00:01
This is the problem one and first you have to understand what is the intrinsic semiconductor the semiconductor which have no impurity at room temperature that means pure form of the semiconductor is known in the intrinsic semiconductor and it requires that is no impurity in pure semiconductor or you can say that the pure form of the same is known as the intrinsic semiconductor and in the intensive semiconductor the number of electron and the number of holes are equals that is n is equal to p is equal to and here ni is known in the intrinsic carry concentration and in the energy band diagram if you see the energy levels of ac and this is av and the fermi energy level in integer lie exactly in the mid of the energy band gap this is efi and suppose this is the energy band gap eg then this efi will be at eg by 2 these are the some properties of the increasing semiconductor and if this n is greater than n i then it becomes the n type semiconductor or if p is greater than an then it becomes the p -type semiconductor or we can define if the fermi energy level is greater than of this efi then it is n -type semiconductor and if ef is less than efi that means p -type or sorry ef is less than efi then it is p -type semiconductor normally the conductivity of the intrinsic semiconductor is low the conductivity electrical conductivity of intensive semiconductor semiconductor is very low so hence we dove the semiconductor by some impurities and that semiconductor becomes the extensive semiconductor so that means the semiconductor without any impurity is known as the intensive semiconductor and the second part of the question is drive this relation for the carry concentration in intensive semiconductor that mean drive the n i derivation for intrinsic carry concentration in order to derive this carry concentration that is n i we should know the expression for the electron concentration and the whole concentration in thermal equilibrium let us first derive the electron concentration in the conduction band and then holes in the valence band and the number of electron in the conduction band are detomined by an e is equal to density of allowed state in the conduction band and denoted by the gce and the probability of the state which can be occupied by the electron this is ffe and this is known the firm in drug distribution function let us assume this is the equation number one now the number of electron in the conduction band in the range de is here no represent the number of electron in thermal equilibrium is the band that is ec2 infinitive and this gce and this value and this value ffe and d .e.
05:28
So we are finding the number of electron in the conduction band that means from the edge ec to infinite and here the value of the gce is it is 4 pi 2 mn star here mn star is the effective mass of the electron 3 by 2 upon h cube e minus e c here is the energy of the electron in the conduction band half and this f f f fd and the value this f f f f e is it is known as the fermi drug distribution function and it is one upon one plus exponential e minus e f f f f bt and let us us assume the e minus e c that the energy of the electron and this ac this difference is greater than kt then we can say that the e minus ef will also greater than 3 is kt and this approximation is known as the boltzman approximation that means then in that case this ff becomes one upon this one can be neglected in that case it will be exponential e minus e f by kb t or we can write a exponential minus e .f by kb t now this equation becomes here we will put this value here and this is constant it can be taken out from the integration it is a it will be that is an no is equal to 4 pi 2 m star power 3 by 2 upon h our 3 is the blank constant here ec to infinitive e minus ac power 1 by 2 exponential international minus e minus ef by kbt and d.
09:09
Let this equation is to solve this equation to integration let us assume e minus e c by k b t is equal to some variable that is y from here it will be e is equal to e plus k b t y and d e is equal to this ec is equal to constant it will be 0 plus kb t and d y find the integration limit when e is equal to ec let's find limit here so e is equal to e c in that case e minus cc that is y is equal to 0 when e is equal to infinitive in that case y will be infinitive now substitute these all values in the equation 2 then an o is equal to 4 pi 2 mn star 3 by 2 upon h power 3 this this will be 0 2 infinite and this e minus ec is equal to kbt power half kbt power half and y power half and this exponential minus this e can be placed by this ec plus kb t by that is minus ec plus kbt y is the value of e minus e c plus k b t y is the value of e minus minus e f divided by kb t and d is equal to d is equal to kb t d y from here this kb t a power and this kvt is one power that kbt power three by two that is an o is equal to four pi two mn star three by two upon hq into k b t power 3 by 2 and in the integration it is 0 to infinitive it is y power of and this can be written as ec minus ea by kb t minus ec bt minus e f by kb t and this portion came written as the expression minus kb t and kb t to kb t you can see that is exponational minus y and d here this value is also constant it can be taken out from the integration that is n -no is equal to this whole value into exponational minus e -c minus ef by kbt and this integration is 0 to infinity y power half exponential minus y dy and this is the gamma function integration and the value of this will be gamma 3 by 2 and the value will be 1 by 2 square root of pi hence this n node will be n no is equal to this whole constant that is 4 pi it is 2 m and star power 3 by 2 kb t power 3 by 2 and divide by h cube this is here into 1 by 2 square root of pi exponential minus ec minus e f divided by k bt let's rearrange this whole and write down again and no is equal to you can see that this pi power is also 3 by 2 the combined whole the three power it is 2 mn star k bt pi also inside the bracket kb t and if i write this h square then it will also h square and this whole power is 3 by 2 and this will be 4 and this is exponential minus ec minus ef by kbt.
16:32
If i write this a constant that it will be n0 is equal to nc exponential minus ef by kbt.
16:56
And here n c is the pseudo constant or effective density of states this is the electron concentration in the conduction band in thermal equilibrium similarly we can find the value the whole concentration in the valence band nv exponential minus e f minus evi v divided by k b t and when you drive this relation you have to consider this f p e is equal to one minus electron that is f f f e and solve the integration as same in the previous calculation so these are the whole concentration in the balance vector now we have to drive the electron intrinsic carry concentration as we know the number of electron and number of holes are equal in the intrinsic semiconductor that is n i if i take product of these two from two equations suppose that is this is equation this is three this is three this is four if i take the product of these to o then n o dot p is equal to n i dot n i that is n i square is equal to n o dot p o that is n c nv exponential minus ec minus ef by kbt dot exponential minus ef minus ev by kbt.
19:59
Or rearrange this one that is n -i square is equal to nc and b exponential minus ec plus ef minus ef and this will be plus eb by kbt and this ef and this ef to ef cancel then here you will get nc nv exponential minus e xonational minus e c b by kv t and this.
21:06
Is the energy difference between the conduction in the balance band and it is represented by the energy band gap of the semi -character that is eg so it is nc n b exponential minus a g by kb and n i is equal to nc n b square root exponent exponential minus a g by 2 k b t now if we substitute the value of the nc and v, here you can see the value of the ncn and there is 2 mn and the 2 mp star for in case of the holes...