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In
the previous lecture we developed a model
for the transistor for FET, that model is
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applicable and will be used for the analysis
of amplifiers, which are constructed using
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either junction field effect transistors or
MOSFETs. So, the analysis, which we shall
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be currently carrying, that is applicable
for MOSFET circuits, as well as, for JFET
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circuits. We will recall, let us recall, that
the bipolar transistor can be used in three
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configurations and that gives common emitter
circuit, common collector circuit and common
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base circuit. In the same way, a field effect
transistor can be used in three configurations
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and we have three kinds of amplifiers; three
types of amplifiers.
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The common source amplifier, that is, CS,
common source, CS, like C amplifier in BJT,
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so similarly here, common source, CS amplifier,
and then, we have common drain, that is, CD
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amplifiers and then, we have common gate,
that is, CG amplifiers. We will study all
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these three amplifier circuits and by study
we mean, that we are going to derive expressions,
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analytical expressions for the characteristic
parameters for the amplifier.
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And you will recall from our previous study,
that characteristic impedance for an amplifier,
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these are for example, voltage gain, voltage
gain, input impedance of the amplifier. These
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two are most important parameters for the,
which will decide the performance of the circuit.
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And then, there are some other considerations,
like what is the output impedance, which is
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important in some cases.
Now, out of these three common source amplifiers,
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in junction field effect transistors or in
MOSFETs, this is most widely used. This is
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like common emitter circuit for BJT in FETs
and MOSFETs, the most commonly used, most
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widely used amplifier is the common source
amplifier. Then, we will study common drain
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amplifier, which is like a and that CC, common
collector or emitter follower, this is known
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as source follower and it has the similar
application as the emitter follower for matching
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purposes, so it is used as a buffer amplifier.
And common gate amplifier is like common base
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amplifier and this has very limited applications
for the simple reason that the input impedance
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of common gate amplifier is very low; it is
few hundred ohms only, so it has limited applications.
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We go one-by-one for the study of these amplifiers.
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First, we take common source amplifier, common
source amplifier, which for example, the JFET
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can be drawn like this.
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This
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is the circuit, here we connect the biasing
source, the battery, V DD and this is R D,
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this is R S, you are familiar with these resistances
and this is R G, and this is the input signal
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v in, this is to be amplified and the amplified
signal is taken at the output here. This is
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the common source amplifier and here, the
input is given between gate and source and
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output is taken from, from drain and source.
Now, this is the coupling capacitor and here,
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this is, C 1 is coupling capacitor, coupling
capacitor and C 2 is by-pass capacitor. If
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we do not use this C 2 by-pass capacitor,
then there will be drop, ac drop across this
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resistance and hence, the gain, the output
voltage, which is available, that will fall.
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This point we will discuss again little later.
Now, first we should draw ac equivalent circuit
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for this actual circuit. This is the actual
common source amplifier circuit, which makes
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use of these three resistances, two capacitances,
the input signal and the output is taken here
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and we draw the ac equivalent of that. For
that, let us remind you we have talked all
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these points when we talked about the analysis
of bipolar transistor amplifier analysis,
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but again, briefly we talk about them, that
for drawing ac equivalent, first thing is,
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that dc voltage sources are grounded; dc voltage
sources are grounded.
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Second point is that these coupling and by-pass
capacitors, they are taken as short at the
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frequency for which this circuit has been
designed. We choose these capacitors such
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that, that the impedances offered by them
at the frequency of interest negligible and
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hence, they can be taken as short. So, capacitors
C 1 and C 2 are taken as short, short circuited.
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So, keeping these points in mind the equivalent
ac equivalent circuit is this.
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This is the ac equivalent of circuit in figure
1, this is figure 1, this is figure 2. So,
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this is ac equivalent of circuit figure 1.
Here, once we take this as short, so this
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is shorted; once we consider this by-pass
capacitor as short, so this resistance is
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shorted. So, zero resistance, this is like
that and this is grounded, so this is like
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this and output is taken against this resistor
R d, this is the ac equivalent. Now, we will
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replace this transistor, this FET or MOSFET,
whatever is, by its, by its model. So, what
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we get is this.
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This is the gate terminal, this is the input,
v in, and this is the transistor in the circuit.
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In the ac model we have replaced by its ac
equivalent circuit, we have replaced by the
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model and this is drain. So, and this is the
current source, which is equal to g m into
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v in and here, because this resistance is
very high, so, v i, v in is equal to v gs.
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You will recall, you will recall, that this
current source in the model we have taken
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as g m into v gs, but here we have shown it
as g m v in because of this and this is true,
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true because R G is very high. With this,
now we can draw these resistances are in parallel,
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with this there may be additional resistance,
for example, R L, the load resistance or another
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stage, which is connected in series with this
amplifier, then what will be the input impedance
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of this next stage, that will be taken here.
So, all these resistances are in parallel.
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So, actually, this can be replaced by one
resistance r D and which is r DS in parallel
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with R D in parallel with R L. One thing,
which is very frequently used in this analysis
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in, in, at many places in electronic circuits,
when two resistances, one high resistance,
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one low resistance, when they are connected
in parallel, then the effective resistance
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of the combination is closer, in fact, smaller
than the smaller resistance.
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Take just one example, 100 k resistance and
1 k resistance, if they are connected in parallel,
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then what will be, you apply, that parallel
resistance rule and find out the effective
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value of the resistance. This will be lesser
than 1 k, but close to 1 k. So, here this
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resistance is highest, which will have least
effect on this. So, this will be very close
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to R D. So, this model, they, all these resistances
we can replace by R D, where R D will be equal
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to this. Also, R G is very high resistance,
this we have seen, 50 k, 100 k, so this can
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also be dropped from this figure.
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And then, this model becomes this.
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This is the model, which we are going to use
and remember, the current source is g m into
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v in, v in is the input voltage. Then, the
drain current i d, that is, this, this current,
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current source and see the direction, so current
is flowing i d and this i d, the drain current
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in this circuit and this is equal to g m into
v in.
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And look at the direction, the direction of
current here, this is this way and the particular
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direction, which we have taken for v out is
just opposite, that is, current normally flows
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from higher potential to lower potential,
but this is flowing this way and hence, it
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is in order to write v out equal to minus
this resistance r D into i d. We substitute
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for i d from this equation v out is minus
g m r D v in and from here, the voltage gain,
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the voltage gain, A v is equal to v out, output
voltage, ac output voltage divided by ac input
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voltage. So, v in and this is equal to minus
g m into r D.
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This negative sign simply says, that there
is a phase reversal in this amplifier, meaning,
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when input will be maximum positive, this
is varying signal, this is varying signal.
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When input signal is maximum positive, the
output will be maximum negative and so on,
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that is the meaning of phase reversal. So,
this sign simply indicates phase inversion
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or reversal phase inversion, like common emitter
BJT amplifier, that also has phase inversion.
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So, similarly, it has phase inversion here.
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And the magnitude of voltage gain A v is simply,
g m into r D in a, just to give an example,
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if g m, g m is 4000 micro , that is the unit
in which g m is written 4000 micro, and let,
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let r D, that is, all these resistances in
the amplifier, the parallel combination of
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this, for example, gives r D to be equal to
4 kilo ohms, then according to this relation
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the voltage gain is 4 into 10 to power minus
3, that is, g m into r D, which is 4 k 4 into
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10 to power 3. So, voltage gain is simply
16.
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This is a simple expression, which expresses
the voltage gain of the amplifier. Now, in
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the circuit we have taken, that R s is bypassed
by the by-pass capacitor C 2. In certain circuits
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we do not by-pass, we do not by-pass this
resistance R s. In that case, as I said, in
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the beginning there will be ac drop across
this resistance also. So, the drop will fall
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or in other words, the gain will fall for
the same input. If this is not bypassed we
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get lesser output because the gain will fall.
So, in that case, if, if resistance R s is
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not, is not bypassed, is not bypassed, in
that case this gain will be reduced and this
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is equal to g m r D by 1 plus g m into R s
here. And once we bypass it, then this R s
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becomes 0 and we return to this expression.
So, this is about the voltage gain.
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And then, the input impedance, input impedance
of the amplifier; input impedance of the amplifier.
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We can see from, for example, from this model
that input impedance, if we measure at the
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input terminals between gate and source, this
will come out to be equal to R G and R G is
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very high. So, input impedance Z i, this is
equal to R G, this is very high. Normally,
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R G will have range, 100 kilo ohms to 1000
or more kilo ohms. So, input impedance is
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very high.
So, we can summarize, that CS amplifier, it
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has high voltage gain, high voltage gain and
high, rather, very high, high input impedance
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and it has a phase reversal, phase inversion.
These are the characteristics of a CS amplifier.
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One thing I may mention here, that voltage
gain for MOSFET or junction field effect transistor,
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if we compare with a bipolar transistor it
is much less, but there are other benefits.
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So, FETs are very widely used and this is
not very high input, not very high gain. This
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can always be compensated by adding another
stage, amplifying stage, which is not a problem
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at all.
If two stages, each having gain ten, for example,
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if two stages are cascaded, are connected
in series where individually each stage has
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a gain of ten, then the two stages will have
a combined gain of ten into ten; that means,
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one hundred. So, this is no problem.
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Then, we go for the next amplifier, that is,
common drain amplifier, common drain amplifier.
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Let us first draw the circuit.
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This is the circuit; this is common drain
amplifier circuit. Few, few features may be
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noted. One is that drain terminal is directly
connected to the dc source. When we draw the
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ac equivalent, then this is to be grounded.
As we have talked, that ac equivalent circuits
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require, that dc voltage sources have to be
grounded. So, we grounded, so that means,
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that drain is at ac ground. This is one point.
Second point is that output we are taking
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at the load connected at the source. So, the
2nd point to be noted is, that load is connected
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at source. So, the output that is, that means,
output is taken from source with respect to
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ground; here, source with respect to ground.
This is, this is the only amplifier in which
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the load is connected at the source and output
is drawn. Now, we will draw the equivalent
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of this, this is, these are the coupling capacitors
C 1 and C 2 and we draw the ac equivalent.
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AC equivalent of this figure and of common
drain amplifier and this is, this is the drain
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terminal at AC ground, this is the AC equivalent
circuit, this is the input v s and this is
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G, this is R g and this is v in, this is g
m into v gs and then, R ds R s and R L. This
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is the circuit, this is drain, this is source
and this we can, as we have done earlier,
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we can replace, this resistance is r g of
course, these three resistances are in parallel
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and that can be replaced by say R L. So, R
L will be r ds in parallel with R s in parallel
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with R L and normally, this is very high resistance.
So, these, this R L effectively, will be very
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close to R s in parallel with R L. So, we
can replace these all three resistances by
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this.
And another interesting point is that the
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gate terminal with respect to ground, this
is at v in, this is the gate, gate at v 1
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with respect to ground and source with, with
respect to ground is at v out, and source
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is at v out with respect to ground. So, what
will be v gs?
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If gate with respect to ground is at v in
and this is, source is at v out, then v gs
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will be v in minus v out, v gs is equal to
v in minus v out, and let us call this equation
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1. Now, what will be v out? I can draw another
circuit by just replacing these three resistances
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by one. So, in that case it will be, this
is R L and here, this is R G, this is g m
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v gs and this is
i d, this is plus minus v out, and from here
we can write v out. From the circuit, obviously,
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this current when, when passes through this
effective ac impedance R L, that will produce
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the output. So, i d into r L. Now, i d is
equal to this.
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Since i d is equal to g m v gs, so we have
v out equal to g m r L into v gs and v gs
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from equation 1 we have this. So, we write
like that, putting for v gs from equation
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1, we have v out g m r L into v in minus v
out
or v out 1 plus g m into r L equal to g m
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into r L into v in. From here, we can write
for the voltage gain. The mathematics is very
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simple, so simple, that it is almost very
straightforward. So, the voltage gain
A v, A v by definition is v out by v in and
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this is equal to g m r L by 1 plus g m r L.
One thing let me make clear that here the
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signal is v s, but here we are showing it
as v in because there is a small resistance
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here R s, source resistance. So, sometimes
there is a small drop across this resistance
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that makes v s not exactly equal to v gs.
So, this is taken as v s and here, this is
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going to the circuit v in because there is
a, there may be some impedance, which is normally
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associated with the ac source at the input,
and there may be some drop across this. So,
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so, that is why the two have been taken as
different. Anyway, so this is the expression
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for the voltage gain.
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g m into r L, normally is much higher than
1, that is, A v we have obtained as g m r
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L 1 plus g m r L. Normally, usually g m into
r L is very large as compared to 1, so 1 can
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be dropped, then A v will be very close to
unity. When A v is 1, that implies, that there
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is no voltage gain, there is no voltage gain,
it, it is, whatever is the input, same appears
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at the output, but there will be some power
gain. Now, so this is this.
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Now, here, the input impedance we find out,
gain we have seen, similarly for the emitter
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follower the gain was close to 1. And as I
said, that this is used as a buffer between
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stages and hence, same is here. The input
impedance, input impedance, when we measure
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because this resistance is normally very small
as compared to r g, so it is the input impedance
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is high, z in is equal to R G and R G is very
high. Input impedance is very high, gain is
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1, but let us see the output impedance.
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The output impedance, look at this circuit,
this was the common drain amplifier. Now,
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here and its equivalent we draw here. Now,
as a rule, whenever output impedance is to
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be measured, then we disconnect r L, this
is the general practice and always this is
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done, that r L is to be taken out. Then, what
will be the impedance, which we will get at
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the output? That means, between source and
drain and this is simply, these two resistances,
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this is the drain source resistance and this
is the source resistance, so the two in parallel.
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Therefore, the output impedance, z out, this
is equal to r ds in parallel with R s. And
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as I said, as a practice, as a practice, r
L is always removed, disconnected
from the circuit
to measure Z out. So, we are left with these
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two resistances in which this resistance is
very high. So, Z out is very closely equal
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to R s and source resistance normally is kept
low.
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R s normally may be 200 or 400 ohms, therefore
the output impedance only for this circuit
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is very low, input impedance is very high,
output impedance is low. So, this amplifier
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can be used to act as a buffer between very
high impedance on one side and low impedance
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on the other side. Low impedance will match
with the output impedance here and the high
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input impedance will match at the other point.
So, this is that.
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In summary of CD, common drain amplifier,
which is also known as source follower, source
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follower, here the gain is close to 1, but
actually, it is less than 1 and input impedance
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very high, R G and output impedance, R s,
low and it is used for buffer purposes.
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The last one is the common gate amplifier.
The common gate amplifier, it is different
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from the earlier two, which we have studied,
the common source and common drain and the
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reasons will be very clear. The input impedance
here is extremely low, the circuit we can
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draw like this. Just directly I have come
to the AC equivalent by grounding it because
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we have done it quite few times, so you understand.
This is the input signal, this is grounded,
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this is source, this is drain and this is
gate, this is the common gate, gate is common
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between input and output. Output is taken,
which means drain and ground and this is the
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effective value of r L, which we have talked
two times at least in the two circuits.
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Here, the v out, v out is i d into r L and
i d is, i d is g m into v gs, which is, and
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v gs is same, you can see this voltage here,
v gs is same as v in, v in. So, this we can
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write, v out as equal to g m v in into r L
or the voltage gain A v is v out v in and
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that is equal to g m into r L, which is high,
which is high. And now, we come to the input
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impedance.
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The input impedance
in this circuit, it is not difficult to realize,
that input current is the same as the current
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between drain and source, that is, i d. If
we see the circuit, the, when we discussed
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about this current flows between drain and
source, so it is not difficult at all to see
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that i in is equal to i d and i d, if we replace
this by the simple model, then i d will be
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equal to g m into v gs and this is equal to
v g in g m into v in
because gate source voltage is the same as
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this, we just discussed this. Therefore, Z
in is equal to v in by i in, this is i in
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and this is simply equal to 1 by g m, g m.
So, this is very low, very low input impedance.
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And let g m be equal to, as we have taken
earlier, 4000 micro, then Z in will be 1 by
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4 into 10 to power minus 3, we have to put
in. So, 4000 micro is 1 by, that is, 4 into
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10 to power minus 3 and this is equal to 250
ohms, very low impedance. The output impedance
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is high, the input impedance is low and so,
this amplifier has limited applications. This
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is equivalent to common base amplifier in
the BJT. So, this way we finish the analysis
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of the amplifying circuits.
The most, whenever we want to use the amplifying
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circuit using a MOSFET or a junction filed
effect transistor, then we make use of common
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source amplifier and that common drain amplifier
is used as a buffer.
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Whenever we have here one stage, which is
having high output impedance, high z out and
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then, here this is some other device, some
other circuit where this is having low impedance,
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they cannot be connected together because
of the mismatch. The most of the power will
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be reflected and very little will pass on
to this circuit. So, we require a buffer here
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and this common drain amplifier, which has
high impedance, that will match with this
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and low impedance here will match with this.
So, this is the buffer amplifier very widely
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used in these circuits. So, this was the analysis
and we have completed that.
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Next thing, which we will be taking, which
is very important as far as the circuits are
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concerned, many time, the, the many times
the MOSFET circuits particularly, because
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in integrated circuits very widely used are
the MOSFETs. So, MOSFETs or even junction
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field effect transistors, but that is normally
results for discrete devices. So, MOSFETs
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can be connected as resistors as capacitors,
that means, whenever a resistance load is
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required, that instead of having a resistance
there we can use a properly connected MOSFET.
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When we connect a simple resistance that is
called, for example, this resistance and here
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is a MOSFET, this resistance R D for example,
this is called passive resistance, passive
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load, passive load, and this consumes power,
this consumes power. And when this resistance
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is replaced by an active device, such as another
MOSFET properly connected, then what we have?
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Active load, active load, so how to connect
the MOSFETs as resistors and capacitors, this
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is we are going to take next.