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Today, we will start with the topic on industrial
drive system. In this industrial drive system,
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in this course, we will be mostly concentrating
on the electric drive part. So, we will be
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talking about the electric drive part. A basic
definition of an electric drive system can
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be an electric drive is defined as a form
of machine equipment designed to convert electric
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energy into mechanical energy and provide
electrical control of this process. So let
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us say, let us write that one, electric drive.
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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
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energy designed to convert electric energy
into mechanical energy, energy into mechanical
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energy. So, in an electric drive, what we
are doing? We are trying to convert electrical
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energy into mechanical energy and provide
electrical control of this process or electronic
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control of this process.
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That means when we convert electrical energy
into mechanical energy; so the load, the output
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system required in a particular fashion. So,
we require a process, a necessary control
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of the system. So, all the controls nowadays
we do it with electronics control. So, we
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will say, to convert electrical energy into
mechanical energy and provide electrical control
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of this process electrical control of this
process; this is very essential. You are converting
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electrical energy into mechanical energy with
a necessary control. So, this is the basic
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definition of an electric drive system.
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Now, this definition, I have taken from a
very famous Russian book by Prof. Chilyken
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and I feel that is a very good definition
of an electric drive system. Now, based on
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this definition, let us form a blocks schematic
so that we can clearly understand what is
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the conversion process? What are the blocks
inwards? And, where the control is? For control,
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what are the inputs required? So, basic block
diagram let us write about.
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Now, the basic block diagram for an electric
drive system may look like this; so, we are
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converting electrical energy in the mechanical
energy. So, we should have the source, what
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we call the power source. Power source, we
will put it like this; this is the power source,
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this is the power source.
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Now, this power source, our basic power sources
are our mains; the voltages and currents will
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be in AC. Sometimes, the power source can
be a battery where the voltage and current
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are DC; sometimes we can have solar power.
So power source, the output of the power source
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should go to a power convertor. Why?
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This power convertor, the output of the power
convertor brings the power source to the required
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level; the voltage level, current level, frequency
level last required by the load. So, we require
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a power convertor where the power source can
be controlled and modified or changed AC to
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DC, whatever it is; so, power convertor. We
will come to each block soon. So, power convertor.
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We are converting electrical energy to mechanical
energy; the source and the power convertor
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convert to the desired level and for any mechanical
energy, we should require a motor. In electrical
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drive system, a motor is a part of... and
motor will be connected to the load. This
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is the basic open load block schematic of
an electric drive; we have the power source,
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power convertor will convert to the required
level and give to the motor, motor will give
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the required speed and power to the load.
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So, in our electric drive system, what we
are controlling? Conversion of the electrical
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energy to mechanical energy means the source
is electrical and the load is mechanical.
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So, we are converting into electrical energy
mechanical energy that means basically speed
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and the torque.
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Now, in our basic definition, we should give
the necessary control of this process. So,
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let us draw the block diagram like this, controlled.
Controller - that is our electronics controller;
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now controller output will control what? We
do not have any control of the power source,
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power source is what is available to us given
to us and load is suggested by the customer,
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what he wants.
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So, once we select the motor, then all the
control; motor parameters we cannot change,
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we have to control the whole thing through
our power convertor. So, controller output
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will mainly go to the power convertor. So,
to take the necessary control action, controller
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requires some inputs. So, as many as inputs
are always good for a good controller but
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the more input means more devices, more sensing
equipments are required. Then the system will
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become very costly. So, the controller input
depends on a cost effective design, we will
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limit to what is their minimum required.
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So, controller input can be from power source.
Why from power source? Suppose, if for AC,
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our mains, we are using as the power source;
sometimes the voltage can fluctuate, frequency
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can fluctuate, then the power convertor should
control it such that the motor gets the required
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steady voltage and current irrespective of
the fluctuation in the power source. So, there
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is an input from the power source is required.
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We can also have input from the motor; speed
can be one input, motor voltages and current
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can be one input. Suppose, due to some sudden
change in load, motor drives very large current
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which is beyond its requirement; suddenly
controller can, we can control that one through
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our power convertor. Power convertor, through
power convertor we can control the motor input
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parameters such that the current can be controlled
or sometimes the load require more torque,
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it is rotating. So, motor should drive more
current. So, this can be controlled through
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our power convertor. So, the basic inputs,
we can have power source and from the motor
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load and the power convertor. This is the
basic block diagram of an electric drive system.
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Now, in our course, we will be mostly studying
the power convertor side and the controller.
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Why? because, as a designed engineer, we have
only access to these two. We can only control
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and we can select and choose these two and
motor is selected for decided base from the
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load requirement. Once these are selected,
these are fixed that means power source, motor
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and the load. Then any control parameter,
we can only control through these power convertor
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and the controller. This is only in our control.
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Now, let us again go to further definition
of our power convertor, whatever is a conversion
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process happening here. Let us take the power
source. Power source can be, we can basically,
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we can we can specify as DC and AC. So, we
will go to the next page.
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Depending on the power source available; either
battery or mains, power source can be basically
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define as basic form DC or AC. Now, our output
of the convertor also we can very basic with
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the basic form of output, we can say either
DC or in AC. So, here also there is a power
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convertor; DC or AC. Now, we should have conversion,
the power convertor convert the power source,
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a DC voltage of one level should convert to
a voltage level to a different voltage level
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as required by the motor. So, there is a DC
to DC conversion. This has to be achieved
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through our power convertor.
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So, basic DC to DC power convertor; you know
it, it is a 1, it is at SMPS - switched more
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power supply, SMPS - switch more power sup
supply, one example. Then AC to DC that is
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our mains that means AC power source with
alternating voltage and current, we are converting
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to a DC voltages and current. The basic power
convertor of this one, we can call as the
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rectifier.
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But if you see in electrical, as an electronic
engineer or electrical engineer; you know
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how to rectify? If you have the diode, a three
phases or single phase rectification is possible.
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But we will not be this is not the required
rectifier what we want. Why? We require control
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of the process. So, what we require here into
is a controlled rectification, like using
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thyristors, phase control convertors; so one
example. So, it is controlled rectifier, this
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is controlled rectifier that is AC to DC.
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Then the next conversion; this is power source,
this is on the load side or the motor side,
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DC to AC that means a DC source battery or
a rectifier output that means what is available
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to us is DC source but we want AC. One example
is UPS that is basically invertors. These
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types of convertors are called, power convertors
are called invertors, these are called invertors.
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We will be studying in the due course, the
basic control of the invertors, various invertors
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topologies for speed control applications,
we will be concentrating more on that one
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later. Then another one AC to AC. That means
voltage, current or frequency of one into
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another voltage and current and frequency
of different level. These are basically in
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old books; we can see one class of convertors
is called cycloconvertors, one example. But
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we will not be studying about this one now.
We will be mainly concentrating the AC to
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DC and the DC to AC convertor in our main
course.
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Now, after going through the classification
and block, we set our control of the whole
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processing through the controller. That means
we have to design the controller such that
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the system will quickly adapt to the change
without much translates. So, controller design
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is very important. Nowadays, simulation tools
are available, so we can have a mathematical
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study and we can simulate and we can tool
the controller before a hardware implementation
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is started.
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Why? That will say lot of time and money;
otherwise we have to learn from the failures.
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The hardware if keep on failure fail and leave
be if we learn from that process, the cost
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of the whole system will go up. So, the controller
design and the controller tuning, before I
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control the implementation; a controller design
and tuning can be done through a simulation
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study.
For a simulation study, the whole system should
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be modeled. That means equivalent mathematical
representation of the system is required.
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That means the representation for the power
source, power source basically we can AC or
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DC, we can do it; then the power converter,
then the motor. So, a mathematical model of
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the motor required based on, for our controller
design.
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As I told you before, the whole process is
converting electrical energy into mechanical
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energy, mechanical energy through our; what
is the output of the load? The mechanical
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energy, the torque and the speed, from there
we are generating the energy. So for motor,
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a basic input output relation, a torque speed
characteristic is required torque speed characteristic
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required. That means various speeds, what
are the torques required? Then for the torque;
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what are the parameters we have to control
at the input side of the motor? That is the
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voltage and current. For that voltage and
control, corresponding controller has to be
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initiated for the power converter.
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Now, the basic load that is now the mechanical
load; we can basically divide the torque speed
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characteristic into four category. Let us
take the torque speed characteristic of the
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load.
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One; torque independent of speed. Let us write
the graph, power speed characteristic of this
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load. The graph, this is our speed that is
speed in electrical radiance per second and
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this is the load torque. So, torque is independent
of speed. So, the curve will be, the speed
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torque characteristic will be like this; steady.
That means irrespective of the speed, it will
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always generate constant load. So, what are
the examples for this one?
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Example; low speed hoisting, that is cranes
during hoisting during hoisting that is one,
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then machine tools for feed mechanism; these
are some of the examples. Many examples you
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can find in many text books and these are
some examples. Then two, linear rising characteristics;
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what you mean by linear rising torque speed
characteristic? What do you mean by linear
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rising torque speed characteristics?
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That means whenever the speed changes, torque
will also change proportionally, this is one
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this is our second example. The torque speed
characteristic for a linear rising low torque
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is like this. What is the example? Let us
take a separately excited DC generated connected
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to a load of constant ohmic resistance that
is example. Separately excited DC generator
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connected to a load of constant ohmic resistance
that means a separately excited generator;
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the generator that means machine is working
as a generator and the generator is feeding
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a load of constant ohmic resistance. Then
let us see, the torque speed characteristic
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is linearly rising.
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Now, we know torque is equal to P by omega;
omega is radiance per second. The torque generated,
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the power developed by the generator divided
by the speed give the torque. Now, what is
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the power generator by the separately excited
generator? P is equal to that is the temporary
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voltage into current I. Also, we know that
this V, the voltage generated by the excited
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separately excited DC machine DC proportional
to speed that is proportionality constant
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K into omega. So, V is proportional to k into
omega.
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Why? We want to bring the basic relation between
the torque and the speed. V is equal to K
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omega and what is the current given or taken
by the load from the generator? I is equal
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to V by R, V is equal to proportional to speed
K omega divide by the, this is our constant
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ohmic resistance R, K omega by R. So, we got
I. Now, what is the power? In this, this is
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the basic equation. In the basic equation,
we require the power. Power is equal to P
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is equal to V into I, V into I is equal to
that we can substitute V and I; this will
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be equal to when you substitute K square omega
square divided by R. Now, let us substitute
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this one to this equation. Then what will
happen? Torque is equal to K square by this
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is a constant for a machine, this load is
also constant; so the whole thing, we can
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represent by a constant K1 another constant
K1
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So, this will be equal to K1 omega square,
PV we have substituted here, these part we
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have substituted here divided by omega. Now,
this will be equal to omega square goes, this
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will be equal to, torque will be proportional
to omega. So, as the speed increases, torque
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also will increase. Power will increase, torque
will also increase. So, this is the equation,
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linearly rising characteristic.
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Some simple examples, I am telling here. Now,
let us take here nonlinear rising characteristic.
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Let us go to the next page. Nonlinear rising
characteristic; now the third example is nonlinear
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rising that is parabolic characteristic.
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Nonlinear rising torque speed characteristic;
torque speed, torque versus speed
characteristic that is parabolic. How it will
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look, nonlinear rising? The speed let us again
draw the axis, this is our omega speed, this
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is our torque. So, it will be something like
this, third characteristic. Previously, the
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constant load speed characteristic that was
one, linear rising was like this, this is
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the third one. Third one is this one.
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Most of the final loads belong to this one
that means torque will be proportional to
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omega square. Most of the loads will be most
of the load final loads will come under this.
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You should see; constant, linear rising, then
nonlinear rising, then the next one will be
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for nonlinear falling characteristics, nonlinear
falling torque speed characteristic.
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Are we to look. So, nonlinear; here let us
take that is torque is
proportional to 1 by omega. This is nonlinear
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rising, nonlinear falling characteristic that
is something like this; this is nonlinear
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falling characteristics, torque is proportional
to omega. That means or we can again say T
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into omega is a constant.
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Let us take, one example, rolling mill. In
a rolling mill, the paper comes out from the
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mill, the constant speed and it will be a
thin roll to a drum. So, paper is the radius
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r. As the paper gets rolled on to the drum,
the radius will keep on increase. Now, the
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paper is coming from the mill with the constant
velocity and we should take care of the speed
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of this drum has to be adjusted that we have
to keep constant tension, we have to keep
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a constant tension such that the paper should
get rolled on to the drum properly. So, what
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is the torque speed characteristic it shows?
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Now, if you see here, the speed of rotation,
what is the speed of rotation? Speed of rotation
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is distanced by the speed B. So, speed of
rotation
speed of rotation is equal to 2 pi r divided
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by V, this is the speed of rotation. This
is second, period in seconds. Now, what we
200
00:30:26,490 --> 00:30:31,630
are attempting with this equation? We have
to get a torque speed that is torque and the
201
00:30:31,630 --> 00:30:35,600
speed that is the angular speed omega; we
have to keep a relation.
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00:30:35,600 --> 00:30:42,600
Now, what is frequency here? f is equal to
that is the frequency is equal to 1 by T,
203
00:30:47,809 --> 00:30:54,809
1 by T is equal to V by 2 pi r; frequency
we got. Once frequency we got, we know omega.
204
00:31:00,320 --> 00:31:07,320
Omega is equal to 2 pi f. So, from the speed
V, this V, we can get omega -- 2 pi f. Now,
205
00:31:14,600 --> 00:31:21,020
what is the power exerted by the torque by
the drive?
206
00:31:21,020 --> 00:31:26,120
Power exerted by the drive, see we have to
give a constant torque opposing force here.
207
00:31:26,120 --> 00:31:33,120
This is force f. So, what is the power? Power
is equal to torque is equal to the power exerted
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00:31:38,450 --> 00:31:45,450
by the drive p is will be equal to that is
F into r; r is the radial distance, f is the
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00:31:48,940 --> 00:31:55,940
tangential force. The torque, this will give
the torque, this is the torque f into r, torque
210
00:31:59,250 --> 00:32:06,250
into, what is omega? Omega is equal to 2 pi
f, 2 pi f is equal to v by r, from here v
211
00:32:12,090 --> 00:32:13,710
by r.
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00:32:13,710 --> 00:32:20,710
So, these two we will cancel, F and v; these
are constant for this one, so power is constant.
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00:32:24,230 --> 00:32:31,230
That means torque into omega that is p is
equal to p is constant, torque into power
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00:32:33,980 --> 00:32:39,550
is equal to that is p constant that is this
will give constant wholes power load. These
215
00:32:39,550 --> 00:32:46,240
are some of the some of the basic torque speed
characteristic.
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00:32:46,240 --> 00:32:53,240
Now, once we got the torque speed characteristics,
let us take the combined system. This is torque
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00:32:57,350 --> 00:33:01,100
speed characteristic of the complete load.
Now, in a combined system, there is a driving
218
00:33:01,100 --> 00:33:08,100
torque and the load torque. For speed control
this should be matching. So, how that driving
219
00:33:09,679 --> 00:33:16,679
torque has to be controlled the meet the load
torque? That is our basic control purpose.
220
00:33:16,840 --> 00:33:23,840
So, let us take the combined system. Now,
let us go to the next page again. There is
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00:33:30,850 --> 00:33:37,850
a combined load torque characteristics, combined
torque speed. That is torque speed
torque speed characteristic.
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00:34:06,490 --> 00:34:13,490
Let us try, draw the axis; this is omega,
this is T. Now, the load torque that is not
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00:34:17,489 --> 00:34:24,249
in your control, load torque will keep on
varying and for a stable operating speed or
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00:34:24,249 --> 00:34:30,789
speed in speed control system, the drive torque
should match the load torque. So, let us take
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00:34:30,789 --> 00:34:35,239
a load torque a typical load or something
like this that I have put into under negative
226
00:34:35,239 --> 00:34:41,629
axis because this is the opposing force, opposing
torque, this I will put load torque.
227
00:34:41,629 --> 00:34:48,629
Now, let us take our motor, the speed torque
carries of our motor; let it be something
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00:34:49,379 --> 00:34:56,379
like this. That is our motor driving torque
for the motor. Now, in our speed control system,
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00:35:01,910 --> 00:35:08,910
suppose we have to operate, the system will
come to a stable state where the load torques
230
00:35:10,259 --> 00:35:17,259
approximately this point, load torque will
be equal to the driving torque. But the load
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00:35:20,410 --> 00:35:27,410
torque can vary that is not in our control.
Load torque can vary means suppose the speed
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00:35:27,569 --> 00:35:31,170
varies, load torque goes here. What will happen?
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00:35:31,170 --> 00:35:37,900
Load torque goes here means speed has changed.
Now, this is the load torque demand by the
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00:35:37,900 --> 00:35:44,900
load. But according the motor characteristics,
the driving torque that is Td is less than
235
00:35:48,039 --> 00:35:54,960
the load torque now. So, what will happen?
System will slow down and come to this point.
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00:35:54,960 --> 00:36:01,960
Now, let the load torque decrease. In this
case, what will happen at this point?
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00:36:05,549 --> 00:36:10,650
The load torque is less than the driving torque;
so this will accelerate it and it will come
238
00:36:10,650 --> 00:36:16,180
to the stable point. But in most of the cases,
this is for a typical load speed characteristics;
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00:36:16,180 --> 00:36:23,180
in most of the cases, it will not come to
the stable system so that depending on the
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00:36:24,839 --> 00:36:29,470
load torque requirement; our motor characteristic
torque speed characteristic, we have to control
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00:36:29,470 --> 00:36:32,809
we have to modify through our power converter.
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00:36:32,809 --> 00:36:39,809
So, our whole control block, the purpose of
the control block is to dynamically control
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00:36:40,690 --> 00:36:47,190
the torque speed characteristic so that the
driving torque will be always equal to the
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00:36:47,190 --> 00:36:53,960
load torque as demand by the load for a particular
speed. So, such cases, we require a close
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00:36:53,960 --> 00:37:00,960
loop control system. Now, as I told you before,
see I have so far I have told you TL - load
246
00:37:04,089 --> 00:37:11,089
torque. So, load torque basically, we can
define as or the driving torque Pd is equal
247
00:37:15,089 --> 00:37:22,089
to the load torque Td sorry this is Td, this
is TL plus if the speed is constant, driving
248
00:37:25,400 --> 00:37:27,950
torque will be equal to load torque.
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00:37:27,950 --> 00:37:34,109
Suppose, during acceleration and deceleration;
there is more load, there is a torque change
250
00:37:34,109 --> 00:37:41,109
requirement or torque demand from the load.
So, that can be explained as J into d omega
251
00:37:44,339 --> 00:37:51,339
by dt. J into d omega by dt, this approximately
J is the moment of inertia. So, whenever acceleration
252
00:37:57,509 --> 00:38:02,180
deceleration is there, the total driving torque
should be equal to the constant load torque
253
00:38:02,180 --> 00:38:04,380
plus the acceleration and deceleration.
254
00:38:04,380 --> 00:38:11,380
Now, how to find out for a particular load
system, how to find out the J? There are various
255
00:38:12,509 --> 00:38:19,509
techniques available in literature. I will
take one from the famous Leonard's book.
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00:38:20,839 --> 00:38:27,839
What we have to do? The complete system under
no load, you system machine to the full speed,
257
00:38:44,150 --> 00:38:51,150
drive the system to the full speed, then full
speed and find out the sorry this is omega
258
00:39:01,559 --> 00:39:08,559
0, this is ML, this is our type two axis.
So, drive the system for various speeds for
259
00:39:11,880 --> 00:39:18,880
operation. 5, 6 points you take it and tie
the system and find out the load torque required.
260
00:39:20,769 --> 00:39:27,460
So, you plot the load torque; you got the
ML.
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00:39:27,460 --> 00:39:34,349
Now, run the system to the full speed and
switch off the power supply. Then what will
262
00:39:34,349 --> 00:39:41,349
happen? The whole system will decelerate.
So, during the deceleration, what will happen?
263
00:39:42,390 --> 00:39:49,390
The driving torque is 0, so the decelerating
torque generated by d omega by dt will be
264
00:39:53,519 --> 00:40:00,519
will be opposed that will be opposed by the
friction and windage loss. So, slowly full
265
00:40:00,789 --> 00:40:07,460
system will decelerate. So, at various point,
measure omega; you note the time and measure
266
00:40:07,460 --> 00:40:14,460
the omega. At various point, measure the time
and omega and approximately at this point
267
00:40:19,819 --> 00:40:22,809
and time, you can find out d omega by dt.
268
00:40:22,809 --> 00:40:29,809
So, d omega by dt we know, the system is under
no load at various speeds. So, we know at
269
00:40:32,910 --> 00:40:39,910
various speed, what is the driving torque.
So, this driving torque at the no load torque,
270
00:40:43,509 --> 00:40:50,509
so if you write the equation J into d omega
by dt plus ML is equal to 0 because we are
271
00:40:55,210 --> 00:40:58,229
switched on the system and the system is decelerating.
272
00:40:58,229 --> 00:41:05,229
So, at various speed, J into d omega by decelerating
torque is the no load torque of the system
273
00:41:07,289 --> 00:41:13,979
that we can get from this one. So, from this
equation; d omega by dt we know, ML we know,
274
00:41:13,979 --> 00:41:20,979
we can approximately we can find out J. This
is the one way of doing the thing. There are
275
00:41:21,729 --> 00:41:26,210
other techniques also available; this is the
one simple way of finding out J. So, once
276
00:41:26,210 --> 00:41:31,700
you know the J, once the speed; then what
we know? We now, approximately we can find
277
00:41:31,700 --> 00:41:35,930
out the torque speed characteristic of the
system during acceleration also.
278
00:41:35,930 --> 00:41:42,930
Now, in this class, in this lecture, we have
just basically; the basic definition of the
279
00:41:44,940 --> 00:41:51,940
electric drive is introduced and basic block
diagram is given. Then depending on the voltage
280
00:41:52,229 --> 00:41:57,509
source, whether AC DC and the output requirement,
we have classified the power converter where
281
00:41:57,509 --> 00:42:02,630
conversion process into various four category.
Then the torque speed category of the machine,
282
00:42:02,630 --> 00:42:08,279
we have various categories -- rising, constant,
nonlinear rising, nonlinear falling, we have
283
00:42:08,279 --> 00:42:10,170
defined with typical example.
284
00:42:10,170 --> 00:42:16,839
Then we talked about the combined characteristics.
Combined characteristics, what we found out?
285
00:42:16,839 --> 00:42:23,839
It is required to modify the motor, low torque
characteristics through our power converter
286
00:42:25,380 --> 00:42:31,039
such that depending on the load variation
that is the TL, always we have to keep the
287
00:42:31,039 --> 00:42:38,039
driving torque equal to the load torque. So,
there is a close loop control system is required.
288
00:42:39,619 --> 00:42:46,619
The driving torque, the load driving torque
is not only the steady; note in the acceleration
289
00:42:46,989 --> 00:42:52,549
and deceleration also we have to give the
transient power also to the system so that
290
00:42:52,549 --> 00:42:56,160
the transient power acceleration and deceleration
depend on the moment of inertia of the system
291
00:42:56,160 --> 00:43:03,099
and typically, how to find out the moment
of inertia, a typical system is also we explained.
292
00:43:03,099 --> 00:43:10,009
Now, let us go to the some of the basic power
converter configuration so because we require
293
00:43:10,009 --> 00:43:15,329
the power converter. Once you select the power
converter, then only we can choose the control
294
00:43:15,329 --> 00:43:21,920
function. Then once we have studied some of
the power converters; first we will study
295
00:43:21,920 --> 00:43:27,589
the power converters required for DC modern
drive applications, then we will go for the
296
00:43:27,589 --> 00:43:33,519
inverters that the power converters required
for AC motor drive application, then how you
297
00:43:33,519 --> 00:43:40,519
design the closed control for DC motor and
AC motor will be discussed during this course.
298
00:43:42,319 --> 00:43:47,329
For the DC motor, we will be mostly talking
about the separately excited DC machine because
299
00:43:47,329 --> 00:43:53,249
we will not have much time for all other cases;
separately excited DC motor that is widely
300
00:43:53,249 --> 00:43:59,739
used for motor drive applications. Then the
AC motor drive lectures, we will be talking
301
00:43:59,739 --> 00:44:06,039
about induction motor drive, mainly induction
motor drive applications. During steady state,
302
00:44:06,039 --> 00:44:13,039
dynamic equivalence circuit model we will
study, then the symbol v by f control, then
303
00:44:13,390 --> 00:44:18,039
the high dynamic and the controller like field
oriented controller for high dynamic performer
304
00:44:18,039 --> 00:44:20,130
application we will be studied later.
305
00:44:20,130 --> 00:44:27,130
Now, let us go to the basic power converter
configuration. Let us go to the next page.
306
00:44:32,130 --> 00:44:39,130
Here, what we will be talking about is the
power converter that is AC to DC converter
307
00:44:45,779 --> 00:44:52,779
for DC drive applications. So, control rectification
is a must for this one. Let us study some
308
00:44:59,559 --> 00:45:06,559
of the basic power converter configuration.
So, single phase, let us take a typical converter,
309
00:45:11,309 --> 00:45:17,420
single phase, before coming to the three phase
control rectification; let us talk about single
310
00:45:17,420 --> 00:45:24,420
phase AC to DC power converter.
311
00:45:39,239 --> 00:45:45,920
The most basic single phase power converter
is, the schematic is like this, let us represent
312
00:45:45,920 --> 00:45:52,920
our mains like this; this is our controlling
device, this is the thyristor, then you have
313
00:45:57,779 --> 00:46:04,229
the load. Load, I will simply represent it
like this, it is inductive with back EMF,
314
00:46:04,229 --> 00:46:10,920
it can be there. So, we will have inductive
plus back EMF alone that is the motor can
315
00:46:10,920 --> 00:46:15,359
be represented like this. Then, it will go
like this.
316
00:46:15,359 --> 00:46:22,359
Now, how do you control the output DC that
is DC this part? So, at the appropriate point
317
00:46:27,400 --> 00:46:32,460
of the mains, we will be turning on the device
and we will transfer the voltage to the load.
318
00:46:32,460 --> 00:46:38,450
So, by controlling the firing angle of the
thyristor, we can control the output voltage.
319
00:46:38,450 --> 00:46:42,650
How it can be controlled? Let us see.
320
00:46:42,650 --> 00:46:49,650
This is our V0. What is the nature of V0 for
a single phase with a single thyristor controller?
321
00:46:54,940 --> 00:47:01,940
How it looks? This is our mains. We will not
be whenever, for our present analysis, we
322
00:47:09,249 --> 00:47:16,049
will assume the control of the devices our
instantaneous; we will not worry about the
323
00:47:16,049 --> 00:47:20,670
switching types involved in this one. So,
the moment you give the gate pulse, thyristor
324
00:47:20,670 --> 00:47:26,930
will be turned on and when the AC mains is
getting when the thyristor when across the
325
00:47:26,930 --> 00:47:31,839
thyristor reverse voltage is applied, the
thyristor will switch off immediately.
326
00:47:31,839 --> 00:47:38,839
So, let us say at this point, we are giving
the firing angle alpha. Firing angle we represent
327
00:47:39,440 --> 00:47:45,249
as alpha, alpha with respect to this point,
the starting point of our mains, here, zero
328
00:47:45,249 --> 00:47:51,069
crossing. Now, if the load is resistive that
is the simple and the easy load if the load
329
00:47:51,069 --> 00:47:58,069
is resistive; what will happen? During this
portion during this portion, the voltage will
330
00:47:59,519 --> 00:48:06,519
come across the load. During this portion,
the voltage will come across the load and
331
00:48:07,099 --> 00:48:12,680
if it is purely resistive load, then the current
will also be proportional instantaneously
332
00:48:12,680 --> 00:48:19,489
proportional to the voltage before but pure
resistive load is only a text book answer;
333
00:48:19,489 --> 00:48:25,190
we will not have a pure resistive load, you
will have resistive as well as inductive.
334
00:48:25,190 --> 00:48:32,190
So pear, you will know in this class will
be talking about mostly on inductive loads
335
00:48:32,430 --> 00:48:39,430
and with back EMF load also so. Now, with
inductive load, what will happen? So, before
336
00:48:40,200 --> 00:48:45,109
coming to the inductive load, let us say how
for the resistive load; how the current, it
337
00:48:45,109 --> 00:48:52,109
will be? Current will be in this form, exactly
this will be.
338
00:48:53,630 --> 00:49:00,630
This is our Vin; during this portion, this
any value of the current I0 will be equal
339
00:49:02,809 --> 00:49:09,809
to Vin divided by R. But during the half cycle,
the moment the voltage go negative; what will
340
00:49:14,219 --> 00:49:19,450
happen? There is a reverse voltage across
the thyristor, anode and cathode. The thyristor
341
00:49:19,450 --> 00:49:21,130
will be immediately switched off.
342
00:49:21,130 --> 00:49:28,049
So, during this portion that output voltage
is, there is instantaneous voltage across
343
00:49:28,049 --> 00:49:34,869
the load is zero for a resistive load. Then
next half cycle, again this will appear. This
344
00:49:34,869 --> 00:49:41,869
is the current, this is the voltage. So, we
will see a pulsed wave form will be coming
345
00:49:44,769 --> 00:49:51,729
across the load. This pulsed waveform, it
will have the DC value and the AC value also
346
00:49:51,729 --> 00:49:54,329
will be there, high frequency AC value will
be there.
347
00:49:54,329 --> 00:50:01,329
So, we will not talk about but we are worried
more about the DC value. So, what is the DC
348
00:50:02,210 --> 00:50:09,130
value? That is V0, average value of this pulsed
wave form that can be easily found out. V0
349
00:50:09,130 --> 00:50:16,130
alpha will be 1 by 2 phi that average value
integrating alpha 2 pi. So, this is our pi.
350
00:50:23,940 --> 00:50:29,269
So, alpha to pi we will integrate and divide
by the P rate, you will get the average value.
351
00:50:29,269 --> 00:50:36,269
So, this will be equal to Em sin omega t divided
by dm, this we are integrating.
352
00:50:39,910 --> 00:50:46,910
So, Em sin omega t if you integrate, what
will happen? You will get minus cos omega
353
00:50:49,039 --> 00:50:56,039
t integrating limit is alpha to pi. So, when
you substitute this one, finally the V0 alpha
354
00:50:56,519 --> 00:51:03,519
will be alpha is equal to Em by 2 pi into
1 plus cos alpha. Here, alpha can be varied
355
00:51:13,049 --> 00:51:20,049
from 0 to pi. When alpha is equal to pi, what
will happen? Cos pi that is minus 1, output
356
00:51:20,339 --> 00:51:23,479
voltage will be 0, the DC value will be 0.
357
00:51:23,479 --> 00:51:30,479
When alpha is equal to 0, alpha is equal to
0; what will be the output? Alpha is equal
358
00:51:33,719 --> 00:51:40,719
to 0, the output will be alpha is equal to
0, V0 is equal to Em by pi, this is the output
359
00:51:50,420 --> 00:51:57,420
and alpha is equal to 0 means we are firing
the thyristor at the zero crossing, the positive
360
00:51:59,079 --> 00:52:06,079
zero crossing. What it says? It says, this
can be replaced by a diode. So, output DC
361
00:52:09,210 --> 00:52:16,210
voltage for a single phase half range rectifier
with a single diode is equal to Em by pi.
362
00:52:21,380 --> 00:52:28,140
But if you see here, during the negative part,
we are not using, we are not drawing any power
363
00:52:28,140 --> 00:52:32,359
from the mains. That means even though power
source is there, we are not fully utilizing
364
00:52:32,359 --> 00:52:33,349
the power source.
365
00:52:33,349 --> 00:52:38,059
So, even though in the text book, these type
of AC to DC converter using diode or thyristor
366
00:52:38,059 --> 00:52:44,499
are available. This is only for the understanding
that one, how to get the DC value. So, this
367
00:52:44,499 --> 00:52:49,549
value what we talked about, this is the DC
value but along with this value because of
368
00:52:49,549 --> 00:52:55,130
this variation, we have the repel content
also will be there. So, that repel content,
369
00:52:55,130 --> 00:52:59,700
the repel voltage will produce its own repel
current. That has to be suppressed by the
370
00:52:59,700 --> 00:53:03,509
load. But we will not be putting any extra
filters to suppress this one.
371
00:53:03,509 --> 00:53:10,509
To put filter means it is a cost effective
proposition. So, inductive load; as the frequency
372
00:53:11,989 --> 00:53:16,229
high frequency of the voltage waveform, there
the impedance of the load that is l omega
373
00:53:16,229 --> 00:53:21,130
will increase and the voltage divided by l
omega, the current will the harmonic current
374
00:53:21,130 --> 00:53:26,839
can be suppressed. So, harmonic current, the
amplitude harmonic currents can be suppress
375
00:53:26,839 --> 00:53:32,680
only by through the motor inductance. But
what we are interested is the DC part.
376
00:53:32,680 --> 00:53:39,680
So, this value gives the DC value. So, this
will be another it will be used with, never
377
00:53:40,660 --> 00:53:46,130
use for any applications. But for understanding,
already it is available in text book. So,
378
00:53:46,130 --> 00:53:51,349
whole analysis, what we have done is for a
resistive load. So, this is for resistive
379
00:53:51,349 --> 00:53:57,369
load with the firing angle control. Now, let
us see, instead of resistive load what if
380
00:53:57,369 --> 00:54:01,059
we put an inductive load? Let us go to the
next page.
381
00:54:01,059 --> 00:54:08,059
In the case of inductive loads, again let
us draw the block schematic of the AC to DC
382
00:54:08,289 --> 00:54:15,289
converter with the firing angle alpha, alpha
can be varied from 0 to pi. Now, we have the
383
00:54:16,259 --> 00:54:23,259
highly inductive load. If we use here; what
are the type of repel voltage appearing across
384
00:54:26,710 --> 00:54:33,710
this one - V0 (t)? That is the repel, this
BDC plus the repel. Here, this is our Vm.
385
00:54:35,420 --> 00:54:42,420
If you see here, this is our voltage wave
form, alpha is fired here, firing angle is
386
00:54:51,059 --> 00:54:56,609
fired from here; now what happens? Because
of the inductance previously when the resistive
387
00:54:56,609 --> 00:55:02,559
load was there; the wave form, this repel,
the current wave form across the resistance
388
00:55:02,559 --> 00:55:08,219
was the nature of the current variation was
exactly same like the voltage variation.
389
00:55:08,219 --> 00:55:15,219
But here, in this case, what will happen;
the current wave form, current wave form will
390
00:55:15,400 --> 00:55:22,400
be starting at this point because currently
the inductance cannot change instantaneously.
391
00:55:24,489 --> 00:55:31,489
So, what will happen? The current will vary
like this. At this point, when the voltage
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00:55:37,140 --> 00:55:42,819
is equal to 0, there is a current in the inductance.
So, thyristor cannot switch on, current through
393
00:55:42,819 --> 00:55:45,779
the inductance cannot be changed instantaneously.
394
00:55:45,779 --> 00:55:52,779
So, what will happen? Voltage will go negative.
Voltage will go negative, then thyristor is
395
00:55:53,920 --> 00:56:00,329
conducting; so all voltage has to be suppressed
across the inductive load. What will happen?
396
00:56:00,329 --> 00:56:04,950
At this point, inductance, the rate of change
of current, it will go negative so that the
397
00:56:04,950 --> 00:56:10,019
voltage become instead of positive negative
here, it will become positive or negative.
398
00:56:10,019 --> 00:56:12,959
Here it have the input voltage has changed.
399
00:56:12,959 --> 00:56:17,930
So, this rate of change of current will, the
load will take care of such that the rate
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00:56:17,930 --> 00:56:24,369
of change of current is just sufficient to
forward bias the thyristor here and current
401
00:56:24,369 --> 00:56:29,160
will keep on continuing and this current will
be from the stored energy. Till the stored
402
00:56:29,160 --> 00:56:34,529
energy is released, current will keep on coming.
So, what is the disadvantage here? During
403
00:56:34,529 --> 00:56:39,200
the negative portion upto this point also,
the voltage was appearing to the load.
404
00:56:39,200 --> 00:56:46,200
Now, compared to the previous equation; firing
angle and DC is we cannot apply here because
405
00:56:47,630 --> 00:56:52,239
the negative portion also has come here. So,
the previous equation of alpha with respect
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00:56:52,239 --> 00:56:59,239
to the output DC with respect to firing angle
is no more valid here because the thyristor
407
00:57:00,519 --> 00:57:03,420
is conducting in the negative direction also.
408
00:57:03,420 --> 00:57:10,420
Then how to avoid this negative portion? At
the same time, we have to give a free varying
409
00:57:11,719 --> 00:57:15,410
path for the inductance. This is during this,
it is called sorry during the current conduction,
410
00:57:15,410 --> 00:57:19,789
the negative portion also we should give an
alternative path for the inductance. At the
411
00:57:19,789 --> 00:57:25,269
same, this negative voltage should not appear
across the load. So, than the previous equation
412
00:57:25,269 --> 00:57:32,269
that is V0 alpha is equal to Em by 2 pi into
cos ... one plus cos alpha is valid. How to
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00:57:36,150 --> 00:57:38,849
come to that work; we will study in the next
class.