Welcome to this presentation. In this lecture we will derive the conductivity of semiconductors. Metals are unipolar.
There is only one type of charge carrier in metals that is electron thus it is unipolar. In case of semiconductors we have two types of charge carriers electrons and holes thus it is bipolar. If E is the electric field applied across the semiconductor, the electrons and holes move in opposite direction because of their opposite sign.
But the current due to them will be in the same direction. So let me write down this points. When electric field E is applied, electrons and holes are drifted in the opposite direction.
Electrons and holes are drifted in opposite direction. But the current due to them will be in the same direction. Now let's say the drift velocity of electron.
Let us consider the case of electron and if Vd is the drift velocity. for electron Then we can say that the drift velocity Vd is directly proportional to the applied electric field intensity. And if I remove this proportionality sign, I'm going to have a constant mu.
Mu is called mobility. Mu is called mobility and it is constant. The mobility of a material is constant. This is... mobility and it is constant.
Let us find out the unit for mobility. The unit for mobility mu is equal to VD by E. VD is velocity.
hence having unit meter per second E is electric field intensity and in terms of potential difference I can write it as volts per meter so we have the unit we have the unit of mu as meter square volt seconds so this is the unit for mobility and if we talk about the current density then we can easily calculate the current density because we know the current i equals to n e a v d and is the number of charge carriers e is the charge of the charge charge carriers that is the charge of one electron. A is the area of cross section, Vd is drift velocity. Now current density J is given as I per unit cross section area.
So we will divide both the sides by A and eventually we are going to get NeVd and we have Vd equal to Mu E. So J current density is equal to NeVd. density equals to Ne mu E. This portion of right hand side is represented as sigma and sigma is conductivity.
Sigma is conductivity and this is also the property of the material that we are using so we can write J equals to Sigma E and this is nothing but Ohm's law This is Ohm's law. Now in case of semiconductor materials the whole movement is exactly like electron movement. It is like positive charge movement. The charge is equal to the electron that is 1.6.
The charge of electron is 1.6 into 10 to the power minus 19 coulomb. This is charge of one electron. It can be represented by Q, capital Q or E. It doesn't matter whatever representation you are using.
I am using E to represent the charge of one electron. In some books they will use Q. So don't get confused in this representations. And the positive charge is also going to have the same value of charge that is 1.6 into 10 to the power minus 19 coulomb.
The charge is equal to electron with different mass. But the mass is different and we have to talk about two types of mobilities because of this reason. This mu is the mobility of electron but the mobility of hole will be different because it is having different mass. I can say mobility is inversely proportional to the effective mass. The mobility is inversely proportional to effective mass and the effective mass of hole.
The effective mass of hole is greater as compared to the effective mass of electron. So we can say that the mobility of hole is lesser as compared to the mobility of electron. Now we will include the mobility of hole as well.
So we are going to have J equals to. Mu n is the mobility of electron. Let's represent this by mu n and We have to include the mobility of holes So we are going to have the concentration of holes and then mobility of holes and here we will have E Capital E. I have just modified this relation that we have derived n is the concentration of electron mu is the mobility of electron. I have just included the concentration of holes and the mobility of holes. Now we will compare this relation with Ohm's law and we can find out conductivity sigma by comparison and sigma is equal to small n mu subscript n plus small p mu p e.
So this is the mobility. in case of semiconductors and we are going to use this formula in our numerical problem so this is very important you may write it down now we can move to our numerical problem the mobilities of free electron and holes in a pure germanium are 0.38 and 0.18 meter square volt second find the value of intrinsic conductivity assume ni equals equals to 2.5 into 10 to the power minus 19 per meter cube at room temperature. We have to find out conductivity, intrinsic conductivity.
The conductivity is intrinsic because in question it is given a pure germanium. Pure germanium is intrinsic semiconductor because there is no impurity and hence the conductivity that we are going to obtain after the calculations is intrinsic. So let's move to the solution part.
0.38 is the mobility of free electron and I have already told you the mobility of electron is going to be higher as compared to the mobility of hole and you can see 0.38 is the mobility of electron and 0.18 is the mobility of holes. 0.18 is smaller as compared to 0.38. So mu n is 0.38 meter square volt seconds mu p is 0.18 meter square volt second.
This is given in the problem and the value for ni is 2.5 into 10 to the power 19 per meter cube. If you remember the past few presentations I explained you in case of intrinsic semiconductors the electron concentration and the hole concentration is equal to intrinsic. concentration.
So we have the values of n and p equals to ni. So we have everything now and we can just put all these values in this formula and we will have our conductivity. So let's do it. Sigma equals to mu n is 0.38 mu p is 0.18 and and p both are 2.5 into 10 to the power 19. So let's take it common and we will have 2.5 into 10 to the power 19 multiplied with the charge of electron that is 1.6 into 10 to the power minus 19 and the unit for conductivity is going to be ohm meter raised to power minus 1. So we have to do a simple calculation here.
to get our conductivity. I'm going to use my calculator for this purpose. Okay, 0.38 plus 0.18, 0.56, 0.56, 2.5 multiplied by 1.6 will give us 4 into 4, 10 to the power 19 and 10 to the power minus 19 will give us 10 to the power 0. So finally we will have 2.24 ohm meter raised to power minus 1 as our conductivity. So this is the answer and we will discuss more complicated problems in the next presentation. This is all for this presentation.
See you in the next one.