 This is the 30th lecture and the topic for today is BJT biasing and introduction to power amplifiers. We had discussed last time a BJT biasing circuit in which the emitter was connected directly to ground, the collector was connected through R sub C to plus V C C a supply and the base current was supplied by means of a resistance connected directly to V sub C C and this resistance we had decided to call R sub B and we had seen that if due to any reason the current in the collector I sub C if due to any reason I sub C increases then the process is such that after several intermediate steps I sub C further increases and this goes on in a circle till what you get is a thermal run away, it is as if the transistor cannot tolerate or accept any more heat and therefore it runs away from its responsibility which means that it gets damaged, it gets burnt out, thermal run away and the remedy for this that we had suggested last time and had terminated the class at that point was that you introduce to this circuit a couple of other resistances namely you introduce a resistance from here to ground and another resistance at the emitter lead. This is the most common circuit that is used for biasing a transistor, this resistance which is connected to the emitter lead is called RE, the resistance that is connected from the base to ground is R1 and this resistance is re-designated as R2 and this is the standard terminology I sub C is the collector current, I sub B is the base current and the total circuit is called a self-bias circuit, self-bias circuit. Let me draw this again and explain how this helps in getting the bias stabilized, you have a emitter resistance RE, a collector resistance R sub C, this goes to plus VCC plus VCC because it is an NPN transistor, if it is PNP then it would be a negative supply, then you have a resistance R1 here, no R2 and the resistance from here to ground is R1, this is the circuit and this current, this current is I sub C, this current is I sub B, this current obviously by KCL is I sub C plus I sub B alright and the total circuit is known as the self-bias circuit and to understand why and how the collector current stabilizes, you assume that due to some reason I sub C increases, if I sub C increases then so does I sub C plus I sub B and therefore the voltage drop across RE if we call this V sub E increases, if V sub E increases then naturally VBE shall decrease, VBE shall decrease and if the diode voltage decreases then you know the current shall also decrease and therefore I sub C increase means VE increase which means VBE decrease, VBE decrease and VBE decrease means I sub B decreases and I sub B decreases shall mean that the collector current which is beta times approximately beta times I sub B shall decrease and therefore the increasing tendency of I sub C shall be arrested and therefore I sub C shall be stabilized, qualitatively this is the explanation of the bias stabilization due to the self-bias circuit, quantitatively, quantitatively in order to calculate the values of the resistances and what values of resistances should be put to get a proper operating point, the Q point as we call it with stabilization of the operating point irrespective of changes of temperature, irrespective of replacement of transistor, this was the main cause, one of the main causes why the previous circuit could not be could not be used because that had kept IB constant and therefore if we replace the transistor, the beta of the transistor as you know can have a spread of about 1 is to 6, irrespective of whether the transistor is replaced, whether temperature increases or decreases, I sub C, I sub C should remain a constant. Let us see analytically how this happens. In the process you should remember, I will keep this circuit in the in the background plus V C C, R E, R 1, R 2, R C, this is I C and this is IB and this is IB plus IC, as many times as you draw it the better. Now to to analyze this quantitatively, the the transistor, the transistor must be replaced by an equivalent circuit and the transistor simply the base to emitter simply means that you require the base to emitter is a junction, is a diode and we assume that it is an ideal diode in series with a battery of V BE which is approximately 0.7 volt for silicon and 0.3 volt for germanium and therefore the transistor is equivalent to the following. It is equivalent to an ideal diode in series with V BE alright, this is the base and this is the emitter, base emitter, this is the collector and under this condition you know that if the base current is IB then I sub C, the collector current is simply beta IB plus beta plus 1 ICBO. The collector current obviously is independent of V CE alright and therefore the collector, at the collector simply there are 2 current sources, there are 2 current sources, one of them is beta IB, one of them is beta IB and the other is beta plus 1 ICBO okay. This is the circuit, this is the equivalent circuit that we shall use for quantitative analysis. For quantitative analysis we shall also use a simplification, you see this branch, this branch consisting of R2 and R1 and VCC can be separated that is you can include another VCC here, it does not matter as far as analysis is concerned. So at this R2 is connected to VCC then connected to base and R1 and if we look to the left of this line we can replace this by a Thevenin equivalent and the Thevenin equivalent would be a battery VCC multiplied by Thevenin equivalent voltage source is the open circuit voltage multiplied by R1 divided by R1 plus R2, is that clear? No, well let me draw it again, you have a VCC then an R2 and then R1 and this is this goes to the base, we are looking to the left of these 2 lines from base to whatever the point is let us call it G, we are looking to the left of this line. So what we have is we have a battery VCC then we have an R1 and we have an R2 and it is these 2 points that we are looking into for Thevenin equivalence and therefore this is simply equivalent to a Thevenin voltage let us call this VBB in series with a resistance which is parallel combination of R1 and R2 we call this resistance as capital R subscript capital B, this is R1 parallel R2 and VBB obviously is VCC multiplied by R1 divided by R1 plus R2. Now this as you will see greatly simplifies the analysis, if we substitute this in the original circuit then my equivalent circuit becomes, let me draw it on a separate sheet, the equivalent circuit becomes I have a VBB then I have R sub B then the transistor RE then R sub C plus VCC and this current, this current is I sub C and this current is I sub B therefore also this current is I sub C plus I sub B. Now if I write KVL equation around this loop KVL equation then I get VBB is equal to I sub B RB the drop in RB plus the drop from base to emitter plus VBE plus VBE plus the drop across RE which is I sub C plus I sub B RE. This is one equation in which the 2 currents I sub C and I sub B are unknown but we know that I sub C and I sub B are connected by the relation I sub C equal to beta I sub B plus beta plus 1 I sub CBO and therefore and therefore I can eliminate I sub B from here, our purpose is to obtain I sub C and to see how stable the collector current is. So what I do is I replace in this equation the 1st equation I replace I sub B by I sub B equal to I sub C minus beta plus 1 I CBO divided by beta all right. I replace this here and also here. Then I can simplify the algebra and obtain an expression for the collector current I sub C. The result after this algebra which I shall skip is as follows. Let me write down the complete expression. The expression is I sub C equals to VBB minus VBE and you can see where this comes from VBB minus VBE that means I take it to the left hand side and then I shall have a term containing I CBO. I CBO comes from here and also from here. The term is beta plus 1 divided by beta I CBO then RB plus RE. Write down the expression because it is a very significant expression although a bit long but we shall investigate we shall go very deep into this expression to be able to understand how and why I sub C is stabilized. Why is it called a stabilized bias circuit? The denominator is RE plus RB plus RE divided by beta. This is the expression for I sub C which contains its dependence on temperature through 3 things. What are these? In the denominator it is RE or it is there is no RC at all. This is correct RE plus RB plus RE by beta. This is correct okay. Now I want you to notice I want you to qualitatively first understand how I sub C how I sub C depends on 3 temperature dependent quantities. One is VBE. As you know VBE decreases at the rate of 2.5 millivolts per degree C. This is VBE decreases at the rate of 2.5 millivolts per degree C and therefore VBE is a temperature dependent quantity. Beta also increases with temperature almost linearly. Beta is proportional to temperature alright. Almost linearly and I sub CBO increases it doubles it exponentially. It doubles for every 10 degree centigrade rise of temperature. I sub CBO doubles for every 10 degrees C rise which means that if I have let us say 2 temperatures T1 and T2 look at this modeling very simple modeling. If I have 2 temperatures T1 and T2 where T2 is greater than T1 then I sub CBO2 that is at the increased temperature shall be I sub CBO1 that is the decreased temperature multiplied by 2 to the power T2 minus T1 divided by 10. Is that okay? This is a very interesting way of writing the expression. This is approximate but we can work with this. Is this clear? That if T2 minus T1 is 10 then I sub CBO1 is multiplied by a factor of 2 which is in keeping with this relationship that is I sub CBO doubles for every 10 degree centigrade rise in temperature alright. It also can be affected besides temperature even if you keep it within a temperature stabilized enclosure it can be affected by a replacement of transistor. Transistor goes bad in an amplifier so you pull it out and put in another transistor of the same type but you know beta, beta has a large spread. It can vary by as large a ratio as 1 is to 6. If the original transistor was a beta 30 the new one may have a beta of 180 but even then you do not want I sub C to change. You want I sub C to be stabilized and therefore we must examine this expression as to how this stabilization occurs. Before I take up the question of stabilization let me illustrate with an example as to how the things change with temperature utilizing this expression. Suppose we have a transistor circuit I will keep this expression in view is it is it visible okay right. Suppose we have a transistor we have a transistor circuit in which the beta well it is a silicon transistor and beta varies between 30 and 180 as I said 1 is to 6 alright a silicon transistor. Vbe the temperature is such that Vbe varies between 0.5 millivolt and let us say 0.9 millivolt it is a silicon transistor but with temperature Vbe can vary between 0.5 and 0.9 and I sub Cbo can vary between let us say 1 nanoamperes nano is 10 to the minus 9 to let us say 10 nanoamperes as much as 1 is to 10 that means the temperature may rise it will double for every 10 degree centigrade and therefore this is a large range of temperature it is 2 to the power what is the temperature difference for which the ratio can be 10. 10 is 2 to the power 3 is 8 so it is slightly less than 4 slightly less than 4 2 to the power 4 yeah between 8 and 16 okay. What we want to know is what is the worst case variation in I sub C alright obviously then you want to know what the circuit is let us look at the circuit now the circuit is the usual circuit the self bias circuit and let us say the values are given the values are given as 10 K 90 K these are typical values 90 K and then this is 2 K R is 2 K R C is 15 K and V C C is plus 28 volt this is a typical circuit typical bias stabilized circuit to evaluate the worst case currents obviously you require the values of R sub B R sub B is the parallel combination of 90 and 10 K if you recall it is a parallel combination of R 1 and R 2 so it is 9 K and V sub B B V B B is 28 multiplied by 10 divided by 100 so it is 2.8 volt is it okay now let us consider the worst case we recall that I sub C is given by V B B minus V B E plus beta plus 1 divided by beta I C B O R B plus R E divided by R E plus R B plus R E divided by beta the 2 worst cases are the following one is that you put V B E you want the minimum I C you want the worst cases that minimum and the maximum obviously minimum shall occur when V B E is 0.9 all right because it subtracts V B E is 0.9 millivolt beta the smallest value which is 30 now you might ask me why do you take the smallest value because this is 1 beta plus 1 divided by beta and there is a division by beta here. Now smallest is 30 31 divided by 30 and 181 divided by 180 they do not differ much 1 plus 1 over beta and therefore beta basically controls this quantity so you put beta equal to 30 beta equal to 30 and I C B O put the lowest value that will give you the lowest I C is it okay lowest value is 1 nanoamperes you put down these values then I sub C comes as 0.8 milliampere now the other extreme the other extreme is when V B E is 0.5 millivolt let me indicate this in another colour the other extreme is 0.5 millivolt beta is 180 and I C B O is is 10 nanoamperes beta is 180 and under these conditions if you put down the values then I C calculates as 1.1 milliampere and you see the range 0.8 to 1.1 approximately 30 percent is it okay 30 percent but this is not a very good design it can be reduced to much to a much less much lower value by an appropriate design but even a roughly design circuit causes a change of only 30 percent all right. Now after this demonstration of the example let us look at this expression again carefully and see what should be our design steps obviously obviously if you look at the dependence on beta dependence on beta the numerator there is a term containing beta in the denominator also there is a term containing beta and as I said 1 plus 1 over beta when beta changes from 30 to 180 does not make much of a difference because 1 by 30 is 0.03 and 1 by 100 is 1.01 all right. So it does not make much of a difference the dependence on beta there are 2 reasons why it does not make much of a difference what beta is that is because I C B O is also typically a very small quantity and therefore the total quantity here compares negligibly with V B B minus V B E and therefore beta dependence is mostly through the denominator all right and to make this insensitive to beta what does that mean it means that insensitive to transistor replacement all right it is transistor replacement which causes the maximum change in beta. So if I wish to make I C independent of beta what I should do is I should make R E much greater than R B plus R E divided by beta which means that R E should be well beta minus 1 R E see if I multiply beta R E and I bring R E from right to left this should be much greater than R B all right this is the condition for independence to beta and you know beta is much greater than 1 therefore this condition approximates to R B much less than beta times R E because beta is much greater than 1 therefore one of the conditions in a stabilized biasing is that is that this design condition should always be fulfilled R B must be much less than beta R E and in electrical engineering much less than or much greater than always is can safely be taken to be 1 is to 10 and therefore if you know beta you know R E what you choose R B as minimum beta R E divided by 10 which beta should you take the nominal or the worst case or the worst worst minimum case or worst maximum case which minimum all right because you are taking much less than and therefore R B must be much less than beta mean R E which means that you the lowest value of R B that you can take is beta mean R E is the lowest value or the highest value highest value the highest value of R B that you can take would be or let me put it here R B highest is equal to beta mean R E divided by 10 this is a golden rule of thumb and must always be kept near the thumb all right it must always be remembered this is one of the basic design conditions of a BJT biasing circuit. Let us look at the expression again to find out other parameters now suppose suppose I satisfy this suppose I satisfy this condition all right then my expression will simplify to the following it will simplify to V B B minus V B E plus beta plus 1 divided by beta we take as 1 so I sub C B O times R B plus R E is it okay if we compare the previous expression beta plus 1 divided by beta we take as 1 and therefore I C B O R B plus R E and in the denominator we shall have simply R E because that beta term is negligible now how do I reduce my dependence on I C B O I C B O as you know doubles for every 10 degree centigrade rise in temperature temperature. So what I have to do is that this term I C B O R B plus R E must be made very small compared to V B B minus V B E now who determines V B B obviously V C C and therefore the higher the value of V C C the better will be the stability there is also a choice in I C B O you see for silicon transistors I C B O is much smaller as compared to germanium transistors for example a typical silicon transistor as I said I C B O could be 10 nano amperes whereas for germanium transistor I C B O could be as large as 5 micro amperes micro amperes in a relative terms is a small quantity but relative to 10 nano amperes obviously 5 micro amperes is a large quantity 5 times 10 to the minus 6 and 10 times 10 to the minus 9 and this is why silicon transistors are much more preferred than germanium transistor this is one of the reasons that I C B O is much smaller okay now normally in a silicon transistor if you choose your V B B properly then this will be satisfied alright in any case you must see that this inequality is satisfied this is one of the inequalities finally the devil is V B E which decreases at the rate of 2.5 millivolt per degree C how do you reduce the dependence on V B B V B E obviously what you should do is V B E should be much less compared to V B B and how do you obtain this V B B is V C C R 1 by R 1 plus R 2 you must choose your R 1 R 2 and V C C such that this relation is satisfied okay so these are the basic design parameters of a BJT as you can see as far as temperature is concerned as far as temperature is concerned in this expression we have eliminated beta alright in this expression the temperature dependent quantities are 2 one is V B E and the other is I C B O and therefore if temperature changes from capital T to capital T plus delta T alright then what would be the corresponding change in I sub C that is what would be delta I sub C obviously delta I sub C shall be V B B is independent of temperature so minus delta V B E minus delta V B E plus delta I C B O times R B I am intentionally ignoring R E because usually R B is one order higher than R E that is approximately 10 times R E and therefore I am ignoring that this divided by R E this is a simplified expression for calculating the temperature dependence of I sub C for example if you take a silicon transistor then V B E is 0.7 volt and I C B O typically is 10 nanofamperes for a germanium transistor the corresponding figures are 0.3 volt and let us say 5 micro ampere and suppose there is a 50 degree centigrade rise of temperature suppose delta T is 50 degree C then the question is how much does I sub C change delta I sub C contains 2 quantities one is delta V B E and one is delta I C B O due to 50 degree centigrade rise in temperature what will be the factor by which I C B O changes 2 to the 5 that means 32 alright and therefore the new I C B O shall be 32 times 10 nanofamperes 32 times 5 and delta I C B O should be 31 multiplied by these 2 quantities is that clear is that clear why 31 okay so so if I take silicon if I take silicon delta I sub C shall be equal to let us take some specific values of R E O let us take for the same design the design that we have just let us take 2 kilo okay 2 K and what do we what did we take R B 9 K or 9 K alright then delta I C B O for silicon would be 332 times 10 minus 10 so 310 multiplied by 10 to the minus 9 then plus now you see what happens delta V B E V B E actually decreases with rise of temperature so what would be delta V B E it would be delta V B E is my negative but because of minus delta V B E it will be an addition and delta V B E therefore shall be minus 2.5 millivolt multiplied by 50 which is equal to minus 100 and minus 0.125 volt is that correct and then 25 millivolt which is 0.125 so this would be 0.125 I must multiply by 10 to the 3 10 to the 3 and this figure this figure calculates out to 0.05 milliampere approximately do you understand this calculation okay one the other hand one can now say if it is germanium and you proceed to the same calculation delta V B E is the same isn't that right delta V B E is the same it is only delta I C B O which now becomes how much 31 times 5 micro ampere which is 155 micro ampere if you put that delta I C calculates out to 0.75 milliampere and you see the range of variation whereas this is only 0.05 germanium same rise of temperature can cause a large change in ice of sea and this is the reason why silicon is always preferred particularly where the device has to work under extreme temperature conditions all right. We can now summarize the methods the procedure for BJT biasing what are the steps the steps are that the first thing is to do is to choose a Q point how to choose a Q point we shall illustrate later but first choose a Q point what does this mean it means choosing an I sub C the collector current it means choosing a V sub C E all right and if you know the nominal beta then obviously you know I sub B also you know I sub B if you know beta all right this the two these two quantities are enough to specify a Q point this is an additional information that is on the characteristics which characteristic does the Q point lie on you know that the characteristics are a regularly spaced parallel lines depending on the values of beta depending on the values of I sub B I sub B is the parameter and therefore which I sub B line it lies on this is an additional point next next you have to determine at the next you have to go to the emitter you recall that the emitter has a resistance RE and the current through this is I sub C plus I sub B all right if RE is given then of course you know what the drop across this is if RE is not given as is usually the case you arbitrarily assume a value for V E you assume a value for V E suppose your power supply that is given is let us say 12 volt then obviously RE cannot drop how much it cannot drop 12 volts because there would be a VCE there will be a drop in RC and you do not want this drop to be large because you want a large swing in the output voltage typically this is chosen about 20 to 25 percent rule of thumb again there is nothing sacred about it all right 20 to 25 percent and one of the things that textbooks or manufacturer specify is arbitrarily choose V E equal to 3 volt now suppose this transistor is to go into a space instrumentation into a space vehicle where voltages are available are very small maybe 1 volt is the total voltage available then obviously this will not do you will have to change the design appropriately maybe 0.3 volt but for normal operating conditions 12 volt supply or 15 volt supply you assume V E approximately as 3 volts this must be done with care if RE is given you do you need not do this but if RE is not given you assume this to be 20 to 25 percent of the given power supply and then you can calculate RE because you know IC plus IB all right. So the second step would be to determine an RE or V sub E if RE is given then you determine V E if RE is not given then you assume a V E and determine RE all right the next thing is select V CC third step V CC and R sub C select V CC and R sub C now if V CC is given if this is given if the power supply is given then what is R sub C this will be V CC minus I sub C well I am sorry R sub C would be V CC minus V E minus V C E all right because if you recall the transistor is like this this is V CC V CC minus V C E minus V E divided by I sub C so you can determine RC on the other hand if RC is known it might be specified by the by the load the load may be specified then if RC is known then you calculate V CC V CC would be I sub C RC plus V C E plus V E all right this is the third step either you find RC or you find V CC the next step is choose an RB the next step is choose an RB that is R B which is the parallel combination of R 1 and R 2 parallel combination of R 1 and R 2 which means R 1 R 2 divided by R 1 plus R 2 and I have already told you that this should be chosen as beta minimum times RE divided by 10 all right far less than or far greater than 1 is to 10 so you choose an RB all right and once you have chosen an RB why divided by 10 what I want is RB should be much less than beta RE this is the rule of design all right much less than therefore it must satisfy the expression even when beta is beta minimum and much less than or much greater than in electrical engineering is what 1 is to 10 ratio that is enough this is why I divide by 10 you can divide by 20 there is nothing sacred about it all right this is a arbitrary figure but divided by 10 is taken to be specifying much less than or much greater than and the fifth step is after you choose this you know R 1 R 2 you also know that V B B is equal to V CC R 1 divided by R 1 plus R 2 and this would be equal to if you recall the Thevenin equivalent circuit at the base emitter junction this is equal to I sub B RB this are the circuits V B B is equal to IB RB plus V BE plus V E all right you know V BE you know IB you know RB you have chosen it all right and therefore you know 2 things one is R 1 R 2 divided by R 1 plus R 2 and the other thing that you know is R 1 divided by R 1 plus R 2 from these 2 then you find out R 1 and R 2 and this complete the design we shall now have a break for 5 minutes and then continue with the design.