 6th lecture on BJT bias stability continued, in the last lecture 5th one we had discussed this particular bias arrangement which is the most common bias arrangement that is an editor resistor which is bypassed by a capacitor C E plus V C C this is a N P N transistor R sub C and then we have a coupling capacitor C 2 which goes to the load resistance R L this is the output voltage V 0 and at the input the biasing is done through a potential divider 2 resistances R 1 and R 2 then the signal is applied through a capacitor C 1 and a voltage source a signal source V S V S could have an internal resistance R S okay this is the complete C amplifier and the DC part of the circuit the DC part of the circuit is this this is the DC part of the circuit DC does not is not affected by the capacitors this is a coupling capacitor this is a bypass capacitor okay we also took a particular example which we shall carry throughout our discussion in which we have taken V C C as equal to 12 volt I think 12 volt R sub C was 100 ohms R L was 100 ohms R E was also 100 ohms R 1 was 6.8 K and R 2 was 6.8 K we are not specifying R S C 1 C 2 C 3 C 1 C 2 C E because they do not affect our operation at the moment at the moment biostabilization we also discussed that the BJT bias point that is the Q point or the operating point which is the value of I sub C and V sub C E this defines the operating point Q inside a bounded region in the I sub C V C E characteristic the bound occurs due to various reasons let me use different colors one is the P maximum allowable dissipation P max one is this hyperbola second is the breakdown voltage this is V C E O the breakdown voltage that is the voltage beyond which you cannot go without the junction breaking down there is a maximum current limit you cannot draw more than this I sub C max on the lower side on the lower side there is a saturation line saturation line and finally there is a cut off line so it is this region this region within which the Q point has to lie and in order that the amplifier is able to handle a reasonable amount of signal voltage the Q point must lie somewhere in the middle somewhere well inside the region okay if it is close to the cut off then obviously the dynamic range will be very small if it is close to P max well the transistor might break down because this P max is for is manufacturer specification and we know that even if fabricated by the same process steps there are lot of variations between one transistor and another from the same lot and therefore you cannot take that risk of operating near P max you cannot go to V C E O you cannot go close to IC max you cannot go close to set so the Q point has to be somewhere here and we showed that the Q point lies on what is called a DC load line a DC load line where this point is given by V C C okay and this current is given by V C C divided by yes RC plus RE this is plus RE with a certain assumption what is the assumption that beta is much greater than unity this is the DC load line DC L L and in addition we also say that there is an AC load line which passes through the Q point which passes to the Q point that is a higher slope and AC load line is somewhat like this AC L L whose slope is minus 1 over R L dash which is RC parallel okay this is this describes what we did last time now we start from this point we said that not only we want to establish a Q point we would also like to stabilize the Q point we would also like to make sure that if the transistor is replaced by another transistor of the same number or if temperature increases or decreases which causes a change of beta V BE and also I C BO all the three quantities change or if there is a fluctuation in the power supply V C C changes then naturally V BB also changes and there may be a change in the parameters also therefore the Q point not only we want to establish we want to stabilize it and this is what we want to investigate today yes you have a question you have a question do not bite the pen okay we established this relation that I sub B I sub B the base current by taking the left loop that is the base loop we established that this is given by V BB minus V BE minus I C BO beta plus 1 we are not making any approximation beta plus 1 multiplied by RE divided by RB plus beta plus 1 RE this is the value of I sub B that we had established relating I sub B to V BE to I C BO and beta these are the three variable quantities alright you also know from the relation that I sub C equal to IB times beta plus beta plus 1 I C BO this is a fundamental relation yes that is right we are using the additional symbol B because of the conventions okay this is the common emitter and it is used as C BO if it is common base then the current that would be I C BO which is I C BO multiplied by beta plus 1 okay now if I combine this with this that is if I write this IB as equal to I sub C minus beta plus 1 I C BO divided by beta isn't this doesn't this follow from this relation okay now therefore and I ignore the rest of the quantities then this should be equal to this from which you get a relation between I sub C the collector current and the other quantities this algebra after clearing the mess the result is not too complicated what we get is the following I sub C is equal to beta times V B B minus V BE plus I C BO beta plus 1 multiplied by RB plus RE divided by RB plus beta plus 1 RE okay this is one of the coordinates of the Q point and the other coordinate is V CE V CE which is equal to V CC minus I sub C plus RE with that assumption beta much greater than 1 beta much greater than 1 multiplied by I sub C therefore if I can stabilize I sub C I have stabilized V CE also so all we need to do is to focus our attention on this on the stabilization of the collector current against variations in beta beta varies V BE varies and I sub BO varies okay beta varies this varies V CC also varies then you are stuck V BB will vary them well if there are ways of compensating for variations in V CC also but what is preferred is that you use a regulated power supply rather than an unregulated power supply in other words you make V CC constant by using a regulated power supply alright that is relatively easy to take care if V CC is not constant then we make a simple Zener diode regulator and apply that to the transistor circuit so let us look at the variation of I sub C which we shall call delta I C due to a change of beta due to a change of V BE due to a change of I C BO separately there are 3 parameters you see I sub C is a function of 3 variables V BE beta and I C BO so let us find out partial I sub C that is delta I C due to a change of let us say V BE beta and I C BO are held constant then we will investigate change due to beta when V BE and I C BO are held constant okay with 3 factors separately then we shall try to combine alright this will be our procedure actually what we should do is strictly what we should do is that we consider 2 situations what we should do is the following we consider 2 situations in under 1 situation the collector current is I C 1 which has beta equal to beta 1 V BE 1 and I C BO 1 alright this be at room temperature for example and at an increased temperature since all of them vary let them vary to the following quantities I C BO 2 and let the resulting collector current be I C 2 alright then we should have we should find out we should define delta I C as equal to I C 2 minus I C 1 and should find out delta I C as a function of delta beta which is beta 2 minus beta 1 which is simultaneously that is all the factors considered together beta 2 minus beta 1 delta V BE as equal to V BE 2 minus V BE 1 and delta I C BO as equal to I C BO 2 minus I C BO 1 this is what we should do you understand what I mean take the circuit or take the expression I C expression for I C C find out I C 1 find out I C 2 subtract the 2 subtract this the first from the second due to the variation of all the 3 simultaneously and then find out the percentage change by dividing delta I C by I C 1 and multiplying by 100 this will be the percentage change in the collector current this is what you should do but as you can see the expression itself is quite involved and if I if I do that I can get an expression which will fill the whole page or fill the next page also but it is not very meaningful and as engineers we do not want to make life more complicated then what is demanded of the situation okay. So we try to make life simple and we shall instead in one of the tutorial problems we shall work out actually what happens with numerical values that is not too difficult actual value of the percentage change in I C but as far as as far as mathematics is concerned general formula is concerned let us consider variations one at a time then try to combine this is the ingenious approach alright even here we will make a simplification let me write this expression I C equal to beta V B B minus V B E plus I C B O beta plus 1 R B plus R E divided by R B plus beta plus 1 R E one simplification that you can very easily do is the transistor will be useless if beta is not much greater than unity. So we can assume beta much greater than unity and then this expression becomes beta if beta is much greater than unity then one drops out this one drops out and beta can be taken out right therefore what we get is beta V B B minus V B E plus I C B O R B plus R E alright bracket closed beta is a common factor of the two expressions and in the denominator we get R B plus beta R E this is the assumption that beta is much greater than unity alright and we would also we would also do another simplification mainly that when we consider variations with respect to beta or V B E we shall ignore this I C B O term okay otherwise it is an unnecessary complication we shall ignore this term and to make things look logical let us calculate this term as compared to this term in the particular example that we took in the particular example V B B minus V B E was equal to 6 volt 12 6.8 and 6.8 so V B B is 6 volt and V B E was given as 0.6 so this is 5.4 volt on the other hand I C B O R B plus R E is equal to I C B O typically is 0.01 micro amps that is 10 to the minus 8 ampere which is 10 nano ampere this is a reasonable figure for a BJT so I C B O R B plus R E is 10 to the minus 8 multiplied by R B is 6.8 parallel 6.8 so 3400 plus 100 which is 3500 times 10 to the minus 8 that is equal to 35 micro volt agreed which is indeed much less than 5.4 volt therefore compared to this term V B B minus V B E we can ignore this term but as long as we are considering variations with respect to beta or V B E if we are considering variations with respect to I C B O itself obviously we cannot do that agreed we also know that I C B O doubles for every 10 degree centigrade dies in temperature so there may be a temperature at which this term shall become comparable but as far as beta and V B E variations are concerned let us ignore this in other words we take the expression I C as equal to beta V B B minus V B E divided by R B plus beta R E alright so let us consider delta I C when beta changes beta is beta 1 and when beta 1 changes to beta 2 delta I C which is I C 2 minus I C 1 is equal to beta 2 V B B minus V B E is a constant I could as well take this out then beta 2 divided by R B plus beta 2 R E minus beta 1 divided by R B plus beta 1 R E alright this is delta I C due to a change of beta only and what I have to do is to find out delta I C divided by I C 1 if you clear this algebra and divide by I C 1 while this algebraic simplification I leave it to you the expression very simply becomes becomes delta beta which is beta 2 minus beta 1 divided by beta 1 times R B divided by R B plus beta 2 R E let me write this again delta I C divided by I C 1 is equal to delta beta divided by beta 1 R B divided by R B plus beta 2 R E now an observation suppose R E was equal to 0 someone asked in the previous class why use an R E at all suppose R E is made equal to 0 then this term becomes unity which means that a 10 percent variation in beta will cause a 10 percent variation in I C C is it not right in other words 1 to 1 even if beta changes by 1 percent I C will change by 1 percent whereas multiplication by this factor which can be made much less than unity how by making beta 2 R E much greater than R B if we do that then this quantity will be very small and therefore even a large change in beta will cause a small change in I C and that is what stabilization means is it not right even a beta changes by replacement of a transistor which let us say beta is from is was 100 earlier and it is 150 now even if that large change 50 percent change in beta we want to make I C as stable as possible for example in the particular example that we have taken suppose beta changes from 100 to 150 nothing else changes then the right hand side becomes 50 delta beta divided by 150 no I am sorry beta 1 that is 100 okay multiplied by R sub B is 3400 3.4 K divided by R sub B 3400 plus beta 2 which is 150 multiplied by 100 okay R E is 100 ohms so this should be delta I C by I sub C 1 and it calculates out to 9.2 percent how did I get percent multiplied by 100 okay now look at this even a 50 percent change in beta has caused a change of I sub C by a quantity less than 10 percent so the reduction is almost 1 fifth is that clear I can make it larger if I had used a higher value of R E if I used a higher value of R E I could make it larger now if you recall why not why do not you make R E as large as possible okay that is a problem let us look at this problem the problem is the following this is R E okay and this is V C C this R E goes to ground and you know V C E V C E is equal to V C C minus I sub C R C plus R E if you keep I sub C a constant and increase R E naturally you will have to increase V C C and there is a limit to which V C C can be increased transistors basically are low voltage devices you do not want to use a supply which is more than 12 volts if you use a 100 volts supply yes it can be done but the 100 volts supply itself is a very bulky thing right so R E cannot be increased indefinitely if we could do that there is one more one more problem so I think we use a 100 volts supply R E is sufficiently large due to some reason if R E gets shorted due to some reason due to mishandling what will happen the transistor will blow up because it will exceed V C E O is that correct that high voltage shall come across the transistor it will burn it off therefore it is risky business and R E there is a limit the limit well one should always try to make beta R E we do not know what under what condition beta R E under all conditions should be much greater than R B and as you know 1 is to 10 is electrical engineers much greater and therefore the thumb rule is beta R E equal to R B 10 times R E if I make it in this particular case well R B is 34 3.4 K so this should be 34 K and therefore R E needed would be if beta is 100 it would be 3.4 K well it is not too bad but you have to increase your I beg your pardon 0.34 K that is simply 340 ohms we used to 100 ohm resistor and 340 ohms can jolly well be used without without increasing the power supply to a large value okay this is the story about beta let me also mention that beta usually increases with increasing temperature and the thumb rule is for every 100 degree C beta changes by 50 percent approximately 50 percent alright this is the thumb rule and therefore this change could as well be due to a change of temperature from 25 to 125 okay we will take an example of complete example at the end now let us look at the passive change in I C due to a change in V B E our expression is beta V B B minus V B E simplified expression divided by R sub B plus beta R E now our change is only in this quantity and if you follow the same procedure that is I C 1 V B E 1 I C 2 V B E 2 subtract one from the other and divide by I C 1 it is very easy to see that the expression shall be delta I C by I C 1 shall be equal to minus delta V B E divided by V B B minus V B E this comes from the division by I C 1 is very simple simple to do this algebra which I shall express as minus delta V B E divided by V B E alright which V B E is this one or two one that is correct and therefore I will use this one multiplied by multiplied by V B E 1 and intentionally introduce this term to express percentage change divided by V B B minus V B E 1 alright and once again you see that if this term is unity if this term was not there if this term was not there then 10 percent change in V B E would have cost 10 percent change in I sub C however because of the presence of this term which is less than unity because V B E is typically let us say 6 volt and V B E 1 is 10 times less 0.6 therefore this quantity is less than unity and therefore a 10 percent change in V B E will cause a much less change in I sub C and as you can see this sensitivity to V B E can be reduced drastically if you could make V B E V B B minus V B E 1 much greater than V B E 1 which means that V B B E should be much greater than twice V B E 1 okay. So if you could make if you could make V B E 1 is 0.6 if you could make V B E approximately 12 okay then we could have a reasonable stabilisation of the operating point but the situation that I have shown you is not too bad as you can see if I take that example in our example it is Delta I C by I C 1 is equal to minus Delta V E suppose there is 100 degree centigrade rise in temperature that we take this. Suppose temperature goes from 25 degree centigrade to 125 degree centigrade then V B E changes from V B E 1 to V B E 1 minus or plus minus it decreases so it is minus how much 2.5 millivolt per degree centigrade therefore minus 25 millivolt alright is that okay 2.5 millivolt per 250 millivolt yes okay 250 millivolt therefore Delta I C by I sub C 1 would be equal to minus Delta V B E which is 250 millivolt that is 0.25 alright divided by V B E 6 minus 0.6 agreed this is the expression and this comes out as 4.6 percent so it is not too bad 4.6 percent okay finally we consider the change in I C I sub C due to a change in I C B O and here we have no escape we will have to use the expression which contains I C B O alright otherwise otherwise we cannot we have ignored the variation our approximate expression is V B B minus V B E plus I C B O times R B plus R E again I C B O may have changed due to a rise of temperature alright divided by R B plus beta R E we do the same thing we take I C 1 V I C B O 1 I C 2 corresponding to I C B O 2 subtract one from the other and do this algebra then Delta I C by I sub C 1 keeping all other parameters constant this algebra simplifies to the following beta Delta I C B O times R B plus R E divided by beta V B B minus V B E can you tell me why this comes it comes because of division by I C 1 right plus I C B O times R B plus R E now we have taken care of the variation of I C B O I C B O 1 thank you this is important it should be I C B O 1 this is Delta I C B O this is perfectly alright now we have taken care of the variation due to I C B O and therefore at this stage when we are numerically calculating can we wonderful beta does get cancelled that simplifies the expression further yes very correct now at this stage yes in the first case when we consider the variation is I C due to variation due to beta correct and that is the expression we have taken simplifies the expression in which we assume that beta is much greater than really if beta is somewhere near unity it is not greater than 10 okay if beta is somewhere near unity we will throw the transistor in the dust plate if it freezes if the circuit is taken to enter ticker and due to temperature beta goes down to near unity we will not the circuit will not work and therefore we will use a beta is we will use a transistor with beta was 1000 for example we will use a super beta transistor okay so beta much greater than 1 unless that is so the circuit is not useful there are ways even if you have a long transistor with a beta of let us say 10 I can increase it to 100 by making a circuit connection we will see the circuit connections later we will have to use two transistors in what is called a Darlington connection and that increases beta 200 beta 1 beta 2 because we have multiplied we will see the circuits later okay now in this expression as I say once I have achieved the expression now I can I can ignore this term compared to this term I have already done so I have already taken account of ICBO variation this term will be very small compared to the first term so we ignore this and therefore my expression becomes delta ICBO divided by not the way of writing this expression divided by VBB minus VBE divided by RB plus RE okay if the percentage change now no longer are you dividing delta ICBO by ICBO 1 do you notice this it is not required okay number 2 so it is not a percentage change it is the absolute change the absolute change in ICBO reflects itself in terms of the percentage change except for a reduction factor that is division by this quantity naturally we want this quantity to be as large as possible as large as possible means we have to make RB plus RE as small as possible this is contradictory we wanted RE to be made as large as possible and VBE of course we can make large compared to VBB we can increase VBB but nevertheless whatever the circuit this is the factor by which the percentage change is reduced that is an absolute change in ICBO causes a percentage change in I sub C which has a reduction factor of this and we should aim at making this quantity as large as possible but as I said this is contradictory to previous assumption and life is a measure of compromise from this circuit is no exception all right here also you have to make compromises now let us take an example suppose ICBO the same example is 0.01 micro ampere at 25 C and suppose the temperature increases to 100 125 let us say 100 degree centigrade rise in temperature causes a delta ICBO which is equal to 10 to the power I beg your pardon 2 to the power 10 and considering delta ICBO so 2 to the power 10 minus 1 multiplied by 10 to the minus 8 so many amperes which comes which calculates out as 10.23 micro ampere 10 to the power minus 8 10.23 micro ampere therefore delta IC by I sub C1 is equal to 10.23 micro ampere okay 10.23 micro ampere divided by 6 minus 0.6 divided by RB plus RE is 3400 plus 100 we should actually like this is 10 to the minus 8 because this is going to be dimensionless quantity right 10 to the minus 6 I beg your pardon this calculates out to 0.66 percent only okay 0.66 percent only therefore the effect of ICBO variation is not very drastic unless the temperature goes very high to to furnace temperatures for example okay 1000 sub degree. So ICBO is a harmless quantity as far as ordinary discrete transistor circuits are concerned but I must caution you that it plays havoc in low current circuits I if I sub C is in the order of micro ampere okay low current circuits are the are the are required in integrated circuits in integrated circuits you want to make the volume as small as possible and therefore the heat dissipation is a problem so you want to pass as little a current as possible so that there is as little a dissipation as possible okay there ICBO change causes havoc there are ways of reducing the effect of ICBO in both circuits also which will come to at the appropriate time now I consider what I considered was delta IC by I sub C1 due to variations in individual parameters keeping the other two constant as you know in the case of a multivariable function you can talk of partial differentiations and then if you want to find the total change then you shall have to combine the three and therefore if that is permitted here that is if the changes are not very drastic are not very large then I can write delta IC by IC1 due to a change in all the three parameters simultaneously as the sum of D3 delta beta by beta 1 RB divided by RB plus beta 2 RE which was due to beta variation plus delta VBE divided by VBB minus VBE plus delta ICBO multiplied by RB plus RE divided by VBB minus VBE okay please close this firmly in the process due to the disturbance I have made a mistake can you point out the mistake that is correct wonderful if I take the same example that is temperature going from 25 to 125 if I take the same example the rule of thumb is that you can make a combination if the individual percentage changes mind this is a rule of thumb individual percentage changes are less than 10 percent if you know the 10 percent then you cannot do this then you shall to go back to the original circuit consider the changed parameters recalculate the current and calculate delta IC by IC1 is the point clear if the individual changes if each of these changes is less than 10 percent then you can combine in our particular example they are less than 10 percent the highest was 9.2 percent and therefore in our in our example delta IC by IC1 this calculates out as you remember what was what was IC1 IC1 IC1 was 40 million if you remember the 2 point was 40 million and 4 volts okay 40 million multiplied by the percentage change that is 9.3 percent plus 4.6 percent plus 0.66 percent the positive sign here occurs because of a negative sign here minus delta V B E okay and this comes out as 5 point no I beg your pardon now it has been it has been taken care of here how did we calculate 9.3 percent it has been taken care of this comes out as 58 million here if I am 9 times 9 plus 4 13 yes think about it we have taken care of this in calculating this percentage I am glad you insist otherwise we made a mistake okay so this whole quantity is delta IC you are absolutely correct okay fine so how much is this 13 look 13.9 0.58 million here okay let us see 40 multiplied by approximately 14 divided by 100 you see this change now delta IC that is a change is 5.8 million here which means that the current is now 45.8 million here instead of 40 okay and this was done by calculating the individual changes and adding them up if we have done it exactly that is by taking the 3 quantities beta 1 V B E 1 IC B E 1 and beta 2 V B E 2 IC B E 2 then it turns out delta I sub C exact this has been calculated as 5.7 million here so we are not too much of okay we are not too much of this procedure therefore is a valid procedure alright yes please no no because this quantity minus V B E divided by V B E plus V B E you see V B E decreases and then the delta V B E itself is negative that is why we have to take this okay is the point taken yes yes same point right so all the variation should be added and the particular variation in carrier sign whether it is positive or negative depending on whether it is increases or decreases that is why the second variation subtracted but because of the expression what was the expression V B E minus V B E IC was proportional to this and therefore delta IC is proportional to minus delta V B E you see if you can say that these 3 factors even though one decreases with temperature they conspire to accentuate the error this is always the case it is a it is a manifestation of Marfield's law if something can go wrong it will here also things go wrong and even if one decreases they conspire so that the error is accentuated this is always the case and you cannot beat it will beat it only to a certain extent all you could do was to limit the change to 5.8 milliampere in 40 it is 6 milliampere okay so it is not too bad now an alternative BJT biasing circuit is as follows an alternative in which R E is not used R E is made equal to 0 is like it is V C C R sub C you have this is an alternative BJT biasing circuit you have the transistor which goes to ground there is no resistance R E but the base well this is a signal source with an RS VS coupling capacitor C 1 there is a coupling capacitor C 2 R L the base requires a biasing now we cannot do this biasing we cannot do a potential divided biasing because then the stability will be very poor is it not that R E cannot be made equal to 0 we have to think of something else what is thought of is this the base is supplied from here that is from V C E rather than V C C through a resistance alright so the base current for example can you this resistance is let us denoted by R B if this is I B then this is I B correct the base current is supplied from V C E rather than V C C one could argue why not V C C why not V C C no that is not the cause the cause is something different there will be no stabilization if it is supplied from here let me tell you why the stabilization occurs yes now what is I B I B is V C E minus V B E divided by R B if I C increases due to some reason V C decreases and therefore I B decreases and I C also decreases let me repeat this if I C increases then V C E decreases this causes I B to decrease and this causes I C to decrease on the other hand if you have taken the supply from V C C then such a thing would not have occurred agreed qualitatively you can see the stabilization alright now there is one problem though yeah this is the source resistance signal source resistance it does not come into the DC biasing there is a problem you see because of the occurrence of R B there is a there is a DC feedback that is what is at the current here is part of the current is divided here so from output to input there is a feedback we do not want this feedback at AC at signal currents we do not want this feedback so what do I do in parallel with R B shall we use a capacitor in parallel with R B let us see what danger or in series come on come on make up your mind and tell me what you want to do in parallel okay suppose we use a capacitor then the whole signal will go here for the signal this is a short and therefore the input signal goes over here and this voltage this voltage shall be equal to the input signal the transistor becomes redundant useless we cannot do that no what is done is the following it is a very ingenious solution and suggested by the problem is that if I have an R B here then not only not only there is a DC feedback there is an AC feedback also part of the signal current also flows to R B which you do not want. Sir what is the signal current? What is the signal current here? Signal current if the signal is there an incremental voltage here this will cause an incremental current here we want the whole of this current to pass through the load we do not want a die pass we do not want a D 2 alright we do not want anybody to share the signal current we want the whole signal current to go to the load so what we do is the following instead of straight R B we split R B into 2 parts R B 1 and R B 2 okay we split R B into R B 1 and R B 2 and from the middle point we connect a capacitor a die pass capacitor to ground okay then part of the signal current still flows like this but the signal current cannot come to the base that is there is no AC feedback we have reduced the AC feedback not only that the signal current cannot go to the output because of the capacitor so this is one of the solutions alright and this circuit is called a collector feedback biasing and you can show this will be one of the tutorial problems that if you follow the same type of analysis then delta IC by IC 1 is delta beta by beta 1 R B divided by R B plus beta 2 R E this is due to beta variation minus delta V B E divided by V C C minus V B E in the previous case it was V B B now it is V C C that is the only change plus delta IC B O R B plus R C divided by V C C minus V B E you can show that for this circuit in which R E is absent this is the total expression for percentage change in I sub C and if you notice carefully I made a mistake again this is R C if you notice carefully all that changes is that V B B is replaced by V C C here also and R E is replaced by R C so this circuit is equally effective in stabilizing the operating point except for that modification that one extra capacitor is needed to bypass the middle point of R B 1 and R B 2 in the next lecture in the next lecture we shall consider F A T biostabilization okay.