductors. These will be used

in all electronic devices. In today’s class we have learnt about intrinsic as well as

extrinsic semiconductors. How extrinsic semiconductors are formed from intrinsic semiconductors by

doping them with p-type or n-type materials and we have also seen the characteristics

or properties of these extrinsic semiconductors. This semiconductor is used as a backbone in

all semiconductor devices and circuits. Let us try one example.

Consider an intrinsic silicon bar of cross section 5 centimeter square and length 0.5

centimeter at room temperature 300 degree Kelvin. An average field of 20 volt per centimeter

is applied across the ends of the silicon bar. In part a, calculate number i, electron

and hole component of current density; number ii, total current in the bar and number iii,

resistivity of the bar. Part b of the problem is if now donor impurity to the extent of

1 part in 10 to the power 8 atoms of silicon is added find the density of minority carriers

and the resistivity. The data given to you are electron mobility is given as 1400 centimeter

square per volt second, hole mobility is 450 centimeter square per volt second, intrinsic

carrier concentration of silicon at room temperature 300 degree Kelvin is given as 1.5 into 10

to the power 10 per centimeter cube and number of silicon atoms which are doped per meter

cube is 4.99 into 10 to the power 28. This last data will be required for the part

b of the problem. Let us first of all do part a of the problem. It is given that it is an

intrinsic semiconductor. The hole concentration and electron concentration is same which is

the intrinsic concentration and it is given as 1.5 into 10 to the power 10 per centimeter

cube. Electron mobility is given as 1400 centimeter square per volt second. Hole mobility is given

450 centimeter square per volt second and charge of an electron is not given but you

should know this value which is 1.6 into 10 to the power -19 coulomb. With this data we

will now find out the current density first. As you know the current density J is given

by n mu n q into E plus p mu p q into E where E is the applied electric field and all the

other data are given to you. This current density has 2 components one due to the electron

and other due to the hole. Electron component of current density will be the first part;

this part will be the electron component of current density.

You just put down the values and you know that electron concentration is nothing but

intrinsic concentration. So it will be ni mu n q into E. Now substitute the values which

are known to you. Each one we will substitute. ni is 1.5 into 10 to the power 10 per cm cube

and electron mobility is 1400 centimeter square per volt second. q charge of an electron is

1.6 into 10 to the power -19 coulomb and the applied electrical field is 20 volt per cm.

All these are in cm that you must be careful because centimeter is the unit in all the

data. If you calculate this value then we are getting 67.2 micro ampere per centimeter

square. Part one of the problem a, is now 67.2 micro ampere per centimeter square.

Now in part 2, hole component of current density has to be found out and that is the second

part of the equation J which is p mu p q into E and that can be replaced by ni mu p q E

because ni is the intrinsic concentration that is equal to the hole concentration as

well as electron concentration in intrinsic semiconductor. Substituting each of these

values ni is 1.5 into 10 to the power 10. Then mu p is 450, q is 1.6 into 10 to the

power -19 and E is 20. We now calculate this and the result is 21.6 micro ampere per cm

square.

That is the hole component of current density. Third part of the problem says that you have

to find out the total current in the silicon bar. In order to find out the total current

you have to multiply the current density J by the cross sectional area. The total current

in the bar is equal to the current density into the cross sectional area of the bar.

Current density is the total current density due to the holes as well as electrons. We

have to sum up these two 67.2 plus 21.6 micro ampere per cm square and multiply it by the

5 cm square cross sectional area and that gives the result as 444 micro ampere which

is the total current and the next part of the problem is resistivity. This is part iii

and this is part ii.

Part iii is resistivity of the silicon bar. Resistivity is 1 upon conductivity and conductivity,

sigma is n mu n plus p mu p into q. You substitute all these values which we already have because

n and p is nothing but ni. ni can be taken out; ni into q then mu n plus mu p which will

give you the resistivity and that is equal to 22.52 into 10 to the power 4 ohm centimeter.

Now coming to the part b of the problem in which you have to find the resistivity of

the doped semiconductor as well as minority carrier concentration. If you look into the

problem it is doped and the doping impurity is an n-type of doping material. Donor impurity

is 1 part in 10 to the power 8 atoms of silicon. In part b of the problem you are doping silicon

with donor atom. The concentration of the donor atoms ND will have to be found out and

we have been told that the doping is 1 in 10 to the power 8 atoms of silicon. We also

know and it is given in the problem that number of silicon atoms per meter cube is 4.99 into

10 to the power 28. We are doping 1 in 10 to the power 8 silicon atoms. What will be

the donor concentration? Donor concentration will be 4.99 into 10 to the power 28 divided

by 10 to the power 8. This value is 4.99 into 10 to the power 20 atoms per meter cube. You

have to be careful. This is given in meter cube. This concentration is given in meter

cube. But other data in part of the problem were in centimeter cube. Be careful about

this and you have to divide wherever necessary to have conformity in the units. What will

be this majority carrier density nn which is almost equal to ND in n-type semiconductor?

We can write it as 4.99 into 10 to the power 20 atoms per meter cube.

Now we have to find out minority carrier density which is asked. Minority carrier here will

be holes. Minority carrier density that is hole density we have to find out and we know

from our previous discussion that pn, the minority carrier density in n-type of material

is ni square divided by ND. When you replace you have to see because ni in our example,

if you look into the problem is given as 1.5 into 10 to the power 10 per cm cube. But here

you see that ND is per meter cube. So we have to bring conformity by multiplying or dividing

accordingly. Now I will find out this hole carrier density that is minority carrier density.

That is 1.5 into 10 to the power 10 per cm cube is given divided by 4.99 into 10 to the

power 20. But this way it will be wrong if I do like this because this is in per meter

cube. So I have to bring this numerator to the same unit as in the denominator. I will

have to bring this centimeter cube to meter cube. So 10 to the power -2 we will have to

multiply, to get it into meter. So it will be 1.5 into 10 to the power 10 divided by

10 to the power -2 into 3; that becomes 10 to the power -6. Then it will be in per meter

cube. There is a square and in the denominator it is 4.99 into 10 to the power 20. So all

are now in per meter cube here in the denominator also.

If I calculate this, then I get this value as 4.51 into 10 to the power 11 holes per

meter cube because I have brought it into meter cube. The hole concentration now is

4.51 into 10 to the power 11 holes per meter cube.

Next part I have to find out the resistivity. Resistivity is 1 upon conductivity. You just

replace these values. Sigma is equal to nn mu n plus pn mu p into q. Here nn electron

concentration is 4.99 into 10 to the power 20 into mobility of electron is 1400. Electron

mobility was 1400 cm square per volt second. So I will convert everything to meter units.

I will have to multiply 10 to the power -4 with 1400 plus the hole concentration I have

found out 4.51 into 10 to the power 11 per meter cube into the mobility of holes is 450.

Again I will multiply by 10 to the power -4 to bring it into meter square unit. The whole

thing I’ll have to multiply by q which is 1.6 into 10 to the power -19 and this calculation

gives me the whole denominator will be 11.1777. So I will get 0.08946 ohm meter.

Now I can convert it to ohm centimeter just to see the difference how the resistivity

changes when you are doping an intrinsic semiconductor. If I convert it to centimeter multiplying

by 10 to the power 2 then I get it as 8.946 ohm centimeter. You compare what was the resistivity

when you had only intrinsic semiconductor and that intrinsic semiconductor had a resistivity

we have done in part a of the problem and that was 22.52 into 10 to the power 4 ohm

centimeter that was of the intrinsic material. Now when we doped it with a material then

that doping material is donor so that doping has brought down this resistivity to 8.946

ohm centimeter. If I find out how much ratio it has brought it down dividing this by 8.946

ohm centimeter, then I find that after doping 25,173.26 times the resistivity of the material

has been brought down.

This effect of doping an intrinsic semiconductor by an n-type or a doping agent is very important

because it has brought down the resistivity means it has increased or enhanced the conductivity

so much. This example illustrated the effect of doping an intrinsic semiconductor by an

external n-type impurity atom that we have learnt today.