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We
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continue our discussion on power amplifiers,
just to remind you few points in power amplifiers,
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because almost the whole region active region
is used, because of large current and voltage
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swings. So, non-linearity’s are required
and are there are present, and because of
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this non-linearity’s the analysis which
we did in the small signal case based on some
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fixed parameters like current gain beta and
so on, that does not prove to be accurate.
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So, graphical analysis is done and for the
graphical analysis as we were discussing that
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load line is very useful, and ac and dc load
line are different, because the impedance
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seen by the by the collector is different
in the two cases.
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And when we were discussing the a c load line,
then the components of a c load line; these
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were i c saturation is I CQ plus V CEQ by
r c plus r E. And V C E cutoff is V CEQ plus
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i CEQ r c plus r E. This two expressions we
got and they represent the true points of
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exchange points of the load line this is the
saturation point this is the cutoff point
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and this two expressions were obtained. Now,
with this what we have said is
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this is i c and this is V CE and this is the
Q point, then
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this is how this is the operating point value
I CQ and this voltage here this is V CEQ and
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this current will vary across this operating
point from this region to this region similarly,
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the voltage will vary.
From this region to this region so this is
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from center to the peak is V CEQ and this
point
this point is V c c which is equal to actually
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because , for the central point V twice of
V CEQ this common this is minimum voltage
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this is maximum voltage similarly, here this
is the minimum current i c maximum and i c
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minimum. And the minimum and maximum voltages
are here this is minimum voltage this is maximum
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voltage. We can write here V C E minimum and
this point is i CE maximum
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and this we are going to use this concept
of ac load line we are going to use for the
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analysis. Now actually, we start the analysis
of a class A power amplifier and for that
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we find out
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What is the maximum output power output ac
power? We are concern with ac power and efficiency
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and efficiency of class A amplifier. To carry
the analysis, we do some simplifications and
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these simplifications are that i c minimum
this 0 starts from here and this is slightly
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actually more than 0. So, we take it as 0;
i c minimum we take a 0; so the first simplification
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or approximation is that i c minimum, we take
at 0 and this maximum current we take slightly
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above this point which is actually the
twice of I CQ point just ever i c set we take.
So, i c minimum we take as 0 and i c max this
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points will become again clear and the approximation
which they will be bringing to the analysis
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is just few percent which is tolerable in
all electronic circuits and this i c max is
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taken as 2 I CQ.
And actually, this we are taking as i c set;
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and similarly, minimum voltage this voltage
is V CE minimum we are taking it at 0 and
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look here this was the maximum voltage which
we are taking at this point. So, this approximation
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brings V CE
minimum as 0 and V CE maximum as two V CEQ.
These simplifications approximations, they
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are just very near to the real situation and
the analysis because of this, simplification
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becomes very simple just we will see that,
the analysis becomes very simple. So, what
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actually with these simplification is mean
that, the whole region of the output characteristics
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we have slightly extended little towards the
cutoff that also we have included in the active
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region and the portion a small portion in
the saturation region that is all now becomes
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the part of the active region.
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So, the whole the active region has been slightly
extended because of these approximations and
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what we get is this here, and this is V CE
and this is i c this is the saturation point
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saturation point this is the cutoff point
and this is the Q point and for the optimum
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use maximum power delivery to the load we
will have to use this whole extending from
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0 to this point which is V cc and since this
is V CEQ so this is double of that because
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it is in the center similarly, this is I CQ
so this point is 0; this is in the center
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since this becomes twice I CQ. And now here
is and similarly, here
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this will be the current i c.
Now, this is the peak value of the voltage
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V p, this is V p; this is equal to V CEQ similarly,
here same and here this is the peak value
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of current
peak value of the current peak value or peak
value current is this. Now, once we know this
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peak values this is the ac current in the
output circuit this is the voltage ac voltage
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then the output ac power because our objective
was what was the maximum output power in class
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A amplifier and what is the efficiency conversion
efficiency of class A amplifier. So, the output
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ac power this is we write P o for output power
so P o is rms voltage into rms current product
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of rms voltage to rms current. Now, from this
figure we see that the peak current peak current
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I p; this is I p peak current; I p this is
equal to I CQ because this is I CQ. And
the peak voltage from here peak voltage V
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p as here V CEQ therefore, we said this are
the peak value rms values.
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We can write we can write the expression for
the output current. output power the output
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power P o therefore, is V CEQ by root 2 into
I CQ by root 2 therefore, in class A operation
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the maximum power is half of the product of
V CEQ I CQ by 2; this is the maximum value,
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so that will depend on what is the current
here, what is the voltage here and depending
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on how much power you want you will have to
use a transistor of that power for example,
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if this current here is a 2 milli amperes
and this is a 5 volts then, 5 into 2 amperes
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10 watts this product divided by 2. So, the
maximum output power will be 5 watts.
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So, this expression we have to keep in mind
and remember that we have derived for the
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central Q and when the whole region of the
active characteristic active region of the
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characteristic is been used. So, this is a
the maximum power maximum output power for
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optimum design amplifier design that means,
central Q and swing is in the whole region.
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Now, what is the conversion efficiency, the
next thing we are going to have is the conversion
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efficiency. Conversion efficiency and as i
said in the beginning by conversion efficiency
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we mean what fraction of a dc power which
we are supplying to the circuit is available
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as a c. so, because after all from where the
a c power will come as for example, we say
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the a c power but from where will it come
it comes from the dc power. So, eta the normal
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efficiency is written and eta it comes from
the dc power, so eta the normal efficiency
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is written in eta and it is expressed in percent.
So, this is average a
c power delivered to the load and this is
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the average dc power drawn by the circuit.
As a conventional in this dc power we consider
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only the power which is taken at the collector
circuit. So, this power only taken at collector
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that means we are ignoring we ignore the small.
Power which is taken by biasing registers
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we ignore power consumption by
biasing registers always when we take the
conversion efficiency then we talk every ac
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power deliver to the load by average dc power
taken by the collector circuit. So, this we
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will calculate and calculation is a very simple
that the dc source voltage source here, this
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is V cc which is of course, equal to 2 V c
cube.
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So, dc power delivered to the circuit is the
voltage is V cc and the current drawn is I
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CQ that is the current and the absence of
the signal, it draws the current I CQ and
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the voltage which we are applying is V c c
and this becomes p dc is equal to 2 V CEQ
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into I CQ; this is because V cc is equal to
2 V CEQ. So, this is the dc current and this
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is the a c current a c power, dc power the
ratio of the two is V CEQ I CQ by 2 and then
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divide by this. So, 1 by 2 V CEQ and into
I CQ and when we are interested as i said
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normally, they are expressed in percent multiplied
by 100. So, this cancels and this is 100 by
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4; so 25 percent. Conversion efficiency in
Class A operation r c coupled the circuit
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which we took in the beginning say r c coupled
circuit, and for the r c coupled class A coupled
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amplifier efficiencies 25 percent that means
for getting 10 watts of power in class A operation
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we will have to supply 40 watts of dc power.
So, this is the efficiency and mind it, this
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is the best possible and for the optimum design
by the optimum design again I repeat that
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we have taken the operating point at the center
of the load line and the whole region of the
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active region is been is been utilized. Now,
in fact we can find out the condition whether
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the design of the circuit is really optimum
or not. For that there is a very simple test
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that for the optimum result as we are seen
that i c set this is equal to 2 I CQ, this
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point i c set is 2 I CQ. Now, in the equation
which we derived here, in the beginning that
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i c set is equal to this now if we put this
as 2 I CQ we get a condition and that condition
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is we remember if i c set is 2 I CQ.
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So, i c set equal to 2 I CQ which is equal
to I CQ plus V CEQ by r c plus r E. And from
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here we get a simple relation that r c plus
r E should be equal to V CEQ by I CQ. If this
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condition is satisfied, then we can directly
use this expression what does this condition
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say this condition says that the effective
a c impedance seen by the collector plus the
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effective impedance seen by the emitter if
you sum up the two this should come equal
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to the ratio of the voltage v.
CEQ at the Q point and the current I CQ, if
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this condition is satisfied then, we can directly
use this expression and the efficiency will
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fine maximum is 25 percent. Or if you are
giving just any circuit, you find out your
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what is V cQ and I CQ and use this exact values
in the expression then we will get the efficiency
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for that circuit. So, this is about the r
c couple class A amplifier power amplifier,
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and just to remind you r c couple this is
r and this is this was load this is r c plus
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V cc. So r c coupling very widely used but
for that the efficiency is 25 percent but
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then, the question comes can we increase this
efficiency. For r c coupled maximum theoretical
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efficiency is 25 percent, if we replace r
c coupling by instead of r c couple load if
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we take a transformer coupled load then the
efficiency can be increased to 50 percent.
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You realize one point that even in the absence
of the ac signal because the quotient current
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the operating point current it is I CQ and
the product is the voltage which will be existing
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this current this much power even in the absence
of the ac signal continues to be consumed
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by the circuit and this is actually, this
can saved. If this resistance is replaced
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by the primary of a transformer then, this
much power which is half of the dc power you
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remember dc power ever, we have seen is twice
of this only. So, half power we are wasting
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in this R c.
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And this can be saved but then, we move to
transformer coupled
transformer coupled class A amplifier.
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Now, here that this is a simple design this
is the primary of the transformer, and this
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is the load R L. Where we are using the capacitor
coupled load and this capacitor and this load
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can be coupled through the transformer also.
Where this is the number of primary terms
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and this is n s secondary terms, and the effective
load R L prime reflected load will be n p
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by n s, where n p is the number of terms in
the primary; n s the number of terms in the
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secondary, this is squared into R L; this
will be the effective load. The dc resistance
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dc resistance of the primary is almost 0.
So, the dc load line will be vertical see
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here.
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These are the characteristics so this is here,
this is the Q point and this is dc load line,
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because resistance dc resistance is almost
0 and you remember that the slope. Slope is
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minus 1 by R L or r c whatever now this is
close to 0. So, slope is infinity that means,
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it is vertical and of course, this is i c;
this is V CE and this is the ac load line,
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where this is twice I CQ and this is twice
of now. We can use directly V cc as Q point.
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So, this will be twice V cc and this will
give like that across this point there will
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be this kind of variation and this is V cc;
in this case, V cc is V CEQ.
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And the a c power as we have done ever this
was V CEQ I CQ by 2 and because these are
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equal so, this can be written as V cc into
I CQ by 2. And I CQ is I CQ is equal to V
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cc by the effective load R L prime and dc
power; this is so the ac power becomes we
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substitute for i c P o maximum is V cc into
I CQ and for I CQ we write V cc by R L prime
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and two this two. So, this is the for the
transformer coupled class A amplifier this
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is the maximum output power.
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And dc power is V CEQ by i c into I CQ but
as i said V cQ now we have taken earlier in
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the earlier case half of the power was being
taken by the coil by the resistance r c, but
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now that resistance has been replaced with
almost 0 resistance of the primary of the
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transformer hence we are saving that much
consumption. So, this is equal to P dc is
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V cc into I CQ and I CQ; we replace by the
expression V cc by R L prime. So, this is
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equal to V cc square R L prime this is dc
power ac power the ratio is p 0 p dc into
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100 and this is V cc square R L prime R L
prime by V cc square into 100. So, this is
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eta comes out to be 50 percent. For transformer
coupled class A amplifier efficiency will
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be close to 50 in practice 45 percent 48 percent
will be there, and this transformer also saves
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the capacitor effect that means this will
not permit dc to be coupled to the load as
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this capacitor does through this capacitor
dc cannot pass.
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So, in the load only the ac operates same
is true here because through transformer through
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this transformer the dc will be Checked plus
this relation effective load this also has
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the advantage because by adjusting this ratio
we can get the max load for maximum transfer
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with the load. So, these are the advantages,
but is still the use of transformer coupled
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class A amplifier is a limited restricted
because the transformer is bulky and frequency
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responds if it has to be good then these transformers
are expansive. Now, the last thing about what
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is the requirement how much power the transistor
device transistor has to dissipate. Because
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in power amplifiers if you transistor is capable
of dissipating 5 watts then obviously, you
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cannot use you cannot get power 10 watt from
these transistors that high current will burn
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the transistors.
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So, this is power dissipation by the device
by the device the transistor, and this is
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written as p d dissipation max. Now, in the
operation we have seen that actually a c currents
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and voltages they will vary the currents will
vary from 0 to 2 CQ similarly, the voltage
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will vary therefore, the dissipation will
keep on varying the dissipation will be highest
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here and least here but it can be shown that
maximum dissipation occurs in the absence
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of the signal that means at the Q point. So,
the P D max is actually the P D at Q point
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and which is equal to V CEQ into I CQ you
remember that P o max was half of it so V
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CEQ I CQ by 2; so there from these two equations
we can write that P D max is equal to twice
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P o very important relation for class A operation.
And it says that, if output power is say 10
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watt then 10 into 2 that means 20 watts the
device should be capable of dissipating 20
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watts that means, double of what you are getting
as at the output. This is another thing which
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is not very favorable in the case of class
A operation. For 50 watts of output power,
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we require transistor which can dissipate
100 watts. So, that is a big power and the
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cost of the transistor will be much higher.
So, this is all about class A operation just
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in summary for r c coupled load the efficiency
of class A operation is 25 percent and but
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out of all class A, class B, and class C,
amplifiers class A gives the most distortion
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free output.
So, that is the point and the efficiency of
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course, can be increased in transformer coupled
amplifier and maximum power to be dissipated
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by the device is twice that of the output
power available. Now, that finishes our class
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A operation.
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We now go for class
B amplifiers, and a particular amplifier which
is actually most widely used that is the one
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which is used that is class B push pull amplifier
push pull amplifier. We said that, in class
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B operation the operating point is chosen
at cutoff point that means here. This is the
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a c load line this is the saturation point
saturation point and this is the cutoff point
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this is i c; this is V CE; and this is cutoff
point and this is also Q point the Q point.
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And see what will happen that means that current
is 0 current is 0 at operating point current
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is 0 at operating point. Now, how we get this?
This means that, the emitter junctions initially
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will have no biasing voltage they will be
half, but emitter junction emitter base junction
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gets forward bias from the input signal here
when we apply the input. I will show you this
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will become clear when we take the actual
circuit, but you just understand that the
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emitter base junction will get forward bias
by for example, by the by the part of the
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input signal and then it will move to the
active region and conduction will occur.
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So let us see this way actually, this is the
circuit this part will go towards the cutoff.
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It will not apply only these portions will
appear in the current. This is current and
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this will appear like this
these half’s will forward bias the the the
input signal is here and depending on the
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device either this will forward bias the emitter
base junction or this you will see soon. So,
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this portions will appear in the output current
while they will be absent like here it is
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absent blank only the output will contain
pulse is like that, they are not useful as
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I said in the beginning.
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So, we use two transistors these are complimentary
transistors one is n p n transistor n p n
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transistor here n p n. So, n p n and the other
is p n p
transistor and that is this is p; this is
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p this is n. So, they are connected what is
called push pull amplifier push pull connection
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very special connection push pull class B
amplifier. And this is this simplified form
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actual forms we will draw little later, but
to understand the principle
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these are the two transistors this is n p
n and this is p n p. This is p n p and their
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emitters are connected together and to the
common emitter the load is connected. So this
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is actually, the connection of the transistors
are being used because we are taking output
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from the load connected to the emitter. So,
it is emitter follower emitter follower circuit
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which is most widely used this can be constructed
with the common emitter configuration also,
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but most common is this.
Now, you will understand what i said n p n
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transistor now input signal is this. This
is V i; and this is 0 volt; and this is time
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n p n meats the base meets positive pulse
positive voltage for forward bias. So, this
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portion will forward bias the n p n junction
at the same time, so reverse bias the emitter
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of the p n p transistor hence, p n p will
be non conducting for this part and the upper
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transistor the n p n transistor will conduct.
Therefore, we will get the output this is
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V out; and this is time amplified output you
will get because of n p n. In the same way,
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this negative pulse will forward bias the
emitter junction of p n p. So, and this will
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conduct and this voltage will reverse bias
n p n. This will not come, so the other half’s
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will appear because of p n p this portion
from n p n similarly, this will continue and
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the lower portions will be coming from p n
p; this is push pull action and we get a complete
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wave form from the sinusoidal input wave form
and the two transistors have be have to be
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used in the push pull connection this is push
pull action.
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I briefly repeat that we take the operating
point at cutoff and if we use just one transistor
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then the portion of the input signal will
forward bias the emitter and only that part
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will appear not the other. So, that will not
be of much use as in this case. So, we make
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use of the circuit in a different form we
notify the circuit use push pull connection
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and make use of two transistors and they give
this complete output because , the upper half
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will forward bias n p n transistor and lower
half will forward bias the p n p transistor
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and hence complete wave forms we will get.
We are continuing with the analysis of this,
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but before that we should see that in class
B operation signal distortion occurs actually,
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And there are two types of signal distortion;
signal distortion in class B operation; in
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class B amplifier.
There are two types of distortions which occur
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and this is two types one is non-linear distortion,
which is also called the harmonic distortion
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harmonic
in which the the the frequencies in the output
will be multiples other components of you
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can see will also with there for example,
if omega t is the input frequency omega then,
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2 omega, 3 omega, 4 omega; these are the harmonics,
which may also appear in the output and the
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reason for this distortion is the non-linearity
and we will see that in a well symmetrical
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design two half’s of the push pull amplifier
when they are made symmetric then actually,
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this is largely taken care of largely taken
care of and the other kind of distortion is
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called cross over cross over distortion; this
occurs for a different region and this will
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amount for a distortion which can be corrected
by moving the operating point not exactly
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at the cutoff, but slightly above and this
is called current. We will see this and how
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it is provided and then this distortion can
be taken care of.
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But, once we move the operating point from
the center slightly above in the active region
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then, it is not a pure B operation. In that
case, it will be called A B operation because
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, this cross over distortion is a important
and we can get read of by choosing slightly
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different operating point. So, almost all
class B amplifiers push pull amplifiers are
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actually they are operated A B, AB, A B amplifiers
and this will be seen this will be seen. Now,
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the distortions can be tackled and we will
finally see that, the outputs are largely
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free from all kinds of distortions and we
will continue the analysis. The efficiency
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of class B amplifier is very high, its more
than 75 percent. Practical efficiency is also
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met 70 percent or so which is much higher.