# Lecture – 1 Electric Drive

drive system. In this industrial drive system, in this course, we will be mostly concentrating
on the electric drive part. So, we will be talking about the electric drive part. A basic
definition of an electric drive system can be an electric drive is defined as a form
of machine equipment designed to convert electric energy into mechanical energy and provide
electrical control of this process. So let us say, let us write that one, electric drive. An electric drive is defined
electric drive is defined as a form of as
a form of machine equipment
machine equipment designed to convert electric energy designed to convert electric energy
into mechanical energy, energy into mechanical energy. So, in an electric drive, what we
are doing? We are trying to convert electrical energy into mechanical energy and provide
electrical control of this process or electronic control of this process. That means when we convert electrical energy
into mechanical energy; so the load, the output system required in a particular fashion. So,
we require a process, a necessary control of the system. So, all the controls nowadays
we do it with electronics control. So, we will say, to convert electrical energy into
mechanical energy and provide electrical control of this process electrical control of this
process; this is very essential. You are converting electrical energy into mechanical energy with
a necessary control. So, this is the basic definition of an electric drive system. Now, this definition, I have taken from a
very famous Russian book by Prof. Chilyken and I feel that is a very good definition
of an electric drive system. Now, based on this definition, let us form a blocks schematic
so that we can clearly understand what is the conversion process? What are the blocks
inwards? And, where the control is? For control, what are the inputs required? So, basic block
diagram let us write about. Now, the basic block diagram for an electric
drive system may look like this; so, we are converting electrical energy in the mechanical
energy. So, we should have the source, what we call the power source. Power source, we
will put it like this; this is the power source, this is the power source. Now, this power source, our basic power sources
are our mains; the voltages and currents will be in AC. Sometimes, the power source can
be a battery where the voltage and current are DC; sometimes we can have solar power.
So power source, the output of the power source should go to a power convertor. Why? This power convertor, the output of the power
convertor brings the power source to the required level; the voltage level, current level, frequency
level last required by the load. So, we require a power convertor where the power source can
be controlled and modified or changed AC to DC, whatever it is; so, power convertor. We
will come to each block soon. So, power convertor. We are converting electrical energy to mechanical
energy; the source and the power convertor convert to the desired level and for any mechanical
energy, we should require a motor. In electrical drive system, a motor is a part of… and
motor will be connected to the load. This is the basic open load block schematic of
an electric drive; we have the power source, power convertor will convert to the required
level and give to the motor, motor will give the required speed and power to the load. So, in our electric drive system, what we
are controlling? Conversion of the electrical energy to mechanical energy means the source
is electrical and the load is mechanical. So, we are converting into electrical energy
mechanical energy that means basically speed and the torque. Now, in our basic definition, we should give
the necessary control of this process. So, let us draw the block diagram like this, controlled.
Controller – that is our electronics controller; now controller output will control what? We
do not have any control of the power source, power source is what is available to us given
to us and load is suggested by the customer, what he wants. So, once we select the motor, then all the
control; motor parameters we cannot change, we have to control the whole thing through
our power convertor. So, controller output will mainly go to the power convertor. So,
to take the necessary control action, controller requires some inputs. So, as many as inputs
are always good for a good controller but the more input means more devices, more sensing
equipments are required. Then the system will become very costly. So, the controller input
depends on a cost effective design, we will limit to what is their minimum required. So, controller input can be from power source.
Why from power source? Suppose, if for AC, our mains, we are using as the power source;
sometimes the voltage can fluctuate, frequency can fluctuate, then the power convertor should
control it such that the motor gets the required steady voltage and current irrespective of
the fluctuation in the power source. So, there is an input from the power source is required. We can also have input from the motor; speed
can be one input, motor voltages and current can be one input. Suppose, due to some sudden
change in load, motor drives very large current which is beyond its requirement; suddenly
controller can, we can control that one through our power convertor. Power convertor, through
power convertor we can control the motor input parameters such that the current can be controlled
or sometimes the load require more torque, it is rotating. So, motor should drive more
current. So, this can be controlled through our power convertor. So, the basic inputs,
we can have power source and from the motor load and the power convertor. This is the
basic block diagram of an electric drive system. Now, in our course, we will be mostly studying
the power convertor side and the controller. Why? because, as a designed engineer, we have
only access to these two. We can only control and we can select and choose these two and
motor is selected for decided base from the load requirement. Once these are selected,
these are fixed that means power source, motor and the load. Then any control parameter,
we can only control through these power convertor and the controller. This is only in our control. Now, let us again go to further definition
of our power convertor, whatever is a conversion process happening here. Let us take the power
source. Power source can be, we can basically, we can we can specify as DC and AC. So, we
will go to the next page. Depending on the power source available; either
battery or mains, power source can be basically define as basic form DC or AC. Now, our output
of the convertor also we can very basic with the basic form of output, we can say either
DC or in AC. So, here also there is a power convertor; DC or AC. Now, we should have conversion,
the power convertor convert the power source, a DC voltage of one level should convert to
a voltage level to a different voltage level as required by the motor. So, there is a DC
to DC conversion. This has to be achieved through our power convertor. So, basic DC to DC power convertor; you know
it, it is a 1, it is at SMPS – switched more power supply, SMPS – switch more power sup
supply, one example. Then AC to DC that is our mains that means AC power source with
alternating voltage and current, we are converting to a DC voltages and current. The basic power
convertor of this one, we can call as the rectifier. But if you see in electrical, as an electronic
engineer or electrical engineer; you know how to rectify? If you have the diode, a three
phases or single phase rectification is possible. But we will not be this is not the required
rectifier what we want. Why? We require control of the process. So, what we require here into
is a controlled rectification, like using thyristors, phase control convertors; so one
example. So, it is controlled rectifier, this is controlled rectifier that is AC to DC. Then the next conversion; this is power source,
this is on the load side or the motor side, DC to AC that means a DC source battery or
a rectifier output that means what is available to us is DC source but we want AC. One example
is UPS that is basically invertors. These types of convertors are called, power convertors
are called invertors, these are called invertors. We will be studying in the due course, the
basic control of the invertors, various invertors topologies for speed control applications,
we will be concentrating more on that one later. Then another one AC to AC. That means
voltage, current or frequency of one into another voltage and current and frequency
of different level. These are basically in old books; we can see one class of convertors
We will be mainly concentrating the AC to DC and the DC to AC convertor in our main
course. Now, after going through the classification
and block, we set our control of the whole processing through the controller. That means
we have to design the controller such that the system will quickly adapt to the change
without much translates. So, controller design is very important. Nowadays, simulation tools
are available, so we can have a mathematical study and we can simulate and we can tool
the controller before a hardware implementation is started. Why? That will say lot of time and money;
otherwise we have to learn from the failures. The hardware if keep on failure fail and leave
be if we learn from that process, the cost of the whole system will go up. So, the controller
design and the controller tuning, before I control the implementation; a controller design
and tuning can be done through a simulation study.
For a simulation study, the whole system should be modeled. That means equivalent mathematical
representation of the system is required. That means the representation for the power
source, power source basically we can AC or DC, we can do it; then the power converter,
then the motor. So, a mathematical model of the motor required based on, for our controller
design. As I told you before, the whole process is
converting electrical energy into mechanical energy, mechanical energy through our; what
is the output of the load? The mechanical energy, the torque and the speed, from there
we are generating the energy. So for motor, a basic input output relation, a torque speed
characteristic is required torque speed characteristic required. That means various speeds, what
are the torques required? Then for the torque; what are the parameters we have to control
at the input side of the motor? That is the voltage and current. For that voltage and
control, corresponding controller has to be initiated for the power converter. Now, the basic load that is now the mechanical
load; we can basically divide the torque speed characteristic into four category. Let us
take the torque speed characteristic of the load. One; torque independent of speed. Let us write
the graph, power speed characteristic of this load. The graph, this is our speed that is
speed in electrical radiance per second and this is the load torque. So, torque is independent
of speed. So, the curve will be, the speed torque characteristic will be like this; steady.
That means irrespective of the speed, it will always generate constant load. So, what are
the examples for this one? Example; low speed hoisting, that is cranes
during hoisting during hoisting that is one, then machine tools for feed mechanism; these
are some of the examples. Many examples you can find in many text books and these are
some examples. Then two, linear rising characteristics; what you mean by linear rising torque speed
characteristic? What do you mean by linear rising torque speed characteristics? That means whenever the speed changes, torque
will also change proportionally, this is one this is our second example. The torque speed
characteristic for a linear rising low torque is like this. What is the example? Let us
take a separately excited DC generated connected to a load of constant ohmic resistance that
is example. Separately excited DC generator connected to a load of constant ohmic resistance
that means a separately excited generator; the generator that means machine is working
as a generator and the generator is feeding a load of constant ohmic resistance. Then
let us see, the torque speed characteristic is linearly rising. Now, we know torque is equal to P by omega;
omega is radiance per second. The torque generated, the power developed by the generator divided
by the speed give the torque. Now, what is the power generator by the separately excited
generator? P is equal to that is the temporary voltage into current I. Also, we know that
this V, the voltage generated by the excited separately excited DC machine DC proportional
to speed that is proportionality constant K into omega. So, V is proportional to k into
omega. Why? We want to bring the basic relation between
the torque and the speed. V is equal to K omega and what is the current given or taken
by the load from the generator? I is equal to V by R, V is equal to proportional to speed
K omega divide by the, this is our constant ohmic resistance R, K omega by R. So, we got
I. Now, what is the power? In this, this is the basic equation. In the basic equation,
we require the power. Power is equal to P is equal to V into I, V into I is equal to
that we can substitute V and I; this will be equal to when you substitute K square omega
square divided by R. Now, let us substitute this one to this equation. Then what will
happen? Torque is equal to K square by this is a constant for a machine, this load is
also constant; so the whole thing, we can represent by a constant K1 another constant
K1 So, this will be equal to K1 omega square,
PV we have substituted here, these part we have substituted here divided by omega. Now,
this will be equal to omega square goes, this will be equal to, torque will be proportional
to omega. So, as the speed increases, torque also will increase. Power will increase, torque
will also increase. So, this is the equation, linearly rising characteristic. Some simple examples, I am telling here. Now,
let us take here nonlinear rising characteristic. Let us go to the next page. Nonlinear rising
characteristic; now the third example is nonlinear rising that is parabolic characteristic. Nonlinear rising torque speed characteristic;
torque speed, torque versus speed
characteristic that is parabolic. How it will look, nonlinear rising? The speed let us again
draw the axis, this is our omega speed, this is our torque. So, it will be something like
this, third characteristic. Previously, the constant load speed characteristic that was
one, linear rising was like this, this is the third one. Third one is this one. Most of the final loads belong to this one
that means torque will be proportional to omega square. Most of the loads will be most
of the load final loads will come under this. You should see; constant, linear rising, then
nonlinear rising, then the next one will be for nonlinear falling characteristics, nonlinear
falling torque speed characteristic. Are we to look. So, nonlinear; here let us
take that is torque is
proportional to 1 by omega. This is nonlinear rising, nonlinear falling characteristic that
is something like this; this is nonlinear falling characteristics, torque is proportional
to omega. That means or we can again say T into omega is a constant. Let us take, one example, rolling mill. In
a rolling mill, the paper comes out from the mill, the constant speed and it will be a
thin roll to a drum. So, paper is the radius r. As the paper gets rolled on to the drum,
the radius will keep on increase. Now, the paper is coming from the mill with the constant
velocity and we should take care of the speed of this drum has to be adjusted that we have
to keep constant tension, we have to keep a constant tension such that the paper should
get rolled on to the drum properly. So, what is the torque speed characteristic it shows? Now, if you see here, the speed of rotation,
what is the speed of rotation? Speed of rotation is distanced by the speed B. So, speed of
rotation
speed of rotation is equal to 2 pi r divided by V, this is the speed of rotation. This
is second, period in seconds. Now, what we are attempting with this equation? We have
to get a torque speed that is torque and the speed that is the angular speed omega; we
have to keep a relation. Now, what is frequency here? f is equal to
that is the frequency is equal to 1 by T, 1 by T is equal to V by 2 pi r; frequency
we got. Once frequency we got, we know omega. Omega is equal to 2 pi f. So, from the speed
V, this V, we can get omega — 2 pi f. Now, what is the power exerted by the torque by
the drive? Power exerted by the drive, see we have to
give a constant torque opposing force here. This is force f. So, what is the power? Power
is equal to torque is equal to the power exerted by the drive p is will be equal to that is
F into r; r is the radial distance, f is the tangential force. The torque, this will give
the torque, this is the torque f into r, torque into, what is omega? Omega is equal to 2 pi
f, 2 pi f is equal to v by r, from here v by r. So, these two we will cancel, F and v; these
are constant for this one, so power is constant. That means torque into omega that is p is
equal to p is constant, torque into power is equal to that is p constant that is this
will give constant wholes power load. These are some of the some of the basic torque speed
characteristic. Now, once we got the torque speed characteristics,
let us take the combined system. This is torque speed characteristic of the complete load.
Now, in a combined system, there is a driving torque and the load torque. For speed control
this should be matching. So, how that driving torque has to be controlled the meet the load
torque? That is our basic control purpose. So, let us take the combined system. Now,
let us go to the next page again. There is a combined load torque characteristics, combined
torque speed. That is torque speed
torque speed characteristic. Let us try, draw the axis; this is omega,
this is T. Now, the load torque that is not in your control, load torque will keep on
varying and for a stable operating speed or speed in speed control system, the drive torque
should match the load torque. So, let us take a load torque a typical load or something
like this that I have put into under negative axis because this is the opposing force, opposing
torque, this I will put load torque. Now, let us take our motor, the speed torque
carries of our motor; let it be something like this. That is our motor driving torque
for the motor. Now, in our speed control system, suppose we have to operate, the system will
come to a stable state where the load torques approximately this point, load torque will
be equal to the driving torque. But the load torque can vary that is not in our control.
Load torque can vary means suppose the speed varies, load torque goes here. What will happen? Load torque goes here means speed has changed.
Now, this is the load torque demand by the load. But according the motor characteristics,
the driving torque that is Td is less than the load torque now. So, what will happen?
System will slow down and come to this point. Now, let the load torque decrease. In this
case, what will happen at this point? The load torque is less than the driving torque;
so this will accelerate it and it will come to the stable point. But in most of the cases,
this is for a typical load speed characteristics; in most of the cases, it will not come to
the stable system so that depending on the load torque requirement; our motor characteristic
torque speed characteristic, we have to control we have to modify through our power converter. So, our whole control block, the purpose of
the control block is to dynamically control the torque speed characteristic so that the
driving torque will be always equal to the load torque as demand by the load for a particular
speed. So, such cases, we require a close loop control system. Now, as I told you before,
see I have so far I have told you TL – load torque. So, load torque basically, we can
define as or the driving torque Pd is equal to the load torque Td sorry this is Td, this
is TL plus if the speed is constant, driving torque will be equal to load torque. Suppose, during acceleration and deceleration;
there is more load, there is a torque change requirement or torque demand from the load.
So, that can be explained as J into d omega by dt. J into d omega by dt, this approximately
J is the moment of inertia. So, whenever acceleration deceleration is there, the total driving torque
should be equal to the constant load torque plus the acceleration and deceleration. Now, how to find out for a particular load
system, how to find out the J? There are various techniques available in literature. I will
take one from the famous Leonard’s book. What we have to do? The complete system under
no load, you system machine to the full speed, drive the system to the full speed, then full
speed and find out the sorry this is omega 0, this is ML, this is our type two axis.
So, drive the system for various speeds for operation. 5, 6 points you take it and tie
the system and find out the load torque required. So, you plot the load torque; you got the
ML. Now, run the system to the full speed and
switch off the power supply. Then what will happen? The whole system will decelerate.
So, during the deceleration, what will happen? The driving torque is 0, so the decelerating
torque generated by d omega by dt will be will be opposed that will be opposed by the
friction and windage loss. So, slowly full system will decelerate. So, at various point,
measure omega; you note the time and measure the omega. At various point, measure the time
and omega and approximately at this point and time, you can find out d omega by dt. So, d omega by dt we know, the system is under
no load at various speeds. So, we know at various speed, what is the driving torque.
So, this driving torque at the no load torque, so if you write the equation J into d omega
by dt plus ML is equal to 0 because we are switched on the system and the system is decelerating. So, at various speed, J into d omega by decelerating
torque is the no load torque of the system that we can get from this one. So, from this
equation; d omega by dt we know, ML we know, we can approximately we can find out J. This
is the one way of doing the thing. There are other techniques also available; this is the
one simple way of finding out J. So, once you know the J, once the speed; then what
we know? We now, approximately we can find out the torque speed characteristic of the
system during acceleration also. Now, in this class, in this lecture, we have
just basically; the basic definition of the electric drive is introduced and basic block
diagram is given. Then depending on the voltage source, whether AC DC and the output requirement,
we have classified the power converter where conversion process into various four category.
Then the torque speed category of the machine, we have various categories — rising, constant,
nonlinear rising, nonlinear falling, we have defined with typical example. Then we talked about the combined characteristics.
Combined characteristics, what we found out? It is required to modify the motor, low torque
characteristics through our power converter such that depending on the load variation
that is the TL, always we have to keep the driving torque equal to the load torque. So,
there is a close loop control system is required. The driving torque, the load driving torque
is not only the steady; note in the acceleration and deceleration also we have to give the
transient power also to the system so that the transient power acceleration and deceleration
depend on the moment of inertia of the system and typically, how to find out the moment
of inertia, a typical system is also we explained. Now, let us go to the some of the basic power
converter configuration so because we require the power converter. Once you select the power
converter, then only we can choose the control function. Then once we have studied some of
the power converters; first we will study the power converters required for DC modern
drive applications, then we will go for the inverters that the power converters required
for AC motor drive application, then how you design the closed control for DC motor and
AC motor will be discussed during this course. For the DC motor, we will be mostly talking
about the separately excited DC machine because we will not have much time for all other cases;
separately excited DC motor that is widely used for motor drive applications. Then the
AC motor drive lectures, we will be talking about induction motor drive, mainly induction
motor drive applications. During steady state, dynamic equivalence circuit model we will
study, then the symbol v by f control, then the high dynamic and the controller like field
oriented controller for high dynamic performer application we will be studied later. Now, let us go to the basic power converter
configuration. Let us go to the next page. Here, what we will be talking about is the
power converter that is AC to DC converter for DC drive applications. So, control rectification
is a must for this one. Let us study some of the basic power converter configuration.
So, single phase, let us take a typical converter, single phase, before coming to the three phase
control rectification; let us talk about single phase AC to DC power converter. The most basic single phase power converter
is, the schematic is like this, let us represent our mains like this; this is our controlling
device, this is the thyristor, then you have the load. Load, I will simply represent it
like this, it is inductive with back EMF, it can be there. So, we will have inductive
plus back EMF alone that is the motor can be represented like this. Then, it will go
like this. Now, how do you control the output DC that
is DC this part? So, at the appropriate point of the mains, we will be turning on the device
and we will transfer the voltage to the load. So, by controlling the firing angle of the
thyristor, we can control the output voltage. How it can be controlled? Let us see. This is our V0. What is the nature of V0 for
a single phase with a single thyristor controller? How it looks? This is our mains. We will not
be whenever, for our present analysis, we will assume the control of the devices our
instantaneous; we will not worry about the switching types involved in this one. So,
the moment you give the gate pulse, thyristor will be turned on and when the AC mains is
getting when the thyristor when across the thyristor reverse voltage is applied, the
thyristor will switch off immediately. So, let us say at this point, we are giving
the firing angle alpha. Firing angle we represent as alpha, alpha with respect to this point,
the starting point of our mains, here, zero crossing. Now, if the load is resistive that
is the simple and the easy load if the load is resistive; what will happen? During this
portion during this portion, the voltage will come across the load. During this portion,
the voltage will come across the load and if it is purely resistive load, then the current
will also be proportional instantaneously proportional to the voltage before but pure
resistive load is only a text book answer; we will not have a pure resistive load, you
will have resistive as well as inductive. So pear, you will know in this class will
inductive load, what will happen? So, before coming to the inductive load, let us say how
for the resistive load; how the current, it will be? Current will be in this form, exactly
this will be. This is our Vin; during this portion, this
any value of the current I0 will be equal to Vin divided by R. But during the half cycle,
the moment the voltage go negative; what will happen? There is a reverse voltage across
the thyristor, anode and cathode. The thyristor will be immediately switched off. So, during this portion that output voltage
is, there is instantaneous voltage across the load is zero for a resistive load. Then
next half cycle, again this will appear. This is the current, this is the voltage. So, we
will see a pulsed wave form will be coming across the load. This pulsed waveform, it
will have the DC value and the AC value also will be there, high frequency AC value will
be there. So, we will not talk about but we are worried
more about the DC value. So, what is the DC value? That is V0, average value of this pulsed
wave form that can be easily found out. V0 alpha will be 1 by 2 phi that average value
integrating alpha 2 pi. So, this is our pi. So, alpha to pi we will integrate and divide
by the P rate, you will get the average value. So, this will be equal to Em sin omega t divided
by dm, this we are integrating. So, Em sin omega t if you integrate, what
will happen? You will get minus cos omega t integrating limit is alpha to pi. So, when
you substitute this one, finally the V0 alpha will be alpha is equal to Em by 2 pi into
1 plus cos alpha. Here, alpha can be varied from 0 to pi. When alpha is equal to pi, what
will happen? Cos pi that is minus 1, output voltage will be 0, the DC value will be 0. When alpha is equal to 0, alpha is equal to
0; what will be the output? Alpha is equal to 0, the output will be alpha is equal to
0, V0 is equal to Em by pi, this is the output and alpha is equal to 0 means we are firing
the thyristor at the zero crossing, the positive zero crossing. What it says? It says, this
can be replaced by a diode. So, output DC voltage for a single phase half range rectifier
with a single diode is equal to Em by pi. But if you see here, during the negative part,
we are not using, we are not drawing any power from the mains. That means even though power
source is there, we are not fully utilizing the power source. So, even though in the text book, these type
of AC to DC converter using diode or thyristor are available. This is only for the understanding
that one, how to get the DC value. So, this value what we talked about, this is the DC
value but along with this value because of this variation, we have the repel content
also will be there. So, that repel content, the repel voltage will produce its own repel
current. That has to be suppressed by the load. But we will not be putting any extra
filters to suppress this one. To put filter means it is a cost effective
proposition. So, inductive load; as the frequency high frequency of the voltage waveform, there
the impedance of the load that is l omega will increase and the voltage divided by l
omega, the current will the harmonic current can be suppressed. So, harmonic current, the
amplitude harmonic currents can be suppress only by through the motor inductance. But
what we are interested is the DC part. So, this value gives the DC value. So, this
will be another it will be used with, never use for any applications. But for understanding,
already it is available in text book. So, whole analysis, what we have done is for a
resistive load. So, this is for resistive load with the firing angle control. Now, let
us see, instead of resistive load what if we put an inductive load? Let us go to the
next page. In the case of inductive loads, again let
us draw the block schematic of the AC to DC converter with the firing angle alpha, alpha
can be varied from 0 to pi. Now, we have the highly inductive load. If we use here; what
are the type of repel voltage appearing across this one – V0 (t)? That is the repel, this
BDC plus the repel. Here, this is our Vm. If you see here, this is our voltage wave
form, alpha is fired here, firing angle is fired from here; now what happens? Because
of the inductance previously when the resistive load was there; the wave form, this repel,
the current wave form across the resistance was the nature of the current variation was
exactly same like the voltage variation. But here, in this case, what will happen;
the current wave form, current wave form will be starting at this point because currently
the inductance cannot change instantaneously. So, what will happen? The current will vary
like this. At this point, when the voltage is equal to 0, there is a current in the inductance.
So, thyristor cannot switch on, current through the inductance cannot be changed instantaneously. So, what will happen? Voltage will go negative.
Voltage will go negative, then thyristor is conducting; so all voltage has to be suppressed
across the inductive load. What will happen? At this point, inductance, the rate of change
of current, it will go negative so that the voltage become instead of positive negative
here, it will become positive or negative. Here it have the input voltage has changed. So, this rate of change of current will, the
load will take care of such that the rate of change of current is just sufficient to
forward bias the thyristor here and current will keep on continuing and this current will
be from the stored energy. Till the stored energy is released, current will keep on coming.
So, what is the disadvantage here? During the negative portion upto this point also,
the voltage was appearing to the load. Now, compared to the previous equation; firing
angle and DC is we cannot apply here because the negative portion also has come here. So,
the previous equation of alpha with respect to the output DC with respect to firing angle
is no more valid here because the thyristor is conducting in the negative direction also. Then how to avoid this negative portion? At
the same time, we have to give a free varying path for the inductance. This is during this,
it is called sorry during the current conduction, the negative portion also we should give an
alternative path for the inductance. At the same, this negative voltage should not appear
across the load. So, than the previous equation that is V0 alpha is equal to Em by 2 pi into
cos … one plus cos alpha is valid. How to come to that work; we will study in the next
class.

## 33 comments on “Lecture – 1 Electric Drive”

1. CalculoTV says:

Excelente.

2. Michel AlSharidah says:

This is a great series of lectures and I owe Prof. Gopakumar big thanks and gratitude for his excellent explanation and presentation of the material. I would hope that there is more after lecture 37 in the near future. Thank you again!

3. Shivang Srivastava says:

great effort … this should be publiscised so that students across india can benefit from the lectures. please have more of these. thank you professor gopakumar

4. Beck ApplianceRepairOfTucson says:

with teachers like you Indian student will beat any other nation soon….great…thanks…keep it up!!!!
Calif.USA

5. Saurabh says:

Thanks Prof. Gopakumar and the uploader of this video..

while listening to him, i get the same tingling feeling like while listening to Bob Ross

7. Pranav Choudhary says:

thanx for these vedios…these r awesome…

8. Kasturi Gon says:

thanks prof .We hail from a village near Asansol and with almost no library so this is a great help

9. Vikas B A says:

i wish i get into IIT bombay. All these profs come from there. Thanks prof.

10. This is Pakistan says:

which device he is using for writing. what is the technical name of it. the pen is special or there is some thing else. also which MS office product he is using ???

11. hatvikah vlogger says:

thank you very much sir

12. babita nanda says:

excellent sir..

13. Sreejith M.G says:

Nice lecture

14. Drew Larson says:

great lecture though the sound quality and prof Gs accent make it hard to hear at times.

15. M says:

Thank you for publishing such a great series of lectures.
(As a friendly comment, the resolution of videos may be improved.)

16. akshay bhor says:

great ………..

17. Dhruv Parmar says:

good lectures but can we have more support and written resources(notes) for study
thank you

18. Lalit Patnaik says:

You can follow latest updates from Power Electronics Lab, Department of Electronic Systems Engg (formerly CEDT) IISc at the lab twitter handle @iiscpowerlab

19. Tarunaanand Singh says:

it a good to give those who did not reach that place but want to learn by best all must try it

20. jakkam girish says:

i wish it was in 360p or more when prof writes it is not so clear

21. soumya sharma says:

its so amazing how complex things get simpler by a good teacher. not only the thoughts are restricted to the four walls, they seem to influence people a thousand miles away.
i dont know you sir, but indirectly you've helped so many students like me out there. cheers to the uploader as well!

22. Deodatta Deshmukh says:

Excellent Concept Delivery

23. Vijay tel says:

very bad quality video but very good lecture

24. CHANDRESH KUMAR says:

There is a mistake in power equation at 24.52 minute….

25. Amsavalli A says:

sound in power electronic .it is highly useful my students

26. Arif Mughal says:

I am not satisfied…

27. a k says:

poor lecture..lacks depth lot of time consuming wont recommend

28. Raj Dipak says:

It's great and thanks a lot for this lecture.

29. Robin Maikle says:

Need a circuit solution? Why not use this Spot: 'Circuit Solver' by Phasor Systems on Google Play.

30. eDrive says:

Nice, but I can't understand this indian dialect…

31. Sourav Dhar says:

Stupid teacher.