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Lecture -15 Front-End Ac to Dc Converter – Simulation study

Lecture -15 Front-End Ac to Dc Converter – Simulation study


In the last two three classes, we were talking about
ac to dc converter, front end ac to dc converter with unity power factor. So, we start with
how we can make use of this converter so that we can have an ac to dc converter with a PWM
control and we do not have to resort high very high frequency PWM, without resorting
to very high frequency PWM switching and using two converters and with a phase shifted sine
and triangle modulating signals appropriately some of the lower harmonics we could suppress. So, the harmonics we were able to suppress
was the harmonics at the fc and its side bands and then the harmonics and the side bands
of 2 fc and we were using an fc of around 11 times the mains. Now, then we discussed
how that inductance we can design and the output capacitor how we can design, then how
we will go for the PWM signal, sine triangle you have drawn on the wave forms, then how
we can draw a closed loop schematic of the whole control scheme, then how we can design
our controller; all those things we have designed for a particular converter input specification. Now, let us see, now all the design part is
over; let us say our design is correct or not. So, we will go to our simulation now. So, if you see here, this is our front end
converter. So, this is our input reference; input reference when we give 10 volts, we
want our output voltage of 2800 volts. Now, this output we are giving a feed back here
that is we are sensing the output dc and feed back here and we are controlling it. So, before coming to the whole controller,
let us first see whether it is working, let us start it quickly; then we will come to
the various controllers. So, the various ports are available; this is our total dc link current,
here we can observe here, here we can see separate output dc voltage, here we can see
the output voltage, total transformer input primary current and the primary reference
voltage. And here, we can see the total output VDC means output dc voltage, then the input
voltage, then the two converters due to PWM that current and the total current. So, let
us start it now and let us see how it works, we will open this one. See, if you see your static, this is the capacitor
voltage, this is the starting transient, it will stabilise soon. If you see here, this
red line that is this one, this is our input mains, sinusoidal representation. Then we
can see the three sets of current here. This violet current, this is the total transformer
input current that is this one and these two are the two converter currents. So, if you
can see the two converter currents it has more ripple than the primary side. Another
thing you should notice that one is this power is from the source ac side to the dc side
with the nearly unity power factor. If you see here, it is nearly unity power factor
and our capacitor voltage is nearly 2800 voltage. Let us, so this X axis I put some lower time
scheme so that I want to see the repel and the current. Now, let us see more number of cycles in a
cycle; instead of 0.2, I will make it here, apply and refer. So, if you see here, see the capacitor repel;
so we have designed for a delta v of 5%, so it is within the 5% thing and here is the
mains voltage run and the two converters and the transformer primary curve. So, transformer
primary current if you see, the repel is highly negligible. Now here, I want to see, quickly
I want to reverse it; let us see whether we can do it. Simulation, I pause it; now, I will try to
change the load resistance. Load resistance we have represented as a resistive load, 1
by r; this is the current feedback. So, to make it regeneration load, I will make it
negative that is how we can make it the regenerative quickly, regeneration. So, load is made negative, apply, then simulation,
continue. Let us see our converter block here. What
happened to this one? See, immediately it reverses because in the
same screen it has happened, see immediately it reverses, you can see here. During the
power reversal, the transient slowly it would slide up. See, there is fluctuation in the
capacitor voltage but it stabilises. Now, if you see here, the power flow is negative,
opposite, in the reversed direction. Now, it will stabilise soon; see capacitor voltage
is stabilised and input control was stabilised within few cycle, one or two cycle. So, everything
is stabilised, quickly powered. Here if you see, it is the input voltage and
the currents are nearly 180 degree phase shifted that means power flow is opposite direction.
Now, so that our convertor model is working; now only fine tuning you can do so that these
transients however we can control it. Now, let us stop this one and let us go to our
model so that a detail study of the model, how we can model it, we will get a field for
it. As I told, so this is our constant 10 volt
that is our input reference, this is Matlab Simulink block, you can get this block and
if u click here, the constant value we can mention here. So, we can measure 10 volts. So this, in Matlab we can extract, we can
stimulate the whole thing based on the actual voltages but here if you see, our controller
will be in the real practice, it will be on the plus or minus or plus or minus of 10 volts
plus or minus 15 voltage level, then through the convertor we boost up the output voltage.
So, to get the say one to one correspondence with the thing, we will make the block 10
volt and we are giving the feedback here. This feedback is taken from our VC. How we
get the VC? That is this side, Vdc will come to that one,
this we will, we have taken that one, from there we will step down the voltage that is
this one that is the KV; we will step down so that when 2800 volts comes, here it will
be 10 volt. So, any change in the capacitor voltage due to sudden loading or anything,
it will act here or whenever if we want to get 2800 volt, suppose you want the output
dc only 2000; we can control this one, correspondingly feedback will take care of that one. But here
we want of front end ac to dc fixed dc. Now, this is our PI controller. PI controller; see P plus I, so last class
I taught, we discussed about the PI controller implementation using analog circuits. There,
I mentioned output has to be controlled that means here the output well within, suppose
it is analog operational amplifier, it has to be well within the plus 12 minus 12; so
to be safe, we will make it plus 10 and minus 10. So here, we give the saturation limit
output controller and at the same time, the last time we said the feedback operational
amplifier with the set values maximum minimum. Here, if you see here, we can set upper limit
and minus limit block, so easy for simulation. Then PI controller that capacitor saturation
I told, so here also there is an integrated … integrator. Here if you take integrator, saturation limit
we can specify here so that we have specified here then sum here. So, because of this gain,
the output also we should limit it that is why this output that is this is the output
is the one here, we limit to plus 10 and minus 10. Now, there is another feedback is coming here,
this is called feed forward. It is coming from the current, output load current. See,
this PI control, this is coming from the output load current; we are putting a load, for the
simulation we are putting a resistive load, this output dc multiply 1 by r gives the load
current, again gain, current gain we will give it here. So, these current gain plus this one when
it gives here; so here also, this gain you should properly do it so that the current
total reference should be always plus or minus 10 volt level.
Now, if you see, this is called feed forward but this we are not talked before. If you
see here, this PI controller; we want this PI controller to change whenever the only
the capacitor changes or whenever our reference change only, we want this PI controller should
act. But if you see, sudden load changes, there is a fluctuation in the capacitor voltage.
That can load change will reflect here, then it will gives a required current reference.
So, instead of going through this output loop, if we can send the output current, natural
practice and then the power balance input power is equal to output power. So, we know the input voltage, peak value,
rms value, sinusoidal we know it, output dc is also known; so from that one, for any sudden
load changing or change in load, what is our required input current we can feed forward?
So, that we can give it, so this is the total current reference. This current reference,
we want unity power factor. This gives for a sinusoidal for the sinusoidal current; what
is the peak value? Now, we will generate a sine wave. This sine wave is in natural practice, we
have to derive from the mains so that it should have the and here we are generating with an
amplitude one, so we want only the sine function one and this frequency – 2 pi into 60 that
is our main frequency. Then this slight phase lead is adjusted here. As I told here, their
phase lead we can adjust such that that is our input mains frequency and this due to,
the current which is reference current and feedback current has come to the current loop
and we are dealing with sinusoidal current, there is a delay will be there. That delay
can be compensated by giving a leads to the reference here that is what we have given;
that with trial and error, we can adjust it so that we will get nearly unity power factor. So, it has come here and we are multiplying
that one. So, our final current reference is here. Now, current feedback; so if you
see here, the current feedback is taken from here, 1 by r. This current feedback, there
are two converters; two convertors will give total is 1 and is 2 input current and this
is our, so two converter should equal current. So, this is the total load current. This total
current we can take it and divide by the current gain and we can feed it here. Then, current control loop; controller we
said we want only we require only a proportional gain that is ls, here ls is there, l is the
inductance. Then t, t is equal to 2 tr we know it; is a Ki current feedback gain, g
is the converter gain. So, this we can put it here. So, if you take the gain factor, we can put
this gain here that is what is specified here; so, gain is given here. Then there is a saturation
limit here. So, here also saturation, we are always saturation
limit, we are making plus or minus 10; this will come here. To this one, minus vg; this
is coming from our VS (t) that is our reference voltage. So, this minus vs vg v this one with
the saturation, we are giving to the converter as the input. See, here also, this also act
as the feed forward. If any change in fluctuation, VS (t) happens. If it is not here, then it
will reflect on the capacitor and then capacitor due to current and then only current loop
will adjust. Instead of that one, we can through a feed
forward, we can give it here that is from the standard back converter equation VS (t)
Vr, VS (t) minus or VS (t) plus l into di by dt is equal to Vr (t), we can give it here.
Now, this also, this is our converter. Convertor we have, if you see here, there
are two convertor; this input is our Vr (t) reference, this reference is compared with
the triangle waveform. So here, triangle waveform we can generate;
first we will generate a sine wave. Sine wave multiplied by, the frequency is
not 60, 16 multiplied by 11. So, this is 2 pi into 660. This triangle waveform, we will
pass through a sine, this arc triangle trigonometric tree function. So, this will automatically,
if you give arc sine if you give, it is a triangle waveform which is in phase with the
sine waveform. So, the phase shift will not be there. So, that is why I think, we would not have
a triangle waveform which is in synchronised with our mains. So, arc term it comes, it
will arc term means it is angle, it will go to the pi value plus minus pi. So, corresponding
gain if you do it her, gain multiplicant you can get the triangle waveform which close
to whatever the value plus or minus 10 or plus or minus 1; whatever we want here. Then this is the logic we will put it here;
in this logic if you see, its output one for positive input that means when the error is
negative that is the sine is greater than the triangle value or negative that is minus
1 when the sine is less than the triangle value. So, this is the one which we can give
it here. So, we will get a PWM waveform. This PWM waveform going from plus 1 to minus
1 multiplied by Vdc by 2 that is our Vdc, we know it. So, Vdc by 2 plus or minus 1,
you will get the pole voltage that is for phase A leg or this is phase B leg waveform
here. So, if you see here, this side it will be plus or minus 1 only that multiplied by
this one, you will get actual voltage here. Maybe, we can connect this one to the input
side and we can observe that one whether it is working. Now, let us take the axis property
also. If it is here, it is going plus or minus 1
but if we are multiplying, we have taken from here; so we have to increase the amplitude. So here, we will change the amplitude, we
want minus 3000, here plus or minus Vdc by 2, so 2800. So, we will make it as around
half of that one, so approximately 2000 and plus 2000. Now, we will run the system, we
will see what is happening in one leg. So, if you see, this is the starting, throughout
the capacitor has to build up, quickly a capacitor is building, then the pole voltage is a waveform
is that starting transient because we initiated. So, once it stabilises, it will go into plus
Vdc by 2, minus Vdc by 2; this is for one converter.
Now, we will stop this one. So, same way, we have the other converter. Here in this converter compared to this one,
as I told before the sine waveform; this also 2 pi 60 and phase shift we have … See, here we are making a 180 degree phase
shift at triangular waveform that is a phase we are given a pi here. So, compared to previous
one and this one, you have phase shift of 180 degree will be there. Again, r tan we
will use it this one and with gain, you will get triangle waveform. Maybe, we can observe how this triangle waveform
looks like. You will copy this one here and two waveforms
maybe we can give it here. So, you require a multiplier block. This is the triangle wave form we are seeing
here which is 180 degree phase shifted. Maybe, we can slightly expand it so that we
can see a better waveform of 0.08. See, the waveforms are opposite each other.
So this scale, we will again adjust it here. So, that will give you, we will make it 20
here. So, two triangles as I told; this is for the
converter one, this is for the converter two and this should be 180 degree phase shifted.
So, that is taken here as I told here. Now, let us see, for the converter two; converter
two we are using the triangle waveform which is 90 degree phase shifted compared to this
phase A but these two are again 180 degree phase shifted. So, let us see, whether this
is 90 degree phase shifted. So here, if we go here, we have already pi
by two that is a 90 degree phase shift we have entered here. So, that is true. Now, let us see whether simulation it is happening
here. So, we will remove this block and you take this one here. So, these two are for the converter one phase
A leg, phase B leg triangle and these for the converter to phase A and phase B. So,
for this converter, the triangle waveform should be 180 degree phase shifted, should
be 90 degree phase shifted. Let us see now. See, these two are 90 degree phase shifted
and so these two waveforms, so converter one and converter two, triangle waveform 11 times
but 180 degree phase shifted A and B; for converter one and convert two, phase A leg
is 90 degree phase shifted. That is what we are seeing it now. Now, let us stop and let us see whether these
two are 180 degree; this is 90 degree and again 180 degree here. So, let us see that
one. So, we will drag this one, we will delete
this one from here and drag it there and then check it. So, with one scope this is called
mugs block with we can multiplex, we can two channels or three channels we can put to a
scope. So, let us take this one here, this one here, I will take this one here, one is
here and this one here. So, let us see, this one would be 180 degree phase shifted. So,
this is also 180 degree phase shifted. So, why want to do this one? As I told before,
these two are 180 degree phase shifted and the switching leg is, logic for this converter
is interchanged; for this one, sine is greater than the triangle, top switch is on and here
the sine is greater on the triangle, bottom switch is on. So, this we can take care of this one, output,
input relation here by specifying here. Now, by doing this one, all the harmonics at the
fc and side of fc can be can be suppressed from the converter itself. So, for this converter
as well as this converter, fc and side bands will be absent and the side band at fc, 2
fc and its side bands for this converter and the current generated due to that one, due
to this converter and this converter are there in opposite. That we saw that the repels are
opposite so that that the primary side, it will get suppressed and primary will have
the harmonics only at side bands of 4 fc. This simulation result will come to the last
stage after going through these all simulation blocks. So, this is clear, this we have understood
it. Now, I will stop this one, then this the converter;
so converter will have the switches, here the switches are assumed as ideal, so whenever
the sine is greater than immediately top switch is on plus 1 minus 1, then multiplied by Vdc
by 2, you will get plus Vdc by 2 minus Vdc by 2. So, here you will get the converter
1 and converter 2 voltages; so slightly shifting we will see. This is the converter 1 output,
this is the converter 1 output, this is the converter 2 output, these are the Vr (t) waveforms.
Let us see what is this one here? See, this is the converter output waveform
here, Vr (t) here, this will show the current waveform. So converter 1, the PWM waveform
is like this; it is going to 2800 by 2 tops and minus Vdc by 2 it goes and this suddenly
you see here, this is due to the capacitor fluctuation.
So, we are trying to similar task process possible to actual one. So, capacitor simulation, we said delta v
of plus or minus 5%, so we are doing that one. So now, it is stabilized. So, we have
started that one. So, now it is more or less constant. So, now let us see our converter,
current waveform for the converter 1. See, how to generate the current waveform here. See, current waveform we can generate We are getting from the PWM waveform; from
the PWM waveforms, plus Vdc to minus Vdc by 2, 1 by SL integral; here we will get the
current waveform, 1 by SL. So, this is the one; 1 by SL, this is the
one, this way you will it, 1 by SL, 1 by l. So, here is the integral, 1 by L and integral,
so while integrating we found that the dc offset error. So, to cancel that one, you
get the feedback here. So, this will give the complete motor current. If you see here,
simulation, converter current, it will be like this. See, this is the converter current, one converter
current, goes to the starting. So, it is now, slowly it will stabilise now. So starting,
this is your starting only, slowly it will stabilise, the converter current will get
stabilised, this is the converter current. Now, the second converter current, second
converter current will be, so this converter current is second converter current is here
that is for the first converter, this is for the second converter; so the Vab, so Vr, actual
Vr minus VS (t) 1 by SL that is this one, this is the VS (t) part minus one. So, that will give 1 by SL. So, here plus
or minus that that depends on how you take it; you can take it Vr (t) is equal to VS
(t) plus L into di by dt or the other way, whatever way you can take it but plus or minus
is not important here. So, this current comes here. Now, current we got it. So, from this
one, how do you find out the converter Id current? So, if you see here, these converters, there
are phase leg and phase b are there. So, these converter current, rectified current comes
at the dc side. So, here we will be multiplying by the plus or minus the one output here,
plus or minus one output here. So, if you see here, so the converter output; so plus
or minus, when the top switch is on, isk will be going to other side. During the negative,
when the negative sign is on, if the current is negative, it will go to the reflector on
the other side. So, the converter current if you see here, it will be like this. So, this is the converter output current. It is going negative ripple, 0 to negative.
So, this is because we have made regeneration, load is negative here. So, load we will make it positive here, then
we will get the current flow from ac to dc side. So, now if you see here the Id, see,
it is going to the positive side, quickly it changes. So, its transient is very minimum. So, this
is the Id current. Now, this Id current, these are converter
1, this is converter 2, now we have only shown the total current. If you see, if you want
to see the individual current, you can switch off this one and remove this one, individual
current we can put it maybe two currents, individual current. See, this is individual current, so the pulses,
it is starting, every time we switch off and shut, its starting. So, the individual current,
that is coming towards the Id side that is coming out of the converter, converter if
you say dc and the repel will be there, repel will part will go through the capacitor and
as I told there is another frequency. See, this capacitor fluctuation, we have designed
with two times the input frequency that is visible here. So, apart from the high frequency
repel, there is a two times frequency is there, this current. So, the capacitor repel has
to be decided for this two times that we are done be before for this PWM switching frequency;
this is for one converter. So, this one, we are giving to 1 by sc, so that will give your
capacitor voltage. Then divide by 1 by r that is our resistance. So, this is the block diagram. Now, let us see the primary, let us say our
VS1, total current Vdc and the transformer primary voltage. This is the starting and the starting here
is assumed, capacitor is not charged that is why heavier transient. So, simulation we
can do it. So, this is the transformer primary current, this is the mains voltage and this
is the capacitance, dc link capacitor voltage. So, here we can see, it is nearly unity power
factor, nearly unity power factor. Now let us see, let us reduce, more number of cycles
let us bring it here. So, that we can, I will make it 0.3; 0.1,
I will make it. So, more number of cycles you can see here.
So, this is capacitor repel is visible here. Now, I want to pause this one, pause and I
will suddenly change the load to minus. This is minus, applied that is our screen,
this is our screen and I say simulation continue. See, immediately power flow is power flow
is reversed quickly. So, this is the transient during the power flow reversal. See, quickly
power flow is reversed; this is the primary current, this is the primary voltage. Now,
let us see the total current and total voltage here that is here
that is this one. So, this is the one, this is the total current;
we can expand and see the repel clearly. See, during the X axis we can do it. So, we
will make it 0 to 10 times, we will increase it; see the repel. As i told, see because this opposition from
the two converters, that makes the harmonics at the second fc, side bands of second fc
get suppressed at the primary side. So, the primary repel will be much less, you can see
here; these are the individual converter currents. Now, let us see again here, increase the scale
here and see. The current reversal we can see it here; instead
of 0.02, I will make 0.1 here. So, this is the one. Now again, load, sudden
load change; I want to forward power flow. So, I will pause this one simulation, pause
it and I will go to this one, change the load to positive. That is the way we are simulating the forward
power flow and the reverse power flow here and simulation continue. Let us see. See, it quickly changes, very fast change
with nearly unity power factor. Previously, it was reverse power flow, now it is forward
power flow. See, quickly it changed, this is the one. See, here also if you see here, see it has
quickly changed. Here, two converter currents, we showed. So, this way, we can, before fabricating,
before final design of the converter, controller we can tune it like this. So, our assumption
and theories are correct. Now, let us see the effective waveforms here
it is coming out, whatever we predicted is coming. So, we will again go back to our design
quickly and come to our simulation result. So, I have shown this one to get a feeling
for you that so all the design with engineer approximation using the correct engineering
approximation, we can model the system, find out the transfer function and from the transfer
function, we can appropriate controlled design we can initiate and we can arrive at the controller.
Once that controller parameter is also designed, we can go for a simulation study and we can
use a various control blocks here and we use the simulation study here. So, this is the converter waveform. When the
converter is positive, the current will be going through the Id; the negative side, it
is the reflected one it will come. So, this is by multiplying the plus part and minus
part, plus 1 and minus 1 by the iS (t) current, the reflector Vdc rectified current, we can
get it. So, that is the Id; that Id, i showed you that is this one, this Id. It has a two times the main frequency component
is there. So, your capacitor has to be designed for that one. So, we have designed the capacitor,
capacitor repel also we have seen; now let us see, again quickly we go back to our fftb;
both is coming. So, design of the simulation front end converter;
we said, input voltage is equal to 1432 Vrms last class we have seen and output dc at 2800
volts and the switching frequency that is the triangle frequency 60 into 11 times. Now,
peak input voltage that is 2025 voltage, input frequency is 60 hertz, rated load current
500 amperes, assumed efficiency of 98 %; rated rms input current we can find out, equating
the input power to the output power with efficiency. So, this is the input current we can get.
Our rated peak current is this much. Assumed and modulation index, we set 0.8,
we will not go to the peak value of that. Our modulating is Vr (t); we do not want to
the peak value of the triangle waveform to avoid in practical case, to avoid another
switching pulses. So, modulation touch will only get 0.8. So, that means if 2800 volts
is our output dc voltage, our Vr (t) that is our Vab reference maximum we can go upto
0.8, 22 2240. So, that with controller gain, we have limited
that one. So, maximum rms converter input voltage, then peak repel delta v of 5%, that
we found 140 volts. So, this is the thing. This is the converter we have used, this is
the current, this is the current. See, this current is the current coming here. The one;
how to get this one Id? When this switch is on, this current will come here. Then these
two switches are on, whatever the current coming to this left that is if your taking
this current, the negative of that one will go here that is why we have taken the Vab,
this one positive side, Vab positive one, negative one multiplied by the current value
that is we are show showing the rectification; we will get the idrm Id current here that
is how 2 Id current we have stimulated here, then this is the capacitor. Then, also we found out that for this limb,
A limb and B limb of converter 1, the triangle waveform is 180 degree phase shifted and we
are using the same sign wave but the switching logic for this one is ((Ulta -reversed)) that
is whenever sine is greater than triangle when we are switching on this one here with
the 180 degree phase shifted triangle, we are switching on this one. So, for the fundamental
to get added and all the carrier frequency at fc and its side bands, get suppressed cancel.
This is true here also. So, this current waveform and this current waveform will have the harmonics
at 2 times the side bands 2 fc and it appears at the side bands of 2 times the carrier frequency. Now, by 90 degree phase shifted between the
triangle waveform of this one and this one, and this one and this one: two times, so 2
into 90, approximately 2 into 9, 180 degree. So here, these currents, the flux generated
by these two currents get cancelled and it will not reflect here. So, the transformer
primary current will get only the harmonics at 4 fc and its side bands; this, from the
simulation study we have found that one. So now, these are the current designs, see
the unity power factor, how to do that one so that we will generate the is in phase with
Vs, that we have done and we have given a slight phase lead to take care of the converter
delay because we are doing the sinusoidal currents here. Now, how to design the inductance? So, this
is very easy. From this equation, Vs is equal to Vs (t), Vs (t) plus the L di by dt is equal
to Vr. So, from this equation, Vr is the root that is it is a triangle, right angled triangle.
From there, we can find out Vr, L we got approximately this value here. Then converter gain G into 1 plus STr, triangle
we found for the 60 times, what is the time period, gain for a 10 volts, it is 280, we
can give that G. Then the current loop design; for a first
order lag equating that thing, we got the current, we require only a simple gain loop
here and the first order lag here. So, we found the current response is something
like this and assuming … factor from the control system design, from the standard text
book detail, zeta is equal to 0.707, we can design or we can find out the T. So, T is
equal to, that T initially we put it as a first order lag we want, the whole system
we refer it to the first order lag that is the equation. So, T we found, it is equal
to 2 Tr with this …, Tr is our triangle period. Then Ki, current line, that gain that
gain we have found out. Then these currents loop; see this current
loop is coming inside a voltage loop, the dynamics of the voltage loop is much slower
than this one. So, all the S square terms in the current loop we can, this part, we
neglect it. Then when we are considering this loop within the voltage loop and the current
transfer function, we approximate like this that is here. Then, we this one insert into
the voltage loop and would put PI controller. So, this is the final loop, once again I will
do it; this is the reference, this is the feedback. So, final loop, we get it. So here,
what you have to find out? The Kn and the Tn. Then we use the principle of modulus hugging,
closely transfer function. See, the purpose of this close loop is input should follow
the output as fast as possible. So, for wide frequency range, the ratio, the modulus of
the output by input should be equal to 1. So, how we can choose Kn and Tn such that
for wide frequency range, the modulus will be equal to 1? So, we studied that one and we got this equation
and we found this is third order system. So, for order systems, so b0 should be a0 and
b1 should be a1, one condition; so low frequency, this will be the one. And, for higher frequency, we found out the
modulus and got this equation should be 0. So, final transfer function with modulus or
with optimisation approach, what we have adopted, it will be like this; from this one, we found
out the gains Kn and Tn equating that one. This is the one we are using in our blocks;
the final block is like this. So, this is the one we have, this block we
have used for simulation. So, this is the reference, this is the feedback, this is the
voltage gain, then the PI controller we have selected the value and the PI controller has
a limit plus or minus 10, then this is the feed forward resistance. If you send the load
current, we can give a feed forward here such that any time load changes, it should not
the current loop should not act or need not act through the capacitor the output voltage
loop because any change in, if this is feed forward is not there, need not be or if feed
forward loop is not there, it reflects on the capacitor voltage and the correct reference
comes through this one otherwise quickly it can act through this way. Then we are multiplying this one with current
reference and a phase lead; current reference comes, then the current feedback. So, here
two converters are there; converter 1, converter 2 and each converter should supply equal power
and equal currents. So, both currents we can add and we can take the total current. So,
we do not require two different current loop voltage loops. So, one current loop and the corresponding
references, we can feed it to the converter. So, this is the converter 1 and converter
2. So, converter minus our input sign voltage 1 by SL will give our current waveform. This
current waveform rectified through the mains, it will appear as the Id1, Id1 we found it
has the repel content as well as two times the input frequency. This current one charged
with 1 by SC, you will get the output voltage and this output voltage, we will be feeding
here; this is the block we have simulated here. So, that is the MATLAB block we have
simulated previously. Now, how to find out the capacitor value delta
V? Again from, two times the input frequency. So, that is why we got this value, we got
approximately 5000 and we are using a standard value of 10,000 for our simulation. Then the simulation result; see this is the
one converter waveform, this is the simulation result that we have seen now. So, this is
the converter voltage. If you see here, this is the repel due to the capacitor, capacitor
repel two times the mains frequency. So, we can see that one; this frequency is not based
on our carrier frequency PWM frequency, it is depends on the input mains frequency, capacitor
voltage. And see, the large scale, the capacitor, delta
V 5% of our 2800 voltage. So, this is our capacitor voltage and these
are the, this is the mains, this is the main waveform what is shown here and inside this
one is the large one is the current waveform, primary current and these two are the two
converter current waveform. So, this is power flow from the source to the load with a unity
power factor. So, clear waveform is visible here. Now, this is, see you have two converter waveforms.
If you observe the waveform here, measure the waveform and the converter side that is
your transformer primary, here, this waveform will be the reflected waveform from the two
secondaries. So, what we do to get a feel for the transformer?
And, second harmonic will not be here. Harmonic at the harmonics at the second fc, 2 fc side
bands will not be there. So, if you see here, the primary side voltage will be, will be
like this. This is the primary side voltage. Now again, now we are suddenly quickly we
are doing the power flow reversal is there. So here, quickly we paused the waveform, made
the resistance negative and how to, see, but the natural practice this much will not be
there but we want to show that dynamics is very fast here. So, our controller shows also
accuracy of our controller design. See, quickly power flow reverses. So, power
flow changes, so there is a slight repel from the voltage waveform so that transient quickly
settles within two cycles, one cycle or two cycles it settles here; it is visible there.
So, this also shows our controlled designer, accuracy; our approach is correct. Now, the simulation I could not show because
it was not quickly, suddenly removing the load. So here, I suddenly removed the load.
If you see here, when the load is high, repel is high. So, maximum repel we are limiting
within plus or minus 5%. Now, the load current is slightly reduced. So, quickly changes.
So, capacitors repel also reduced but there is transient that quickly adjust, the transient
is very less. This also shows during transient conditions not only power forward for reverse
flow our controller design approach is correct. So, sudden change in load that is also correct. Then no load operation, this one shows. This
may be no load operation, we have transformer primary current will be there, magnetising
current. So, this is only what we show is or we have the … very no loads operation.
Now, let us go to the FFT. See, FFT of the current waveform IS1 converter
1, if you see, you have the fundamental and the harmonics, 11 times, 11 times it is somewhere
here that is the kind of figure; there all the harmonics are suppressed. Then the next
higher order harmonics happens at this side; 2 fc and the side bands and 3 fc also it is
cancelled, then 4 fc it happens; this is for converter 1. This is for the converter 2 but this is only
amplitude, there is a phase difference is there. So, their phase difference is 180 degree. So, if primary if you go primary current that
is suppressed and you have the harmonic side bands at 4 fc and side bands only. So, amplitude
is much less. So, this is also visible from the voltage
waveform that summed voltage waveform that I showed here. See, this is the equivalent
transformer voltage that means we are summing the primary Vr1 and Vr2. So, this is the voltage,
this happens at the 4 of it sides in the harmonics. So, what is the advantage here? Even though
we are using a carrier frequency of 11 fc; if it is fc, it is of 11 fc somewhere here.
But we could shift our high amplitude harmonics to this side not by resorting to high frequency
ft, 4 times fm, fm is here now, by properly doing a phase shift at PWM control, sine triangle
PWM control so that without resorting high frequency PWM switching, we could suppress
all the lowered harmonics and the harmonic amplitudes are at high frequency side; this
is the voltage. So, the impedance L, line impedance by L omega
is very high for these high frequency voltages. So, current is highly suppressed. So, you
have the fundamental and the other current. So, this is the simulation study. So, the
approach what I want to say; instead of blindly using P PI controller, there is a way to select
the P PI controller. That depends on the system transfer function and its order. So, first model with the symbol approaches
engineering model, we like the way, converter we approximated as the first order lack. So
same way, you arrive at the transfer function, various transfer function then link the blocks
and get the closed loop. Once the closed loop system is there, you start from the inner
loop; you start from the current loop. Then from the current loop, we try to initially
start with the PI controller, then try to cancel the pole zero cancellation; it was
not working, damping factor was missing. So, any disturbance comes, system will oscillate. So, with the first order, for a first order
lag, how to design the controller? Then we found current loop for the symbol because
the whole system goes through only the converter. That converter is represented as the first
order lag. So, we require a simple gain controller we require; we found that one and our approximation
is correct that is also proved from our simulation study and the various dynamic conditions,
results at the various dynamic conditions. Then we found that current loop is acting
very fast and the high frequency, so the current loop can be approximated to the first order
lag by removing the S square terms because the next we are inserting this current loop
into the voltage loop where the transients are the frequency of the current loop is much
lower than the current loop corresponds. So, we we represent as the first order lag and
then from the voltage loop, then design our PI controller of the voltage loop here. Then the whole system we modelled and simulated
and we got the result. So, this approach we can use for other, for not only for front
end ac to dc convertor, even for dc motor control or even for induction motor control
which we will be studying subsequently and for our not only V by f control, for our high
performance application like vector control, we can design same approach. So, when it when we come to that level, this
design approaches we will only mention it because already we have talked about that
one, we will use this standard approaches to make use of our controller dc. With this
one we will stop and now we have so far studied various controlled rectification; ac to dc
you have taken dc to dc converter, then ac to dc converter with unity power factor, so
now we can control the voltage with good efficiency good transient performance. Now, making use of this one, we can go for
a dc motor speed control. Here, we will be talking about the separately excited speed
control. Then how the speed of the dc motor can be controlled based on the same approach
what we have used; we will study in the next few classes.

Comments (1)

  1. @OSRcook ahaha..same here dude..I can't understand the one explaining!..LOL!

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