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We were talking about the distortions in the
output signal in the class B operation, and
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we said that there are two types of distortions.
One is two types of distortions in class B
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operation: one was the non-linear, which is
also called harmonic -
harmonic distortion, and the other one is
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cross over distortion. Distortion implies
these two kinds; for example, in the first
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one the signal frequency as well as its harmonics
that is if omega is the angular frequency
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of the input signal, then at the output other
than omega frequencies like 2 omega, 3 omega,
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4 omega the harmonics will also be present,
they are distorted, and they are distortions.
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And many applications will not except this
distortion, the other one is cross over distortion.
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Now, we not only talk about these two distortions
in details, but also the remedies; that means
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class B operation finally we will see is free
from both these kinds of a distortions. So,
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first we study the non-linear
or harmonic distortion, we know that the current
and voltage swings involved with the power
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amplifiers are large.
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Now, let us look at the trans conductance
curve, dynamic trans conductance curve
this is the output current and this is the
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voltage the input voltage now see here then
the whole region of this characteristics will
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be involved when the output current swing
for example has to be high, so let us look
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here; see the different regions of the input
voltage they will give rise to different asymmetric
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output currents, and this is the net distortion,
this is the dynamic transfer curve, in transfer
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curve 1 property, 1 parameter of output port
is involved and the other parameter is from
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the other port, here output current versus
input voltage that is transfer curve and it
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is non-linear, and we are going to use the
whole part of it and this is bound to give
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the distortion and as a result of this non-linearity
other than the signal frequency the multiples
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the harmonics will also be present,
Now, we analyze it and we will see that the
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output finally in the push pull case then
there is a symmetrical design then the output
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will be free from even harmonics like the
biggest distortion will be from if omega is
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the input frequency then 2 omega the second
harmonic will be have an largest magnitude
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amplitude of distortion that will be absent
not only that although even harmonics will
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be absent finally, and odd as the number the
3 omega 5 omega they will be there, but their
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contributions will be negligibly small, so
we continue with the simple analysis of these
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this distortion caused by this non-linearity.
For linear curve for example like this, this
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is the linear curve here it is i c and this
is a vBE we can represent the for this linear
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case ic the collector current this is ic this
is equal to K which is a constant that means
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k is proportional to the voltage this is the
linear case where k is constant k constant
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and this is for linear characteristics but
our curve is like this and this we can approximate
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for the simple analysis we can take it parabolic
across the operating point, So, taking the
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curve as parabolic we can write this ic as
k1, vBE plus k2, vBE square this is the equation
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00:07:27,830 --> 00:07:32,680
parabolic, equation of a parabola.
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00:07:32,680 --> 00:07:39,680
Now, let the input signal we take as a sinusoidal
and it has the form the input signal vBE is
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vm the maximum amplitude cos omega t where
omega is the angular frequency of the signal
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so this is the equation of the input sinusoidal
signal, then if we put this in this equation
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then we get ic is k1, vm, cos omega t plus
k2. k1, k2 are constants vm square cos square
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omega t, now this cos square omega t term
can be written in a different form using the
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trigonometric relation and that relation is
cos square omega t is equal to half plus half
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cos 2 omega t, so we can replace that and
then ic will be any dc component present that
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is taken care by this term, plus these constants
gamma 0 plus gamma 1 cos omega t plus gamma
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2 cos 2 omega t this is the expansion,
Now, for this in the push pull as we have
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talked that there are 2 transistors and the
input signals are differing in phase by pi,
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the upper one responds the positive half of
the input signal the lower half responds to
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the negative half, so writing the input for
the second transistor as 2 equal to vm cos
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omega t plus pi, since now we are using 2
transistors to this we can say 1 this is for
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upper transistor and for lower one we are
writing, so we will get another equation ic2
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in the similar way a simple equation and this
ic1 for the upper half is ic2, gamma 0 minus
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gamma 1, cos omega t plus gamma 2, cos 2,
omega t, and if the final because both these
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currents are flowing through the common load.
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Then net current in the common load will be
net current in the common load of the amplifier
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will be proportional to the difference i will
be proportional to the difference of these
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2 currents, and therefore we can write i will
be equal to some constant X this is equal
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to X, ic1 minus ic2 so that will be gamma
1, cos omega t, so we are seeing that second
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harmonic is completely absent and only pure
signal frequency is there all harmonics are
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absent, but if we take a general case we took
approximately parabolic shape of the v of
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the trans conductance curve, but in general
case when Taylor series is used then actually
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this i we will find 2X into gamma1, cos omega
t plus gamma3, cos3 omega t and so on,
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In fact all the odd modes are present, but
as I said as the order of the harmonic increases
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the amplitude falls, So, the contribution
of gamma3 will be there but it is very small
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and gamma5 will the cos that 5omega term that
will be still a smaller, so this way and here
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we see the which take consideration of the
dc currents, for the symmetric case have cancelled
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in this expression we have cancelled these
capital c1, c2 the dc current, and this expression
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which has been written this is for the symmetric
design which is always carried out for push
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pull amplifier, So, in symmetric design in
symmetric design of push pull amplifiers in
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the above analysis we have taken ic1 equal
to ic2 which is true for symmetric design
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so this is what we conclude from here.
The simple analysis shows that harmonic distortions
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are there but the biggest contribution in
the distortion will come from the second harmonic
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which is absent in this because of the design
of the push pull amplifier and in fact all
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the even harmonic are absent only odd harmonics
are there but their amplitudes are negligibly
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a small then the cross over distortion.
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The next 1 is a cross over distortion, in
class B operation the cut off condition is
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achieved by keeping the Q point at the cut
off, Q point operating point is kept at cut
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off point, let us and at this point the emitter
base junction emitter base junction of the
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transistor has or both have 0 bias, look at
this design and then we will be able to explain
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that these are the 2 transistors Q1 and Q2,
Q1 is n p n type and the other 1 is a p n
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p and both are biased by a single source V
CC, and output is taken at the common load
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Rl, if the input signal is like this; this
is the input signal where this is voltage
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vi and this is omega t or time, now in strictly
speaking in the B operation the emitter base
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junction is at 0 potential, so and we remember
that if the devices are of silicon then this
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transistor will not come to conduction state
unless the emitter base junction is forward
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biased by 0.7 volts that is required.
In this case that means when the signal goes
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from 0 then first 0.7 milli volts will be
used to forward biased only after 0.7 volts
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the conduction will start, similarly in the
lower case up to 0.7 this transistor Q2 will
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not conduct, any nut shell here when this
voltage is of 0.7 volts for a silicon device
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this transistor will not conduct for this
part, similarly in the lower 1 this part of
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0.7 it will not conduct, the 2 transistors
will not conduct this 0.7 and this will amount
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2 distortion in this fashion.
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Output will be in little execrator format
it will be like
this is V out and this is omega t, but in
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practice also the outputs are like this, they
are not the exact replica of the input signal,
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So, this distortion is known as cross over
distortion, and this can be taken care of
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how to take care of that our input output
curve is like this, IB, this is base current
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and this is VBE and this is the input curve,
the strict B operation will imply to take
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the operating point here, instead of here
we take a little we shift our operating point
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from here to here so this becomes the operating
point, and this becomes the base current at
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the operating point,
Correspondingly this is the load line and
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the this is ic and this is vCE, at for class
B operation the operating point at exactly
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cut off, but now we have moved so that the
quotient current is not 0, here when this
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is done this is the Q has been shifted to
this, this will take care of that 0.7 bais
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if that means this voltage this is 0.7 volts
after we choose the operating point here,
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then these transistors are ready to conduct
right from the beginning because 0.7 volt
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is already have been provided, similarly here
0.7 has been provided, so the output will
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be the way we expect replica of the input
it will be only much higher in power.
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Now, in shifting this the conductance is now
more than for 180 degrees conduction angle
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little more, and this operation is class AB,
why we do class AB operation because we want
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to get rid of the distortion which is called
cross over distortion how we can take corrective
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measures by shifting the operating point from
ICQ 0 to a little bit value which may be 5
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percent of the maximum value or even less
and this operation is called class AB operation,
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and this current is small a small collector
current collector current ICQ is known as
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trickle current, trickle current
so this trickle current provides the forward
bias required and both the transistors are
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in the exact conduction state.
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Now, briefly we take biasing of class AB push
pull amplifiers biasing, there are 2 methods
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for biasing one is voltage divider, voltage
divider
bias and the other one is diode bias, now
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this is the complete design of class B push
pull amplifier, Q1 is this is voltage divider
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bias and here this resistance R1, this is
also R1, this is R2 this is R2, and this R1
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and R2 they have been chosen such that the
voltage developed across here is vBE for this
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and vBE for this transistor, that means 2VBE
so we have done voltage divider bias in the
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normally small signal amplifiers by very simple
means we can calculate this resistance R2
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which is to be used, and Q1 is of npn type
and Q2 is pnp type this is the push pull.
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circuit which is very widely used,
And for the analysis purposes only half of
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the circuit can be taken because they are
symmetric, so normally half is good enough
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to be taken into consideration, this voltage
is VCE1, similarly this voltage will be VCE2
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this current is Ic1 and here it will be Ic2
and Ic1 is equal to Ic2 is equal to Ic, and
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VCE 1 is equal to VCE2 which is equal to VCE,
and VCE1 plus VCE2 this is equal to V CC,
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and hence VCE1 alone VCE 1 this is V CC half,
if this is 10 volts 5 volts will be dropped
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here 5 volts will be dropped here that is
the meaning this is showing, and so this is
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very simple design, and as I said we can take
just upper half and we can calculate R2, same
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R2 we have to use we have to take R2 such
that the voltage this voltage should be 0.7
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volts, we can find out and this is the voltage
divider bias.
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There is a problem in the voltage divider
bias if you look at the curve it is very sharp
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in the just one condition here, therefore
a small changes in voltage VBE, ic trans conductance
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curve here the changes are very fast, and
remember this thing and when we were talking
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of BJT
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We said that voltage VBE is temperature dependent,
with the it changes 2.5 milli volts per degree
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centigrade. So, this is for silicon transistor
700 milli volts, by if the temperature rises
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by 20 degrees then 50 milli volts change will
be there, and 50 milli volts will amount to
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a large variation in this current here, therefore
many times the resistance R2 is replaced by
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the diode and then we call what is called
diode bias. The characteristics of this diode
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are same as the characteristics of the emitter
base junction of the 2 transistors, there
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is one thing that the pair of these 2 transistors
along with these 2 diodes is available in
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the market for different power ranges for
5 watts, 10 watts, 50 watts.
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The all the 4 pieces these 2 transistors complementary
transistors 1 npn other pnp and these 2 diodes
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they are sold in a single pack and that makes
life little easy, So, this is R, this is R,
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this is diode 1, this is diode 2, and this
is the common route and we take the output
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here this is plus V CC, this is the design
of the diode bias, any changes of temperature
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will change VBE1 also VBE2 because the characteristics
of these transistors are same as of these
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00:30:17,080 --> 00:30:22,580
junction base emitter junctions are same as
these diodes, So, same changes will occur
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00:30:22,580 --> 00:30:29,179
in these diodes, they are and that will take
corrective measures and the operating point
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00:30:29,179 --> 00:30:36,179
will not shift because of this variation,
temperature variation in particular, so this
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is about the diode bias.
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The next thing which we take are power relations
and efficiency; efficiency for class B operation,
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this analysis is strictly for B operation
that means we are neglecting the small effect
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of the trickle current which was required
to eliminate cross over distortion, by neglecting
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by taking this trickle current is 0, we make
the analysis much simple, and accuracy as
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00:31:47,140 --> 00:31:54,140
I said few percent accuracy variations are
always acceptable in electronics. So, not
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a big deal so taking the operating point taking
Q point at cut off
that is neglecting trickle current.
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Then our this is the situation this is the
ac load line and this is the operating point
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here this is ic this is vCE and
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00:33:15,289 --> 00:33:22,289
this is t and this is vCE and the corresponding
currents
this is ic set this is ac load line and this
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00:33:39,779 --> 00:33:46,779
is Q point this is 0 this is VCEQ and the
current will flow only for these half’s,
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00:33:52,549 --> 00:33:59,549
and it will be ic like that, and we will take
now that ac load line components you remember
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00:34:14,000 --> 00:34:21,000
when we were talking about in the beginning
of this module we spoke about the ac load
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00:34:22,740 --> 00:34:29,740
line, the 2 equations which we got the ac
components or the
components of a c load line, we got two expressions
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00:34:42,940 --> 00:34:49,940
ic set was ICQ plus VCEQ by effective collector
and effective emitter resistance, and similarly
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00:34:59,900 --> 00:35:06,900
we got vCE at cut off equal to VCEQ plus ICQ
into rC plus rE, this 2 equations we got when
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00:35:19,869 --> 00:35:26,869
we were talking about the ac load line, in
this present case because the operating point
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00:35:29,719 --> 00:35:35,799
we have taken at cut off and that is small
trickle current we have neglected for the
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00:35:35,799 --> 00:35:42,799
simplicity of the analysis, so ICQ we are
taking as 0, So, if ic u is kept 0 here in
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these 2 equations
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00:35:47,380 --> 00:35:54,380
they are reduced to but ICQ is 0 that is in
B operation,
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Therefore ic set is equal to VCEQ by rC plus
rE, and similarly when we substitute ICQ 0
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in the voltage cut off at cut off equation
we get it simply VCEQ, So, this these 2 equations
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will help us in arriving at the maximum ac
output power, further as earlier we have shown
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that VCEQ is equal too, because the total
voltage V CC which we are applying here this
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00:37:10,579 --> 00:37:17,579
is half appears here and half appears here,
So, VCEQ is equal to V CC by 2 this is when
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00:37:24,579 --> 00:37:31,579
we use single source voltage source, we can
have a choice instead of biasing both transistors
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00:37:34,589 --> 00:37:41,589
with a single source a positive voltage can
be given at this collector in negative voltage
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00:37:43,910 --> 00:37:50,910
we can give here, So, that this negative voltage
will bias this collector in the reverse, and
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the upper one will be reverse bias by plus,
so in that so this is single source when single
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00:38:02,049 --> 00:38:09,049
dc source is used when 2 sources will be used
then VCEQ this is just for the sake of knowledge,
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00:38:13,789 --> 00:38:20,789
we are using only one battery so for as this
is true but other choices applicable that
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00:38:21,359 --> 00:38:28,200
one voltage plus here, negative voltage here
to reverse bias this collector in that case
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this VCEQ will be equal to V CC whatever we
use so then we can write the expression the
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00:38:39,559 --> 00:38:40,749
peak a c voltage.
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00:38:40,749 --> 00:38:47,749
Let us this figure we should note this voltage
here is VCEQ so this voltage, this is VCEQ,
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00:39:01,150 --> 00:39:06,309
and this current is ic set current because
this point is 0 ic current.
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So, the peak ac output voltage is VCEQ, and
peak a c output current this is ic set saturation,
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then the maximum ac output power will be the
maximum ac output power in class B amplifier
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this will be Po max maximum is equal to Vrms
into Irms, and voltage rms this is the peak
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value and hence this is VCEQ by root 2 this
is ic set ic set by root 2 so this is Po maximum
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is equal to half VCEQ into ic set since VCEQ
is V CC by 2 so Po max is equal t V CC into
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ic set by 4, this is a important relation
and this will we know the value of ic saturation
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and what V CC single dc source we are using
to bias both the transistors, we know the
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magnitudes of these 2 parameters we can find
out what will be the maximum output ac power
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available from the push pull amplifier.
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And then we can go for efficiency; and the
efficiency this is convergence efficiency,
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that what fraction of the dc is available
as ac, so convergence efficiency and this
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is we write as eta and this is equal to Po
that ac output power by dc output power, and
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when it is to be expressed as a percent then
we multiply this by hundred and this we have
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already find out the maximum output power,
what will be the Pdc this is the maximum dc
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power which will be supplied to the circuit.
The voltage is VCEQ and this is ICQ, now one
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thing is very important to remember that ICQ
is 0 because we have chosen the operating
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point at the cutoff point so in the absence
of the signal the ICQ is 0.
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The transistors will come to the conduction
state by the upper transistor will conduct
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because of the positive half, when positive
half input signal is there then the transistor
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conducts, and when it conducts then we will
have it draws dc power from the source, similarly;
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the lower transistor will be conducting for
the negative half’s of the input signal
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and then that transistor will consume the
dc power and the current this ic the current
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will flow, now so we can find out what is
on an average what is the value of the collector
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current which is drawn from the battery, and
this you might have done the average over
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the period by using integration, so ICQ in
half wave rectifiers this analysis has been
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done.
But, I just do it here; ICQ 1 by 2 pi 0, 2
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pi ic set sin omega t, d omega t, we have
taken the signal as sinusoidal and the current
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which flows is this is the maximum current,
and this is that sin part we solve it and
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we find that ICQ is ic set by pi, therefore
Pdc
for 1 transistor is VCEQ, ICQ which is VCEQ
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00:46:21,039 --> 00:46:28,039
now we substitute for ICQ so ic set by pi,
When similarly there will be a expression
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for the second transistor
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and therefore Pdc for transistors 1 and 2
that is for Q1 and Q2 both, Pdc this will
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be double of Pdc 1 which we have calculated,
and this is 2VCQ into ic set by pi, but 2VCEQ
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is equal to V CC, therefore Pdc is equal to
V CC, ic set by pi, this is the expression
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for the dc power which is taken from the dc
power source, and we have already obtained
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00:48:22,869 --> 00:48:29,869
the expression for the ac power which will
be available at the load.
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Therefore, the efficiency eta in percent this
will be P0, Pdc into 100, and this is we substitute
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the values V CC, ic set by 4 into pi by V
CC, ic set into 100. So, they cancel and we
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00:49:31,660 --> 00:49:38,660
are left with pi by 4 into 100, and that comes
out to be eta comes out to be 78.5 percent,
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00:49:50,769 --> 00:49:57,769
very high efficiency. So, this is of course,
the theoretical highest possible efficiency
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00:49:59,029 --> 00:50:06,029
78.5 percent when the transistors are operating
in class B as a class B amplifier. This efficiency
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for the class A power amplifier when the load
was the coupled it was 25 percent, here it
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is 78.5 percent which is quite high very high,
and that is the reason one of the important
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reasons that class B push pull amplifiers
are very widely used,
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All the almost all public attires systems
are class in fact A B operation because distortion
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has to be completely taken care of, So, A
B operation and there the efficiency will
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00:51:05,609 --> 00:51:12,609
slightly fall so the practical efficiency
will be close to seventy or so, So, this is
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this, the last thing about class B operation;
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We can say and that is about power dissipation
that means how much power the transistors
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are suppose to dissipate the cost of the transistor
will vary a transistor which can dissipate
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00:51:51,019 --> 00:51:57,809
say 2 watts of power and 5 watts of power
they will be quite different, 5 watts power
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00:51:57,809 --> 00:52:04,809
dissipation transistor will be quite high
in this case in class B operation, this power
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00:52:07,809 --> 00:52:14,809
dissipation is very small the P D dissipation
power dissipation, this is 1 5th of Po, output
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00:52:32,029 --> 00:52:39,029
power is 5 times of P D or P D is 1 5 th of
Po. This is another point which goes in favor
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00:52:54,549 --> 00:53:01,549
of class A class B push pull amplifiers, for
if we want output power as say 50 watts or
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00:53:07,269 --> 00:53:14,269
100 watts, let us say Po is 100 watts then
we need transistors which can dissipate only
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00:53:16,069 --> 00:53:23,069
20 watts of power, So, this is a big game
with game, with class B power amplifiers.
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00:53:27,930 --> 00:53:34,930
We have been discussing this these power amplifiers
with BJT circuits, in fact MOSFETS can also
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00:53:41,599 --> 00:53:48,599
be used instead of 2 transistors BJT we can
use MOSFETS, and they are a special fat devices,
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fat transistors where there is a V grove in
the device and these are called V type FETS
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00:54:07,329 --> 00:54:14,329
V FETS, these can be used and the power amplifiers
this class B amplifier can be constructed
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00:54:26,420 --> 00:54:33,420
instead of by B J T we can use this.
Specially designed field effect transistors
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also, and another thing that we took a emitter
follower circuit for the for the push pull
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00:54:51,660 --> 00:54:58,660
amplifier, we can use the common emitter circuit
also, and the design is quite simple, and
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00:55:04,009 --> 00:55:11,009
in that case normally the transformer coupling
will be used at the output, So, many times
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00:55:11,839 --> 00:55:18,839
the push pull amplifier is constructed using
2 transistors in common emitter, and the output
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00:55:24,179 --> 00:55:31,179
is coupled to the load through a transformer
coupling. So, this is all about class B operation,
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00:55:38,420 --> 00:55:45,420
class B power amplifiers and the most popular
one of this was push pull amplifiers.