Mod-03 Lec-34 AC filters for grid connected inverters


Welcome to class 39 of topics in power electronics
and distributed generation. We have been discussing filter design, and we have
been looking at an LCL filter and the constraints that we have been looking at in
the LCL filter are the ripple injection into the
grid. The amount of reactive power that is drawn by the filter and the DC base voltage
that is required. If you have very large value of the filter inductance the amount of
reactive power required would be more, you also have consideration of what is a pass
band and stop band frequencies that you would like for the filter. Then you are also
looking at what would lead to a very efficient filter in terms of what value of parameters
for L would reduce to the power loss in the filter. And based on this one can come up
with the preliminary values for L, L 1, L 2 and C. And we saw that L 1 is equal to L
2 is equal to the total L divided by 2 would be
a suitable design and the C would then be related to the resonant frequency and the
value of the inductance that is used. So, what is shown over here is a 3 phase power
converter with a LCL filter; in the solid line you have the configuration for a 3 phase
3 wire system. And in the if you add the .dotted lines; where you have the connection
to the DC base being point. And to the output to the neutral then you have a filter
for a 3 phase 4 wire system. So, just if you look at the ideal LCL filter there are possibilities
of resonant that can happen such an LCL filter; one possibility might be when
say you are your inverter is not switching. So,
you could consider your L 1 side to be open and you could have a resonance between
your L 2 and C; and if the filter is still connected to the grid.
. So, in that particular situation essentially
the resonant frequency in this particular case
would be 1 by square root of L 2 C. And at the resonance you have this particular
condition for the impedance in the circuit. But this particular scenario might to some
extend can lead to over voltages on the capacitor but you have the anti parallel diodes of
the switch; which can actually absorb some of the energy also you have the possibility
that if you have such a response if you have a upstream contactor or a circuit breaker;
if that is open then the excitation to such a
cycle would be prevented. So, it depends on how you are operating the power converter.
Another possibility for resonance is when you when the power converter is operating
as a voltage source inverter; in which case what
you would have is a your L p which is essentially the parallel combination of L
1 and L 2. And this particular impedance condition being infinite would lead would
lead to operation of this filter at resonance; this resonance can be excited both from your
inverter side and also from the grid side. .And you can have over currents in the filter
and you could have say voltages across the capacitors or the filter components. This
particular configuration cannot be prevent cannot be mitigated the opening of a upstream
breaker. Because you are actually operating the power converter and you are
interested in sending power exchange between your converter and the grid. So, you
need to actually damp this particular resonance effectively.
. So, if you consider the excitations that can
happen for the resonance; it can be both from the power converter side or it can be from
the grid side. If you look at the power converter side the primary excitation is at
the fundamental frequency and the switching frequency. However you have non idealities
of the inverter such as a dead time, the on state voltage drops, the turn on the turn
off delays etcetera; which introduce additional frequencies in the output of the inverter;
you also have deter in the control timings, you
have contestation of the in the control. So, many such factors can introduce noise at the
output of the power converter which can actually need to a excitation.
However, whatever comes from the power converter is to a large extend controlled by
the designer. If you have excitations from the grid side it could be because of harmonics
from connected loads, it could be from operation of neighboring converters which are
injecting voltages at the input of the filter; you could have non idealities of the
equipment. For example, you have a transformer over here and the non idealities of the .non-linear b h loops or power resonance in
the transformer can actually introduce harmonics from the grid side. You could also
have upstream contactors or breakers which are cycling and introduce frequencies
which can excite the filter; these external excitations from the grid side is not under
our control it depends on what is connected to
the external side. And is important to address the resonance and ensure that you do not
lead have overloading of the filter. Because of over currents or higher voltages appearing
across the components; which can actually lead to damage of inverter component or
tripping of the inverter. So, if you look at the resonance in a in such a LCL filter
one way of addressing the resonance is by introducing
a resistor into the network. . So, as soon as you introduce a resistor in
a network your power dissipation goes up. And
objective of a passive damping is to reduce the quality factor or Q F with minimum
power loss. So, in this particular where you have the resistive damping we are
considering adding a single resistor to the LCL filter. .. And, a single resistor can be introduced in
multiple ways. So, if you have an LCL filter you could think about say adding resistor
in series with the inductor. So, this could be
one possibility; so in this particular case whatever power flow is happening between a
power converter and the grid will introduce power loss in this resistor. So, you would
have a increase in losses in this particular resistor. So, another possibility of introducing
a resistor damping element could be say in series with the capacitor; in which case the
fundamental power flow between the inverter and the grid will not affect the power loss
in the resistor to a large extent; you would of course have your ripple current going
through your capacitive branch. And the ripple current would now introduce losses in the
damping resistor also the capacitor would draw some fundamental reactive current which
would also cost losses in the resistor. So, if you look at the configuration between
putting a resistor in series with the inductor or a resistor in series with a capacitor;
the one with the resistor in series with capacitor would lead to lower losses for a given level
of damping. And you can see that in this particular case you have your LCL filter plus
one component which is essentially the damping resistor; if you now think about one
way in which you could reduce the losses in the damping resistor is to provide a path
for your ripple current to flow which is in parallel with damping resistor. So, by connecting
say your resistor C 1 parallel with the damping branch or essentially what you have
done is you have split your capacitor C into 2 branches C 1 and C D. And you connect your
resistor in series with one branch and the .capacitor C 1 is directly connected across
you are as the LCL filter. So, your ripple current essentially would primarily go through
L 1 which reduces the power loss in the damping resistor.
Here, your complexity has gone up so you have your LCL filter plus 2 components. So,
you would expect the power loss to come down but the number of components have
actually gone up. If you look at then case of the S C R L damping you have a LCL filter
plus 3 components; you have L 1, L 2 and C 1. Then, you have 3 components C D, R D
and L D. So, you can see that the complexity is increasing as you go from a simple
resistive damping to the split capacitor resistive damping to the split capacitor R L
damping. So, if you look at the as you increase the
number of components your complexity of the circuit goes up also often the cost of circuit
would go up; if you have more components. So, your it does not mean that for all situations
you have to go in for the most complex network for lower power levels; where say
for example you might be operating at power levels of less than a kilo watt. Then, you
might go in for essentially the direct resistive damping. And as the power level increases
you would go in for say for example you are looking at the order of 10 kilo watts maybe
you would go for split capacitor damp resistive damping. And if you are talking
about 100 of kilo watts to mega watt power levels then the added complexity would not
matter that much. But the reduction in power loss would be significant; so you would go
for the more complex forms of damping. So, in each of the situations you will have to
look how to select your R D, L D, etcetera. So,
we will start with the simple resistive damping in which case if you do a transfer
function analysis; you would see that the transfer function is relatively straight forward. .. If you look at your capacitor voltage to a
inverter voltage transfer function essentially you would have S L 2 divided by 1 plus R d
C S divided by S L 1 plus S L 2 1 plus R d C S plus L p C S square; in
this particular case L p is the parallel combination of L 1 and
L 2. The natural resonant frequency of this particular transfer function can be obtained
in a straight forward manner; where the numerator
is the second order 1 0 at the origin another based on the R d C time constant;
the denominator again we have pull at the origin. And then you have a second order transfer
function from which one can easily find out the location of the pores.
The undammed natural resonant frequency of this filter is omega r is 1 by square root
of L p C; and you could obtain an approximate
expression for the quality factor at this particular frequency omega r. So, you could
define your quality factor Q F for this particular circuit as V c (s) divided by a
V i (s) at s is equal to j omega r divided by V c
(s) divided by V i (s) at s going to the origin. So, you could evaluate this in a straight
forward manner and you get an expression where you have 1 plus square root of L p by
C divided by R d the whole square. So, you can see that when R d is extremely small you
have a very high quality factor; which means that you are filter can oscillate very easily
when or your added value takes on a larger number; then your quality factor comes down
essentially damping out your resonance. .. So, if you have a more complex passive damping
circuit; such as this split capacitor R L damping or a split capacitor R damping one
can actually do a analysis by looking at the model of for the filter. And one particular
model might be a if you look at is state space model of such a filter. And for the state
space model we will consider one possible set of
state variables might be the inverter side current, the grid side current, the voltage
across the capacitor C 1, the voltage across the
capacitor C D and the current through the inductor L D; the inputs to this particular
model could be the excitation from the inverter side and the excitation from the grid side.
So, if you take inverter side excitation it essentially your p w l voltage; if you are
considering the line to D C base voltage you are
looking at excitation of either plus V D C by 2 or minus V D C by 2; from the grid side
if you are assuming the grid voltage is a pure
sin wave at the fundamental frequency. So, you have that particular frequency component
at say 50 hertz; for all other frequencies essentially the voltage source
can be considered as short circuit for your analysis. And looking back at the inverter
again we assume the inverter to be a voltage source; again assuming that this is voltage
source your ignoring the control related impedances that can be seen in a inverter.
But often the control bandwidths of the inverter is lower than the resonant frequency
and approximating it as a voltage source is a good approximation. .. So, if you look at the state space model of
this S split capacitor R L damped LCL filter; you have a equation state equations of the
form x dot is equal to A x plus B u; y is equal
to C x plus D u you are a matrix is 5 into 5 matrix; the 5 states in this particular
model you have 2 inputs. So, your b matrix is of
this particular form you are inputs are your inverter voltage and grid voltage. And your
state variables are given by is selected to be
these 5 variables one particular output of interest is essentially a your current in
the damping, resistor; the current in the damping
resistor is good information. Because essentially the power loss in the damping
resistor is given by your i R d square R D gives
you your power loss then R 2. So, again keep in mind the objective of a passive damping
network is to reduce your power minimize your power loss and achieve the lowest
possible, lowest quality factor in a circuit. Again, the definition of the quality factor
we will take it as the peak value of your transfer
function. .. So, here the transfer function that is plotted
is the capacitor voltage by the into grid inverter voltage. So, you are looking at essentially
this particular ratio of the peak value at the resonant frequency to the peak value
at D C to be your quality factor. And if your quality factor is close to 1; it means that
you’re not having any peaks over there exciting resonance or essentially you are not amplifying
any input when you are at the particular resonant frequency.
So, there are actually a couple of ways in which you could look at the quality factor
of such a S C R L damped a passive damping network.
And one is to again use this particular definition of the quality factor
in this particular ratio; again the assumption that
your resonant frequencies stays at omega r; where omega r is equal to 1 by root L p C
is actually not exact. Because you know that
adding passive damping components to the circuit will actually shift your resonant
frequency to some extent. So, this is an approximation but still you could make use
of that to evaluate what your quality factor is. .. And keeping that in mind one can define the
quality factor in this particular manner; the
transfer function V c (s) by V i is actually now a ratio of a polynomial; it is third order
in the numerator, it is fifty order in the denominator.
So, it is not easy to directly simplify it to get the roots of such of a polynomial.
So, these approximations help in giving you ah
comparatively insights into what the quality factor is.
So, assuming we will make a couple of assumptions; one is we will define the term K to
be the ratio of R D by L D; R D by L D is actually if you look at the passive damping
network is the corner frequency of essentially this R D, L D branch. And if you look at
the ratio of that particular corner frequency to the fundamental frequency; you have the
term K which gives you a feel for what the how the R D, L D filter time constant should
be placed with respect to your fundamental frequency; we also saw in filter design that
taking L, L 1 is equal to L 2 to be equal to L by 2; it would be a suitable design
guideline. Also we will see that taking C 1 to be equal to C D to be which would be
C by 2 would actually also lead to a fairly good
design. And we will also see that it selecting R
D to be square root of L by C would also be a good design choice.
So, with these assumptions in this particular transfer function we could plug in and then
take this particular ratio and simplify the quality factor. And it comes out to be a fairly
simple expression twice the square root of this term this quadratic term plus 1. And
again .we are assuming the resonant frequency has
not shifted they are still at square root of L p
C. So, from this particular expression we can
see that a couple of things are possible; one is
by a appropriate selection of the R D by L D term; it might be possible to make this
particular term equal to 1 which means that your quality factor would have a value close
to 2. So, by suitably selecting this particular L D you might be able to make your quality
factor get close to 2. Another thing that it shows is there is a selection of K; which
can be related to your omega and or fundamental frequency
which is your main frequency at which you are trying to reduce the excitation
the flows through the R D branch to actually reduce your quality factor.
And, this normalization term K is given as R D by L D into one by your fundamental
frequency. Note that this is an approximate way of a looking at the quality factor; you
can have a alternate method of looking at the quality factor; which would be by directly
going to a numerical approach you plot the boarding plot gain plot of this particular
transfer function. See where your peak of your resonant peak of your transfer function
is and evaluate this particular peak value to
your D C value ratio that would give you a exact value for your quality factor. So, that
would be numerical as this gives you an approximate field for what the quality factor
is. Now, that we have a framework for evaluating the quality factor.
. .The next item of importance is to evaluate
the power loss in this damping filter network. And the power loss in the damping filter again
primarily involves the excitation; that the main excitation of this filter is the 2 frequencies
which would be the fundamental frequency and the resonant frequency. So,
your fundamental frequency excitation can then be evaluated by assuming that in your
LCL filter; we can assume that at fundamental frequency the inverter voltage
is close to 1 per unit also your grid voltage is
close to 1 per unit. And we can actually find out what would be the current that flows
through this particular resistor; in this R D at the fundamental frequency from a direct
phase of analysis. And, using a phase of analysis you could actually
then get an expression for your fundamental current flowing through the branch.
And we have a power which is equal to the re of V i conjugate. And V is 1 per unit
and evaluating your i conjugate in terms of the expressions of that particular branch;
you can get a simplified expression as a power of the fundamental frequency on a per unit
basis is C D, R D; C D square R D divided by
K minus C D R D the whole square plus 1; where these terms are again in per unit as
form. So, one can see from this expression that
as K takes on a large values your power loss through the at fundamental frequency can actually
come down. Because you have a quadratic term the denominator. So, as this
particular term increases your power loss can
actually come down; the second component that is to be evaluated is essentially the
power loss because of your ripple frequency. So, what P r i indicating the power loss in
the damping network due to the ripple component, due to the switching action of the
power converter. And this can be evaluated using the state space model of the power
converter. .. So, if you look at this state space model;
the model of the power converter of the filter is
essentially x is equal to A x plus B u and y is equal to i R D. And this is essentially
equal to V c minus V d by R D. So, in this particular
case if we include the winding resistance; you could have generate a matrix this invertible.
And your inputs are u is essentially V i and V g and we have V i is equal to plus V
d c by 2 or t belonging to the duration 0 comma T O N; where t O N comma T s w. So,
this particular duration is your T off and if you look at your grid side voltage; we
will take our grid voltage to be equal to V d c
into d minus 0.5; where d belongs to the range 0 to 1.
So, this means that your assuming that at your inverter voltages exactly balanced by
your grid side voltage. So, you are assuming a
cautious steady state at each duty cycle of operation of the power converter with the
LCL filter. So, you could then obtain a solution for the state equations. .. So, you would have you could write your X
at your switching interval is e to the power a
T s w times your X naught plus integral 0 to d T s w; which is your t on into the power
of A T s w minus tau b u 1 of tau d tau plus
d T s w to T s w into the power of A T s w minus tau b u 2 of tau d tau again you want
is essentially your plus V d c by 2 comma V
g. And V 2 is essentially minus V d c by 2 comma V g. And if the system is in steady
state it means that at the end of the switching interval at T s w; you get back to the same
point where you started so this would be equal to your X naught. And given that your e
matrix is invertible; we can actually solve this particular equation to obtain what your
X naught is. .. So, once you know your X naught your initial
condition then you could use that to evaluate your current in your damping branch;
numerically at points at n points between 0 and T s w. And this is this is then used
to evaluate your current R M S current ripple during the duration T s w. So, essentially
what you are doing is if your time is proceeding like this 0 T s w to T s w will
be further down; essentially your dividing it
into n points. And using the state evolution equation you know your value of X naught
at. So, you know your X naught at this particular point; you can use that to evaluate your
state variables at all the subsequent points. Because you know your excitation you know
your initial conditions you can solve your dynamic equation of the filter and essentially
the output it is evaluated is your i R D. So, once you have your i R D evaluated at
these points; you could then generate your R
M S currents over a switching interval over T s w using n sub intervals in T s w. And
you have i R D R M S over T s w is summation over
the n points of the square and you average it; and this is for a given duty cycle.
So as your duty cycle is waiving between zero and one this particular i R D R M S would
vary over the different switching periods different T s w’s; as one goes over the
fundamental cycle. .. So, if you look at a fundamental cycle you
have i R d r m s over a fundamental cycle to
be equal to the square root of 1 by T; where T is 20 milli seconds of j is equal to 0 to
p; i d r m s square over the duration switching
the duration T s w over all the p points times T s w. So, essentially your P is equal to
T divided by T s w; where T is 20 milli seconds for 50 hertz and T s w depends on your switching
frequency that you are using for your particular filter. And essentially once you
have your i R d r m s your power loss at your ripple frequency is i R d r m s square times
R d. So, you could actually look at this particular
evaluation. So, essentially what you are doing is now that you know your r m s current
over a switching interval what you are doing is your now looking at 0 T s w to T
s w over the overall fundamental cycle T. And
looking at the r m s the power loss and the ripple frequency from the r m s loss through
this particular damping branch. And the total loss power loss in this damping resistor is
because of your fundamental frequency power loss and the ripple frequency power loss.
So, essentially what you are trying to do is to minimize your quality factor while keeping
your power loss in this particular total power loss to be a small value. So, that is
essentially the overall frame framework. And now we have a method for evaluating both
the quality factor and the power loss could actually go about looking at how to actually
select the components of the LCL filter. .. So, what is shown over here; if you look at
the power loss over the different switching durations, over different switching T s w’s.
And as a function of your duty cycle d what is plotted over here is i Rd r m s. So, you
can see that the losses in the damping branch is
maximum for a duty ratio d. But because you are as you are going over your fundamental
sine wave your duty cycle is sweeping a value depending on your modulation depth; it
may be between say 0.1 to 0.9 or 0.2 to 0.8 depending on your modulation depth or the
inverter. And then you evaluate the r m s components of over the different switching
situations and that particular value is what is plotted over here i r m s in R d over the
fundamental cycle. So, this is over a switching duration T s
w which is now a function of d. So, if you then
take the overall r m s we will then get a value of the current which is lower than the
peak the worst case being at a duty cycle point
five for a 3 phase 4 wire system; if you do again a r m s evaluation for a 3 phase 3 wire
system the power loss in the damping branch is further reduced. So, if you are
doing a 3 phase 3 wire design adopting this particular approach to give a consecutive
value for the design; in the sense that the actual
loss is would be lower than what you would have if you are thinking analyzing as a 3
phase 4 wire power converter. .. So, to actually then proceed with the design
we will look at an example design; the example that we would choose is a 40 kilo
watt power converter with a line to neutral of
voltage 240 volts. So, given the power level and the voltage you can then calculate your
base current; from which now you had base voltage and base current you know what
your impedance base is. And then you can calculate your this inductance and base
capacitance and your base frequency is essentially 50 hertz or in radiance to 50 radiant’s
per second. Now, the use of the base per unitized notation
for a power filter is actually useful because you want to compare different filtering
up rage. And this looking at in milli Henry and micro Farads you would not make
the comparison directly easy to compare whereas doing a comparison on a permanent
basis can lead to a simpler and easier comparison. .. So, for the design that we are considering
looking at a switching frequency of close to 10
kilo hertz. And a resonant frequency of 1 kilo hertz you are taking considering 3 possible
damping operations of just the R damping; then the split capacitor resistive damping.
And the split capacitor R L damping and we will take the value of C to be 0.25 per unit.
And in case it is split we will consider C 1 is equal to C d is equal to C by 2. So,
it is just 1.25 the value of R d is selected to get quality
factor of close to 3. And if we could see that under this particular condition both
the S C R damping and the S C R L damping it
is possible to meet the given attenuation requirement into of this case minus 65 d b.
Whereas in the R damping because of the damping resistor position you are not able to
get that level of attenuation. And, we will see that from the transfer function
why that is the case you can also see that in case of the R damping the power loss is
about 1.5 per unit. So, if you are talking about
a 1 kilo watt power converter, you are talking about a power dissipation of about 15
watts which might be reasonable to handle. Whereas, if you are not now talking about
a mega watt converter you would this loss might
be much higher; you are talking about is of 15 watts, 15 kilo watts.
So, going to higher more complex damping network would lead to a lower much lower
power loss of 0.07 percent. So, you can see that as the complexity of the damping .network is increased, your power dissipation
in your damping circuit is actually coming down.
. So, to look at the recent why I your one when
you have your one S C R damping what is shown over here is a plot of ideal LCL filter.
So, this is the ideal LCL filter and what is
shown over here is your resistive damping. So, at low frequencies the transfer function
between your grid current to your inverter X voltage is falling off at 20 D B per decade;
for the resistive damping at the higher frequencies the role off is at minus 40 d b per
decade whereas, for the ideal LCL filter it is actually minus 60 d b per decade. So, if
you are switching frequency is 10 kilo hertz;
you are talking about a lower attenuation of this
particular S C R damping network compared to the ideal LCL filter.
If you look at the case of your S C R L, S C R and S C R L type of damping network
they both give a similar level slightly lesser damping than your ideal LCL filter. But the
attenuation at high frequency is at minus 60 d b per decade. So, that is the reason
why this particular case you are not able to achieve
the decide damp level of attenuation with the R damping. But you could get a higher
attenuation with the S C R and the S C R L damping. .. So, one approach to look at the selection
of the parameters of the damping S C R L damping say LCL filter is to look at multi
parameter optimization. So, the objective of
the optimization would be the same to minimize your quality factor and to minimize
your power loss. So, you can then look at a multi parameter optimize optimization
approach; what gives you minimum values gives you the range of L D, R D, C D etcetera
that gives you puts you at the minimum. But often the output of such an optimization
engine may not be intuitive the optimization engine will give you particular design. But
it might not give you the intuition behind why the particular design gives you lower
value of losses. So, the approach that we would take is actually
to think of the more complex S C R L damping as adding one complexity at a time
to the ideal LCL filter. And you use that to
get a insight to what exactly one is trying to achieve by the passive damping network.
So, essentially you are adding one component to and simultaneously looking at you
factor performance factor such as your quality factor and your power loss and the
implication of adding one component at a time. So, you take your LCL filter converted to
into a S C R damp circuit and then convert that into a S C R L damp LC L filter. .. So, again the filter values that we are starting
off with is with L 1 is equal to L 2 0.02 per
unit in our particular design example. And C to be 0.25 per unit again based on the
resonant frequency constraints that we have chosen in our design example.
(Refer Slide Time: 47: 46) So, a starting point for the design would
be to consider how to split C into C 1 and C d.
And one could consider C 1 and C d to be in a ratio of C d equal to a c times C 1 and
also to C 1 plus C d to be equal to C. So, the
total value of the capacitance is kept at C and
they split between C 1 and C d in the ratio of C d is equal to C d by C 1 is equal to
a c. .. So, for example if you are value of so if
you are a c is equal to 0l it means that C is equal
to C 1 because your C d would be 0. So, C is equal to C 1 which again would correspond
to the condition where you have no damping. Because if C d is 0 and C is equal to C 1
your essentially your damping circuits damping branch is open circuit. So, if you look at
the other end of this extreme case a c effect takes on a very large value this implies that
C d is 10 into 0; this implies that your circuit would simply to essentially your R
damping. So, when a c takes on a very large value essentially your C 1 would be 0; so
essentially your circuit would simplify to your R damping.
So, what it means is if you look at a the trade off in this particular case essentially
a c having a small value would mean that essentially
you do not have any damping, your quality factor is very large. And a c taking
on a large value essentially you have R resistive damping. So, your quality factor
actually comes down if you look at the knee of
this curve at a value greater than 1; you do not have a significant decrement in the
value of a a c. So, region of a c greater than 1
might be considered a suitable design such that
you would get a quality factor which is less than 3.
So, selecting a value of a c greater than or equal to 1 would give a quality factor
less than 3; the other thing that you could see is as
you are a c value is increased essentially the
losses in the damping circuit goes up, because as a c is increased essentially the capacitor
C d is being increased. So, your losses in the damping branch goes up; also you can see .that as your switching frequency is reduced
you have essentially increased losses in your damping resistor branch which is to expected.
Because your switching frequency is getting closer and closer to your resonant
frequency. So, as the switching frequency is reduced
your power loss is going up; power loss is going up as your switching frequency is reduced.
So, this is looking at your power loss as
a function of a c for different switching frequencies; what is interesting to note is
if you plot your minimum quality factor versus
your factor a c; the minimum quality factor actually order minimum quality factor curves
by one on top of the other; which means that the quality factor is independent of
your switching frequency. If R d is selected to
actually in such a manner that minimizes your quality factor term.
This also implies that a design point of say if you say guideline a c is equal to 1.So,
selecting a c is equal to 1 which means that as C 1 is equal to C d is equal to C by 2
would be a good design selection irrespective of your switching frequency; whether it is
2.5 kilo hertz or 9.10 kilo hertz of what would be in a range of switching frequencies
that are commonly used in power converter design;
this particular selection of dividing C 1 to
C d would actually lead to a good design. . So, then the next thing one can look at is
you could once you have actually plotted your taken your C 1is equal to C d. Then, you could
actually try varying your R d we initially saw that R d is equal to square root of L
by C might be a reasonable design choice to .validate that we could vary R d in a neighborhood
from 0.5 or in this particular case 0.6 root L by C to twice root L by C. And you
can look at the variation of the location as
your R d is varied from a small value to a large value. And you can see that the point
at which you get loose damping because lines
of constant damping is in this particular orientation. And as your damping is improved
you will get closer and closer to your real axis.
So, you can see that the point at which your you get just
damping is when R d is equal to square root of L by C; when you are a c takes
a value of 1. So, making use of your state space model of the system you can actually
look at your pole locations and see what value of R d gives you your best damping.
And this confirms that selecting R d of L by C
gives you a good value of damping of you are a S C R damped network.
. So, if you look at then a good starting point
for your design choosing C 1 is equal to C d
is equal to C by 2 and L 1 is equal to L 2 would be a
good starting point. So, here the values are given in physical per unit values
and physical values; the resonant frequency is taken as 20 which 20 times your 50; which
is your fundamental frequency would be 1
kilo hertz and your switching frequency in this particular case is10 kilo hertz. So,
this gives you a initial design point for the LCL
damped a split capacitor resistive damped LCL filter. So, in the next class we will look at how we could
then add the term L d to
such a S C R damped network; again in manner which gives
you a design that minimizes .your quality factor. And reduces the power loss in the damping filter branches
simultaneously. Thank you. .

1 thought on “Mod-03 Lec-34 AC filters for grid connected inverters

  1. Hi everyone.knows any to tell me,what if i run my spwm inverter,with any type of load,without low pass filter to the output?,whatt the matter????? must be a filter anyway? and if yes,why? thank you.

Leave a Reply

Your email address will not be published. Required fields are marked *