# Module – 4 Lecture – 4 Op-Amp Applications Part -2

In the last classes we have studied about
the application of op-amp and we have seen application of op-amp in various arithmetic
circuits like logarithmic, exponential as well as precision half wave rectifiers, etc. Today we will study about the application
of op-amp in control sources. What is a control source? Control source means we can control the output
with a control input. That is suppose we want to vary the output
current of a circuit by varying the input voltage or suppose we want to vary the output
voltage by varying the input voltage or suppose we want to vary the output current by varying
the input current; so the different examples of control sources are there. Here the source means a voltage or a current
source and that voltage or current source which we will get are in fact obtained by
input voltage or current sources. A typical example which we earlier studied
was that of a transistor. If we consider a transistor, we get at the
output a control source or a dependent current source because if we consider a common emitter
transistor, the output current is the collector current and that collector current IC is equal
to beta times of the base current. This example is nothing but that of a current
controlled current source. Here we are controlling the output current
IC by controlling the input current IV. Similarly using op-amp also we can have the
control sources and that we will discuss today and the control sources are having various
configurations like voltage controlled voltage source. It is called VCVS, then voltage controlled
current source, VCCS it is in short called and current controlled voltage source that
is CCVS or current controlled current source that is CCCS. These different types of control sources we
can have by using op-amp. Let us take a schematic diagram of a voltage
controlled voltage source or VCVS. If we consider a block of a VCVS, the input
is Vi that we are giving. We have to give an input and as it is controlled
by voltage, input will be a voltage and that is Vi and the voltage source at the output
or the control source which we will get at the output is also voltage and that is VO. The main thing here is that VO is controlled
by Vi. That is we will get VO, which is controlled
or affected by Vi and that is equal to k times Vi we are writing in order to express that
relationship between output voltage and input voltage. Here this k which is a constant, it will relate
the input voltage and output voltage. If we vary this input voltage, output voltage
here will accordingly vary. This is a block diagram for representation
of an ideal voltage controlled voltage source. Here we are using the term ideal because we
are not considering the practical parameters which exist. For example we know in an op-amp there will
be input impedance, there will be output impedance. But in this ideal consideration, we are not
considering those or we are simply ignoring those and an ideal op-amp consideration still
we are using ignoring the input impedance and output impedance. Let us consider a practical circuit. This circuit is a very familiar circuit. We used it when we were discussing about the
inverting op-amp; inverting amplifier when we discussed earlier, the same circuit configuration
we have used. Input voltage Vi is having an input resistance
which is connected that is R1 and the feedback resistance is Rf and the non-inverting terminal
is grounded. What will be the output voltage VO and from
our earlier study we already know VO equal to minus Rf by R1 into Vi and Rf by R1 for
a circuit once we design it, suppose we are not varying it, we can name it as k. k being
the ratio between Rf and R1 that is the feedback resistance and the input resistance and that
k is relating the input voltage with output voltage. That means we are controlling the output voltage
VO with an input voltage Vi. The output and input both are voltage sources. This is an example of a controlled voltage
source and this voltage source which is controlled is by another voltage source only at the input. Similarly the non-inverting op-amp example
which we are discussing earlier is also an example of voltage controlled voltage source. Because if we see in the circuit we are connecting
a voltage Vi at the non-inverting terminal and there is a resistance RC. This RC resistance is there but it is not
going to affect the overall expression of the output voltage VO. Because if we want to find out what is the
output voltage VO, VO is here and if we see the side where Vi is connected, this RC is
there, but there no current into the op-amp. If we name this current, which we have flowing
through this Rf, the same current will flow through this R1 and this example and the non-inverting
terminal op-amp which we were discussing earlier does not have any difference in analysis because
this presence of RC is not going to affect anything. What will be the output voltage VO? VO is again 1 plus Rf by R1 into Vi only. This example is also an example of a voltage
controlled voltage source. The output voltage VO is controlled by the
input voltage Vi. Another control source is voltage controlled
current source and in short it is known as VCCS. As the name suggests voltage controlled current
source means the output source will be the current source, but it will be controlled
by an input voltage. Here is the circuit of a voltage controlled
current source. We are having a voltage at the input which
is Vi and if we now denote the current flowing in this resistance R, this is Ii. We are naming it Ii. The same current will also flow through the
resistance RL because there cannot be any flow in this op-amp. Current flow cannot enter in the op-amp; whatever
current is flowing from this Vi must flow through this RL. So the current IO is same as Ii and what is
Ii? This is nothing but Vi by R only because we
will go by this loop. So Vi minus zero by R that is Ii and it can
be written as 1 by R into Vi; this 1 by R, let us name by k. So what do we get at the output? Here, the current IO we are getting from Vi
with a relation that is k times of Vi or k is equal to 1 by R or 1 by R times Vi is equal
to IO. If we vary this Vi, the output current will
also vary. Another example of voltage controlled current
source can be this example where we are having a Vi at the non-inverting terminal and we
are having resistances R and RL connected in this manner, RL being the feedback resistance. This circuit and the earlier circuit (Refer
Slide Time: 9:32) have a difference. Where is the difference? Here we are having a RL in this case which
is floating. Floating means it is not grounded. If you see both the sides, this side is connected
to R and this side is connected which is the output voltage VO. This point is not ground; so, this is called
a floating resistance or floating load resistance RL. But here in this circuit when the load is
floating we will have this control source, which is same as the example which we were
considering earlier which is also nothing but a non-inverting op-amp circuit. We will discuss a circuit which is having
a floating resistance and a circuit which will have a non floating resistance; in both
the cases we can use actually a voltage controlled current source. In this circuit we are having the output current
IO which is effected or which is controlled by the input voltage source Vi. The voltage controlled current source that
we are considering right now, finds application in electronic voltmeters. If we consider electronic voltmeters circuit,
electronic voltmeter is basically a voltmeter which should be able to measure even very
small voltages and if we want to find the measurement for a voltage we must have a considerable
deflection in the meter and we may also want that a full scale deflection should occur
for a specified voltage which is to be measured. In those cases, we will use a voltage controlled
current source like the example which we discussed just now. Here these examples show a moving coil meter,
the basic component in an electronic voltmeter and this coil resistance is RL. We want to measure a voltage which is V. We are using it with an op-amp that means
we are not using this electronic voltmeter alone. This moving coil meter, which we are having
for measurement we are not simply using it and measuring any voltage, we are combining
it with an op-amp. The reason for this combination with an op-amp
is that we want to have a specified deflection in this meter with respect to a particular
voltage being applied at the input. Suppose we are measuring a millivolt range
of a voltage and we want to have a full scale deflection in the moving coil meter which
will be basically dependent upon the current flowing through it. The current which will flow if we denote by
I that is the current which is flowing through this moving coil meter having the resistance
RL. This current should produce that much of deflection. Suppose if we want to have a deflection of
100 milliampere for a millivoltage range which is specified then how we can achieve this,
is dependent upon how much resistance we will connect at R, because this circuit and the
circuit (Refer Slide Time: 11:30) which you discussed which is voltage controlled current
source are typically same because there we were using a feedback resistance which is
RL. This RL is now the resistance of the coil
resistance of the moving coil meter. Here the voltage which is to be measured is
given at the input, non-inverting terminal. The current which is flowing through this
coil or the moving coil meter is I. So this current is equal to, from the earlier
consideration we have seen, Vi by R. R is the key element which will decide the
current and in order to have a specified value of current for a specified amount of voltage
at the input, we must connect that R which will give our requirement. This is a very important example of application
of voltage controlled current source and here one thing that is achieved by combining it
with an op-amp is that the voltmeter input resistance is almost infinite we are getting
and that is very useful. Also another factor is that the scaling factor
will only depend upon the resistances because here whatever the output current must be that
is decided by the resistance and that is why we use this type of VCCS in electronic voltmeters
where we can even measure very small millivolt range of the voltage very accurately. If we consider this example a voltage controlled
current source, here we are having a same VCCS that is voltage controlled current source
which is the similar example as the earlier case. But there is a difference here. In this case we see that the load resistance
which is connected is grounded. In the earlier cases, here (Refer Slide Time:
11:30) as well as here also (Refer Slide Time: 9:32), the load resistance is not grounded,
but here the load resistance is grounded. This is also an example of a voltage controlled
current source. The output current IO is the current of interest. That is we want to find out what will be this
IO? But our input is a voltage which is Vi. In order to know what is IO let us first assume
one thing, before finally analyzing the circuit, that R3 by R1 the ratio between these two
resistances is equal to R4 by R2 that is maintained and if this resistance ratio is maintained
then we can show that IO which will be finally flowing at the output is equal to minus Vi
by R2, Vi being the input voltage. We can show that the current at the output
IO is equal to minus Vi by R2 if R3 by R1 is equal to R4 by R2. Let us do this example. We will try to show that IO is equal to minus
Vi by R2. Drawing this circuit again, let us name the
currents which are flowing. The current which is flowing through R1 is
i1 and the same current i1 will also flow through R3 that is also i1 because there cannot
be any flow of current into the op-amp which is the condition or which is the assumption
which we have been following, following the ideal op-amp consideration. What will be the current IO, we have to find
and for that let us name one node voltage V1 which is here. Let us name this node voltage as V1. If this is V1, this node voltage is also V1
because of this op-amp being ideal. This voltage will be also V1 that means this
voltage is also V1. This point and this point are same. We now apply Kirchhoff’s current law, KCL
at this node V1. One current is IO, which is outgoing and let
us also assume the current which is flowing through R2 that is also outgoing and the current
which is flowing through RO is say incoming. This again is our assumption. You can also assume it as outgoing or this
is incoming as you wish because ultimately we will get same expression only if we follow
Kirchhoff’s current law. Let us name it as the current here is say
i2. This is i3. If we apply Kirchhoff’s law at V1, we get
the sum of the incoming current is equal to the sum of the outgoing currents. Here the outgoing currents are i2 plus IO
that is equal to the incoming current i3. What is i2? i2 current is nothing but V1 by R2 and what
is i3? We can find out i3 because i3 is nothing but
the difference in potential between this point and this point which is VO minus V1 divided
by R4. Using this result that is V1 by R2 plus IO
these are the outgoing currents, we have denoted like this and equal to incoming current which
is equal to VO minus V1 by R4; that is the Kirchhoff’s current law being applied at
node V1. You will further simplify because we know
what is V1? If we find out V1 from this input side, what
is V1? This voltage is nothing but Vi minus this
drop which is i1 into R1. So V1 equal to Vi minus i1 into R1; we will
use that here. Instead of writing V1, we will write Vi minus
i1 R1 by R2 plus let us say keep IO, as it is because we want to find out IO and from
the figure what is VO? We can write VO in terms of V1 as V1 minus
i1 R3; this drop when you subtract from V1 that will be VO. That is VO can be written as V1 minus i1 into
R3 by R4 minus V1 by R4 let us separate. We are keeping V1 in all the expression and
if we now separate out some of the terms, Vi by R2 minus i1 into R1 by R2 plus IO equal
to again V1 can be written as Vi minus i1 R1 by R minus i1 R3 by i1 R3 by R4 minus again
V1 we can write Vi minus i1 R1 be careful about these signs plus and minus. This is the expression. Now we write as Vi by R2 plus IO and transfer
all the other terms to the right side. If I transfer these terms to the right side
that means i1 into R1 by R2 will be positive plus what we will have here is Vi by R4 minus
i1 R1 by R4 minus i1 R3 by R4 minus Vi by R4 plus i1 R1 by R4. If you look into this expression, there will
be some terms which will cancel out; this one and this one Vi by R4 and this Vi by R4
minus will cancel out and i1 R1 by R4 and minus i1 R1 by R4 will cancel out. So we are left with these two terms in the
right side, which is equal to i1 into R1 by R2 minus R3 by R4 and that is equal to left
side which is Vi R2 plus IO. Now look into the right side because we are
using an assumption which is given to us that is R3 by R1 is equal to R4 by R2. We will use that; R3 by R4 is equal to R1
by R2. Using this expression we will now have these
two cancel out, so these two will cancel out. We will get here zero that means Vi by R2
plus IO equal to zero. If the right hand side of this Vi by R2 plus
IO is zero that means we get IO equal to minus Vi by R2. That means we can show that the controlled
source current or current source which we will get at the output which is IO that is
equal to minus 1 by R2 into input voltage Vi. Our result is now obtained which you have
been asked to get by using this law or by using this relation actually. This example was an example of a voltage controlled
current source. Another source which is a controlled source
is current controlled voltage source. So far we have discussed about voltage controlled
voltage source as well as voltage controlled current source. Now we will discuss current controlled voltage
source and current controlled current source. Current controlled voltage source or CCVS,
as it is named, is having this diagram which is an ideal block diagram form of a current
controlled voltage source. Here the input is a current. Ii is current and at the output we are getting
a voltage which is VO and VO is a voltage which will be controlled by the input current
Ii according to the law k times of Ii; k is actually a constant. For example we have this circuit where we
are having an input current Ii and the voltage at the output if we want to find out what
is this VO? VO is found out by flowing in this loop starting
from ground which is zero minus IO into RL is equal to VO or IO we can just write as
Ii. Because IO is equal to Ii there cannot be
any current flow in the op-amp. So zero minus Ii into RL is equal to VO. So that means VO is equal to minus Ii into
RL. Here the k which we are writing here is this
resistance RL. This example is a very simple example of the
current controlled voltage source. Basically in the example which we have shown
here, input is having a current source. But if we think about a current source, basically
a current source is having actually an infinite resistance ideally, in parallel, and as larger
value that parallel resistance has, it is more and more towards an ideal current source. Because we know that along with the voltage
and current there will be series resistance in voltage source and current source will
have a parallel resistance. An ideal voltage source is when the series
resistance is zero and an ideal current source is when the parallel resistance is infinite. But practically we never get an infinite parallel
resistance with a current source and that is why even if we want that the input current
should pass on to the load wholly, that means 100% of the current available from the current
source must be available for the input circuit, that does not happen because a part is lost
as the current through the parallel resistance which is along with it. This is a current source. Basically a practical current source will
have this resistance, RS. We apply directly a current source in a circuit. In this circuit whatever we have shown, it
is a current source just applied in this circuit; but then the current source is having a resistance
that was not shown in this case. But because of this fact that it has a resistance
which is parallel to it and which is to be avoided and because we want that the resistance
should not draw any current so that the whole source current must be available in the circuit,
we are combining it with an op-amp. What will we gain? We now see what part of the current will flow
through this resistance? In this resistance, this point is the negative
or inverting terminal of the op-amp and its positive is grounded. We are having an op-amp. These two points are having the same ground
voltage. This voltage is also ground voltage and at
the other point of this source resistance, this is also ground. There cannot be any flow of current through
this resistance RS because both the points are at ground because of the connection through
an op-amp. Our aim that there should not be any current
flow in the parallel resistance of the current source is being achieved and so the current
will be flowing wholly through this R and so the output voltage we will get minus R
into IS. This is an example of the current controlled
voltage source because we are getting an output voltage VO which is controlled by the current
source IS. Now current controlled current source is another
example of control source driven by an input current and here the output is a current. Earlier case we got a current controlled voltage
source but here the current controlled current source we are getting. This is the ideal block diagram representation
of this current controlled current source. This is the model of the current controlled
current source. Here this is the dependent current source
at the output which we obtain from the input by this relation that IO is equal to k times
of Ii. Let us now see the practical implementation
of a current controlled current source. This is an example of a current controlled
current source. Here we are having an input current and we
have to find out what will be the output current IO? There is this op-amp and positive or non-inverting
terminal is grounded and we are having resistive network having the resistances R1, R2 and
RL. The current if we denote at the input as I1,
this I1 will flow through R1 as there cannot be a current in the op-amp and at this point,
one part of the current will be say I2 and the other current is IO. So I1 is basically IO plus I2. So IO is I1 minus I2. This current I2 we can find out. If I name this voltage as V, the current I2
is nothing but V by R2; this is I2. Now what is V? Suppose we use this loop; this current is
flowing. This point is plus, this point is minus. So V plus I1 R1 is equal to zero; because
I am coming from this point downwards, minus plus is raising voltage, so V plus I1 into
R1 is equal to zero. This is ground. From here I get what is V? V is equal to minus I1 R1. I am replacing this minus I1 R1 here. In place of I2 I am writing minus I1 R1 divided
by R2. That is the expression for I2. Simplifying further, taking I1 common I get
1 plus R1 by R2. So IO is equal to 1 plus R1 by R2 into I1. This can be written as k. This is the ratio R1 by R2 plus 1. This is a constant. That means we get this expression of this
IO as k times of I1 and this is the form of a controlled source which is a current source
driven by input current. This is an example of a current controlled
current source, CCCS. We are controlling the output current by an
input current. Example of control sources we have taken up
today. We have seen different forms of control sources;
voltage controlled voltage source, voltage controlled current source, current controlled
voltage source and current controlled current source. Let us now try to solve one or two examples
which are based on these op-amp circuits. Using that ideal op-amp consideration, now
let us find out a voltage VO given in a circuit like this. We have an op-amp and we are having a source
applied to it say minus plus, minus plus 2 volt and this resistance, if this one is 20
kilo ohm, VO is to be found out and the positive terminal is grounded but the point here is
connected to this positive terminal and this resistance is 10 kilo ohm. Note this circuit using this op-amp. We have to find what will be the value of
VO? In all analyses we use ideal op-amp consideration;
that means we will use that property of ideal op-amp. In order to find out VO, let us name this
node voltage as VX. What is VX? If we come down from this point to ground
VX minus 2 volt is equal to zero. So VX is equal to 2 volt and let us name the
current as i, which flows through 20 k. I am showing the direction like this; intuitively
I am assuming that this is flowing to ground, so it will be in this direction. But there is no hard and fast rule. We can take the direction from left to right
also; that will not harm. We now find out what is I? I, current can be found out if we come from
the VX to ground through this resistance. The current i, which is flowing will not go
through this part because then it will have to enter the op-amp. So it will flow down like this. This current i and this current i, which flows
through k, 10 k is same. So now I can write VX minus 10 k into i minus
zero equal to zero. So what is i can be found out; VX by 10 k
ohm and we know VX is 2 volt. So 2 volt by 10 kilo ohm is the value of this
current. That means it is 0.2 milliampere. We got the current i is equal to 0.2 milliampere. We are interested in finding out VO. How we can find out VO? VO can be found out if we flow from this point
VO minus 20 into i minus 2 volt is equal to zero. The voltage VO can be found out using this
expression or KVL; VO minus i into 20 this drop minus 2 volt to ground if we come. VO is equal to 2 plus i into 20; 20 is in
kilo ohm and i we have found out in milliampere. So it will be in volt only; 0.2 milliampere
into 20, so 2 plus 4 which is 6 volt is the voltage VO. This is an example of a simple circuit. Similarly we can proceed to find out actually
different parameters in a circuit. Another example let us do to find out the
voltage in this circuit. If we have a circuit having say a resistance
is connected here which is 10 kilo ohm, another resistance is here which is also 10 kilo ohm
then this resistance is 20 kilo ohm. There is a source or a battery 2 volt. This point is ground and another source is
in the opposite direction of polarity. This is 3 volt. Another resistance is there in the feedback
part which is having 20 k ohm. We have to find out what is this voltage VO? To solve this example, let us name the node
voltages as v1 here. If this is v1, this is also v1 according to
the ideal op-amp assumptions or ideal op-amp properties and let us name the current also. This current is i let us name it and the current
which is flowing through this 10 k ohm must be also equal to i. We have to find out what is VO and for that
let us first find out what is this node voltage V1? If you see
this circuit, we want to find out the voltage at this point. The other point is grounded. This is a circuit where we can find out the
voltage V1 simply by following the voltage division law. So V1 is equal to 2 into 20 by 20 plus 10;
kilo ohm I am omitting because all are in the same unit. So we get 2 into 20 by 30. That means we are getting a 4 by 3 volt at
this node which is same as this voltage. V1 is equal to 4 by 3 volt. We can now apply the Kirchhoff’s current
law at this node V1. If we apply Kirchhoff’s current law, at
this node V1 the current incoming is i, which is equal to VO minus V1 divided by 20 k and
that is equal to the current outgoing which is equal to V1 minus -3 because this polarity
we have to be careful. This is minus, this is plus, so if I find
out the potential difference between these two points, this point is grounded and this
point, V1 minus this voltage but this voltage is -3; so V1 plus 3 divided by 10 k that is
the outgoing current. This expression can be applied here at V1
of the Kirchhoff’s current law at node V1. KCL, Kirchhoff’s current law we are applying
at node, which has a voltage V1. So that gives us the incoming current i is
equal to the outgoing current, same current but we will find out this expression from
the node voltages VO minus V1 by 20 k. Omitting this k because both sides will have
the same unit equal to as I have said V1 minus -3 that is V1 plus 3 by 10 and we already
have known what is the value of V1. V1 is equal to 4 by 3 volt. So this is 2; so from this relation we get
VO minus V1 equal to 2 times V1 plus 6 or what is V1 can be found out? VO is equal to 3 times V1 plus 6. So replacing this value of V1 by 4 by 3 what
we get at the output is 10 volt. The output voltage VO is 10 volt that we have
got here in this example following the principal of ideal op-amp. In another example let us try to find out
the power in a circuit. For example we have a circuit having an op-amp
here. We are having this battery here minus plus
minus plus 4 volt. 4 volt battery is applied in the feedback
part and we are having in the non-inverting terminal, a resistance which is 30 kilo ohm
and another resistance is connected which is 20 kilo ohm and here there is a resistance
which is 10 kilo ohm to ground. This is the circuit. We want to find out the output voltage. Find VO as well as the power supplied by the
4 volt source. In order to find out the power supplied by
the 4 volt source, we must also know the current flowing through it. Because power is equal to voltage into current,
we must know the current which flows. Let us name it by i. So we will have to find out the current through
it because then only we can find out what is the power supplied by the 4 volt source. Power supplied by the 4 volt source will be
4 into i. So i we need to know. In order to solve this following similarly
let us name this node voltage here as v1 which is same as this point voltage which is V1
only; these two voltages are same. We now find out what is? We can find out V1 in one way just by finding
out this voltage which is that portion of the voltage available from VO. We do not know VO but then we will be solving
following this relation. What is V1? VO into 30 k divided by 30 plus 20 k; because
if we consider this as a source this is ground, this is the part. The voltage here into this resistance by this
resistance plus this resistance; this is the voltage division principal we are applying. We are interested to know this voltage. We are finding out VO into 30 by 30 plus 20;
so 3 by 5, 30 by 50 is 3 by 5. So 3 by 5 VO is equal to V1. We have found this voltage, here as well here. This is the same voltage 3 by 5 into VO and
so this can be written in another way VO minus 4 volt is equal to again V1 that is equal
to VO minus 4 volt. Using these two relations we can find what
is VO? So VO minus 4 is equal to 3 by 5 VO or we
can find what is VO? VO is equal to VO into 1 minus 3 by 5 equal
to 4 VO into 2 by 5 is equal to 4. So VO equal to 10 volt. We have known 10 volt is VO. If we know VO, then V1 we can find out. V1 is nothing but VO minus 4; the voltage
here is VO minus 4. 10 minus 4 is 6 volt. We know what is V1 and so we can now find
out what is the current flowing i? i, this current is nothing but the same current;
this current is same current i, so V1 by 10 k. V1 is 6 volt so i is equal to V1 by 10 k;
6 by 10 is 0.6 milliampere which is i. The current which flows here is 0.6 milliampere. 0.6 milliampere is the current which is flowing;
i is equal to 0.6 milliampere, the current we know. We know the voltage which is 4 volt; so, we
know the power supplied by the battery is 4 volt. The power supplied by the 4 volt source equal
to 4 volt multiplied by i and 4 volt multiplied by i, we have found out to be 0.6 milliampere
and that gives us 4 into 0.6 which is 2.4 milliwatts; volt into ampere is watts. So miiliampere into volt is milliwatt. That means we have found out the power supplied
by the 4 volt source which is 2.4 milliwatt and for that we have to find out the current
flowing through that circuit or we have to know the current supplied by the source 4
volt actually. In this example we have seen that power can
also be found out if we know the current as well as the voltage. These are some of the examples which we have
solved today by following the same analysis of op-amp being ideal. In the whole analysis or in all the examples
we have solved till now we are following only the ideal op-amp consideration. In order to solve such problems we have to
know the ideal op-amp consideration only and then we can apply Kirchhoff’s voltage law
as well as current law and we can find out the required quantities whether it is voltage
or current in the circuit. In today’s discussions we have seen the
control sources being built up using op-amp and the four different variations of the control
sources we have studied today. These four variations of control sources are
voltage controlled voltage source, voltage controlled current source, current controlled
voltage source and current controlled current source and these are extensively used. Mainly in instrumentation circuits also you
will find applications of these control sources and also we have seen how we can solve numerical
examples using ideal op-amp considerations and the KVL, KCL, etc to arrive at the required
solutions.

## 4 thoughts on “Module – 4 Lecture – 4 Op-Amp Applications Part -2”

1. Abdela Mohammed says:

show you tube

2. Dheeraj Kumar Raghuvanshi says:

With due regards … Its really tough to understand when the quality of the aid used is so poor. When the professor is pointing on certain points on the circuit its difficult to understand which point she is talking about as the pointer is invisible. I guess the video not taking the output from the device she is working rather its capturing the the screen by a camera.

3. Vamshi Arakatla says:

in the problem given in the lecture i don't think the solution was right because negative terminal of an opamp is usually considered as a virtual ground as positive is grounded and if it is so why is it considered the same current in the upper loop too.
if any one gets the logic please help me. For all those thanks in advance.

4. Vamshi Arakatla says:

The below comment corresponds to first example in the lecture.