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We will start today, power electronic circuits
and equipments. The, depending upon functions,
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the entire power electronic circuits or equipments
can be classified into 5 groups. The first
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1, they are known as the converters. The input
voltage is AC, output voltage is DC. When
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I am saying output voltage is DC, it means
that average value of the output voltage is
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finite, not a necessarily a constant value.
The second 1 is inverters: wherein, the input
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is DC, output is AC. What comes to your mind
when I am saying output is AC? Is it a pure
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sinusoid? No, when I am saying when the output
is AC, it means that average value of the
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output voltage is 0, not necessarily a sin
wave.
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Third one: input is DC, output is also DC.
Output voltage can be varied, it could be
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either choppers or switch mode power supplies
or DC to DC converters or DC transformers.
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The fifth one, the fourth one: input is AC,
output is also AC but the frequency of the
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output voltage is same as the input, whereas,
the magnitude is changing. These are known
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as phase controllers. They are often used
in fan regulators, you might have seen. A
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small regulator, just by changing the knob
you are varying the speed. They are basically
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phase controllers.
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The last 1 input is AC, output is also AC.
Both, voltage and frequency are variable.
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They are cycloconverters. A new group of power
electronic equipments, they are also known
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as matrix convertors. So, last 1 is cycloconverters
and matrix converters. We will study each
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1 of them in detail.
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First 1, converters: what are the basic assumptions
that are we making? All the devices and the
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circuit components are ideal. In other words,
when the device is conducting, voltage across
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the device is 0 and when the device is open,
then the current flowing through the device
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is 0 and when I am saying so, passive elements,
L is a pure inductor, C means a pure capacitor
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and input is a pure sin wave. The first one
is diode circuits: a low power, a 3 ampere,
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600 volt diode. 2 element device: anode, cathode.
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A 40 ampere, 600 volt device, what exactly
they mean? I will tell you later. Anode, cathode:
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these are, directly this device cannot be
used in a circuit, this has to be mounted
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on a heat sink. This is a heat sink. It has
to be mounted on a heat sink.
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So cathode, anode: body is anode, this is
a cathode. It is a 2 element device. These
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are known as the uncontrolled device: hence
the name, uncontrolled rectification. Let
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us see, what are the various parameters?
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Alpha is the angle at which diode starts conducting:
beta, angle at which diode stops conducting.
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So, what is the conduction angle? It is beta
minus alpha. All these angles, alpha and beta
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are measured with respect to the positive
0 crossing.
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Half wave rectifier, there is only 1 diode
connected in between the source and the load.
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We will consider first, load to be a purely
resistance. Input is an ideal sin, here, a
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pure sin here. Device is ideal, so diode starts
conducting at the positive 0 crossing. When
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the diode starts conducting, output voltage
V0 is same as the input voltage. Load is purely
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resistive: we know that current and voltage
are in phase. KVL says, VI is equal to I into
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R or Vm sin omega t is equal to I into R.
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So current is also sinusoidal, reaches peak
when omega t is equal to pi by 2. At omega
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t is equal to pi, voltage becomes 0, current
also become 0, diode turns off. So, alpha
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is 0, beta is equal to pi radians, duration
for which diode conducts in each cycle is
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pi radians. What happens in the negative half?
There is no current flowing in the circuit.
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Current became 0 at pi, anode potential becomes
negative, diode cannot conduct.
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So in the entire half, diode is in blocking
mode. So, input voltage comes across the diode.
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So, the entire voltage, the diode has to block
or the voltage rating of the diode should
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be a minimum, should be Vm. So, if an input
is 230 volts, the peak up, the input will
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be 230 into root 2. That is around 310 volts
or so. So, when I have said 40 ampere, 600
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volt diode, it means that it can carry 40
amperes and it can block 600 volts.
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In other words, peak reverse voltage that
can be applied theoretically is of the order
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of 600 volts.
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See, these are the various equations, simple
equations that I have written. What is the
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average value of the output voltage? I said,
AC to DC converter, output voltage has a finite
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DC value. The average value is given by Vm
by pi. Output voltages are not a constant
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DC, it has, it is varying from, output voltage
are varying from 0 to Vm and to 0, in 0 to
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pi radians and becomes continues to remain
0 till pi to 2 pi.
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So average value, if I integrate it over the
cycle, it comes to be Vm by pi. It also has
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other AC component. If I write a Fourier series
of this, is a Vm function, so has an average
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component and other V1 frequency components.
So, what is the RMS value?
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RMS value is Vm by 2. I said, output voltage
is not a constant DC, it has an AC component.
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The measure of AC components in the output
voltage, they are known as the ripple voltage,
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it is given by this expression. What is the
ripple factor? Ripple factor is measure of
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the AC components in the output voltage to
the DC value. It is as high as 1.21.
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We have discussed about the resistive load.
Now, we will replace the resistive load by
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a pure inductor. Input is a sinusoid, again
diode starts conducting at the positive 0
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crossing. When the diode starts conducting,
the circuit equation is Vi is equal to V0
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that is equal to L di by dt. So, as the input
voltages increases, current also will increase.
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Will it reach peak at omega t is equal to
pi by 2, as in the case of resistive load?
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No, it will reach peak when the input voltage
is 0 or it will reach its maximum when omega
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t is equal to pi.
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Why am I saying? See, 1 way to analyze the
circuit is write a differential equation,
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solve it or second one is a, use a graphical
approach. I will use a graphical approach.
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I said Vi is equal to L di by dt, instantaneous
value of the input voltage becomes 0 at omega
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t is equal to pi, so, L di by dt also should
be pi, sorry L di by dt is equal to 0 at omega
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t is equal to pi. So, L di by dt is equal
to 0, it can happen only when di by dt is
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0. So, either it could be positive or positive
maximum or a negative maximum.
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But then, we know that it started, current
started increasing from 0. So, it will reach
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a peak. Current is in the, current can flow
in the only in the positive direction, so
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therefore, current reaches peak at omega t
is equal to pi. Because at this instant, voltage
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across inductor is also 0, input voltage is
also 0. Vi is equal to L di by dt, that is
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equal to 0. So, from 0 to pi, di by dt is
positive, so therefore, L di by dt is also
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positive. So, area under the curve from 0
to pi is the positive L di by dt.
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We know that voltage, average value of the
voltage across the inductor is 0. Current
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reached peak at omega t is equal to pi. From
omega t is equal to pi plus, instantaneous
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value of the input voltage becomes negative.
So, L di by dt also should be negative. So,
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current starts decreasing. Where will it become
0? 0 to pi, it was increasing, reached peak
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at omega t is equal to pi, so it will reach
0 at omega t is equal to 2 pi.
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At that instant, Vi is equal to L di by dt
is equal to 0. So, di by dt should be 0. Positive
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L di by dt should be equal to negative L di
by dt. So, diode conducted for the entire
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2 pi radians. Beta is 2 pi, therefore gamma
is also 2 pi. Average value of the output
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voltage is 0. Diode conducted for 2 pi radians,
but the average value of the output voltage
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is 0.
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I have a question. How that diode started
conducting beyond pi? because, input voltage
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becomes negative. A point to be noted that
it is not the anode potential alone will decide
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the conducting state of that diode, it is
the cathode potential also. If the cathode
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potential is less than the anode potential,
diode has to conduct and diode will conduct.
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So, what happens beyond pi? Input voltage
may be becoming negative, but cathode potential
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which is also equal to L di by dt, current
is also negative because di by dt is negative.
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So, this polarity as this terminal has become
negative and this point has become positive.
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So therefore, diode continuous to conduct
till the current becomes 0.
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So remember, it is not the anode potential
alone that will decide the conducting state
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of the diode, it is the cathode potential
also. Here, because of the inductor, L di
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by dt becomes negative, L di by dt is negative
because, di by dt is negative. So negative,
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positive: this is also negative may be, but
this point is a bit more negative than this.
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A positive voltage or positive voltage of
1 volt also may appear across the diode as
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it keeps conducting.
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See, these are the various equations, i is
equal to omega Vm, we have one minus cos omega
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t, V average is 0, power delivered to the
load V average into I average is 0. Remember,
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diode conducted for 2 pi radians, average
power is 0. In the previous case, diode conducted
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for pi radians, average power is finite, V
average into I average. V average in the previous
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case is Vm by pi.
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Third case, now we will add, we discussed
R, we discussed L, now we will see RL. Again,
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diode starts conducting at the positive 0
crossing. When diode starts conducting, output
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voltage is equal to the input voltage and
the circuit equation is R into I plus L di
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by dt is equal to Vi that is equal to Vm sin
omega t. First order differential equation
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that has to be solved.
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What is the general equation? Sorry, What
is the general solution? Has a sinusoidal
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component and an exponential term. The ... is
given by this term, pi is sin inverse omega
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L by r. What are the initial conditions? When
omega t is equal to 0, i is equal to 0. You
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solve it, this is the general, this is the
solution for the current.
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Load is purely resistive. It reach, current
reach maximum at omega t is equal to pi by
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2. Load is purely inductive, current reach
maximum at omega t is equal to pi. So, if
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the load is RL, current will reach maximum
somewhere in between pi by 2 to pi. Where
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it will reach maximum? It can be determined
by solving the equation, the above equation.
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But, I know that when current is maximum,
voltage across a conductor is 0. So at this
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instant, the value of the input voltage is
Imax into R. Is that okay?
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When current is maximum, the instantaneous
value of the input voltage is Imax into R.
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So, this point, this magnitude could be or
is Imax into R. Beyond that, current becomes,
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current starts decreasing. But then, current
is always positive, it is flowing in this
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direction only, current cannot reverse, this
is a positive direction of current. So, I
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into R is always positive. L dr by dt, now
dr by dt has become negative. So, this point
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becomes negative, this point has become positive.
Current starts decreasing at omega t is equal
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to pi, instantaneous value of the input voltage
is 0. But then, current is still flowing in
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the load. So, I into R, L dr by dt should
be zero, KVL has to hold good.
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Beyond pi, input voltage becomes negative
and since there is an inductor in the load
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circuit and current is decreasing, this current
continues to flow beyond pi, that is because,
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L dr by dt has become negative. Cathode potential
is less than the anode potential, if I assume
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the direct to be a non ideal. If it is an
ideal, cathode potential is equal to the anode
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potential or non ideal diode is, difference
could be of order of 1 volt. So, current continues
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to flow till it becomes 0.
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Till a current becomes 0, input voltage is
equal to the load voltage or V0. Vi is equal
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to V0 till i becomes 0. So, once the diode
turns off, the entire input voltage appears
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across the diode. So, this is the diode wave
form, voltage wave form till it starts conducting
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again in the next ...
What is the average value of the output voltage?
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You need to find out, you have to integrate
it. It is given by Vm by 2 pi into 1 minus
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cos beta.
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What are the observations? We found that when
gamma for a purely resistive load, diode conducts
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for pi by 2 radians. Sorry, diode conducts
for pi radians. When the load is purely inductive,
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diode conducts for 2 pi radians and for a
RL load, diode conducts for durations greater
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than pi. So, gamma increases with an increase
in L. But then, average value of the output
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voltage decreases with L. Why? Because
beyond pi, we are applying a negative voltage
to the load.
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That is why, average value of the output voltage
decreases with increase in gamma beyond pi.
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What is the significance of gamma? Why the
load current should be continuous or why there
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should be a finite current in the load? What
happens?
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Consider a simple case, I have a DC motor,
torque equation is K into phi into IA and
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the equation is J d omega by dt plus B omega
plus TL is equal to Te. Te is the torque developed
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by the motor, load torque, frictional torque,
J is the moment of inertia. So if I neglect
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the frictional torque, d omega by dt is equal
to TE minus TL added by J.
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So at a steady state, d omega by dt is 0.
It possible only if Te is constant, in other
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words, if Te varies with time, d omega by
dt also will vary. When Te will vary with
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time? Either IF or IA or both, if they pulsate,
Te will also pulsate. So, in order to have
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a constant Te, we should have a constant value
of IF and constant value of IA. So, it is
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always desirable to have finite IA, if not
a constant IA.
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Just by increasing L, we found that gamma
may be increasing but then, load voltage,
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the average value of the output voltage decreases.
So just by increasing L, may not be a solution
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for these circuits. What next? Is there a
way out? I think so, there is a way out. What
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is that?
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Consider the circuit, a switch, there are
switch that can be close to 1, 2 or 3. Initially
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I will close it to switch 1. What is the circuit
equation? This is the circuit equation, R
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into i plus L di by dt is equal to V. We have
studied this circuit. After sometime I would
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like to open the switch, either I will close
it to 2 or I will close it to 3.
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What happens if I close it to 2? Voltage applied
to the load is 0. So, current decays in this
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fashion. The circuit equation is Ri plus L
di by dt is equal to zero, forcing function
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is 0, current decays slowly. Whereas, in the
second case, forcing function is negative.
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So, current decays fast. So, by applying a
0 voltage to the load, I have achieved 2 things.
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Current decays slowly, so in other words,
beta increases. But then, I am not applying
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a negative voltage to the load unlike in this
case. So, my average voltage remains constant
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even though my beta is increasing. Whereas
here, since I am applying a negative voltage
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to the load, current decays faster plus the
average voltage comes down.
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So therefore, it is possible to increase beta,
also increase V0. At what cost? Of course,
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nothing comes for free. If you are ready to
spend something, you will get a, there will
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be solution for every problem that you have
created. That is a law of nature. ...
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Consider a case, in the previous case, there
was only 1 diode, I am connecting another
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diode across the load. Positive half, only
D1 can conduct, D2 cannot conduct in the positive
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half. So in the positive half, D2 is blocking
or D2 is off, D1 is conducting. We have this
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equivalent circuit, input voltage, diode D1
R into IR and L. Same way, from current increases
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which is a peak somewhere in between pi by
2 and pi and starts decreasing. At omega t
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is equal to pi plus, this potential becomes
higher than this potential, becomes negative.
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So, if D1 conducts, a positive voltage appears
across D2 which is not possible. So therefore,
191
00:29:11,929 --> 00:29:18,929
D1 turns off, D2 starts conducting. When D2
starts conducting, this point gets connected
192
00:29:23,009 --> 00:29:30,009
to the cathode of D1. So, negative voltage
is appearing across D1. So, diode D1 turns
193
00:29:30,269 --> 00:29:37,269
off or diode D1 cannot conduct. So, our assumption
of, assumption of saying that D2 conducting
194
00:29:41,159 --> 00:29:45,109
in the negative half is proved.
195
00:29:45,109 --> 00:29:52,109
So beyond pi, till the current becomes 0,
D2 conducts. But, I told you that when the
196
00:29:56,519 --> 00:30:03,519
D2 conducts, voltage applied to the load becomes
0. So, rate of decay of current or rate of
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00:30:03,719 --> 00:30:10,719
decrease of the current that is flowing to
the load also reduces, current slowly reduces.
198
00:30:11,749 --> 00:30:18,749
So, average value is same as that of a purely
resistive load: in the sense, output voltage
199
00:30:18,899 --> 00:30:25,899
is equal to input voltage till omega t is
equal to pi and beyond pi, it is 0.
200
00:30:27,889 --> 00:30:34,889
So, average value is same as that of a purely
resistive load. Beta is also increased now.
201
00:30:36,729 --> 00:30:43,379
Here the decay is very slow, it depends on
the time constant and in the negative half
202
00:30:43,379 --> 00:30:50,379
the entire input voltage appears across D1.
So, voltage rating of both these diodes is
203
00:30:50,879 --> 00:30:57,879
same as the peak of the input. So, D1 blocks
in the negative peak and D2 blocks in the
204
00:31:01,389 --> 00:31:08,389
positive peak. So, this D2 diode is also known
as the freewheeling diode. It provides a path
205
00:31:12,129 --> 00:31:19,129
for the current to flow when D1 is off. Load
is freewheeling through D2.
206
00:31:19,690 --> 00:31:26,690
Now let us see, let us discuss about RLE load.
The limit in case, L is 0. So, we have R and
207
00:31:33,219 --> 00:31:40,219
E. Can the diode start conducting at the positive
0 crossing? No, now we need to know, what
208
00:31:46,769 --> 00:31:53,769
exactly is there in the load circuit? If the
load is active, we need to know the load parameters.
209
00:31:54,019 --> 00:31:58,839
If the load is passive and if the current
is discontinuous, we do not need to know,
210
00:31:58,839 --> 00:32:03,499
because diode starts conducting at the positive
0 crossing. Now, even if there is no current
211
00:32:03,499 --> 00:32:08,779
here, cathode potential is E with respect
to this point.
212
00:32:08,779 --> 00:32:15,139
So no current, same potential: this point
and this point are the same potential. So,
213
00:32:15,139 --> 00:32:22,139
E is the potential with respect to E of the
cathode potential with respect to the ground
214
00:32:22,649 --> 00:32:28,489
and anode potential is Vi. So, diode starts
conducting when the instantaneous value of
215
00:32:28,489 --> 00:32:35,489
the input voltage is E and once the diode
starts conducting, the circuit equation is
216
00:32:37,219 --> 00:32:44,219
Vi is equal to R into I plus E or I is equal
to Vi minus E divided by R, purely resistive
217
00:32:48,539 --> 00:32:54,809
circuit, so these are the equations.
218
00:32:54,809 --> 00:33:01,809
When diode is on, output voltage, when the
diode is on, Vi is equal to V0 and when diode
219
00:33:08,529 --> 00:33:15,529
is off, V0 is equal to E. So, from 0 to alpha,
diode was not conducting, output voltage was
220
00:33:23,219 --> 00:33:30,219
V0, diode starts conducting at alpha, Vi becomes
equal to E again at pi minus alpha. So, diode
221
00:33:34,200 --> 00:33:41,200
turns off at this point. Once the diode turns
off, till it starts conducting again, V0 is
222
00:33:46,440 --> 00:33:53,440
equal to E. This is the voltage across, V
minus C is the voltage across R. So, current
223
00:33:53,700 --> 00:34:00,339
is V minus C divided by R. This is the sinusoid,
reaches a peak at omega t is equal to pi by
224
00:34:00,339 --> 00:34:07,339
2. These are the various equations.
225
00:34:14,669 --> 00:34:21,669
Now second case, now we will make R is equal
to 0. We have a diode, inductor and a battery.
226
00:34:32,470 --> 00:34:39,470
Same, as in the case of RE load, diode starts
conducting when instantaneous value of the
227
00:34:43,369 --> 00:34:50,369
input voltage is equal to E, because anode
potential is Vi, cathode potential is E.
228
00:34:52,130 --> 00:34:59,130
So, anode voltage should be higher than or
equal to E for the diode to conduct. Once
229
00:35:01,900 --> 00:35:08,900
the diode starts conducting, the circuit equation
is a L dr by dt plus E is equal to Vi. So,
230
00:35:12,160 --> 00:35:19,160
when diode starts conducting, when the diode
is on, output voltage V0 is same as Vi. So,
231
00:35:25,430 --> 00:35:32,430
difference of input voltage and the battery
voltage appears across L and beyond alpha,
232
00:35:34,350 --> 00:35:41,350
this voltage is increasing. So therefore,
current also starts increasing. Where will
233
00:35:41,730 --> 00:35:48,580
it reach peak? Will it reach maximum when
omega t is equal to pi to 2, as in the case
234
00:35:48,580 --> 00:35:55,580
of resistor? Of course, 1 way to, 1 way to
write a differential equation is to solve
235
00:35:56,350 --> 00:36:03,350
it. But, if I use a graphical approach, I
can say that current will reach maximum when
236
00:36:06,760 --> 00:36:13,760
Vi is equal to E at omega t is equal to pi
minus alpha, because at this instant Vi is
237
00:36:19,670 --> 00:36:26,670
equal to E. So therefore, L dr by dt is equal
to zero, in other words, i is equal to Imax.
238
00:36:27,330 --> 00:36:34,330
So, current started reaching started from
0, reached a peak at omega t is equal to pi
239
00:36:35,460 --> 00:36:42,460
minus alpha and from there onwards, it starts
decreasing. Where will it become zero? Again,
240
00:36:45,360 --> 00:36:52,360
it depends on E and L. It can be determined
by solving the differential equation. But,
241
00:36:55,070 --> 00:37:02,070
I know that average value of the voltage across
inductor is 0. So, this is positive L dr by
242
00:37:05,110 --> 00:37:12,110
dt. So, current become 0 at that instant or
this area should be equal to this area at
243
00:37:20,310 --> 00:37:23,530
that instant, current becomes 0.
244
00:37:23,530 --> 00:37:30,530
So, current continues to flow beyond pi minus
alpha till negative L dr by dt is equal to
245
00:37:33,440 --> 00:37:38,890
positive L dr by dt. So that is the very basic
concept, average voltage across inductor should
246
00:37:38,890 --> 00:37:45,890
be 0. So, diode turns off at beta. In this
case, I have shown that diode continues to
247
00:37:48,920 --> 00:37:55,920
conduct beyond pi, it need not be, it can
happen anywhere beyond pi minus alpha. So,
248
00:37:58,260 --> 00:38:05,260
till the diode is on, output voltage is equal
to the input voltage, load has no role to
249
00:38:06,850 --> 00:38:07,100
play.
250
00:38:07,030 --> 00:38:13,120
Remember, when the diode is conducting, input
voltage and the output voltage are same. It
251
00:38:13,120 --> 00:38:19,540
is only when there is no current and if the
load is active, in other words, if there is
252
00:38:19,540 --> 00:38:26,540
a battery there, we need to know the value
of the battery voltage or output voltage is
253
00:38:28,070 --> 00:38:34,400
equal to the value of E when the diode is
not conducting.
254
00:38:34,400 --> 00:38:41,400
It was not conducting from 0 to alpha and
it stops conducting from beta to again 2 pi
255
00:38:42,990 --> 00:38:49,990
plus alpha. In this region, Vi, this is the
input wave side. So, this is positive L dr
256
00:38:53,720 --> 00:39:00,720
by dt and this is negative L dr by dt. These
are the various equations that I have written.
257
00:39:01,700 --> 00:39:08,700
E Vi is equal to E at omega t is equal to
pi minus alpha, V0 is equal to E for 0 to,
258
00:39:09,870 --> 00:39:16,870
for omega t in between 0 to alpha. These are
the various equations. i is equal to I0 at
259
00:39:23,190 --> 00:39:28,790
beta, when positive L dr by dt is equal to
negative L dr by dt.
260
00:39:28,790 --> 00:39:35,480
Now, what happens if I put a small resistor
RLE? When the load is purely resistive, we
261
00:39:35,480 --> 00:39:42,030
found that current reaches maximum at at omega
t is equal to pi by 2. Of course it starts,
262
00:39:42,030 --> 00:39:49,030
both the cases it started conducting at alpha
where alpha is sin inverse E divided by Vm.
263
00:39:49,790 --> 00:39:56,790
In the case of LE load, sorry in the case
of EL load ,current reach maximum and current
264
00:39:58,870 --> 00:40:05,870
reach peak at omega t is equal to pi minus
alpha. So therefore, for RLE load, current
265
00:40:08,340 --> 00:40:15,280
will reach peak somewhere in between pi by
2 and pi minus alpha. It reaches peak here,
266
00:40:15,280 --> 00:40:22,280
current starts decreasing and it becomes 0
when ... somewhere in between ... it becomes
267
00:40:28,110 --> 00:40:33,900
0. It all depends on R L and E. It is a bit
difficult to determine now, because there
268
00:40:33,900 --> 00:40:40,900
is R is also present. So, till diode is conducting,
Vi is equal to V0, when diode is off, V0 is
269
00:40:45,940 --> 00:40:46,250
equal to E.
270
00:40:46,250 --> 00:40:53,250
So, if the load is highly inductive and has
a very small battery voltage, beta may be
271
00:41:00,170 --> 00:41:06,870
high. Because, if E is low, as E reduces,
alpha reduces, because alpha is sin inverse
272
00:41:06,870 --> 00:41:13,870
E divided by Vm.
As alpha reduces, the time available for the
273
00:41:17,160 --> 00:41:23,300
current to increase also increases. I mean,
because current reaches peak somewhere in
274
00:41:23,300 --> 00:41:30,300
between pi by 2 to pi minus alpha. As alpha
reduces, time for the current to increase
275
00:41:32,030 --> 00:41:39,030
also increases and if I have a substantial
value of L wherein, current cannot change
276
00:41:41,830 --> 00:41:48,830
instantaneously and if I have a low value
of E, this beta can be increased.
277
00:41:50,230 --> 00:41:57,230
But then, beyond pi, we have a same situation,
a negative voltage appearing across the load.
278
00:41:59,260 --> 00:42:05,830
So, the moment the negative voltage appears
across the load, average voltage, average
279
00:42:05,830 --> 00:42:12,390
value of the voltage applied to the load also
decreases plus current decays at a faster
280
00:42:12,390 --> 00:42:19,390
rate. Somehow, I need to stop this. What will
I do? I will connect a freewheeling diode
281
00:42:20,500 --> 00:42:25,180
across the load.
282
00:42:25,180 --> 00:42:32,180
I have connected a freewheeling diode across
the load. Till pi, the behavior of the circuit
283
00:42:40,750 --> 00:42:47,190
is same as before, no change.
284
00:42:47,190 --> 00:42:54,190
The wave forms also, no change from 0 to alpha,
it is the input, from 0 to alpha, it is E
285
00:42:57,780 --> 00:43:04,780
because no current. So, V0 is equal to E.
From alpha to pi, assuming that there is some
286
00:43:12,290 --> 00:43:19,290
current flowing in the circuit, Vi is equal
to V0, in other words, V0 is equal to Vi.
287
00:43:22,560 --> 00:43:29,560
At omega t is equal to pi plus, the freewheeling
diode starts conducting. So, voltage applied
288
00:43:29,780 --> 00:43:36,780
to the load is 0, diode turns off. So, it
does not supply power to the load. But then,
289
00:43:40,920 --> 00:43:47,920
since there is an inductor, the stored energy
in the inductor continuous or stored energy
290
00:43:48,340 --> 00:43:55,340
in the inductor ensures the current to flow
in the load beyond pi.
291
00:43:57,720 --> 00:44:04,720
So at beta, current becomes 0. Once the current
becomes 0, voltage applied to the load becomes
292
00:44:06,020 --> 00:44:13,020
E. So, difference is only from pi to beta,
pi to beta, voltage applied to the load is
293
00:44:14,700 --> 00:44:21,700
0. See, all this circuits can be analyzed
by writing differential equation for various
294
00:44:24,220 --> 00:44:31,220
modes and solving them. I am not using that
approach. You are free to do that, rather
295
00:44:31,830 --> 00:44:38,830
I would encourage you to write those equations
and solve numerically. So, this is what it
296
00:44:41,310 --> 00:44:43,720
is. i starts flowing ...
297
00:44:43,720 --> 00:44:50,720
Now, consider a case, assume that load current
is continuous. Do not ask me what sort of
298
00:45:01,250 --> 00:45:07,780
a load. It could be anything. It could be
anything, I mean, it can be LE type or whatever,
299
00:45:07,780 --> 00:45:14,270
do not bother.
I may have a battery or may not have a battery.
300
00:45:14,270 --> 00:45:19,900
Only thing is current is continuous. So, if
the current is continuous, either D1 should
301
00:45:19,900 --> 00:45:26,900
be on or D2 should be on, at any given time.
In the positive half, we already proved that
302
00:45:26,950 --> 00:45:33,950
D1 conducts and in a negative half, D2 conducts.
So, at any given time, one of the diode is
303
00:45:34,570 --> 00:45:41,570
on: in the positive half, D1is on and in the
negative half, D2 is on.
304
00:45:42,110 --> 00:45:49,110
So, when D1 starts conducting? D1 starts conducting
when omega t is equal to 0 or at the positive
305
00:45:54,850 --> 00:46:01,850
0 crossing, diode starts conducting. So therefore,
if the load is continuous or load current
306
00:46:01,910 --> 00:46:08,600
is continuous, diode starts conducting at
the positive 0 crossing. Diode starts conducting
307
00:46:08,600 --> 00:46:15,600
at the positive 0 crossing and D2 starts conducting
at the negative 0 crossing. So, even if I
308
00:46:17,830 --> 00:46:24,650
have a battery in the load and if the current
is continuous, it does not matter. D1 starts
309
00:46:24,650 --> 00:46:30,240
conducting in the positive 0 crossing, remember,
even if I have an E here, that is because
310
00:46:30,240 --> 00:46:37,240
of L di by dt. Because, once the current has
reached the peak, in a RLE type of load we
311
00:46:39,320 --> 00:46:46,120
found that current reaches peak between pi
by 2 to pi minus alpha and beyond that current
312
00:46:46,120 --> 00:46:47,280
starts decreasing.
313
00:46:47,280 --> 00:46:54,280
So, L di by dt becomes negative. That will
ensure the current to flow. So therefore,
314
00:46:55,620 --> 00:47:02,040
even if I have a battery here and if the current
is continuous, diode starts conducting at
315
00:47:02,040 --> 00:47:07,970
the positive 0 crossing. Please do not say
that you have a battery in the load, so that
316
00:47:07,970 --> 00:47:14,970
is why cannot conduct till alpha. That is
true only when, only when current is discontinuous.
317
00:47:16,750 --> 00:47:23,260
If the current is discontinuous, we found
that cathode potential of the diode is at
318
00:47:23,260 --> 00:47:30,070
E. So, anode potential should be higher than
or equal to E for the diode to conduct. That
319
00:47:30,070 --> 00:47:35,500
is why it starts conducting at alpha which
is equal to sin inverse E divided by Vm.
320
00:47:35,500 --> 00:47:42,500
So, if the current is continuous, D1 starts
conducting at the positive 0 crossing, an
321
00:47:42,870 --> 00:47:49,400
important concept to be remembered. So, these
are the proof, so here is the proof.
322
00:47:49,400 --> 00:47:56,400
So, potential of A is potential of C in the
positive half. I am assuming D2 is on. If
323
00:47:56,440 --> 00:48:03,440
D2 is on, so, what is, when D2 is on, Potential
of B is equal to potential of C. So, in the
324
00:48:06,570 --> 00:48:12,150
positive half, if this occurs in the positive
half, a positive voltage appears across D1.
325
00:48:12,150 --> 00:48:17,930
Because in a positive half, potential of A
is higher than C and if D2 conducts, this
326
00:48:17,930 --> 00:48:24,830
potential gets connected to B. A positive
voltage appears across D1. So, D2 cannot conduct
327
00:48:24,830 --> 00:48:31,830
in the positive half. So again I will repeat,
if i is continuous in the positive half, D1
328
00:48:38,060 --> 00:48:41,350
conducts and negative half, D2 conducts, independent
of type of load.
329
00:48:41,350 --> 00:48:48,350
So, what are the conclusions? First 1 is the
conduction angle or the duration for which
330
00:48:49,780 --> 00:48:56,780
diode conducts increases with load inductance.
Second is for the diode to conduct, voltage
331
00:48:59,560 --> 00:49:06,560
across the diode should be 0. It is not only
the anode potential alone will decide the
332
00:49:10,130 --> 00:49:16,220
conducting state of the diode, it is also
the cathode potential, it is also the cathode
333
00:49:16,220 --> 00:49:23,220
potential will determine whether that diode
can conduct or not.
334
00:49:24,120 --> 00:49:31,120
Use of freewheeling diode increases gamma.
Why it increases? because, it applies a 0
335
00:49:31,530 --> 00:49:38,530
voltage to the load. It applies 0 voltage
to the load, so therefore, decay process also,
336
00:49:38,720 --> 00:49:45,720
current decay process also, decay, the rate
of decay also reduces. But then, voltage applied
337
00:49:46,470 --> 00:49:51,750
to the load is 0. If there was no freewheeling
diode, the voltage applied to the load beyond
338
00:49:51,750 --> 00:49:55,750
pi is negative, average value comes down,
decay process also faster.
339
00:49:55,750 --> 00:50:02,750
So, we are achieving 2 things at the cost
of 1 diode. So, next point is if the current
340
00:50:05,800 --> 00:50:12,800
is discontinuous and if the load is RLE, the
instant at which diodes starts conducting
341
00:50:13,950 --> 00:50:20,950
is given by sin inverse E divided by Vm and
if the current is continuous, alpha is 0.
342
00:50:28,580 --> 00:50:35,580
It is independent of flow. It is because of
L di by dt. These are the important concepts
343
00:50:43,670 --> 00:50:47,290
to be remembered.
344
00:50:47,290 --> 00:50:54,290
If I know the behavior of LC with sinusoidal
excitation, the study of power electronics
345
00:50:59,160 --> 00:51:06,160
is very simple. I would encourage you to use
a graphical approach rather than the analytical
346
00:51:10,330 --> 00:51:16,610
approach. Writing a differential equation,
solving it, anyone can do. If you try to use
347
00:51:16,610 --> 00:51:21,000
a graphical approach, it will give a better
insight rather than writing a differential
348
00:51:21,000 --> 00:51:22,650
equation.
349
00:51:22,650 --> 00:51:29,190
So, what did you do in the entire class? We
just used a simple KVL and 1 equation, Vm
350
00:51:29,190 --> 00:51:34,630
sign omega t is equal to R into i plus L divided
by dt plus E. That is the most complicated
351
00:51:34,630 --> 00:51:38,970
equation and I have, did not take any time
in solving those equations. You can always
352
00:51:38,970 --> 00:51:43,900
solve that equation,
I use a graphical approach. Somehow, I like
353
00:51:43,900 --> 00:51:50,900
it more and I encourage you to use this approach
rather than solving it, solving using a differential
354
00:51:52,730 --> 00:51:53,130
equation.
355
00:51:53,130 --> 00:51:53,920
Thank you.