In yesterday’s class, we made use of the
two port network theory to evaluate the performance
factors of amplifiers, such as, input impedance,
output impedance, voltage gain, current gain
and power gain.
These are the important factors, performance
factors, associated with any amplifier. Taking
for example, Y parameter, I had shown you
how to evaluate the input impedance and asked
you to evaluate the output impedance; and
I evaluated the voltage gain, asked you to
evaluate the current gain and power gain.
Today, let us see whether you have really
understood this. In case you have any difficulty
in evaluating this, we would work out the
whole thing for the Y parameter case now.
I would request you to perform the same calculations
for all other parameters in almost identical
fashion and see what these input impedance,
output impedance and power gain, voltage gain,
current gain are in the respective parameters.
So, I would like you to concentrate on the
method that is adopted here. The same method
must be adopted for the rest of the three
parameters.
So, first, let us see how we can evaluate
the input impedance. So, we take this equation
I i; and Y i is the self-impedance at the
input; this multiplied by V i and Y r; and
you have to replace this V naught by something
that is dependent upon V i so that we can
find out the input impedance, which is Z i;
Z in, which is one over Y in
Z_in=1/Y_in
; and Y in is I i by V i
Y_in=I_i/V_i .
This is the input impedance seen from here.
In fact, if Y r is zero, this is same as Y
i. If Y r is not zero, this is Y r into V
naught; and this V naught has to be replaced
in terms of V i. So, it is necessary for us
to evaluate and find out what is the relationship
with V naught and V i. That we can get from
the second equation.
We know that I naught, I naught is the output
current here
I_o=V_o Y_L
. But the current then will flow this way;
so
?-V?_o Y_L
. So, I naught in any of these circuits is
always equal to minus V naught into Y L. This
is always the case. I naught is known in terms
of V naught and load admittance.
So, because this is assumed as plus, minus,
the current would have flown this way and
I naught is assumed to be opposite, we put
the negative sign. So, substitute this I naught
in the second equation. So, we have
V_o Y_L=V_i Y_f+V_o Y_o
So, this equation is the relationship between
V naught and V i as V naught over V i is nothing
but minus Y f divided by Y naught plus Y L
V_o/V_i =-[Y_f/(Y_o+Y_L )].
This, in the last class, we had very simply
evaluated from this circuit itself. Y f into
V i is the current in this.
This current will flow through the total admittance
which is Y naught plus Y L; the resultant
voltage therefore has to be
V_o/V_i =-[Y_f/(Y_o+Y_L )]
.
That is what we have derived here. So, this
is an important equation for voltage gain
V naught over V i.
Now, once we know V naught in terms of V i,
we can substitute this in the first equation
and get
I_i=V_i Y_i-(Y_r Y_f Y_i)/(Y_o+Y_L )
. This is ... This minus Y f into Y naught
by Y L, minus Y f divided by Y naught plus
Y L, is the term that we are replacing it
with, into V i is V naught. So now, we will
get I i by V i which is nothing but input
admittance of the amplifier, is equal to
I_i/V_i =Y_in=Y_i-(Y_r Y_f)/(Y_o+Y_L )
.
An important relationship, in the sense, that
this relationship is valid, whatever be the
parameter you select, as long as we replace
this parameter with the corresponding parameter.
That, what I mean is, if it is Z parameter,
then Z_in=Z_i-[(Z_f Z_r)/(Z_o+Z_L )]
.
If it is h parameter,
h_in=h_i-[(h_f h_r)/(h_o+h_L )]
. If it is g parameter, g in is bound to be
equal to g i minus g r g f by g naught plus
g L. And therefore, it is immaterial; that
is why I introduced the term immittance. Instead
of calling this impedance or admittance, it
is immaterial what the parameter is. It is
called immittance matrix; and a general term
that is used is p. So, you can just say that
p in is always equal to
I_i/V_i =p_in=p_i-(p_r p_f)/(p_o+p_L )
.
So, now that you know the results in all the
other parameters, I would request you to adopt
the equivalent circuit approach or equations
to derive these relationships and convince
for yourself that this is nd it remains unaltered
in its nature. So, this is as much about input
admittance and voltage gain. This, we had
derived in the last class itself by some other
approach.
We will now see the other parameters, other
important performance factors associated with
the amplifier, which is, I naught over I i,
current gain. Now, this can be again derived
from the equations; but I would like to do
it slightly differently, just as we did it
in the beginning of the class on amplifiers.
Now that we have evaluated V naught over V
i, why not we evaluate I naught over I i,
in terms of V naught over V i? That is the
best way; because, we have solved already
all these equations and therefore we know
V naught over V i. So, V naught is going to
be equal to, we know again, in terms of I
naught, we have got that, is equal to I naught,
-I_o/Y_L .
This is from this equation.
V naught is already known in terms of I naught.
Another important parameter, we also know
V i in terms of I i. What is it?
V_i-I_i/Y_in
, from this. You have already evaluated this;
V i is equal to I i, this goes there, divided
by Y in. Subsequently, therefore, we have
V_o/V_i =Y_in/Y_L ? I_o/I_i
Important relationship, once again. That is,
I can either express the voltage gain in terms
of current gain into Y in by 
Y L; or
I_o/I_i =-V_o/V_i ? Y_o/Y_i
.
Is that clear? So, this is a simple way of
always remembering how to evaluate I naught
over I i, given V naught over V i. We do not
have to start all over again solving the equation.
That can be done independently by solving
the equation, eliminating the other variables,
V naught and V i from the two equations.
So, this is done knowing the relationship
V naught over V i, how to find out I naught
over I i. If you know this, then, the power
gain becomes a very simple factor. Now, one
word of caution. Power, what is power input?
Power gain here, in general, is going to be
output power divided by input power. This
is one way of defining power gain. There are
several other ways. This is the most common
way of defining power gain.
Let us now see how this can be evaluated in
general. We have done this already for amplifiers.
Let us do it for a two port general network.
What is output power? We know that output
power is going to be expressed as V naught
square, that is the output voltage, rms value,
V naught square, divided by R L; or, into,
real part of, what? Y L; because, we are talking
of admittances now
?P_o=V_o?^2 Re(Y_L)
.
Real part of Y L. This is the input power.
This is output power. Input power is, in a
similar fashion, what is it? We can now easily
see. This is output voltage square into real
part of Y L. This should be, input voltage
square, real part of what is it? Y in, which
is already evaluated. Y in is already evaluated.
It is the real part of this that you have
to take in order to find out what the actual
power that is inputted
?P_i=V_i?^2 Re(Y_i)
.
So, power gain therefore is the ratio of these
two; that is, voltage gain V naught over V
i square, V naught over V i square, into real
part of Y L by real part of Y in.
So, the same thing can be derived for any
of the parameters, Z, g or h, by adopting
the similar procedure and similar definitions.
So, in summary, I can say that we have understood
how to derive for any two port network. The
important performance factors of the amplifier;
these are: voltage gain, current gain, power
gain and input impedance and output impedance.
These are the five important parameters associated
with the performance factors.
So, these can be derived in terms of any one
of the other three parameters that we have
left out; Z, g and h. These are going to be
left out as homework problems for you to show
that input immittance is going to take the
nature exactly similar to this.
Now, output admittance, I have not derived
that. We can derive this. Again, I must mention
that when I am evaluating output admittance,
this is something that has not been understood
clearly, I am trying to find out what the
output current is. I will remove the load.
This is... I will excite it by means of a
voltage here, which is V naught. This is not
the voltage which has been generated because
of input. This is a voltage which I am applying
here and finding out what this current is.
That is the definition for output impedance.
V naught by I naught, when I am exciting the
output. And then, this excitation has to be,
this is an independent excitement, this has
to be removed. In this case, obviously, if
Y f is not existent, then, output impedance
is going to be 1 over Y naught straightaway.
This is called the self-admittance,
Y_o
.
Otherwise, it is going to be determined by
this Y f and Y r. Y r is absent; again, the
same thing is going to happen. It is not going
to be dependent upon what is happening at
the... That is, both Y r and Y f, if they
are absent, these are responsible for forward
and reverse transmission.
If something is applied at the output, this
Y r should be present, so that it conveys
it to the input and this input is in turn
bringing it to the output. So, both Y r and
Y f should be present in order to fill something
at the output, when I am exciting at the output.
If Y r is zero, then nothing happens; nothing
happens that is transferred from input to
output. So, this impedance is going to remain
as the output impedance.
Let us now see what it is. Here, the circuit
looks exactly similar to the circuit that
is used for evaluating the input admittance.
If you notice, V naught is what I am applying
here and I have to evaluate I naught. I naught
is given by Y naught V naught, this is the
second equation,
Y_f V_i+V_o Y_o
. This V i, I had to replace in terms of what?
V naught. Then, I will get the relationship
between I naught and V naught.
So, because of applying V naught, there is
something that is happening here. There is
a current that is generated here, which is
Y r into V naught. This is the reverse transmission.
So, Y r into V naught generates a current
here; and because of this, there is going
to be current pumped in this, which will generate
a V i. And what is that V i is what we have
to substitute.
So, V i is generated here because of V naught.
So Y r into V naught is the current. And how
much is the current that is flowing in this?
This current is going to get divided between
this admittance and this admittance, so that,
that admittance that is going to take up current
from this, this is nothing but Y s here, the
source admittance. So, what is the current
that is going there? That is nothing but Y
r divided by Y i plus Y s into... with a negative
sign. This is going to be the voltage at this
point. This is going to be equal to V i.
Y r into V naught is the current through this;
this current is going to flow through the
entire admittance, which is Y i plus Y s;
develop a voltage which is V i. Since the
current direction is this way, the voltage
is going to be plus and minus. That means,
there is a negative sign that is to be put;
and therefore, this is equal to Y naught V
naught minus Y f into Y r 
by Y i.
Is this clear? So,
V_o=Y_o-[(Y_f Y_r)/(Y_i+Y_s )]
. You can now compare this with what we got
for... This is the Y out. This is the Y out.
So, compare this with what we got for input
Y in, is equal to Y i minus Y r Y f by Y naught
plus Y L. Wherever i is there, replace it
with o, wherever o is there, replace it with
i; wherever L was there, replace it with s;
and Y r and Y f also have to be replaced.
But, since they come as a product, it is of
no consequence; it will always remain as Y
r into Y f. So, you can see Y r into Y f product
is going to remain unaltered in this change.
So, Y i minus Y r Y f by Y naught plus Y L
Y_i-[(Y_f Y_r)/(Y_o+Y_L )]
is the input admittance; Y naught minus Y
f by Y r by Y i plus Y s is the output admittance.
And in the immittance matrix, p out ... So,
this is what we can write. p out is going
to be p naught minus p r p f divided by p
i plus p s
p_out=p_o-[(p_f p_r)/(p_i+p_s )]
. This is the general output immittance formula.
Once again, I request you to show that this
is the case in the other three parameters
by adopting either equivalent circuit approach
or solving the simultaneous equations.
Now, let us take an example here to illustrate
what we have learned. For the two port network
shown, determine V naught over V i, I naught
over I i, p naught over p i, R in; that is,
input resistance and output resistance.
Now, the only difference between what we had
earlier worked out in the amplifier case and
this; the earlier amplifiers were all unilateral
what it meant was that Y r or h r or Z r or
g r was zero.
So, transmission was possible only in one
direction, forward direction. Whatever was
happening here was not getting reflected on
to this side. Now, we have put non- idealities
everywhere. So, here is the case where we
have some reverse transmission. Whatever is
happening here is felt here. So, this... what
happens here is felt here; what happens here
is anyway going to be felt here; that is what
we want in amplifier. What is happening here
is to be amplified; but what is happening
here, should not be reflected here.
So, we do not want any reverse transmission
in a good amplifier. But, this is not a good
amplifier; there is some reverse transmission
here. So, under this situation what happens?
Let us consider this. Here, we have a current
source feeding on to the input. Whether it
can be represented as a current source or
otherwise is justified by comparing the impedances.
This is 100 K and this is point 1 K; so this
is definitely a current source drive. There
is no doubt about it.
The source impedance is very high compared
to what it is here. At this point, the control
source, this... whether it is correctly represented
as a current source or not can be again justified
by comparing it with the load, 5 K. So, 5
K is very small compared to its source impedance
of 50 K; so, it is correctly represented as
a current source.
Now, if these are the currents due to output
and input respectively, then this whole representation
can be identified by comparison with what
you have learned with Y parameters. So, this
is what you should be able to do. That is
because, here, I i is equal to V i divided
by point 1 K plus this 10 to power minus 5
times V naught. That is the total that is
the equation, one equation. The other equation
is, I naught is equal to V naught divided
by 50 K plus 10 to the power minus 2 into
V i. These are the two equations that describe
this two port now, which is nothing but Y
parameter representation of the equivalent
circuit.
So obviously, I have to now put down the Y
matrix for this network as... This is the
Y matrix. Y i is going to be 1 over point
1 K; Y r, this is Y i; Y r is 10 to power
minus 5. You can see that this should be a
very small factor in terms of siemens, compared
to what? Because, this is now having a dimension
of siemens. Compared to other parameters,
this parameter, these are all Y parameters;
so, this should be very small. Then only,
I can call it as a very good amplifier. If
it is zero, it is the ideal amplifier.
So now, on this side, Y naught is going to
be 1 over 50 Kilo volts; and Y f is the main
transmission, 10 to power minus 2 siemens.
So all these are siemens. We can see that
this should be really a large factor compared
to this, in any case, so that it is more unilateral.
That means, transmission from here to here
is more than reverse transmission. So, these
are the factors of it.
Now, because of these two factors, there is
interaction between input port and output
port. What happens at the output port is going
to be reflected at the input port. So, the
input impedance is going to be different from
the original impedance of point one K which
is the self-impedance. If this is not there,
if this is zero, it would have been just point
1 K. Because of this now, it is going to be
different form point 1 K. How different it
is going to be is what we are going to now
see.
Let us first evaluate the voltage gain for
this. V naught over V i. This is going to
be therefore Y i, this is Y r, this is Y f,
this is Y naught.
... is going to be minus Y f, divided by 
Y 
i plus Y L. So,
Y_f/(Y_i+Y_l )
.
That is the value, which comes out as minus
10 by 10 point 2, sorry, Y naught, this minus
Y f by Y naught plus Y L.
So, which comes out as... This is therefore,
50. So, how much it comes? So, this is 1 by
50 plus 1 by 5; which is, minus 50 by 1 point
1. Minus 45 point 4. So, this is the voltage
gain.
Next, Y in is going to be Y i minus Y r Y
f divided by Y naught plus Y L. This factor,
we have already evaluated as minus 45 point
45. So, this is Y i. Y i is going to be 1
by point 1 into 10 to power 3, minus Y r Y
f, 45 point 45 into, that is Y f by Y naught
plus Y L; minus of that is, minus 45 point
45, into Y r, which is 10 to power minus 5.
... which is going to be... this is 10 millisiemens
or 10 into 10 to power minus 3, minus, point
4545 into 10 to power minus 3. This is going
to be... how much is it?
9 point ... which is to mean that R in is
equal to 1000 divided by 9 point 54 which
is 104 point 8 ohm.
Compare this with what we had earlier. It
was 100 ohms; now it has become more than
100 ohms. How did it happen? Because, earlier
100 ohms. Why did it become more than 100
ohms? That is, the effect of the feedback
is such that it is now simulating a negative
resistance here, such that, the combination
of this 100 ohms and what is simulated here
is more than 100 ohms.
Because, if it is a positive resistance that
it assimilated, it would have become less
than 100 ohms. So, the effect of the interaction
of the output at the input here is to boost
up the impedance from the original impedance
of 100 ohms. Original impedance without any
feedback would have been just 100 ohms. Because
of this reverse transmission, it has become
more than 100 ohms. Why did it happen? Because,
it is now simulating a negative resistance
across this positive resistance and boosting
up the input impedance.
This is called positive feedback. This is
called positive feedback. So, why is this
impedance more than the self-impedance at
the input port? That is because, it is simulating
a negative resistance and effective resistance
is more than the original impedance.
So, this is called positive feedback. How
do you recognize that it is a positive feedback?
What is the factor that was responsible for
this? We know that it is Y r into Y L divided
by Y naught plus Y L, which is coming additionally.
If this is, assuming that these are positive,
this becomes positive feedback only when Y
r into Y f is positive. So, we can quickly
identify whether it is positive feedback or
negative feedback by seeing the product Y
r into Y f.
Here, 10 to the power minus 2 into 10 to the
power minus 5; that is the factor by which
the effect of output is reflected at the input.
And, it is getting subtracted from the input
admittance. That is, effective admittance
becomes lesser or impedance becomes higher.
So, this is the effect of positive feedback.
We will discuss more about positive feedback
much later; but you should now know that in
an amplifier, this, an ideal amplifier, this
is not at all wanted. If Y r into Y f is negative,
then, it is called negative feedback. That
means, negative, what happens at... then?
You can look at it.
If this is negative, in this case it is not
negative, let us say this is minus or this
is minus; if any one of this is minus, then,
it can be negative feedback. Then, what happens?
This becomes positive; that means, the total
admittance increases or impedance decreases.
So, this is what happens in negative feedback.
In this example, we are giving positive feedback
here; and that is why, the impedance is increasing.
Next, we would like to evaluate I naught over
I i. I naught is nothing but V naught, minus
V naught by R L.
And, I i is nothing but V i divided by R in.
That is right. So, if you know the voltage
gain, which is V naught over V i, which has
already been evaluated as minus 45 point 45,
minus, minus, becomes plus; this into R in;
R in is also evaluated as 104 point 8; that
divided by R L, which is given as 5 into 10
to power 3. This is the current gain.
How much is it? Now, this is a good example...
point 9, where the current gain is less than
1; still it is acting as a power amplifier.
It is giving you voltage gain and it is having
a current gain which is less than 1.
This is quite possible in practical amplifiers;
either it has to give a current gain or voltage
gain so as to give ultimately power gain,
so that it can retain the name of amplifier.
So, what is the power gain? This is current
gain into voltage gain, magnitude of that;
it is 45 point 45 into point 952; which is
43 point 3.
As long as the power gain is greater than
1, it is called an amplifier. This is an active
device; if the power gain is less than 1,
the two port network can be termed as a passive
device, of no use, for amplification purposes.
So, power gain has to be greater than 1 in
order to treat it as an amplifier. Even though
this is giving you less than 1 as current
gain, it is ultimately acting as an amplifier
because power gain is greater than 1.
Now, we have to evaluate only the final thing,
that is R out. So, Y naught Y out is equal
to Y naught minus Y r Y f divided by Y i plus
Y s. This is what we have evaluated earlier.
So, how much is this? Y naught is going to
be 1 over 50 Kilo ohms, minus Y r Y f is same
as 10 to power minus 7, divided by Y i plus
Y s which is 1 by point 1 K plus 1 by 100
K.
We will write it more clearly up somewhere
here. So, Y out is going to be equal to 1
over 50 into 10 to power minus 3 minus; now
this 10 to power 3, 10 to power 3 goes, 10
power minus 4; so, 10 to power minus 4 divided
by 10 point 01 . So, this is going to be 10
to power minus 3 into 1 point 001. Is it correct?
10 to power minus 4; this is 10 to power...
just a minute. 10 to power minus 4 divided
by 10 point 01; this is 10 to power minus
5 into 1 point 001.
So, how much is this now? This is going to
give you 100 by 50 into 10 to power minus
5. That is, 2 into 10 to power minus 5. This
is very nearly 1, into 10 to power minus 5.
So, this is equal to 10 to power minus 5 siemens,
which is therefore going to give you R out,
equal to, 100 K.
Again, you will notice; a very, the same strange
thing that has happened. The self-admittance
at the output is 1 over 50 K; or, the output
impedance, self-impedance is 50 K. But now,
because of interaction from the input, it
has become 100 K. It has increased. That means,
the same effect is there. Because of positive
feedback, the impedance level, both at the
input and output, have gone up. From 50 K,
it has gone up to 100 K. That means, there
has been an effect of negative resistance
on this side also. So, this clearly illustrates
what happens because of this dangerous negative,
that is, reverse transmission factor, which
is causing positive or negative feedback.
Now, we will, let us see that we can purposely
introduce positive or negative feedback, once
we understand the amplifier thoroughly. So,
before understanding the amplifier thoroughly,
we must design amplifiers which have no reverse
transmission. This reverse transmission is
going to play lot of role in later designs;
both in negative feedback amplifiers as well
as oscillators. So, we will preserve this
knowledge for use at a later date. Henceforth,
we will assume that for all our amplifier
discussion, until we again raise it, the reverse
transmission is zero.
I just wanted to bring this about so as to
make it clear to you that this factor is sometimes
a troublesome factor in amplifier design.
An amplifier has to be unilateral. It has
to transmit only from input to output. There
should not be any feedback. It is this feedback
which causes most of the troubles faced by
designers of amplifiers.
So, now that we have discussed two port theory
and amplifiers, let us summarize the entire
thing. Two port theory as applied to amplifiers,
we would like to understand. We said that
there are four basic amplifier configurations.
They are all called controlled sources.
Amplifier, therefore according to network
theory people, are nothing but controlled
sources; and therefore, ideal amplifiers are
the following: voltage controlled voltage
source, current controlled current source,
voltage controlled current source and current
controlled voltage source. These are the four
types of amplifiers that can be designed.
Now, we saw that these two are more basic
than these two because if you have ideal voltage
controlled current source and ideal current
controlled voltage source, by cascading this
with this, we can get voltage controlled voltage
source; by cascading this with this, we can
get current controlled current source. There
is no need for therefore synthesizing these
two blocks at all. If you have these two,
you can obtain ideal amplifiers of all the
four categories. So, these two are more basic
than these two.
This is called voltage amplifier, traditional
voltage amplifier. This is called current
amplifier. This, this is a voltage controlled
current source; this is called trans..., it
is a current source so, admittance amplifier,
or more popularly called, transconductance
amplifier. This one is called transimpedance
amplifier or more popularly called transresistance
amplifier. This is what we have discussed.
Then we said, an ideal voltage amplifier has
a matrix representation, which is, zero, zero,
zero, whatever it is... and has a voltage
ratio which is K, something like that. So,
obviously, this voltage ratio comes only as
basic definition in g matrix. So, g parameter
is the only parameter which can represent
a voltage amplifier.
What does it mean? What it simply means is,
if I try to represent a current amplifier
or a transadmittance amplifier or transimpedance
amplifier in terms of g parameters, these
will go towards infinity; any one of these
three. So, ideal amplifier, these things will
become meaningless. Only in g matrix, this
can be meaningfully represented. And, these
parameters should go to zero; and this is
the only
?(0&0@k&0) ?[g]
So, that defines a voltage controlled voltage
source. If you see this here, this will define
g i which is going to zero; that means, the
impedance is infinity. This g is going towards
zero; this output conductance. So, it is a
voltage source. Now, it is unilateral; so,
no reverse transmission. So, these things
are going towards zero.
Again, this is the representation. Therefore,
for current amplifiers, the dual of this;
that is, only h parameter, it is the only
parameter which is suitable for representation;
no other parameter is suitable for...
I mean you can represent a non-ideal amplifier
by any one of the four; but once you recognize
the source impedance and load impedance, you
will recognize that particular thing as a
particular amplifier. And therefore, you must
adopt the proper parameter for analysis; then
you would know an idea about the magnitude
of these. These will go towards zero and this
is the only parameter which is going to be
realistically large.
Then, transadmittance amplifier. So, you can
see this - voltage controlled current source.
Again, this will be zero. The current source
magnitude I, is going to be represented as
V i into g. So, this is going to be conductance
or admittance. So, this is going to be represented
by Y
[Y]??(0&0@G&0)
The other one is going to be zero, zero, zero,
R; or Z. So this is going to be appropriately
represented by Z
[Z]??(0&0@R&0)
So, this is the summary of what I was trying
to impress upon you by trying to correlate
the two port theory with amplifiers. This,
you must remember as the basic theory necessary
for you to understand the concept of amplification
itself; and these factors being zero, indicate
the amplifier to be unilateral.
That means no what? feedback. So, this, let
us say, p r being zero in any one of the four
indicate the unilateral nature of the amplifier
where no feedback is there. And p r into p
f, the sign of this, indicates whether it
is positive feedback if it is positive; negative
will indicate that it is negative feedback.
More about this feedback, we will learn later.
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