# BJT AMPLIFIER-3

welcome back to basic electronics in the last

lecture we showed that correct biasing of the b j t is crucial in amplification we also

looked at the simple biasing scheme which was found to be very sensitive to the beta

of the transistor we will now look at an improved biasing scheme and show that it is much more

robust with respect to variations in the transistor beta we will then turn our attention to the

second major issue in b j t amplifiers that is adding the signal voltage to the bias voltage

we will illustrate how a coupling capacitor can be used for that purpose so let us get

started here is an improved ah circuit and ah we will

see that ah dependence of i c or v c e on the transistor beta and the circuit is much

bigger than the previous circuit to analyze the circuit let us redraw so what we have

done here is ah instead of showing v c c like that we have drawn that explicitly and let

us check that the circuit is still the same as before from the base we have r one going

to v c c here from the base we have r one going to v c c this node is the ah common

node or ground shown here from the collector we have r c going to v c c same thing here

collector r c and then v c c so these two circuits are actually the same now what we

will do is to look at this part and find the thevenin equivalent circuit what is the thevenin

resistance let us deactivate this independent ah voltage source and then we see r one and

t two in parallel so that is r t h what about v t h the open circuit voltage here so imagine

that nothing is connected here and then this ah voltage is r two divided by r one plus

r two times v c c so that is our v t h so after ah making that transformation we get

ah this circuit let us calculate v t h and r t h v t h is

r two by r one plus r two times v c c r two is two point two k r one is ten k so v t h

is two point two by ten plus two point two times ten volts and that turns out to be one

point eight volts what about r thevenin r thevenin is r one parallel r two so ten k

parallel two point two k and that turns to one point eight k and now we know everything

about circuit here and ah let us now write k v l equation for this loop assuming that

the b j t in active mode so what does k v l say k v l says that this voltage drop v

t h is the sum of three voltage drops r t h times i b the first term here v b e about

point seven volts the second term here and r e times i e the third term here now i e

is bet plus one times i b and therefore we can write k v l as v t h equal to r t h times

i b plus v b e plus beta plus one i b times r e all right

we can now solve this equation for i b and we get i b equal to v t h minus v b e divided

by r t h plus beta plus one times r e now i see the collector current in the active

mode it is simply i b times beta and therefore we get this expression for i c all we need

to do is simply multiply this ah expression here by beta now let us calculate ah the collector

current for beta equal to hundred first and that turns out to be one point zero seven

milliamperes if we increase beta to another value let us say two hundred then i c turns

out to be one point zero eight five milliamperes and we notice that ah there is hardly any

difference between these two numbers for all practical purposes they are the same and this

of course is a big improvement over the previous circuit in which i c was very sensitive to

the value of beta here we have changed beta by a factor of two and see hardly any difference

in the collector current value so this is a big advantage of this configuration and

it is therefore commonly used for amplification having obtained i c equal to one point one

milliamps let us now proceed and ah get the other quantities of interest what about v

e v e is i e times r e and i e is nearly equal to i c because our beta is large enough about

hundred and if beta is ah hundred then alpha is beta divided by beta plus one or hundred

divided by hundred and one that is very close to one and therefore we can say that i c and

i e are nearly the same so we will use for i e one point one mill ampere so that times

one k is one point one volts and that is ah what is indicated over here what about ah

v b v b is v e plus these voltage drop which is point seven so one point one plus point

seven one eight volts what about ah v c v c is v c c minus i c r c v c c is ten i c

is one point one million r c is with three point six k so that turns out to be six six

volts and now we see immediately that the b j t is actually ah at operating in the active

mode n p n p is at one point eight volts n is at six volts so n is higher than p so therefore

the base collector junction is under reverse biased so that tells us that the b j t is

operating under operating in active mode or linear region what about ah this voltage difference

v c e six volts minus one point one so that turns out to be four point nine volts

there is a way to quickly estimate the bias values if we assume that beta is ah very large

ah same finite and what is the implication of large beta the base current i b is i c

over beta and if beta is large then the base current is small so what we are going to do

is ah ignore the base current which is fair if ah beta is large and in that case this

current is zero negligibly small and therefore this ten volts gets divided between r one

and r two ah this is like an open circuit so there is no current there and then we can

ah obtain v b simply by voltage division r two by r one plus r two r two by r one plus

r two times v c c so that turns out to be one point eight ah volts once we get v b we

know that ah the base emitter junction is under ah forward bias so there is a voltage

drop of point seven volts there so one point eight minus point seven that is one point

one volts i e once we get v e ah i e is simply v e divided by r e r one point one milliamp

and if beta is large then alpha is equal to one and i c and i e r then equal so i c is

alpha times i e nearly equal to i e that is one point one milliamp and ah then we can

quickly find v c e v c c minus i c r c time minus i e r e v c c minus this voltage drop

minus this voltage drop that is v c e so that is ten minus three point six k times one point

one milliamp that is i c r c minus one k times one point one milliamp that is i e r e and

that turns out to be five volts so there is a slight difference here ah the earlier v

c e was four point nine when we actually consider the value of ah beta about hundred and if

you consider beta to be very large then it is five volts so four point nine there and

five here not ah a big difference and very often this ah quick estimate of ah the bias

values is very helpful ok we have now found a reasonably good biasing

scheme which is insensitive to ah the transistor beta value and now the next challenge is to

add the signal to the bias so here in this circuit there is no signal anywhere there

are only d c quantities or bias quantities not to that we want to add the signal if we

cannot add the signal ah of course the whole amplifier is useless because ah then it will

not be amplified so the input signal v s of t which is typically taken as a sinusoid for

example needs to be mixed with the desired bias value v b so that the net voltage at

the base this voltage is the d c component which we have already computed in the last

ah slide and in that example it was one point eight volts plus the signal

now the signal we have taken as a sinusoid but in general it need not be a sinusoid for

example we have any [aud/audio] audio application then it is not one signal ah frequency but

ah whole bunch of frequencies with its different fourier components and so on so if you look

at the audio signal in time domain it will look very different from a pure sinusoid but

ah for simplicity of analysis we often take ah v cap sin omega t as the input signal so

let us proceed with that so now the challenge is to mix these two and this can be done by

using the coupling capacitor c b the coupling capacitor here so we have the signal here

v cap sin omega t for example and we have the rest of the amplifier and we connect the

signal to the rest of the to the amplifier with this ah coupling capacitor it is called

coupling capacitor because it couples the signal to the amplifier and how does it work

so let u explain that by considering a simple circuit ah that will illustrate this ah idea

of a coupling capacitor ok let us start with this simple r c circuit

lets leave the b j t aside ah for the moment because it makes life ah more complicated

and ah for this r c circuit we have two sources one is ah d c source v zero and one is sinusoidal

ah source v m sin omega t and we are interested in the solution what do we mean by the solution

currents and the voltages in the circuit in the sinusoidal steady state and we have ah

come across this term earlier what is sinusoidal steady state that is when the exponential

transients are vanished and the exponential transients arise because ah we have a time

constant in this circuit so there will things like e raise to things like minus t by tau

and ah if t is sufficiently large then those terms would go to zero and that is what we

mean by ah the exponential ah transients have vanished so after that has happened each quantity

x of t could be able to it could be this voltage it could be this current is of the form the

constant x zero plus a sinusoid and the sinusoid has an amplitude of ah x m and a phase of

alpha our job now is to find ah x zero and ah x

m and alpha that means the complete solution for x of t in the sinusoidal steady state

there are two ways of going about it one solve the circuit equations directly let us see

how to do that let us say that this ah node voltage is b ah sub a and this is ah total

voltage that is it has constant part as well as a sinusoidal part now this current here

is v a by r one and we are talking about the instantaneous current v a divided by r one

that current there this current is v a minus v zero divided by r two because this node

voltage is with respect to ah this count v zero so v a minus v zero divided by r two

that is this ah second term here and ah this current plus that current must be equal to

the current entering this node ah from the capacitor side and what is that current given

by it is given by c d v c d t what is b c b c is v s with respect to this ground v s

minus v a so d d t of v s minus v a so that is the equation that we get thus one we have

doing it and ah of course we can imagine that it is not easy to ah solve this ah equation

this so therefore there is a second way which we can ah which we might find much easier

and that is to use the d c circuit plus a c circuit approach and ah we will ah see what

this is in the next slides what we are now going to do is to consider

this components one by one and see how they behaved when a combination of b c and sinusoid

is applied let us start with registers we have ah v r the voltage equal to constant

part and ah sinusoid v r cap sin omega t plus alpha and ah here is a new notation that we

have introduced small v capital r is the total instantaneous voltage capital b capital r

is the d c part and small b small r is the sinusoidally varying part that is given by

v r cap sin omega t plus alpha similarly we have current ah given by a constant plus sinusoidally

varying part and ah we are considering the sinusoidal steady state in a circuit and this

register is part of that circuit and we know that for register the total instantaneous

voltage is simply r times the total instantaneous current and ah when we substitute for p r

and i r we get this relationship and ah we can split this into two equations one capita

v capital r times i r this part and ah small v and small r is r times small i small r these

are the d c quantities v r and i r and these sinusoidally varying quantities in other words

we can think of a register in this situation where there is d c and sinusoidal part to

be described by two circuits a d c circuit and an a c circuit in the d c circuit we have

the d c quantities capital v capital r and capital i capital r they are related by this

ah equation which is ah the resistor equation then we have the a c circuit in which we have

small v small r the sinusoidally varying part small i small r again the the sinusoidally

varying current nd they are also related by the same behavioral equations so small v small

r is r times small i small r [vocalized-noise] so in the d c circuit as well as in the a

c circuit the resistor essentially looks the same it behaves like ah a resistor and the

relationship is p equal to r times i where v and i could be either d c quantities or

they could be the sinusoidally varying quantities let us repeat it for ah the capacitor now

so we have again ah total instantaneous voltage a d c part and a sinusoidal part total instantaneous

current a d c part and a sinusoidal part so these are constants and these varies sinusoidally

and ah because ah the capacitor involves ah derivative we have made this alpha and beta

different and ah at this stage we do not know what the relationship between alpha and beta

is but we will find out now the since total instantaneous current

a c d d t of the total instantaneous voltage we substitute for i c like that and for v

c like that and now we can once again split this equation into two the first one relating

the d c quantities i c is equal to c d d d t v c and that is zero because c is a constant

the second equation this small i small c is c d d t of small v small c that is the second

equation and this equation looks pretty much like ah capacitor equation and ah what does

this ah correspond to this is saying that the d c current is zero so that means in the

d c situation the capacitor is an open circuit not surprising and ah in the a c situation

or the sinusoidal situation we have the same equation as ah we would have ah for the total

instantaneous quantities so the capacitor in this situation in the a c situation looks

like a capacitor and can we now figure out what is the relationship

between alpha and beta we can turn this into a facer we can turn that into a facer and

then go through the facer analysis and then we will see that the current leaks the voltage

and therefore they are related by beta equal to alpha plus pi by two what about ah a d

c voltage source the equation is the total instantaneous voltage is a constant and ah

because it is a constant there is no sinusoidally varying part that is zero and so in the d

c circuit we have that constant voltage source and in the a c circuit we have voltage equal

to zero that is a short circuit and the d c current capital i capital s is shown in

the d c circuit and the sinusoidal current of the a c current is shown in the a c circuit

that is small i sub small s let us look at an a c voltage source the equation

is ah the total instantaneous voltage is zero d c plus the sinusoidal part the time varying

ah sinusoid now since ah the d c part is zero the d c circuit simply has v a is equal to

zero which is a short circuit and ah sinusoid is then incorporated in the a c part so the

a c circuit is simply ah ah voltage source an i c source and the i c circuit is a short

circuit once again the d c current is shown in the d c circuit capital i capital s and

the a c or the sinusoidal current is shown in the a c circuit small i sub small s

we can now use the findings ah in our original r c circuit so we replace r two with its ah

d c equivalent which is also registered and the a c equivalent that is also a resistor

similarly for r one what about the capacitor the capacitor as we have seen in the d c circuit

it is an open circuit and in the a c circuit it looks like a capacitor the d c source in

the d c circuit is a d c source and in the a c circuit is a short circuit zero volts

similarly the a c source is a short circuit in the d c circuit and in the a c circuit

it is ah the sinusoidal source so we have split the original circuit into ah d c circuit

and an a c circuit and we know how to get these from the original circuit by systematically

replacing each of the components now let us ah write equations for these two

circuits the d c and a c circuits for the d c circuit we have v a over r one this v

a over r one that current plus v a minus v zero by r two v a minus v zero by r two that

current and they should add upto zero because this is an open circuit remember capital b

sub capital a is a constant ah voltage for the a c circuit we have small v sub small

divided by r one that current plus this current which small v small a divided by r two so

that is the sum of those two currents that current and that current and let us remember

that ah this quantity is ah sinusoidally varying quantity and it has no constant in it now

these two currents the addition of these two must be equal to the capacitor and that is

given by a c d d t v s minus v a so that is the equation we get for the a c circuit if

we add equations one and two we end up with equation three here and what is it saying

it is saying that the d c b a plus the a c b a divided by r one that is the total instantaneous

current plus the d c b a plus a c b a minus v zero divided by r two that is the total

instantaneous current through r two should be equal to c d d t v s minus v a and both

of these are sinusoidal quantities let us compare this equation with what we

had obtained ah earlier directly from the original circuit and that equation is here

this is the total instantaneous quantity this is also the total instantaneous quantity this

is instantaneous v a minus v zero this is also the instantaneous v a minus v zero and

what about ah this ah capacitor current d d t v s minus the instantaneous v a now this

instantaneous v a has two parts one is a constant part and one is the sinusoidal part and the

derivative of the constant part of course is zero and therefore we get the same ah terms

as this one here so in other words equation three and four are actually the same and this

a very ah powerful statement because now instead computing v a directly from ah equation four

we can compute the constant part using equation one the sinusoidal part using equation two

and then then use ah the instantaneous v a as the sum of d c v a and the a c v a we can

do this computation separately and that is big advantage and then just simply add them

to get the total quantity so this is very useful approach splitting the original circuit

into a d c part and then a c part working on the d c circuit and the a c circuit separately

it turns out to be much simpler than the than working on the original circuit and then simply

adding the two solutions to get the total instantaneous ah quantities

to summarize we looked at the robust b j t biasing scheme and showed that it is relatively

insensitive to the beta of the transistor we then discussed the use of a coupling capacitor

to couple the signal voltage to the amplifier we have considered an r c circuit to explain

how a coupling capacitor works in the next class we will consider a b j t amplifier with

the coupling capacitor so see you next time