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We were discussing the method, mathematical
operations through amplifiers, and op amps
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are used in this process, and in fact that
is the reason, that is why these are called
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operation amplifiers. So, the mathematical
operation, which was currently under discussion,
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is summing amplifier is scaling amplifier
and an averaging amplifier. These are three
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different amplifiers, but because they all
the three, can be drawn from the basic amplifier.
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So, that is why I am taking them all under
single heading. Now, these amplifiers can
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be realized by using a inverting amplifier,
inverting op amp or by non inverting amplifier
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equally well. We are taking just one case
that is of inverting amplifier, so how to
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realize a summing amplifier with inverting
configuration. The summing amplifier is the
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one, in which the output is the sum of all
the inputs. I repeat, that summing amplifier
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in is the one which gives a output which is
the sum of all the inputs. This is summing
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amplifier. Now, the inputs can be more than
one, so may be two, three, four, five, six
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or it can be number eight or ten, but we just
to make the diagram simpler. We take three
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inputs and when we use a inverting in configuration;
then this circuit is this. v 1 v 2 v 3 are
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the three voltages, which are to be summed
up and output is here, and this non inverting
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input.
Normally, so far we have been putting at a
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ground potential, by directly grounding it,
but for a better arrangement is, when this
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resistance R x equivalent to the parallel
combination of these resistance is attached.
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It is a still at grown potential, but not
directly, but through a resistance and this
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reduces. It reduces the offset effects; we
have not yet talked about the offset effects.
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We will talk little later about these things,
but they arise, because of the a symmetry
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in the design of the circuit. And even when
the inputs are absent some currents is still
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flow. They are of course, very mild, very
small currents, but still, so those off set
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effects can be reduced drastically, if instead
of directly connecting the non inverting input
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to ground. It is connected through a resistance
equal to the parallel combination of all the
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input resistances. Now here this is at a potential
v b, and the three currents i 1 i 2 and i
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3 they flow here, and from here will flow
i f.
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Now, remembering that v b is at zero potential,
remember we have said that the inverting input
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is virtual ground. So, this is at ground potential.
Hence v b is zero, if this is so, this voltage
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is zero. Here it is v 1, then we can simply
write the expression for the current i 1,
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this is actually speaking v 1 minus v b by
R 1, but because v zero. So, i 1 is simply
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v 1 by R 1. Similarly, we can write for i
2 and i 3. So, i 2 is v 2 R 2, i 3 is v 3
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R 3. These are the currents and now we are
seeing that this is simplified. And the whole
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expression gets simplified by this concept
of virtual ground, and that is why emphasis
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was given to this point, that the inverted
input is a virtual ground. Now, these three
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currents are flowing, there will be a current
here I B, but I B as we are seeing that, because
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the input impedance of the amplifier is extremely
high. So, this I B can be taken as zero.
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I B is equal to zero, we neglect this current.
It is extremely small current it is extremely
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is small current, if the neglect this current
I B is equal to zero, then; obviously, the
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summation of these currents is equal to I
F. Therefore, I 1 plus I 2 plus I 3 this is
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equal to I
F. And we can write for I F, all the three
currents we have written. We can also write
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for I F the current we can write here.
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This three currents we write here. Now, what
will be I F. This is v B minus this terminal
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is at potential v b, which is zero, we will
write, finally zero minus v 0 by R F and this
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is equal to minus v 0 by R F. So, substituting
the values of all the currents in this expression,
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we get v 1 by R 1, which is equal to I 1 plus
v 2 by R 2 plus v 3 by R 3 and this is equal
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to minus v 0 R F, or the output we can write
from here, or v 0 is equal to minus R F R
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1 by v 1 plus R F R 2 v 2 plus R F R 3 v 3.
This is the relation which is we may, say
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the basic relation for this circuit, and then
by choosing. You recall that R F by R 1 is
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the magnitude of the gain by which v 1 will
be amplified. Similarly, R F by R 2 is the
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gain by which v 2 will be amplified, and f
appears at the output and so on. So for now,
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from this basic equation which we call as
first equation, we can now go for summing
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amplifier. Summing amplifier in this equation
one, if we choose R 2 equal to R 2 equal to
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R 3, say this is R.
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Then what we have is this, v 0 is equal to
minus R F by R and v 1 plus v 2 plus v 3.
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If further we choose R F this gain as one.
So, R F further, if R F is taken equal to
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R, then magnitude of
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gain is equal to one, and in that case this
term will be one, and v 0 is simply minus
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v 1 plus v 2 plus v 3; that means, the output
voltage for this summing amplifier under these
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conditions of choosing R 1 equal to R 2 equal
to R 3, and let us say that is R. And R F
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also equal to have that is alright this resistance
is in this circuit are taken as single value
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resistance. So, all these resistance is R
F r 1 r 2 r 3 are taken as a single value
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resisters say 10 k, 20 k and so on. Then the
output will be, the sum of all the input voltages
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with a reverse sign, with a negative sign;
that is sign is not really material, if you
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want positive we can use the sign changer
circuit, after that which we have already
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discussed, and because this minus sign is
coming, because we have taken a inverting
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amplifier. If we take non inverting this sign
automatically will disappear, but as I said
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it is not is significant point. So, this is
the summing circuit.
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Then we can go for is scaling, scaling or
weighed amplifier, in the
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simply summing amplifier, we have seen that
all the inputs are amplified by the same amount,
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choosing all the resistance is as single value
are in R F also of R. Then simple summing
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amplifier results is scaling are weighted
amplifier is the one, in which the inputs,
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each input voltage is amplified differently.
I repeat that in a scaling, a scaling amplifier
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is also called weighted amplifier, in which
the amplification in factor for individual,
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input voltage is weighted, is not same, but
they are different, they are weighted. And
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so then the amplifier is called a scaling
are weighted amplifier. Such amplifiers are
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needed. For example, if we want that the output
should be when v 1 v 2 v 3 are the input voltages,
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then v the output should meet this requirement;
that means, this equation should be satisfied,
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this is equation.
The output should we according to this equation,
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where v 1 v 2 v 3 are the inputs. Now here
same circuit and same fundamental equation,
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that circuit is the same, but here the values
of R 1 R 2 R 3 and R F we have choose differently,
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and we will take help of this basic equation,
which we have written for that amplifier.
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Now, in this case if we choose a particular
value of R F. For example, let R F the feedback
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resister is of the value of twelve kilo ohms.
We can choose 24, but just as an example I
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am taking 12. Then we should choose R 1 is
3 k, 3 kilo ohms. So, that this will be amplified
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by R F by R 1, look here R F by R 1 into v
1. Here we want affecter of four, according
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to the requirement. And if we choose R F at
12 kilo ohms, then R 1 necessarily we will
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have to take at 3, so that if factor of 4
comes.
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And R 2, we are suppose to take as 24 kilo
ohms, because R F by R 2 here, we need is
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point five. So, this will v point five, when
12 kilo ohms divided by 24 kilo ohms. So that
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means, 12 by 24, which is point five. So,
we will get again of point five. And similarly
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R 3 we take as at 8 kilo ohms. These resistance
are whole different R 1 R 2 R 3, they are
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not same they are all different, and they
are different, then R F as well. So, this
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scaling can be done according to this equation,
by choosing this resistances, and then keeping
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that we will get, which will be 4 v 1 point
4 v 2 plus 1 point 5 v 3 with a reverse sign,
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with a minus sign, if this minus sign we already
talked, it is not really much important, it
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can be taken care of without much problem.
So, this is the scaling are weighted amplifiers.
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Then the third one was the averaging amplifier.
Averaging amplifier, what say averaging amplifier.
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Averaging amplifier is the one in off the
inputs. Average of all the inputs put together,
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SO that is averaging amplifier. Now averaging
amplifier again by the same fundamental equation
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we can get. If R 1 is equal to R 2 is equal
to R 3, say R. This is the condition and R
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F by R is taken as 1 by n, this is additional
point, first point is this. This is the second
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point in R F by R should be taken as 1 by
n, where n is
the number of inputs, in the present case
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which is under discussion. There are three
inputs. So, this factor R F by R should be
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equal to, in the
present case under discussion, we should have
R F by R equal to 1 by 3. Now, if we substitute
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R
F by when we take R 1 equal to R 2 equal to
R 3 equal to R. So, here this is common term
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which we can take out, R F by R, and R F by
R in this case, is taken as 3. So, obviously,
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v zero is minus R F by R into v 1, plus v
2, plus v 3. And since this is taken as one,
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so v 0 will be minus v 1 plus v 2 plus v 3
by 3, because R F by R is 1 by 3. So, this
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is the result for averaging amplifier. The
output is the average of all the inputs with
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this sign reversed. So, this is the averaging
amplifier, we continue with these amplifiers,
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we will take other circuits also and we continue,
let us talk of a logarithmic amplifier.
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Logarithmic amplifier is the one, in which
the output is the log of the input. I repeat
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what is logarithmic amplifier. Logarithmic
amplifier is the one in which output is proportional,
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to the log of input, let us say, that is v
i, this is the logarithmic amplifier. Now
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if in the simple inverting amplifier, if we
replace the resistance R F by a diode, what
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will the circuit. This will be the circuit,
v out and this is a D C voltage v s, d c source.
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This is the circuit of a logarithmic amplifier.
Here the I B characteristic of diode, we remember
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that I equal to I 0, exponential v Q by K
T minus 1. This is the equation of diode,
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which describes the current voltage relationship.
This is applied voltage in this the resulting
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current, and here I 0 is the saturation current;
the saturation current. This we are going
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to use and here this K T by q, this is Boltzmann’s
constant, this is temperature in Kelvin and
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this is q we charge electronic charge when.
So, the thermal equivalent of that room temperature,
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the voltage equivalent this is that, the voltage
equivalent of thermal energy at room temperature.
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This comes how to be roughly 26 milli volts
at temperature T equal to 300 degrees Kelvin.
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So, as long as this v applied field, is say
of the order of 500 milli volts or one volt.
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This one will be neglected, because this is
exponential term which will rise very fast.
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And so neglecting that one, we can use the
equation I equal to I 0 exponential v q by
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K T from here, or exponential v q by K T is
equal to I by I 0, or we take log of both
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sides; v equal to K T by q into log I minus
log I zero, and this current I here in this
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circuit. This is I S from the source, this
current and this current is I and this voltage
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across this which will are applying that is
v. And this current the current here, this
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is actually zero, because of very high impedance
of the operational amplifier. The current
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going in the inverting input is zero. So,
taking I as I s equal to v s by R. And further
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neglecting small contribution, this contribution
of this term is negligibly small. So, we neglect
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that, in that case this v will be equal to
minus. It will v 0, because see here. Here
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this is the voltage developed across this,
and because it is inverting. So, with the
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minus sign it will be K T by q into log v
s by R. Thus this is a constant. So, the output
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voltage is proportional to the natural log
of the input v s. Output voltage is proportional,
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to the natural logarithmic of the input voltage.
This is the logarithmic amplifier. Now we
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will go for other mathematical operations,
using these op amp circuits, we go for the
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integrator.
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The integrator, this is also called integration
amplifier. Here the output voltage is integral
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of the input voltage, output voltage is the
integral of the input voltage then the amplifier
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is called, the circuit is called the integrator.
Integrator can be easily realized, by using
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the op amp in the inverting mode, and by replacing
the feedback resistance R F by a capacitor.
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So, the circuit of the integrator, the basic
circuit is this. This is R, this is that cap
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stance, this is the point x. And here we attach
the input signal, and output is taken here
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v out. This is the circuit, R F has been replaced
by the capacitor c. And here this is the current
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i which would be flowing, and here also, because
this current we taking as zero. So, the same
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current will flow here, in the capacitor.
Now, from the simple basic capacitor theory;
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from the basic capacitor theory or the theory
of the capacitor, the voltage developed across
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the capacitor and the charging current have
a relationship; that i is equal to c d v by
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d t. If this is the relationship between the
charging current and the voltage, developed
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across the capacitor, this is simple relationship.
The charging current gives a rate of change
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of voltage, and the capacitor is charged.
Now, a applying this equation in the present
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case, then it will be. Let us say which of
course, this voltage is v 2, which is zero
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of course, but for the sake of the argument,
this is c into d v 2 minus v 0. The two sides
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of the capacitor, one is at a voltage of v
2, and the other and is at v 0 and this is
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d t, but v 2 is 0, as it is virtual ground.
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So, i is equal to minus c d v 0 by d t, and
i this is the same as this i. So, i is; obviously,
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here v i minus v 2 by R, i is equal to v 1
minus v 2 by R, v to v in 0, is equal to v
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i by R. This is i. So, this i we substitute
in this expression, therefore we get v i by
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R, the current is equal to minus c into d
v 0 by d t. or from here we can write d v
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0 by d t is equal to minus 1 by R c into v
i by d t. integrating both sides, we get v
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0. So, this will not be there, when we take
this, this is simply this will be in the case.
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So, v 0 is equal to minus 1 by R c, integral
0 to t, v i d t plus c 1, where c 1 is a integration
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constant, c 1is a integration constant and
represents initial capacitor voltage at t
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equal to 0. And it represents initial voltage
at capacitor at t, the time equal to zero.
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And this we can set c 1 can be set to zero.
In that case v 0 the output voltage will be
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1 by R c to integral v i over d t. This is
the expression for the output voltage, and
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the amplifier, therefore provides an output
voltage, which is proportional to the integral
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of input voltage. Output is proportional this
is a constant R c, time constant of the circuit.
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So, v 0, the output voltage is proportional
to the integral of the input voltage. This
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is the integration and the integrator or a
integration amplifier. Now let us discuss
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further more, and what are the problems in
this basic integrator, and how we take care
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of those things. Now here if the input voltage
is constant, that is v i is a constant, say
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let us write with capital V, then by substituting
that in this equation, we will get v 0 will
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be V t y R c. So, from time to zero to time
t, this will be keeping on changing. So, it
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will be a ramp, the output will be a ramp,
because of this negative sign, it will be
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going this way. This is time and this is the
voltage v zero. So, a constant input voltage
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if it is positive, the ramp will be like this.
If v i is a negative voltage then the ramp
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will go in the other direction, this is t.
So, this will be the ramp. This is v zero
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verses t, and for a square wave input, we
will get a triangular shape output.
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If the input is a
square wave, the output will be triangular;
that means these once. This
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is input in square view, this is time, then
the output, this is constant, so it will be
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a ramp, and then this is also constant, it
will be ramp and that these two ramps will
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be going in a opposite direction. This is
t, so this is the output, a triangular. So,
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if the input is a square wave, output is triangular.
In square wave input will give a triangular
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move.
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And if the input is a sinusoidal signal, let
input voltage be sinusoidal, say v i is a
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zero sin omega t. We put this in the fundamental
equation, in this equation. Then v 0 will
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be minus 1 by R c into a 0 sin omega t d t,
and this v 0 comes out to be a 0, omega R
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c cos omega t. Here this is important that
the maximum amplitude of the output, will
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be is frequency dependent. This is frequency
dependent. Therefore, whatever is the maximum
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output for a signal of a frequency, say one
kilo hertz at two kilo hertz, it will be half.
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Maximum output will be half, so it will be
frequency dependent. Frequency
dependent
maximum output voltage. Now let us talk, we
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continue with the integrated amplifier, that
what are the problems with the basic circuit,
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which we have drawn simply by replacing R
f by a capacitor c.
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Now, there are two problems; problems with
basic integrator, problems are that, for v
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1 when it is zero, the integrator x as in
open circuit for d c. Look at this circuit,
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here when v i is zero, then in this circuit,
the integrator x is a op amp circuit, because
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this capacitor will act as a blocking capacitor,
and it will not permit any d c currents to
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be fed back to the input. I repeat that when
v i is zero capacitor c blocks d c current,
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and from where this current will come. These
are drift currents, drift and off set current.
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I said about offset currents, we will be talking
later, but at least for the time being, we
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can say offset currents are there, because
of a symmetric design of this circuits, which
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are almost impossible to get them in perfect
mast conditions. So, this small current, there
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will be blocked by the capacitor, and these
currents will not be fed back to the inverter,
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but they will continuously charge the capacitor.
And hence the voltage gets developed, and
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then it is amplified and the output voltage
appears, while the input is zero. So, this
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presents of this output voltage, it is a error
voltage and it should not b when input voltage
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is zero, the output voltage should be zero,
but it will be finite, because of the charging
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of the capacitor, and then that voltage is
amplified by the circuit and there will be
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a error voltage, gives error voltage at the
output. Another problem; that in this circuit
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at very low frequencies close to, for example
zero. The gain of this inverting amplifier,
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is impedance of the capacitor divided by R
F by R. R F is now replaced by X c, that is
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the capacitive reactants of the capacitor.
So, the gain is this, and this is close to,
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this is this is very high reactance, because
the capacitor 1 by omega c is the reactance.
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So, omega approach is zero, this reactance
X c approach is infinity, as omega approach
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is zero. So that means, the gain will be very
high for is small frequencies. These problems
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can be taken care, by attaching a register
R F with the capacitor, and that is the practical
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circuit.
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Practical integrator; the circuit is this.
C F is that capacitor, and
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v zero we take here. Now, this R F we will
take care of the problems, which we have discussed,
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that it minimizes. R F minimize is the output
error voltage, and it will also take care
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at very low frequencies, the gain will not
be infinity, which will be at very low frequencies;
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that means omega going to zero. Now, this
is infinity we can neglect, because R F is
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in parallel, and gain will be R F by R which
is finite. And we will see that normally this
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is kept around ten by design, we keep it around
ten. So, this takes care, but it puts a limitation.
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True integration will be there, when the combination
R F and C F produce a capacitive impedance.
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So, there is a restriction that at very low
frequency, this circuit will not be a true
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integrator, and what is that condition we
will just see, but this is the practical circuit
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and inclusion of R F that gives the modification
and the performance.
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Now, we discussed the gain magnitude and frequency
response of
this circuit. Frequency response of the integrator;
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if we take v i, we have done it above as sin
omega t and v 0 is minus 1 by R C F into,
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this is omega a zero, integral of this a zero
into cos omega t. This is input. This is output
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after integration. And if we take the ratio
of the peak voltages, v zero peak and v i
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peak, peak voltages, ratio of peak voltages,
then this is a zero, a zero omega R C F and
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this is 1 by omega R C F. So the gain, this
is gain, gain falls. This is v 0 by v i by
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taking the peaks. This is gain and this is
in d B. this is frequency f, and this is f
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1, and we will see what it is. This will come
out from here. f 1 comes out to be 1 by 2
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00:56:22,310 --> 00:56:29,310
pi R 1 R, let we have taking only R C F. And
this is a slop minus 20 d B fall per decade
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of change of frequency. So, this is the frequency
response, and this is for the basic amplifier,
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basic integrator
and how this is modified for the case of the
practical integrator, we will take further
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and discuss other circuits also.