in the last lecture I introduced transistors in this lecture I will explain how transistor works I also explained the meaning of named BJT BJT B stands for bipolar B stands for bipolar we have bipolar in the name because there is involvement of both type of charge carriers there is involvement of electrons as well as holes in transistor that's why we have bipolar in the name G stands for Junction J stands for Junction two junctions are formed in transistor Junction j1 and Junction j2 two junctions are formed in transistors so we have Junction in the name now the most important thing is the meaning of transistor why we call this three terminal device transistor the name transistor is coined from word transferred resistor transistor is coined from transferred resistor in which we have trans and store I sto R and this two makes the word transistor now what is the meaning of transferred resistor if we talk about active mode if we talk about active mode of operation then we already know Junction j1 is forward biased and Junction j2 is reverse biased when Junction j1 is forward biased it will offer a very low resistance or ideally resistance should be zero when Junction j2 is reverse biased it will offer a very high resistance or ideally the resistance must be infinity so this is what we have in active mode and if we have same current same current flowing then initially it will flow through a low resistance let's say the current is I and after this at output it will flow through high resistance let's say it is capital R and the same current is flowing through both these resistances so somehow we have transferred the low resistance to the high resistance that's why we have the name transferred resistor and from this we got transistor there is very significant use of this thing for example let's say VI is the input voltage and V o is the output voltage we are measuring VI across this small resistance and we are measuring vo across this large resistance so V I is equal to I multiplied by r and vo is equal to I multiplied by capital R the current is same and the input resistance is smaller than the output resistance so we can say that VI is smaller than we owe so we had a weak signal at the input but we have an amplified signal at the output so there is amplification amplification by using the three terminal device all these things will be clear when I explain the working of NPN transistor in active mode this is the NPN transistor so the emitter region is n the base region is P and the collector region is n this is NPN transistor this terminal is emitter terminal this terminal is base terminal and this terminal here is collector terminal and we want to operate this transistor in active mode we want to have proper eight this transistor in active mode and we already know in active mode Junction j1 is forward bias so two forward bias Junction j1 we need to we need to apply a forward bias potential and is connected to the negative terminal and P is connected to the positive terminal so emitter is connected to the negative terminal and base is connected to the positive terminal let's say this forward biasing potential is V EB and we have to reverse bias Junction j2 so n is connected to the positive terminal and P is connected to the negative terminal collector is connected to the positive terminal base is connected to the negative terminal and let's say this reverse biasing potential is we see B now we have Junction j1 forward biased and Junction j2 reverse bias now we will analyze the movement of electrons and holes in this three terminal device and let's say let's say VB is the barrier potential for Junction j1 in Junction j2 when the transistor terminals are open circuited there is no biasing potential in the transistor so VB is the barrier potential for Junction j1 and Junction j2 when there is no biasing potential and we have to analyze what will happen to the barrier potential once we apply the biasing potentials in the transistor this is barrier potential of Junction j2 and it is equal to VB this is barrier potential of Junction j1 and it is also equal to VB now Junction j1 is forward biased after the application of V EB and the barrier potential will now reduce and let's say the new barrier potential is this the new barrier potential the new barrier potential is equal to VB VB minus V EB forward biasing potential on the other hand Junction j2 is reverse biased so barrier potential will increase and the new barrier potential the new barrier potential is equal to VB VB plus VC B plus we see be the reverse biasing potential now we can easily analyze the movement of electrons and holes because we have idea about the barrier potentials because of the reduced barrier potential at Junction j1 the electrons on the N side that is the emitter will cross the junction and move to the base and recombine with the holes in the base there is one very important thing that you always have to keep in your mind base in case of transistor is very small it's very small and it is also lightly lightly domed and because of this because base is thin and it is lightly doped there is very small recombination of electrons from the emitter and most of the electrons from the emitter pass over to the collector I will repeat this point again barrier potential at Junction j1 is reduced so electrons from emitter will cross over the junction and very small amount of electrons from the emitter will recombine in the base because base is very small very thin and it is lightly doped so most of the electrons will cross the junction j2 because they have high velocity and this implies they have high kinetic energy so most of the electrons emitted by the emitter will find themselves in the collector a very few electrons very few electrons will recombine with the holes and this means electrons will move to VEB the positive terminal of the forward biasing potential let's say n number of electrons enter the base and number of electrons enter the base out of which one minus alpha n out of which one minus alpha n electrons combined with the holes in the base and alpha n alpha and electrons move to the collector this is what is happening in the transistor only 2 to 5% 2 to 5% electrons are combined in the base and 95 to 98% electrons are collected in the collector therefore most of the electrons emitted by the emitter moved to the collector so this is what will happen when you forward bias the junction j1 and junction j2 in bipolar Junction transistor there is one more thing that we must not forget and it is the reverse saturation current Junction j2 is reverse biased so there must be reverse saturation current through Junction j2 we have minority charge carriers on n side and we have minority charge carriers on P side on n side minority charge carrier is hole on P side minority charge carrier is electron so this hole will move like this and this electron will move like this so there there is current called as a reverse saturation current or leakage current when the junction j2 is reverse biased I am calling the reverse saturation current reverse saturation current or leakage current or leakage current IC o---- because this current is associated with the collector so there is subscript C this o stands for open circuit we measure this current when emitter terminal when ammeter terminal is open circuited so this is why I am calling this current IC o---- and if I if I want to find out the collector current IC then it is equal to alpha I alpha times the emitter current because n is the number of electrons entering the base and alpha n is the number of electrons moving to the collector so current will be alpha times ie plus IC o---- so this is the value of collector current the next thing is to find out relation between the emitter current base current and collector current for this purpose first we have to find out direction of the three currents you can see electron is moving from left to right so direction of current is like this from right to left this is i.e the emitter current when electrons from emitter recombine with the holes the direction of current will be like this and the current is IB because electron will move in this direction to the positive terminal of the battery so current will flow in opposite direction and to find out collector current we again have to focus on the movement of electrons electrons are moving from base to collector like this in this fashion so current will be like this from right to left and to find out relation between ie IB and IC we have to use Kirchhoff's current law KCl and from KCl we know sum of currents entering is equal to the sum of currents leaving ie is leaving I II is leaving IB and IC are entering so ie is equal to IB plus IC this is what we have in the NPN transistor in active mode these two relations are very important and we will talk more about alpha in the coming presentations this is all for this lecture see you in the next one