Lecture – 17 Power Electronics


In my last lecture I discussed the operation
of 3 pulse converter that is 3 phase half wave rectification. We found that each device conducts for 120
degrees and each phase is also supplying power for 120 degrees. So, each phase supplies power only in half
a cycle for 120 degrees duration. We used 2 such bridges and we got a 3 phase
full wave rectifier and we found that there is 1 common cathode configuration and there
is another common anode configuration. Each device conducts for 120 degrees but then
each phase supplies power for 240 degrees in a cycle, provided, the load current is
continuous. So, in other words, at any given time, 1 phase
is open. It does not supply any power to the bridge. There are only 2 devices are conducting at
a time. 1 device in the upper half, another device
in the lower half and they do not belong to a same phase, remember. So, at any given time, in a 3 phase full wave
rectifier, 1 phase is open and 2 phases are supplying power to the load and there are
totally 6 pulses per cycle, each pulse of 60 degree duration. See remember, each device conducts for 120
degrees but then there are 6 pulses, each of 60 degree duration. Hence, the name 6 pulse converter. So, we have so far you have studied, 1 pulse,
2 pulse, 3 pulse and 6 pulse converter. What are the advantages of increasing number
of pulses? I will just show you a … slide. This is the waveform for half wave ratification. I am assuming, it is a pure resistive load
or may be a half wave rectification using a freewheeling diode, full wave rectifier,
3 pulse converter and a 6 pulse converter. I said this is AC to DC conversion. I would like to have almost a constant value
of DC voltage. So definitely, I would filter this waveform
to get a constant DC. Now, if you compare here, in which case the
filtering requirement is maximum and which case it is minimum? To have a constant DC in half wave rectification,
definitely, the filter requirement is the highest because for 0 to pi, the load voltage
is same as the input voltage and pi to 2 pi it is 0. Now, I need to have a constant DC. So, definitely, my filter requirement here
is more, whereas in 2 pulse converter, there is no 0 voltage duration, there is no period,
wherein, output voltage is 0 here. So, definitely, filtering requirement reduces. But then output voltage changes from 0 to
Vm. What happens in 3 pulse converter? We found that yesterday, there are 3 pulses
and the minimum value of V0 it is proportional to sin 60. It is root 3 by 2 or root 3 by 2 per unit
and this is again the peak value is Vm. For alpha is equal to 0 half 3 phase half
wave rectifier, the voltage output voltage varies from root 3 by 2 to 1. What happens in 6 pulse? We discussed yesterday for alpha is equal
to 0 or that is equivalent to an uncontrolled bridge. We found that this voltage changes from 1.5
per unit to root 3 that is approximately 1.7. So, in 6 pulse converter with half alpha is
equal to 0, the variation in the output voltage is 1.5 to 1.7. So, that is approximately a constant for an
engineer, it, I can say that output voltage remains approximately constant. I may not require any filtering. See, therefore, as the number of pulse increases,
your filtering requirement comes down. The maximum in 1 pulse converter, minimum
in 6 pulse converter. 6 pulse converter, we will see somewhere while
doing DC to AC conversion. We will find that filtering requirement there
is minimum. In some cases, you can do away with this filtering,
even to have a constant DC. Now, what is the expression for the average value of the output
voltage? There are 6 pulses per cycle, each pulse of
60 degrees duration. Alpha is measured with respect to the point
of natural commutation. So therefore, if I need to integrate from
pi by 6 plus alpha to pi by 6 plus alpha to pi by 3 and we found that when D1 or T1 and
D6 or T6 are conducting, output voltage is Vab. So, the magnitude of Vab is root 3 Vm sin
omega t plus 30. So, we will found, if you simplify this, you
will find that it is 3 root 3 by pi into Vm into cos alpha where Vm is peak of the phase
voltage. See, we have taken Van is Vm sin omega t. This is a phase, so it is, V is the RMS value. So, if I substitute for Vm, you will get 2.54
into V into cos alpha where V is the RMS value, alpha is the trigger angle. So, this is also equal to 2 into this expression
and we found that this is the expression for output voltage of a 3 pulse converter. So, average output voltage for a 6 pulse converter
is twice the 3 pulse converter. So, here is the equivalent circuit. It is 2.54 into V into cos alpha. RMS value of the phase voltage is variable
because there is a function of alpha. 2 devices are conducting at a time and here
is the load voltage, this is the load voltage, the current. I told you that each device conducts for 120
degrees. So, if I find the RMS value of the source
current, it is found to be root 3 by 2 into I0. So, source, if alpha is equal to 0, phase
A supplies power from pi by 6 to 5 pi by 6. In the positive half and in the negative half,
it is 210 degrees to 330 degrees. So, each phase in each cycle, power supplied
by the source is of 120 degree duration. So, if I assume the load current to be constant
and ripple free that means load current is constant of value I0. If I find RMS value of the current, source
current, it is found to be this value, root 3 root 2 by 3 into I0, whereas, in a single
phase bridge, if I assume that load current is of constant and ripple free, RMS value
of the source current is same as I0, whereas here, it is different because each phase supplies
power for 240 degrees. So, if I assume a load current of a constant
and ripple free, source current is of a square wave of duration 120 degrees in each cycle. In a single phase case, it was a square wave
of pi radians. So, if I take the harmonic series or if I
write the Fourier series for a single phase bridge, we found that all the triple N harmonics
are present. In other words, all odd terms are present;
sin omega t, sin 3 omega t, sin 5, 7, 9, 11 so on. All odd harmonics are present in a single
phase case, whereas, in a 3 phase, 3 wire system, you cannot have triple N harmonics. I am sure, this result your teacher who was
teaching machines might have told you. So therefore, in a 3 phase, 3 wire system,
you cannot have triple N harmonics. So, if I write the Fourier series, you will
find that there are only 6 N plus or minus 1 harmonics. In other words, there is fundamental sin omega
t. Next is sin 5 omega t, sin 7 omega t, sin
11 omega t, sin 13 omega t. This is also equal to sin 6 N plus or minus
1. N takes a value of 1, 2, 3 so on. So, in a 3 phase, 3 wire system, there are
only 6 N plus or minus 1 harmonics and the fundamental. So, harmonic content is also reduced in a
3 phase. There are no triple N harmonic. The frequency of that harmonics present in
a 3 phase, 3 wire system also increases. What implication it has? We will see some time later. Yesterday, we discussed for alpha is equal
to 0. Today we will study the waveform for alpha
is equal to 30 degrees. Alpha is measured with respect to the point
of natural commutation. So, T1 is the reference for thyristor T1,
this is the reference point. T3 – this is the reference point, T5 – this
is the reference point while device that is connected that is 2, this is the reference
point, 4 – this is the reference point and this is for T6. See, D phase is getting negative. So, at alpha is equal to 30, this is alpha
is 0, alpha is 30, T1 is triggered, G1 is getting a triples. So, prior to point P, T6 is conducting. At P or P is the reference point for 2. So, prior to 2, 6 would conduct in the lower
half, so, 6 was conducting. Now, what is the output voltage? When T1 starts conducting, the positive DC
bus or potential of X is Van and potential of Y is Vbn. Refer the bridge which we have drawn it yesterday. So, Van is proportional to sin 60, mind you,
alpha is measured with respect to this point. But then, we need to find out the instantaneous
value of the phase, the corresponding phase with respect to its 0 crossing. So, at alpha is equal to 30, instantaneous
value of Van is proportional to sin 60 that is root 3 by 2 and Vbn, it is sin 300, 6 was
conducting. So, it is minus root 3 by 2. So, therefore the peak is therefore, the value
of the output voltage, VO is Van minus Vbn is found to be in a root 3. So, with alpha is equal to 0, the peak value
of the output voltage is root 3. So, even at alpha is equal to 30, the peak
value of V0 is still root 3. Now, let us see after 30 degrees, what happens? So, V or somewhere at this point that is P,
we will find that A phase is at the peak that is sin 90, 6 is still conducting, so instantaneous
value of phase B is sin 330 that is minus half. So, output voltage is 1.5. This corresponds to 1.5. Now, 30 degrees beyond P, T2 is getting triggered. Let us see the output voltage just prior to
triggering 2. At that instant, Van is proportional to sin
120 or say 120 minus that is still root 3 by 2. Vbn is approaching 0, may be, 360 minus. Therefore, it is 0. So, the load voltage is Van minus Vbn, it
is proportional to root 3 by 2 itself. So, just prior to triggering T2, output voltage
is root 3 by 2. The moment you trigger T2, now, the load voltage
is Van minus Vcn because T2 starts conducting. T2 is connected to phase C, so load voltage
is Van minus Vcn. Instantaneous value of Vcn is proportional
to sin 60 or sin 240 that is again minus root 3 by 2. Van is 120, again it is root 3 by 2. Output voltage is Van minus Vcn. So, output voltage jumps to root 3, root 3
by 2 minus of minus root 3 by 2, it is root 3 itself. So, peak value of the output voltage even
with alpha is equal to 30 degrees is root 3. But then the minimum value is reducing. It is now root 3 by 2 and now … there is
a sudden jump, root 3 by 2 to root 3 because at this instant you are triggering another
device, incoming device. So, from in this instant, T6 and T1 are conducting
and T1 and T2 and so on. I encourage you to draw the remaining 6 pulses
also. So, what is the effect of increasing alpha
on the power factor or the instance at which the source supplies power? Phase A voltage is becoming positive here,
this is the positive 0 crossing. If alpha is equal to 0, it would have started
supplying the current from this instant. I cannot reduce this period beyond this. So, alpha is 0 is here, so it can, the phase
A can start conducting only at alpha is only at omega t is equal to 30 degrees. Now, I am triggering at omega t is alpha is
equal to 30 degrees that is omega d is equal to 60. So, phase A is open till omega t is equal
to pi by 3 radians. So, the positive crossing of the source current
can be at this instant. If alpha is equal to 0, the positive crossing
could have been somewhere here. Now, with alpha is equal to 60 degrees, the
positive crossing becomes here. In other words, I am delaying the current. In other words, source current started becoming
more and more lagging. If alpha is equal to 0, the 0 crossing of
the source current would have been here, now, it is here. So, this angle is increasing. The angle between the positive 0 crossing
in the voltage and positive 0 crossing of the source current is increasing. In other words, power factor is deteriorating. Remember, as you increase alpha, power factor
deteriorates. Now, let us see what happens with alpha is
equal to 60 degrees? At this instant, T1 is triggered. Prior to T1 or prior to triggering T1, T5
was conducting. In the lower half, in this instant, T6 is
conducting. So, what is the output voltage just prior
to triggering T1? It is Vcn because 5 is conducting, minus Vbn,
6 is conducting in the lower half, upper half 5 is conducting. So, Vcn minus Vbn. What is Vcn? Vcn is 210, proportional to 210 degrees that
is minus half. Vbn, it is 330 degrees. It is again minus half. So, output voltage is Vcn minus Vbn, it is
0. So, now it is touching 0, whereas, in the
both the cases with alpha is equal to 0 and alpha is equal to 30 degrees, it did not touch
the X axis. Now, with 0, with X, alpha is equal to 60
degrees, it is does touching the X axis. Now, at point P or at this instant, you have
triggered T1. Now, output voltage becomes Van minus Vbn. What is Van at this instant? Proportional to sin 90, it is 1, whereas,
Vbn is still at 330 plus. So, it is again minus half. So, VO is 1.5 times the peak value or 1.5
per unit. See the difference, now. For alpha is equal to 0 to 30, we had still
the peak value of root 3. So, alpha increase above 30 degrees, the peak
value comes down. Now, with alpha is equal to 60 degrees, we
found that peak value is 1.5, minimum is touching 0 now. Now, what happens to the output voltage waveform
beyond 30 degrees from there? At this instant, Van is sin 120 that is root
3 by 2. We have not triggered T2, 6 is still conducting. So, Vbn at this instant is 0. So, output voltage is root 3 by 2 itself. Van minus Vbn, Vbn is 0. So, it is root 3 by 2 and what is the output
voltage just prior to triggering T2? It is same as the output voltage just prior
to triggering T1, whatever that happen to T1, it will happen to T2 also, may be, in
the lower half. So, you will find that output voltage is just
touching 0. The moment you trigger T2, output voltage
jumps to 1.5. So, output voltage here varies from 1.5 to
0. Now, it is touching 0 with alpha is equal
to 60 degrees. So, here is the procedure. I would expect you to continue the complete
the remaining 4 pulses also. Now, let us see what happens with alpha is
equal to 90 degrees? Corresponding value of omega t is 120. Van is root 3 by 2, bottom of 6 is conducting,
Vbn is sin 60, it is 0. So, V0 is root 3 by 2. The moment I trigger T1, it jumps to root
3 by 2. What happens after 30 degrees? At this instant, Van is sin 150 that is half
and Vbn is sin 30, it has become positive. So, instantaneous value of Van is same as
instantaneous value of Vbn. So, output voltage is 0. Now, what is the output voltage just prior
to triggering T2? Van is, may be, 180 minus is approximately
equal to 0. Vbn is sin 60 minus. It is approximately, root 3 by 2 itself. So, Van is 0, Vbn is plus root 3 by 2. So therefore, output voltage is Van minus
Vbn. It is minus root 3 by 2. So therefore, for alpha is equal to 0, output
voltage varies from root 3 by 2 to minus root 3 by 2, provided, the current is continuous,
remember. Provided, the current is continuous, output
voltage when alpha is equal to 0 is varies from root 3 by 2 to minus root 3 by 2. So, if I summarize the 3 waveforms, you can
have a peak value of root 3 till alpha is equal to 30 degrees. So, anywhere from 0 to 30, peak value of the
output voltage is the same. For alpha less than 60, V0 is always positive. At 60 degrees, V0, the instantaneous value
or minimum value of V0 becomes 0 and if I increase alpha beyond 60 degrees, minimum
value of V0 now, starts becoming negative. It becomes negative. So, if I ask a question, if the load is purely
resistive, so, what will be the instant or what should be the trigger value of alpha
beyond which current becomes discontinuous? Definitely, it is 60 degrees because at 60
degrees, the minimum value of V0 is 0 that is a voltage appearing across the resistor,
current also will become 0. So, current is just continuous in the case
of a resistive load for alpha is equal to 60 degrees. Above 60 degrees, current becomes discontinuous
and mind you, you cannot use the expression 2.54 into Vm into cos alpha if the current
is discontinuous. That is valid only if the current is continuous. Alpha is equal to 90 degrees, we found that
average value becomes 0. What sort of a of load we are talking about? What sort of a load we are talking about or
in which case average value of the output voltage is 0 but the current is continuous? It can happen only if the load is purely inductive
because average voltage across the inductor is 0. So, what happens is, in this waveform, this
is positive L di by dt and this is negative L di by dt. Current starts from 0, reaches a peak here
because at this instant L di by dt is 0. So, current has to be peak and again comes
down and becomes 0. It is just continuous, just at the boundary
between continuous and discontinuous. For other loads you cannot have a continuous
conduction for alpha is equal to 90. For anywhere, for even if the load is of RLE
type, I into R is always positive, E is positive. So, average value, you cannot have. Average value across the inductor is 0. So, RI plus E should be 0, it is just not
possible. So, load current can be continuous for alpha
is equal 90 is only for an inductive load, purely L load. Last case, I will just increase alpha is equal
to 120. Now, corresponding omega t is equal to 150. So, Van is proportional to sin 150 that is
half. Beyond, in the lower half, T6 is conducting
because we have not triggered 2. 2 is triggered somewhere at this instant. So, 6 is conducting, the potential of Y is
Vbn and potential of X is Van. Instantaneous value of Van is same as instantaneous
value of Vbn. So, output voltage is 0. Vbn is sin 30 and Van is sin 150, output voltage
is 0. Let us see what happens beyond, 30 degrees
beyond from this point? Van is sin 180, 0. Vbn is proportional to sin 60 that is root
3 by 2. Output voltage is Van minus Vbn, it is minus
root 3 by 2. What is the output voltage just prior to triggering
T2? Just prior to triggering T2, output voltage
is still Van minus Vbn. Van has become negative, it is sin 210, 210
minus or so and Vbn is sin 90 minus. So, Vbn is minus 1 approximately, Van is minus
half. So, output voltage is minus 1.5, this peak
is minus 1.5. So, at this point T2 is triggered. Now, the moment T2 is triggered, now output
voltage is Van minus Vcn, sorry Vcn, right, Vcn. Instantaneously, this voltage becomes 0. Instantaneously voltage becomes 0 because
AN is 210, CN is 330. So, both are minus half. So, minus of minus half minus half is plus
half, it is 0. So, for alpha is equal to 120 degrees, if
I summarize, it becomes 0 and always negative, whatever that happened for alpha is equal
60 degrees. Alpha is equal to 60 degrees, we found that
instantaneous value or minimum value of V0 is 0 and it is positive. So, at alpha is equal to 120 degrees, it is
0 and negative. Just the opposite, these are the 2 pulses. See, we have triggered T1 at this instant. The positive value, the positive 0 crossing
of the current is somewhere here. Positive crossing of the phase A voltage is
somewhere here. So, this and this goes on increasing. Power factor deteriorates with increase in
alpha. I am not going to do for alpha is equal to
150 and alpha is equal to 180. Same procedure, you follow it and you will
find that whatever that happened for 30 degrees, it will happen for alpha is equal to 150 but
negative. Whatever that happened for alpha is equal
to 0, it will happen for alpha is equal to 180. So far we drew the waveforms for alpha is
equal to 0, 30, 60, 90 and 120. Suppose, now that you have to go to the lab,
you have been given a 3 phase bridge, there is a load. Now, what value of alpha will you choose? Now, forget about 3 phase. Let us talk about single phase, relatively
easier. Say, assume that we have a DC motor as a load,
RLE type load, DC motor. Motor is stationery, in the sense, motor rotor
is not rotating. What will you choose the alpha to be? In our machines lecture, our teacher has told
that you need to apply, reduce voltage to the DC machine. Full field reduced armature voltage, so that
the current drawn with the motor reduces, also, it accelerates faster because when the
motor is at stand still, back EMF is 0. The current is limited by the armature resistance
itself. Invariably, students say that we should choose
alpha is 90 degrees, if a single phase fully controlled bridge is feeding a DC motor. Why? because, if alpha is equal to 90 degrees,
average value is 0. So, alpha is equal to I am applying a very
small volt, in fact, 0 voltage and go on reducing alpha. Now, my question is can we start at alpha
is equal to 90 degrees? Answer is simply no, why? See the waveform here. The input AC, RLE the load, E is 0. Say, inductance is very small here. I am assuming that there is no filtering inductor
in the DC side. Things, of course, things do change if there
is a filter inductor here. Now, the current is limited by R itself, which
is very small, armature resistance is very small for a motor. Again, armature inductance is also very small. Prior to triggering these devices, there is
no current. So, if you trigger the bridge at alpha is
equal to 90, the instantaneous value of Vn is Vm. That is the peak value. So, your peak voltage, you are applying to
the motor terminals. If it is 230, you will get around 320 or so. 320 volts, current is limited by R plus small
L. It will so happen that a large current will flow and may damage these devices. So therefore, a safe value may be, towards
150 to 160. It is all depends on the parameters of the
machine or the DC side. If I have an inductor, of course, you can
decrease this value. Otherwise, may be, start somewhere near this
point and reduce alpha towards 90 degree. Now, do not say that sir, you are triggering
at 150, average value is negative. Average value becomes negative beyond 90 degrees
only if the current is continuous. You have not even established the current
in the circuit. So, the question of current being continuous
does not arise. So, depending upon the value of R, L, E, current
builds up and it may become 0 because if you keep the same alpha it may become 0 much before
that. So, if you want to accelerate the motor, you
go on reduce the alpha towards 0. What happens for a 3 phase case? Things are not very straight forward here,
why? At any given time, 2 devices are conducting. Change over from 1 device to another device,
1 device in the top and 1 device in the bottom; they do not take place at the same time. So, change over from T5 to T1 takes place
at some omega t. After 60 degrees, change over from 6 to 2
takes place. The second is at any given time, 2 devices
should be on. There is no current, you have not even established
any current in the bridge. You have put on the AC supply, to establish
the current in the bridge, 2 devices have to be triggered. So, your pulse patterns should be such that
2 devices or any pair T1T2, T2T3, T4T4 or whatever, should get trigger pulse simultaneously. So, all the waveform that I have drawn so
far, I have just drew or just shown you only 1 pulse? That scheme may work at steady state, provided,
the current is continuous. I repeat; I have shown you only 1 pulse in
the gate circuit. That may work at steady state, provided, the
current is continuous. So, while starting or if the current has become
0 in between, to re establish the current in the bridge, 2 of them again will be triggered
the same time. So, the pulse pattern should be something
like this, T6T1, T1T2, T2T3. So, I need to have 2 pulses separated by 60
degrees. See here, 6 and 1 are triggered at the same
time. Now, 6 also may be triggered at the same instant
but the moment you triggered 2, say, after 120 degrees or so, from here, 6 will turn
off and 2 will come. So, at any given instant, 2 devices should
be triggered simultaneously. By the way, this pulse should be this is an
ideal case. Do not think that just because we are triggering
again here, 6 will continue to conduct for another 120 degrees, no. See, you are triggering 2 after 60 degrees. See, this duration is 60 degrees. The moment you trigger 2 somewhere here, 6
turns off and 2 starts conducting. So, ideally you need to have 2 pulses which
are separated by 60 degrees. In a single phase case, we found that if the
load is of RLE type, current may become discontinuous. You have triggered the bridge T1 and T2 in
the positive half, they started carrying the current. Now, E is higher than the instantaneous value
of Vi. In that case, we found that di by dt is negative
and it may so happen that current may become 0. Please recall, I have discussed all this. Therefore, we said that instead of having
only 1 pulse, have series of pulses. Same concept is still valid in 3 phase case
because if the load is highly inductive, we know that gate signal should be present till
the current through the device is higher than the latching current. So, gate signal should be present. We know that if the load is highly inductive,
current cannot change instantaneously, a current slowly builds up. So, if there is only a sharp pulse, by the
time IG becomes 0, current would not have increased to, is the latching current. So, SCR may not even turn on. So generally, a series of pulses of duration
70 to 75 degrees is used. I might have said that theoretically, 2 pulses
of 60 degree duration, sorry, 2 pulses which are separated by 60 degrees are required. General practice is; have a series of pulses,
of the total period is of the order of 70 degrees or so is used. What is the maximum value of alpha to start
the bridge in a 3 phase case? We found that for alpha is equal to 120 degrees,
minimum value of V0, in other words, V0 touches 0 and becomes negative. It does not become positive at all. I have not even established the current in
the circuit. Let it be, whatever may be the type of load,
we have not established the current in the circuit. To establish the current, I need to have a
positive voltage across it. If alpha is equal to 120 degrees, it is either
0 or negative. So, you cannot establish any current in the
circuit. So therefore, in a 3 phase bridge, the maximum
value of alpha to start the bridge is or should be less than 120 degrees. It is only then a positive voltage appears
across the load. For alpha less than 120 degrees for a small
duration, a positive voltage appears across it. If the positive voltage appears across the
load, current starts flowing, current builds up, then it can go on reducing alpha depending
upon the load requirement. So, alphamax should be less than 120 degrees,
should be 120 degrees so that a positive voltage appears across the load, voltage across the
load and i starts flowing. Again, I am assuming here that load is passive. Things will change if I have a battery there,
things will change. So, if the load is a passive, alphamax should
be less than 120 degrees. So, here is the sequence. A series of high frequency pulses of 70 to
75 degrees and see, there is an overlap here. This is the overlap, this duration, this is
the overlap. 1 and 6 are triggering at this point or this
is the instant were G1 should be triggered. So, at that instant, G6 still have few pulses
because even if the current has become 0 somewhere this, at this instant, G1 has also started
conducting, G6 sorry at this instant, G1 is getting a trigger pulse and there is a trigger
pulse for 6 also. So, current can establish in the circuit,
so far, so good. On this analysis, we discussed for source
inductance is 0. We are neglecting the source inductance. So, that is why current changes instantaneously
from 0 to the value of load current, the moment you trigger the thyristor and if I consider
a finite source inductance, instantaneous change is not possible and in the single phase
case we found that during commutation period, output voltage is 0. So, during the overlap period, mu, output
voltage is 0. So therefore, there is a net reduction in
the voltage applied to the load. This is due to the source inductance. What happens in a 3 phase case? Will the output voltage is 0 during commutation
period? We will see. Assume that I am, case 1, I am just assuming
an uncontrolled half wave rectification or rectifier. D1 was conducting, D2 starts conducting at
the point of natural commutation, when the positive when the Vn becoming more positive
then Van, for sometime D1 and D2 starts conducting, there is a short here, D3 is still open. So, what is the KVL for this loop? It is Van minus Vbn is 2LSi ds by dt is the
current that is flowing here. Current increases, this current decreases,
iT. So, output voltage is V0 is equal to Van minus
Van minus L dis by dt. So, if I substitute here, V0 is found to be
Van plus Vbn by 2. So remember, in a 3 phase case when the 2
devices are conducting, output voltage is the sum of the 2 voltages, phase voltages
divided by 2. So, just see here, at this instant is the
point of natural commutation. Prior to this point is the load voltage. Whatever it be, whether, it is An or Bn or
Cn, does not matter. At this instant, both the devices have started
conducting. Now, this voltage is this voltage is instantaneous
value of this plus instantaneous value of this divided by 2. So, I need to add up these 2 voltages and
divided by 2, will give you the output voltage. I am assuming mu to be 30 degrees. So, at this instant, overlap is complete,
so outgoing device turns off completely. Now, the output voltage is corresponding to
the instantaneous value of the incoming phase. So, there is a jump. Though I have used diodes, I should have had
a smooth waveform. There is no jump in a uncontrolled rectification. Now, here you find that due to source inductance,
there is reduction in the voltage and there is going to be a jump
in the output voltage. This is for controlled bridge.

7 thoughts on “Lecture – 17 Power Electronics

  1. can u share some info about this lecture so we can easily understand about the lecture topics to b discussed before

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