 we are going to talk about common base and common collector amplifiers, mid-band analysis as usual. The common base amplifier has the base common to both input and output and the circuit looks like this, you have the transistor, now you draw in this fashion because the base is to be common alright, the input is to be applied between the collector, between the emitter and the base and the output is to be taken from the collector to the base, but then the collector must be biased, must be biased, so we attach a resistance R sub C and take it to plus VCC, as usual this is an NPN transistor, then the base also has to be biased, the base current has to be supplied, so what we do is from the base to ground we connect a resistance R sub 2 and from this junction to VCC we connect a resistance R1, it is exactly the same way just orientation has been made different and then the output is to be taken from the collector, so the output goes via a coupling capacitor C2 to R sub L and this is my V0, the Rittman square voltage, the phasor voltage and this is the current I0, the load current, now at the input obviously if I have to connect between between emitter and base a source I cannot make it short, I have to use a resistance okay and this resistance cannot be bypassed right, because if it is bypassed then the source the AC signal cannot be applied, so we must have a coupling capacitor C1 and the source the usual story RS and the source is VS, this is my common base amplifier okay the total circuit, you understand the connections? Once you know how to argue what is to be applied where the circuit can be drawn from logical arguments, you do not have to commit to memory alright, now the biasing as you can see is in the same manner as the common emitter amplifier except for the fact that if the base is like this this will cause a feedback and therefore the base now has to be bypassed, you connect a capacitor C sub D so that it does not cause a feedback, if the base is not bypassed then there will be problems with regard to feedback okay, it will affect the gain, it will affect the input resistance and everything okay, this is the circuit, yes, importance of CB, you see what I want is that the AC signal should be applied between the emitter and the base okay and this cannot be done unless R2 is bypassed, if R2 is not bypassed then obviously the input signal V sub I appears between emitter base plus R2 which causes a feedback because there is some amount of outputs which also shall flow through R2, some amount of output okay, so we have to bypass R2 now and that is why we call this capacitor CB base bypass capacitor, now to draw the equivalent circuit if you follow the sequence we have VS, RS, C1 is a short for mid band and therefore I have RE then I have the emitter terminal, the base is grounded here between the base and this is the collector okay, between the base and the emitter you shall have the Rx and R pi and the voltage across R pi is V pi with polarity this positive here and negative here, we usual usual hybrid pi equivalent circuit you must do it carefully and then this is the this is the internal base, this is the internal base and this is the emitter between emitter and internal base is R pi, this voltage is V pi, between the collector and the emitter we shall have the current source gm V pi from the collector to emitter, you must keep the directions the same, in addition we have this inevitable R0 which I shall show as a dotted connection R0 and we will assume that R0 goes to infinity in order that it does not complicate matters, R0 is of the order of for the example that we have been considering it is 139k and mostly it can be neglected, from the collector from the collector then you have the what do you have, you have RC going to ground and the C2 by the C2 the coupling capacitor acts as a short and therefore you have RL, this is V0 and this is I0 okay, this is the equivalent circuit, what happened to R1? R1 is connected from VCC which is a ground to a point which is ground, so R1 does not affect, neither R1 nor R2 affects the AC operation, is that clear? We have killed RB the parallel combination of R1 and R2 which shows its teeth in the common emitter amplifier so prominently okay, we have killed it. Now this circuit is now to be analyzed, as you can see if we ignore R0 this green part of the circuit then GM V pi this current source, this current flows through the parallel combination of RC and RL and therefore V0 can be immediately written as minus GM RL prime where RL prime is equal to RC parallel RL multiplied by V pi that must of course be there but then you have to find out what is V pi, if you look at the circuit this voltage is VI the input voltage, so V pi can be very simply written as V pi is the result of potential division of VI between R pi and RX but we take negative sign because the polarity of V pi opposes that of VI so V pi is minus VI times R pi divided by R pi plus RX and as you know, yes? RX that is correct RX and as you know RX is negligible compared to R pi so this is approximately equal to minus VI and therefore the voltage gain AV V pi if you substitute minus VI the voltage gain simply becomes GM times RL prime therefore our expression is simplified to AV which is V0 by VI equal to GM RL prime notice that there is no negative sign here in the common emitter circuit again the gain was minus GM RL prime there is no negative sign and therefore V0 and VI are in phase this is the one of the distinct differences between common emitter and common base amplifier that the output is in phase with the input, this is AV now we want to calculate the input resistance input resistance that is the resistance that is faced by the source R sub I obviously the input resistance shall be the parallel combination of RE and whatever VI faces alright so if I call this current as I sub I then I sub I and this current is I1 then I sub I shall be VI by RE plus I1 is that right? I sub I the current fed by the source is equal to VI by RE plus I1 now let us see what I1 is I again refer to this circuit what is the current through this resistance this current we have taken the polarity as this so this current is V pi by R pi this current is V pi by R pi alright and R0 has been ignored this current is GM V pi therefore I1 I1 is simply equal to minus V pi by R pi if I write KCL at the node at this node editor node V pi by R pi minus GM V pi all these 3 currents go towards the node the sum of them is equal to 0 so I1 is this and you can see that this is equal to minus V pi beta plus 1 divided by R pi right and since beta is much larger compared to 1 we can ignore 1 and the beta by R pi simply becomes GM so this is approximately equal to minus GM V pi and since V pi is approximately equal to VI we get this is approximately equal to minus GM VI plus GM VI where did you lose the negative sign? V pi is approximately minus VI therefore therefore my I sub I this equation if I substitute this in this equation I get I sub I as equal to VI by RI plus GM VI alright and therefore RI which is the ratio of VI to II is simply equal to the parallel combination of RE and 1 by GM and as you shall see RE is of the order of a K or more whereas 1 by GM is several tens of ohms for example if GM is 40 millimaw 39 as we have taken because we took Q by KT is 26 millimaw instead of 25 if GM is 40 millimaw then what is 1 by GM it is only 25 ohms is it not right and what is 25 ohms compared to 1 K or what is 1 K compared to 25 ohms shanty effect so this input resistance is of the order of 1 by GM agreed because the shanty effect of RE is negligible this shows the second difference between a common emitter and common base in a common emitter amplifier the input impedance is of the order of R pi alright which is of the order of a K on the other hand in a common base it is of the order of 1 by GM which is a small resistance so CD input resistance is much less compared to CE input resistance alright and you know that the CE input resistance drastically increases if the emitter resistance is unbiased if the emitter resistance is unbiased then CE input impedance becomes R pi plus beta plus 1 times RE of course you can add RX if you showed this is for unbiased emitter resistance common base circuit has the lowest input impedance of all the 3 configurations not only CE also CC as you shall see in a common emitter common collector circuit this is the order of input impedance in a common collector circuit we shall show this is the order of input impedance and therefore common base circuit has the lowest input impedance now the input impedance is low what kind of source would be the most appropriate voltage source or current source current source if it is a voltage source then there will be potential division between the low resistance and the low resistance right whereas if it is a current source high input impedance all the current will go into the common base agreed so CD is more appropriate for a current source drive whereas CE is appropriate for a voltage source drive okay that is the story with regard to the input impedance then of course our current gain A sub I as we have already shown V0 by RL divided by yes VI by RI this is the input current and this is equal to AV because V0 by VI is the voltage gain multiplied by RI divided by RL and you can see by substituting or let substitute AV is GM RL prime RL prime is RL RC divided by RL plus RC and RI is RE parallel 1 by GM which is equal to RE divided by 1 plus GM RE agreed RI the input resistance we have just shown it is a parallel combination of RE and 1 by GM this is the expression RE into 1 by GM plus RE divided by RE plus 1 by GM this is the simplification and you have to divide by RL okay so RL and RL cancel GM RE is usually much greater than 1 and therefore what is left here is 1 by GM agreed GM RE is much greater than 1 because 1 by GM is much less compared to RE okay this we have already demonstrated this is much less than RE so GM RE is much greater than 1 and therefore what will be left is approximately RE divided by GM RE RE RE cancels GM and GM cancel so this is approximately equal to R sub C divided by RL plus R sub C that is the current gain is divided is determined by the load and the collector biasing resistance only it is approximately independent of what the input conditions are okay this is A sub I and what would be the output resistance what is the resistance that is seen by by the load does it require any calculation what I want is this resistance R0 RC because it is shunted by a current source whatever the value of the current source the impedance the internal resistance is infinity and therefore R0 is approximately RC why am I saying approximately because of R0 if R0 is there then this year this is not the calculation it will affect the output impedance so we say the output impedance output impedance R0 is approximately equal to RC R0 is approximately equal to RC now yes why is the input impedance of a dependent current source also infinity any current source the input impedance is infinity the internal resistance is infinity any current source any voltage source whether dependent or independent physically it is not a current source physically a current flows here yes proportional to some current it is flowing some other part of the circuit no but this current is independent of what you connect across it the current g m v pi is not dependent on what you connect across it so it is a current source similarly a voltage source delivers its voltage irrespective of what you connect to it that is the definition and the other quantity a quantity which we often find is AVS which is simply AV multiplied by R i we have found that out divided by R i plus R s and naturally this would be approximately since R i is approximately what 1 by g m this would be AV multiplied by 1 by 1 plus g m R s is that clear and g m R s normally we may be comparable to 1 if it is much larger than 1 then you can ignore the 1 okay R s has not been specified R s could be low it could be a good voltage source if R s is 0 for example AVS is equal to AV so we cannot make this approximation blindly R s depends on what your source is if it is a microphone source then impedance is usually high whereas if it is a if it is a signal taken from a loudspeaker the internal impedance is only 8 ohms agreed so that completes the that completes the analysis of the CB amplifier now if we if we use this for calculating the performance of a CB amplifier whose parameters are the same as that of the CE that we considered earlier what were the values we had considered R 1 and R 2 we shall continue the same example again and again R 1 and R 2 were 220 K R s was taken as 600 ohm the source resistance R c the collector biasing resistance was taken as 2.2 K and the load was taken as 4.7 K the transistor parameters were g m equal to 39 millimole because I sub c was 1 milliampere and this is 1 milliampere divided by 26 millivolts then beta was given so we could calculate R pi as 2.6 K beta the product of the 2 is 100 R x was given as 100 ohm we did not consider this we could there was no problem in considering R x but it affects very slightly R mu we have ignored because it is 13 mega that was the resistance and R 0 was equal to 139 K we have ignored the effect of this also if we substitute these parameters in this calculation then our results are Av is 58.5 which is the same as that of the CE amplifier same no change R sub i approximately 1 by 39 milli ohm and this becomes how much it will be 1 by 39 26 26 we have already calculated this how did we calculate g m what is R e what is R e we did not specify R e R e was 1 K that is right R e is 1 K okay so R i is 26 ohms which is way down it is much lower as compared to the CE amplifier A sub i the current gain is only 0.32 it is much less than CE case CE case it was less than beta but it was 10 in the order of 10 30 or 31 I do not remember what it was but here it is it is less than 1 and it can be it can be seen it is logical because whatever you are feeding to the emitter the collector current cannot be more than that of the emitter so the current gain shall be less than 1 but it is it is much less than 1 is 0.32 only and R 0 is 2.2 K you can calculate what is A vs would be 58.5 multiplied by 1 by approximately 1 plus R s is 600 ohms and 1 by g m is 26 ohms okay so this is indeed greater than much greater than 1 and you can approximate this as 58.5 into 26 divided by 600 the gain is approximately 2.6 right tends to 2.6 because this is approximately 10 any question on this that leads us to the common collector amplifier the third configuration that is possible common collector amplifier CC it is also called the emitter follower and this terminology emitter follower arises because the output if it is a common collector collector is common between input and output so where do you feed the input obviously the where the base alright and where do you take the output the emitter it turns out that emitter voltage closely follows the base voltage and therefore it is called an emitter follower the voltage at the emitter closely follows the voltage at the base which immediately says qualitatively that the voltage gain of the circuit shall be approximately 1 okay it is however always less than 1 it is always less than 1 as we shall show now the the biasing of the circuit yes less than 1 yes because it can be current gain it can be current gain from base to emitter there is a current gain it can give power gain if the current gain is greater than 1 voltage gain is less than 1 the product can be greater than 1 so it can give power gain alright and therefore the circuit is but the major use of the circuit let me let me point out right here if I call this as the emitter follower this is the input and this is the output the emitter follower has the property that the input impedance is very large and therefore and therefore if you connect a voltage source here the output shall be virtually decoupled from the source okay that is suppose you have an amplifier of gain plus 1 agreed which the emitter follower is and if you apply a source here the output does not interact with the input because the source fits into a high impedance the output impedance usually is small in an emitter follower the output impedance is of the order of 1 by g m alright so what it does is it converts a no it isolates the source from the load you can connect any load here which is much greater than 1 by g m alright so the source is isolated from the load and in that sense it is also called a buffer it is also called a buffer alright let me explain why this term buffer we shall be using this term again and again throughout electrical engineering let us see why it is called a buffer suppose you have a source of resistance 1 k and the load which is also 1 k and the load varies no let the load be constant it does not matter let the load be constant okay now if you connect this directly if you connect this directly only half of the voltage will appear here alright and any change that occurs in the source due to any reason will affect the load voltage also on the other hand if you apply it to an emitter follower which is a gain of plus 1 approximately then the input impedance let us say is 100 k the total voltage here can be transferred over here alright not only that the load variations will not affect the source the source variation will not affect the load so that this is a buffer circuit it isolates the load from the source exactly like a transformer I mean it isolates the source from the load that is load changes do not affect the source source changes do not affect the load okay suppose the source internal impedance changes from 1 k to 990 ohms well because the input impedance is large the voltage that appears here is still the same okay on the other hand if the load changes here the input impedance the because the input voltage does not change the output voltage also does not change why not if the load changes because the output impedance of this is low of the other let us say 25 ohms okay so even if the load changes the load voltage remains a constant this should not have happened if this was not there so it is called the buffer we shall see examples of applications of the buffer later alright now the circuit the circuit is the same as the common emitter same as that of the common emitter except that the collector is virtually grounded okay so what I should have done is I can I take the VCC directly to the transistor directly to the transistor I have an RE which I can no longer bypass because this is my load and therefore from RE I take the coupling capacitor C2 and connect to RL this is my V0 and this is my I0 now the base biasing I do exactly the same manner that is I use two resistances R1 and R2 connected to ground and the source is applied through a coupling capacitor source resistance RS and VS this is my common collector circuit I have not reoriented it I should have brought this down turned it upside down and so on I have not done it I have drawn it in the same manner that we are used to I must also mention R1 is connected to VCC yes this I have shown R1 connected to VCC however in order to get a correct biasing in order to get correct you see now the DC load is simply RE is it not right DC load is simply RE there is no RSC DC load formally was RC plus RE now there is no RE suppose in order to bias the transistor at the appropriate viewpoint you have to use a large RE okay which you do not want to because of reasons that you will know later you do not want to use because large RE will also affect the AC load the AC load is the total combination of RE and RL right AC load is the total combination of RE and RL now you do not want to increase RE then what you do is you do use a resistance R sub C here but in order that it does not behave like a common emitter amplifier we have also used this as a common emitter amplifier we took two voltages one from here and one from here and this were called this was called the para phase amplifier two voltages of opposite phase now in order that R sub C does not develop yes no AC load is determined by RE and RL parallel combination and therefore the voltage swing this is the DC load line and this is the AC load line voltage swing will be determined by the parallel combination of RE and RL right so I don't want to change RE you see RL is fixed I may want to decrease the effective AC load because I want a particular voltage swing alright so if needed I should not hesitate to use an R sub C but in order that R sub C does not affect the AC operation I bypass where they connected capacitor ACROSRC or from collector to ground I shall call this C sub C it is a collector bypass emitter and VCC oh collector yes I can do that I can do that but if I have a ground I will connect it to ground why keep one of the terminals of a capacitor charged make it uncharged so we can as well use another resistance in the emitter and bypass that only we can as well use another resistance in the emitter and bypass yes we can do correct what he says is suppose you used another resistance here and bypassed it yes we can do that we can do that also okay that's a good idea so this is the circuit as far as AC is concerned this point is at 0 potential oh instead of RC instead of RC suppose I use a resistance here okay and bypass that alright let me use let me show you suppose you use an R sub E prime here and bypass it then the DC load is still R E plus R E prime whereas AC load is the R E parallel R N so this is an alternative that's right that's right okay now let's see let's look at the AC equivalent circuit of the common collector amplifier we have the RS VS this you should be able to do by looking at the circuit without any further calculation any further thought just looking at the circuit it should become a part of your daily life routine C 1 is a short R 1 and R 2 come in parallel so that becomes R sub B R 1 and R 2 in parallel then this goes to R X and R Pi this R Pi does not go to ground it goes to R E and also to R L this is V 0 and this is I 0 this is the internal base B prime this is the external base B this is the editor E from the collector from the collector we shall have a G and V Pi where V Pi is this voltage G and V Pi and where does this go it goes to ground in addition we shall have an R 0 let's include this R 0 now because it is easy to do so do you see how how it is easy it comes in parallel R E and R 0 so it can be observed now my R L prime let me call this as R E parallel R L parallel R 0 okay then by looking at this circuit can you simplify let me simplify this circuit a little bit VS then things will be obvious if I do this simplification the calculation will almost be done by inspection this is R B then you have R X R Pi what is the current through this what is the current I B but this is in terms of V Pi it is V Pi by R agree once you recognize this current then there is another current here coming from ground I have pulled that wire and taken down this is GM times V Pi and the effective resistance here is R L prime and this is V 0 I had lost I L though isn't it the load current because I absorbed R L but no problem load current is simply V 0 by R L so I am not lost it I had lost it only in the picture I must keep this in my mind V 0 by R L now therefore in this simplified picture one can very easily write the value of V 0 as you see V 0 is the effect of two currents coming over here V Pi by R Pi and GM V Pi these two currents then flow through R L R L prime and therefore V 0 is simply V Pi by R Pi plus GM V Pi R 0 is included in R L prime this multiplied by R L prime now we also require the value of V Pi all right what is V Pi V Pi is the voltage across R Pi this voltage is V Pi how do you find V Pi V Pi this is V I V Pi is the result of potential division of V I between R X R Pi and not R L prime beta plus 1 R L prime the effective voltage here is beta plus 1 times that current multiplied by R L prime is that clear because the current flowing in this is not I B it is I B plus beta I B okay so the effective V Pi shall be R Pi V I divided by R X plus R Pi plus beta plus 1 R L prime the point clear therefore V 0 by V I which is A V the voltage gain I can write this as GM plus 1 by R Pi which comes from here times R L prime into R Pi divided by R Pi plus R X plus beta plus 1 R L prime and if you notice if you notice the numerator when I multiply by R Pi it will be beta plus 1 so they conspire to make it look very nice expression beta plus 1 R L prime divided by R Pi plus R X plus beta plus 1 R L prime now beta plus 1 R L prime normally will be much larger than R Pi plus R X so this is approximately equal to 1 but you must remember it is always less than 1 okay this calculation also has revealed what the input resistance shall be is not this absolutely clear what the input resistance is if you look at the circuit it would be R B in parallel with whatever this faces what does it face R Pi plus R X plus beta plus 1 R L prime and therefore the input resistance I write without any more ado as R B parallel R Pi plus R X plus beta plus 1 R L prime and you can see why this resistance is very large because of this beta plus 1 R L prime okay and R B normally is a large resistance 110 K was the value for our circuit now A sub I we lost I 0 but not from our mind A sub I is A V R I divided by R L and it can be calculated we know the expression for A V A V is approximately 1 so this is simply R I divided by R L can it be greater than 1 A V is approximately 1 so this is approximately R I by R L that means it must be greater because R I is very large compared to R I okay it is much greater than 1 so you can get currently you can get powered in also now to determine the output resistance of the circuit this is not obvious if we look at the original circuit what we have to do is to look from here and calculate R C it is not obvious because there is a V Pi even if we kill V S V Pi shall be there right let us see let us follow blindly the procedure what do we do we kill the input source connect a source at the output and find out the current let us do that then what do I get again let us say this source is V 0 and this current that is I 0 this will connect to first R 0 it will connect to R L not R L R E R L is what is seen this impedance okay R E then we have an R Pi the voltage across which is V Pi plus minus we have an R X this is R X and then we have R B what else R X so one thing we can do is let us parallel these two resistors and call them R S prime call them R S prime so my simplified circuit becomes V 0 I 0 can I combine these two also into an R E prime R E prime is equal to R 0 parallel R E then I have an R Pi let us write it in this way plus minus is the polarity alright is the polarity alright okay then we have a resistance R S prime now you see even if the source has been killed V Pi is not killed V Pi is the potential division of V 0 between R Pi and R S prime one at a time yes input excitation because we want to find out what R L faces what does R L C what does R L C what resistance that is that we are finding the equivalent resistance so we are filling the input source and connecting the source at the outside at the output and finding the current okay now R S prime is R B parallel R B okay so can we write the expression for V Pi down with R X R X okay okay yes what about dependent current source yes sir which current source should be divided by beta plus 1 very good so it go back why you do point here alright in terms of the this side there should have been a current source here G and G Pi okay there should have been a current source here and this should come here agreed G and V Pi this should come here and if you now calculate if you now calculate I 0 you see I 0 consist of 3 components 1 is G and V Pi 1 is V 0 by R E prime this current is V 0 by R E R E prime and the third is this component and this component is V Pi by R Pi so if you write here I 0 would be equal to G m V Pi minus V 0 by R E prime plus V Pi by R Pi is it point clear there are 3 currents and V Pi is related to V 0 how is V Pi related to V 0 it is V 0 multiplied by R Pi divided by R Pi plus R S double prime is there a minus sign is there a minus sign one of our students is claiming there should be a minus sign and I had no alternative but to agree with him there is a minus sign because this polarity does not agree with the polarity of V 0 so what else do we want to know I know I 0 in terms of V Pi and I know V Pi in terms of V 0 and therefore I can calculate now V 0 by I 0 and the expression I skip the algebra is the procedure clear procedure is not clear what I do is I write KCL at this point I 0 equal to G m V Pi that is right okay I write I 0 plus G m V Pi minus V 0 by R E prime then minus V Pi by R Pi so I know I 0 in terms of plus V Pi plus V Pi plus V Pi this is these 3 currents are coming this 3 and this is leaving so this plus this plus this minus this equals to 0 what does this equation contain this contains I 0 V 0 and V Pi substitute for V Pi from this expression therefore you will be left with an expression containing on the I 0 and V 0 and then R 0 is the output resistance and you can show that this is R 0 parallel R E parallel which I had called R E prime parallel R Pi plus R X plus R S parallel R B we have called this as R S prime and the combination is R S double prime we are now putting this explicitly divided by G m R Pi plus 1 or simply beta plus 1 okay this is the expression now it is not as simple as the calculation of output resistance in the case of C E or C D amplifier you can show from practical approximation that this is approximately equal to 1 over G m and the reason is very easy to see you see R S parallel R B plus R S would be less as compared to R Pi and therefore R Pi by beta is 1 by G m 1 by G m is of the order of 25 ohms R B is of the order of a K 40 times that R 0 is 139 K so these fellows they do not have any contribution to make and the output integration is approximately 1 by G m which is independent of any external resistance connection so the emitter follower has the property that it is a high input impedance and the very low output impedance okay we stop here rolling is the process of producing different components by plastic deformation of work material by passing it between rotating rolls the process is called longitudinal rolling if the forming rolls are rotating in the opposite directions and flow of work material is along the length of the job and perpendicular to the central line of the rolls by this process we can produce products like bars of different cross sections steel for construction work flat products like plates and sheets of various thickness can also be produced by this process another version or type of rolling process is transfers rolling let us have a look at a transfers cold rolling machine in this process the forming rolls are rotating in the same direction and the flow of material is either radial or the flow of material is along the axis of the job which is also parallel to the axis of the forming rolls each variety of products like pulleys of different sections rollers for overhead cranes cycle freeway cups gear blanks and gears can be easily produced by this process this machine can roll circular discs of aluminium mild steel and alloy steel of very thickness and diameters the two forming rolls are mounted on the spindles running in the bearing housings of the two blocks one on either side of the job and these blocks are moving in the dovetail slides of the main bed in this process there is less material wastage higher production rate and long to life hence it is much cheaper a process than any other conventional process nobody can deny that we can travel faster much faster than ever before but at what price increased air land and noise pollution increased casualties in accidents rapid declation of petrol in resources automobiles are in fact now recognized as the chief contributors to atmospheric pollution throughout the world emitting huge quantities of lead oxide of sulphur and nitrogen in addition to carbon monoxide and carbon dioxide road accidents also take a heavy toll in India alone about 40,000 people die every year and over a million are injured and these numbers are rapidly increasing year after year and as is common knowledge automobiles consume a major share of petroleum products consequently this endowment of mankind accumulated over millions of years is getting so rapidly depleted that by the year 2000 we would be left with only 40% of the total stock available at the beginning of the 19th century