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I
have been discussed the basics of operational
amplifiers, we continue to study them for
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example, how they can be used, and what are
their circuit applications. First we take
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op amps in open loop, by open loop we mean
that there is no feedback involved in it;
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that means, that is op amp without feedback
op amp without feedback network. How will
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it behave, just let us consider a op amp with
the two inputs as we have said. So, let us
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take that in open loop
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this is non inverting input to which we apply
a input signal and let this inverting input
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b grounded.
Now, this A is the open loop voltage gain
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and v out we take from here R L, and this
is v out. Now this A is the open loop gain,
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always remember open loop gain gain of the
op amp and the 741 op amp, 741 which is a
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general purpose op amp and that we are taking
as the model just a sample we used. So, this
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741 has A around 10 to the power 5 its 100000
is the voltage gain. Now let us see what will
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happen if we apply A for example, a positive
signal v i is plus 1 mill volt let us say
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this is d c.
Now, mathematically what will be the output,
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the output is gain into input that is, this
v out will be A into v i gain is 10 to the
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power 5 ten to the power five and this is
1 milli volt. So, 1 is to 10 to power minus
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3 volt. So, the simple mathematic says that
v out will be 100 volts. Now very fundamental
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thing that in a circuit this v out the output
voltage cannot exceed the maximum d c voltage
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is available or applied to the circuit. Therefore,
this instead of 100 it will saturate at V
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sat output v out that is output we will saturate
at plus V sat.
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And V sat normally will be slightly low,,
but of the order of V c c, the V c c is 15
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volts then plus V sat will be roughly 14 or
14.5 volts. So, it will saturate. So, what
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we will get with the inverter non inverting
signal the output will be like this, this
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is a time, this is a voltage and this is V
sat this is minus V sat. So, if the voltage
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is positive it will just saturate here
and if it is negative voltage here, if this
voltage is negative then without change of
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polarity it will appear n output and it will
saturate A.
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Now, if we apply a sinusoidal signal let v
i b sinusoidal like this, this is the input
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signal and the if this is say of the order
of a milli volt then same thing will happen
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here. It will just saturate at plus V sat
and here it will saturate at minus V sat.
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So, if this is the input the output will be
this. This is the output voltage time plus
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V sat and here this is minus V sat, this is
highly distorted this is not the replica of
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the input. Therefore, does in open and loop
A op amp amplifier is a rarely used almost
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never used as an amplifier because output
will be a kind of square wave for a sinusoidal
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signal.
So, this is highly distorted when we compare
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the two. The same thing will happen when we
apply the input instead of a the non inverting
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input if we apply at the inverting input,
all voltages are always with respect to ground.
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When we say v i this is like this actually
here this is the ground level and this is
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grounded and here we apply a voltage this
is the inverting input and v i and this is
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the output which is a gain with respect to
ground this is v out this is the load. So,
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this is not shown, but in a statement we cover
it that all voltages output and input voltages
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are with respect to ground.
So, here when we apply it the sign will be
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inverted if it is d c if we apply a positive
voltage it will saturate at a minus V sat
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and if it is negative voltage, it will saturate
at plus V sat and we apply a sinusoidal signal
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and if this exceeds the limit A into v i more
than V c c, then it will produce a signal
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at output A square wave with inversion. That
means, the it will be like this, this is time,
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this is minus V sat plus V sat and this is
voltage. So, this is n if we apply use it
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as a differential amplifier.
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Then the difference of the two will be will
appear will again normally, it will saturate.
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Unless the difference is in say micro volts
then of course, amplified signal will appear.
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Now, what how we summarize this? That normally
as an amplifier in open loop op amps are not
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used; sometimes they can be used as comparators
open loop op amps are used as comparators;
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comparators is a circuit which compares two
voltages. So, here for example, we give here
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a, one is the reference this is a reference
voltage.
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And this is the applied voltage, we applied
to be compared applied means the one which
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you want to compare. Now the output when we
take, then as soon as this will exceed this,
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the output will saturate accordingly and a
signal will be available. So, this way the
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three configurations in open loop the inverting
amplifier, non inverting amplifier and as
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a difference amplifier this is a not used
as an amplifier, but one of the applications
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for open loop is as comparators where v 0
will be A v i d and A into v 1 minus v 2,
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where one of these can be a reference voltage.
So, this can be used as an as an comparator,,
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but the use of op amps as an amplifier is
very wide very wide as I said there are a
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large number of applications of op amps. But
most of these applications are with the using
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proper feedback and
the amplifier realization that is construction
of a amplifier is simplest with a op amp.
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We can choose how much gain we will see we
are coming to that that, we can choose how
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much gain for the amplifier is required and
what is the input impedance we want. Then
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we have choose just two, three resistances
of a rap appropriate value they are to be
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attached with the op amp.
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And the design will be extremely simple you
need not to bother about biasing network this
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and that, that is all everything has been
taken care of by the manufacturer and designing
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the op amp. So, we take the applications of
amplifier of op amp as amplifier first. So,
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op amp as amplifier
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op amps as amplifier there are three amplifiers
and all three we will take one by one. First,
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we take non inverting amplifier non inverting
amplifier with feedback and this feedback
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is negative feedback, in amplifiers we always
use negative feedback.
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This point was stated and made clear when
we were discussing a feedback in amplifiers
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negative feedback. First, let us see what
will be the simple circuit and then we analyze
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this amplifier. This is v out and this is
the input
and the feedback is this R F this resistance
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is R F and this is R 1 look at the simplicity
of the circuit and of course, these sources
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are there plus V c c and minus V c c. Even
if we do not show these two d c supplies it
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altimeters it is implied no electron circuit
will work with a without a d c supply.
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Because, d c voltages are needed for biasing
the transistors which is too important for
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biasing is too important for the operation
of any transistor. So, this is the circuit
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and this we analyze and since the input signal
has been connected with the non inverting
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input which is indicated by plus sign. So,
this is a non inverting amplifier with a proper
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feedback now we analyze it and remember A
many times written here, A is the open loop
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gain open loop gain, gain without feedback.
This is the feedback amplifier how much will
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be the gain that we will estimate and evaluate
for this, but this is provided by the manufacturer
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open loop gain.
And for most of the amplifiers particularly
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741 this is A is 100000, 10 to the power 5.
Now the v out this v identify this is v 1
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and this is v 2, now v out this v out simply
this is gain and the difference of two voltages
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v 1 minus v 2, whatever is the voltage here
that is v 1 and whatever is the voltage here
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that is v 2. So, the difference of this will
be will be multiplied with the gain will be
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the output, if we take in this figure, in
the figure in the figure v 1 this v 1 is equal
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to v i this two are identical. So, v 1 is
v i and v 2 v 2 will be the voltage here feedback
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voltage.
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This is we write v f, v f is voltage feedback
to inverting inverting input and this is simple,
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this we have talked several times, but anyway
I do this. Voltage here is v 0 and these are
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two resistance and what will be the voltage
drop here, this is voltage divider circuit,
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which I can drawn like this actually
this is
R F R 1 and whatever is the voltage drop here,
this is v f the feedback voltage and this
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is v 0 see here v 0 and this resistance these
two resistance is they are dividing this voltage.
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So, this is the network. How much will be
v f the current will be v 0 R 1 plus R F,
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this is the current through these resistance
is which are in series and the voltage drop
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across R 1 will be the current, this is a
current actually this is the current and multiplied
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with R 1 becomes the voltage. So, v f is R
1 by R 1 plus R F v 0 simple voltage divider
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circuit and this we have talked several times.
This is the current which will be flowing
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and what will be the voltage drop across this
multiplied with R 1 that is this voltage v
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f. So, we substitute the value of v 1 substituting
the values of v 1.
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And v 2 this v f this is a equal to v 2 the
voltage here with respect to ground is the
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same as voltage v f. So, this is v 2 and v
1 we already have this. So, when we substitute
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v 1 and v 2 we get v 0 is a v 1, v 1 was v
i minus v 2, v 2 is this. So, R 1 R 1 plus
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R F into v 0; so this is simply v 0 equal
to a into v 1 minus v 2, we have substitute
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the substituted the values of v 1 and v 2
and this we can write or we take this term
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here or v 0 is v 0 into 1 plus A R 1, R 1
plus R F this is a v i let us 0 r across multiplication.
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We will give A 0 R 1 plus R F plus A R 1 this
is equal to A v i R 1 plus R F and now form
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here, we can get the close loop gain from
this the expression we will get A with feedback.
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So, normally we write with F or F B with feedback
gain gain with feedback, that is A F is closed
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loop gain closed loop closed loop gain this
is equal to v 0 by v i here input signal was
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v i output v 0. So, ratio of output voltage
to input voltage is the gain this is the op
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amp in open loop and this is the feedback
network.
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So, this whole amplifier is A and amps which
feedback this is amps which feedback for that,
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this is the gain and which from this equation
v 0 by v i, this becomes equal to A R 1 plus
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R F R 1 plus R F plus A R 1, this is the exact
expression we write here exact. This is the
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exact expression for the voltage gain,, but
when we see the practical situation it gets
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drastically simplified and the situation is
that normally and normally means most of the
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time A R 1 is very, very large as compare
to R 1 plus R F, R 1 into A, A is 10 to the
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power 5; so, 100000 times R 1, while this
resistance is made differ early by effect
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of 2 or 3 or 5. So, this is this condition
will be mostly a available and applicable.
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So, this we can or this can be neglected and
then gain which feedback becomes equal to
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A R 1 plus R F by A R 1 or simple 1 plus R
F by R 1. This is the gain of the non inverting
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amplifier with feedback is 1 plus R F by R
1, how simple you choose the gain how much
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you want 10, 20, 30, 50, 100 accordingly choose
these two resistances if for example.
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Now, because this is a amps is a small signal
device. So, resistance is normally will be
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ohms because ratios suppose we want a gain
of 20. L let A F B desired is 20 and for that
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from this expression, we want R F by R 1 ratio
be to 19. Now 19 you can have for example,
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R F you can choose a say 190 into 2; so many
ohms and R 1, because this is to be 20 this
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ratio as to be 20. So, this becomes 380, 380
divided by 20.
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So, 20 ohms and 380 ohms if we take ratio
will be 19, but these resistance is will draw
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it large current and the currents will exceed
the maximum current which is recommended for
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the op amps and op amps will be burn. So,
instead of in ohms these are taken in clones.
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So, we can choose the this in clones and a
we can have this ratio 19 then gain will be
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20. For example, let R F be 19 clones and
R 1 v 1 ohm then gain A F B will be 1 plus
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19 k by 1 k. So, this is 20 gain will be 20
how simple it is to construct a non inverting
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amplifier with a desire gain.
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Now, let us go further to show that non inverting
amplifier with amps is a very close to an
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ideal voltage amplifier. It is an ideal voltage
amplifier and this we can show by using the
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concepts of feedback. Now from this equation
v 2 equal to v f which was equal to voltage
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divider R 1 plus R f into v 0 this is v f,
you remember that in the feedback networks
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1 plus A B, where A B is the close loop gain
and beta A factor. So, beta is v f by v 0
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the fraction which is written 2, this is the
input of this is the feedback network the
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input.
Here was v 0 this v 0 is the input to the
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feedback network and output we are taking
here. So, the gain of of of feedback network
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is output divided by input v f by v 0 which
from here we write as a R 1 by R 1 plus R
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F. This is the the gain of the feedback network
and then using this expression using R 1 R
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1 plus R F equal to B, in the A F the exact
equation, which was A R 1 plus R F R 1 plus
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R F plus A R 1. If we use this then this can
be written as a A F B is A 1 plus A B just
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substitute like this.
And this becomes this we divide by R F actually.
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So, this is A by 1 plus A R 1 R 1 plus R F
which this is B. So, this is this expression
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which was the expression for a voltage amplifier
with the negative feedback and this term A
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B will be very large in comparison to 1. So,
1 can be drop. So, the gain A F B is 1 by
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B that is highly stable. The gain of the feedback
amplifier op amps does not depend on variations
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in a which may vary because of aging or temperature
or noise in the surroundings whatever may
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be the reasons, but the gain of the feedback
amplifier the of the no inverting operational
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amplifier with feedback is stable.
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And it simple depends on the B, the B network
is normally few resistances. So, with this
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expression every quantity will vary with the
you remember we said this which feedback that
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because, the gain is A 1 plus A B and all
qualities will vary either increase or decrease
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by a factor 1 plus A B. If we increase we
have multiply by this factor and if they fall
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will have to divide it. So, this is this amps
here the sacrifices what we are doing that
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gain falls from A to this, the gain of the
non inverting amp with feedback is much reduced
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it is reduced by this factor 1 plus A B and
this factor may be 100.
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So, then the gain will fall by hundred 10
to the power 5 by 100 even 1000. So, if it
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is 1000 this factor is 1000 it will be just
100 may be less. So, now what are the quantities
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we recollect from our discussion in the feedback
chapter which we finished that input impendence,
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input impedance
with feedback that is R i F with feedback
this is R i and it increases by this factors.
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If this is R i is the input without feedback
then this is the input impedance it will increase
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by this factor.
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And this factor is for example, often 1000
10 and more see 1000 times the input impedance
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will increase and the output impedance, in
this case we will reduced the output impedance
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is reduced
is reduced and with feedback R F. If without
feedback it is R o then this 1 plus A B, it
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will be reduced and a ideal or rather a good
voltage amplifier is suppose to have high
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input impedance and low output impedance and
band width will increase because you will
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recall that gain width product is constant
gain voltage gain into band width product
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is a constant for a device in this case gain
falls by a factor 1 plus A B.
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Hence the band width increase by the same
factor 1 plus A B. So, with feedback band
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width band width with feedback this is, if
this is the band width without feedback, then
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it will be this is band width without feedback.
So, we are seeing all through that non inverting
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op amp is a non inverting amp is a an ideal
voltage amplifier, non inverting input op
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amp which is called actually non inverting
op amp is an ideal voltage amplifier. So,
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this is the major things about a non inverting
amplifier.
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Next, we take inverting amplifier and how
we known that whether the amplifier is inverting
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or non inverting simple. Where we are feeding
the input signal as you have seen in this
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case the which we have studied the input signal
was at as to a non inverting amplifier a non
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inverting input. So, it was a non inverting
amplifier. Next is inverting amplifier inverting
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amplifier with feedback
let us see the circuit, the circuit is this,
this is the circuit for inverting amplifier
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just we connect two resistances R F and R
1.
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These two resistance is externally connected
then it makes a inverting amplifier and now
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we investigate the important properties of
this circuit. Here R F this is the feedback
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resistance feedback resistor and R 1 is a
another resistance and we can see that this
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is the voltage shunt voltage shunt feedback
circuit. Now the most important property of
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an amplifier is the voltage gain the other
important property is input impedance and
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so on. So, first we take a voltage gain of
this amplifier.
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Let us see here this current is i i this is
i b and this i f, this is the input current
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which is divided into two parts. So obviously,
summation of currents at this point which
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is v 2 and this is v 1 this is of course,
is 0 because, it has been grounded this terminal
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is at potential v 2 and this is with respect
to ground of course, and this is at v 1, but
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v 1 is 0. So, summing up currents at v 2 gives
i i this is equal to i b plus i f and i b
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you remember that the input impedance of A
the op amp is a extremely large it is in wave
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ohms.
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So, this current is very feeble. So, since
i b is very close to because the input impedance
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of the amplifier is very high. Therefore,
i i is simple equal to i f now what is the
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value of i i and this terminal one terminal
is at the voltage v i the other terminal as
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v 2. So, what is the current flowing through
this ohm law potential difference is v i minus
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v 2 divided by R 1. So, this is this current
this is v i minus v 2 by R 1 this is equal
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to i f I mean i f the two terminals of i f
1 is at voltage v 2 other is at voltage v
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0 output voltage.
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So, this is equal to v 2 minus v 0 by R F
and from here we can what will be the gain.
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The gain of the amplifier, this is equal to
v 0 and two voltages v 1 by v 2,, but v 1
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is 0 because it is grounded v 1 is 0. So,
since v 1 is 0, A is v 0 by v 2 and of course,
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that is a minus sign here, from here v 2 is
equal to minus v 0 by A and this we substitute
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let us call this equation x. In this equation
we substitute this substituting
for v 2 in equation x we get v i minus v i
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minus v 2, but because v 2 is minus v 0 by
A.
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So, this is v 0 by A by R 1 this is equal
to minus v 0 by A minus v 0 by R F or the
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voltage gain with feedback. This is equal
to v 0 by v i
output is v 0 input is v i. So, this is the
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voltage gain of the total amplifier and this
is equal to minus A R F just from here, it
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is simple manipulation and this is R 1 plus
R F plus A R 1 and since A R 1 is very large
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as compare to R 1 plus R F. So, this we can
drop and A F B gain with feedback this simply
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is minus R F by R 1. How simple it is the
expression for the gain of this inverting
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amplifier this of course, sign shows the inversion
as it is we expected, because it is a inverted
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input, so inverting input.
So, hence this sign indicates there and just
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choose the desired ratio R F R 1 required
to obtain the the desired gain. If gain as
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to be 100 choose this 100 k this as 1 k the
gain will be 100 how simple it is. You cannot
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think an simpler amplifier in which the you
choose the gains several parameters will have
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to adjust and several parameters you have
to calculate here simply two resistances you
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have to take of appropriate value R F and
i R i and this will be the the case. So, this
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is the practical this is the exact expression
for the voltage gain which is rarely used.
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Because it does not make sense while this
condition is very much satisfied hence this
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is the practical expression. So, practical
expression for the voltage gain now one very
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important parameter about this amplifier and
that is this expression can be obtained from
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this equation. If we substitute v 2 equal
to 0 the same expression which we have arrived
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here, we get if we substitute v 2 equal to
0 in equation x in equation x, then we will
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get the same here, this equation x was this
v i is v 2 R 1 v 2 v 0 R F.
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And from here if we substitute this as 0,
this is 0 then the gain A F B is equal to
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minus R F by R 1 same expression means sure,
we have a obtained ever that is here we got
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the same thing if this consideration, but
the same expression we get with v 2 equal
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to 0. What is v 2 equal to 0 this is v 2,
v 2 equal to 0 means that v 2 is at the ground
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potential ground potential it is called virtual
ground. The inverting input very important
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inverting input is virtual ground virtual
ground. Why ground because at ground potential
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and why virtual because, actual ground can
take almost any amount of current hundred
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thousand amperes whatever, but here you know
only milli amperes can be observed in the
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device.
So, it is virtual ground this concept remember
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that inverting input is a virtual ground it
is at ground potential this is a great hell
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in simplifying the analysis of more involved
circuit like summing integrator differentiated,
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which we we are going to be studied. So, remember
two things the gain of the inverting amplifier
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is just the ratio R F and R 1, choose these
appropriate values for the desired gain, and
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the inverting input is a virtual ground and
we will continue our discussion. This is very
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important point remember that inverting input
is a virtual ground it is at ground potential.