In this lecture we will develop the basic analytical techniques that allow us to solve for the voltages and currents of any arbitrary switching converter.
Specifically we will employ the small ripple approximation to simplify the equations, and we'll derive the principles of inductor volt second balance and capacitor charge balance that give us the DC equations of the converter circuit.
So if you recall from the previous lectures we discussed the buck converter having a switch and having a low-pass filter that removes the switch in harmonics but allows the DC components or the DC component of the waveform through to the output.
Now, in a practical converter, we can't build a, a perfect low-pass filter. The low-pass filter will have continuation but it won't completely remove the switching harmonics of the waveform. [COUGH]
And therefore, the output voltage of a practical converter might look like this where there's a DC component capital V that as we previously found, is equal to the duty cycle times
the input voltage vg. And then in addition to that, we have some small ripple that is at the switching frequency and its harmonics that, where a small amount of the switching ripple gets through the filter and appears in the output.
And as sketched here, this ripple isactually probably
a lot larger than it would be in practice.
So, in a practical circuit, we'll have
some kind of specification on how
large this switching ripple can be, and
generally it's a very small number.
So in an output voltage of say a volt
or two for a computer power supply,
this ripple might be ten millivolts
or, or some very small number.
so therefore we can write the equation of
the
output voltage VOT, as being equal to
capital V which
[COUGH]
here I'm using the capitals letters to
denote the DC components.
So capital V is the desired DC component
equal to DVG.
And then plus some AC variation, that's
called V ripple of T here, that
is the undesired, switching harmonics that
make
it through the filter to the output
voltage.
[COUGH]
So in a well designed converter, the cell
c filter will have lots of attenuation.
And the ripple will be very small compared
to the DC component.
So we call this the small ripple
approximation, where the,
the ripple is small compared to the
desired DC component.
And under certain circumstances we will
neglect the ripple and
simply approximate the output voltage by
its desired DC component.
Okay, this has the effect of decoupling
the
equations, the differential equations of
the circuit, and making
them very simple to solve.
Okay, so this is called the small ripple
approximation to ignore this component and
simply approximate V of T by capital V.
Okay, brief discussion here.
[COUGH]
as we're
going to
see in a
minute,
the small
ripple
approximation
formally
can be
applied
only to
continuous
wave forms
that have
small
ripple.
So, we don't apply the small ripple
approximation to switched wave forms in
the circuit such as the switch output VS
of T
[NOISE].
And with these being a second order
circuit, in
general, the kinds of wave forms that we
get are
sinusoidal or decaying exponential in
nature.
So, the actual IL of T, when we solve
this circuit might have some wave form
that looks like
this, for example, with some initial value
IL of zero
and then some solution, of the second
order differential equation.
[COUGH]
Now in a well-designed convertor, this low
pass filter has a cutoff frequency
that is very low in frequency relative to
the switching frequency.
What that means is that the time
constants, or the time that it takes
this response, this ringing response, to
happen,
is long compared to the switching period.
So one switching period might be just this
long, for example.
And in the small ripple approximation,
what we're actually doing is
approximating this solution for a short
time with a straight line.
And for that reason we also often will
call
the small ripple approximation instead,
the linear ripple approximation.
And any continuous wave form
for short enough time can be approximated
by a straight line.
And so that is what we're in fact doing.
Of course if it's a discontinuous wave
form such as a switched
wave form then we cannot in general
approximate it with a straight line.
[COUGH]
So we apply the small ripple approximation
as appropriate to inductor
currents and capacitor voltages that have
responses such as this one.
That we can approximate over the switching
period or
a fraction of the switching period for the
straight line.
Okay, we can do a similar thing with the
switch in the second
position as well were the inductor left
side of the inductor then is connected
to ground instead of VG We get another
circuit to solve, and we can
find the, the wave forms for that interval
as well, in a similar manner.
So, let's look at, for example, for the
first interval, finding the inductor
current wave form.
using this small
ripple approximation.
So from solving this circuit, we have a
loop right here that the inductor is
connected in.
And the loop equation for this circuit is
the inductor voltage BL
is equal to the input voltage VG minus the
capacitor voltage.
Here it's called V.
So we get this equation for the loop the
the inductor is connected in.
Okay.
Now, V of T is the output capacitor
voltage, and as we've discussed
so far, we want this capacitor voltage to
have small ripple.
And so V of T again can be expressed as
its DC component, capital
V, plus the small ripple, and here is a
case where we can replace
[INAUDIBLE]
by its DC component capital V to find the
inductor voltage for this interval.
And so this is a good approximation over
the
short time of when the switch is in this
position.
Okay, so this is a small ripple
approximation.
Knowing the inductor voltage we can now
find what the inductor current does.
Using the well known defining equation of
an
inductor, that V is LDIDT in the inductor.
So if we plug this expression for the
approximate inductor
voltage into there, and solve for DIDT, we
get this
equation Which says that the inductor
current changes with a
slope that is given by the voltage over
the inductance,
with the voltage being equal to a constant
value, Vg minus capital V.
Okay?
So, for this first interval, The
inductor current will start at some
initial value,
IL of zero.
And it will increase with the, this slope.
And it will increase with that constant
slope.
Until this first interval is over, and
the, the switch
is changed to position 2, which happens at
time DTS.
[COUGH]
' Kay.
Then, during the second interval, we get
this circuit,
with the left side of the inductor
connected to ground.
And the loop equation now around this loop
where the inductor is
connected, says that the inductor voltage
VL is equal to minus V.
Like this.
We can again use the small ripple
approximation to replace V of T
with its DC component capital V, and
ignore the ripple during this interval.
And we can plug that voltage into the
defining equation of the
inductor again to find the slope of IL
during the second interval.
And what we find then is that the inductor
current for
this interval changes with a slope equal
to minus V over L.
[SOUND]
So we started at some initial value, we
went up
during the first interval and ended at
some point.
[SOUND]
And then during the second interval we now
change with a different slope.
And the slope is negative, assuming the
output voltage is positive.
So we get a slope for this interval of
minus V over L, and the current
goes down, until the switching period is
over, and we end at, at time
TS we end with some final value.
[COUGH]
' Kay, so the small ripple approximation
makes it simple
to, to sketch what the inductor current
wave form looks like.
We have straight lines that have constant
slopes and are easy to solve.
And this greatly simplifies the equations
so that
we don't have to write decaying
exponential type functions.
And have pages of algebra.
So then here's a summary of what the
inductor
current wave voltage and current wave
forms look like.
We found the inductor voltage during each
interval, it was positive
and equal to an approximately constant
value during the first interval.
And during the second interval, it was
equal to an approximately constant
negative value.
what this did was made the inductor
current go up during the first
interval and then go down with a negative
slope during the second interval.
Okay, one thing we can do with this right
off the, right
away is, get an equation that helps us
choose the value of inductance.
You can see that the actual inductor
current wave form has switching ripple.
The current goes up and down with a
traingle, triangular wave form at the
switching frequency.
And on top of that it has an
average value or dc component, which is
the dc value of inductor current.
That we'll call capital I.
To design the inductor, choose an inductor
value, we generally will
try to limit the amount of switching
ripple in the inductor current.
How much ripple we allow is a design
choise, and at least for the
next several weeks we will.
Discuss converters where this ripple is
limited in value to maybe ten or twenty
percent of capital I at full power.
' Kay, so the ripple, we commonly define
the peak to average ripple as delta IL.
And so two delta il is the peak to peak
ripple, which is from here to here.
And from this waveform and knowing the
slope, we can
easily write an equation for how large the
ripple is.
So for example, during the first
sub-interval, with the switch
in position one, the inductor current
changes from here to here.
And the net change is two delta I.
So we can write two delta IL is the
change in the inductor current, and this
is simply equal
to the slope during the first interval.
Vg minus V capital over
L times the length of the first interval
with is DTS.
Okay?
So knowing the slope and the time, we can
find the change.
And we can solve then for delta IL.
Just divide both sides by two.
And we get this expression.
Okay?
it's, this is the equation that is
commonly used to calculate
the value of inductance, and choose an
inductor for the converter.
So we can solve this equation for l.
Push l onto that side and get this
equation.
And from this equation we can solve for
the value
of l that gives a given value of delta I.
Okay, there's another thing we can do
besides finding the value of L.
We would like to find the steady state
voltage and current in the converter.
So, let's consider now what the inductor
current
does during a turn on trangant of the
converter.
[SOUND].
So, for example,
[COUGH].
Let's suppose that we, start the circuit
with the inductor
current equal to zero.
And the output voltage is zero also.
So we'll start at zero with the circuit
turned off.
And at t equals zero, we'll start
switching with So we'll
switch between positions one and two with
some fixed duty cycle.
Okay?
We've already found what happens, the
inductor current
will change with some slope during each
interval.
And so during
the first interval, the very first
switching
period, the inductor current will go up.
It with a slope that we've already found,
is equal to vg minus v over l.
Okay, well initially what v is the
capacitor voltage, vc in fact let's
just call this v, instead of vc.
So the voltage is initially
zero at the output, and our slope during
the first interval is V G over L.
Okay?
So we go up until the end of the
first interval at time DTS, we've reached
some positive current.
[SOUND].
Okay.
What happens to the output voltage?
Well, its inductor current is small, but
it's a little more than zero.
That inductor current flow to the output.
And it will slightly charge the capacitor
voltage up a little.
So V is still small but its a tiny bit
more then zero.
Okay during the second switch interval
with a switching position two, we found
that the conductor current changed with a
slope of minus V over L.
' Kay, so we have a slope here, of
minus V over L.
Well V, the output voltage,
is nearly zero.
so minus V over L is pretty close to zero.
And there's hardly any change in inductor
current during
this second interval because the slope is
nearly zero.
After one switching
period, we end up with the inductor
current at what IL of TS.
A little bit positive.
It's a little greater than 0.
Now we repeat the process.
So, for the second switching period, we'll
go up with a slope of Vg minus v over L.
V is a little positive now, so the slope
during the second interval,
this slope, is a little bit less than that
slope was, last time.
So we don't quite increase as much.
Okay, and then during the, second, with
the switch in position two, for
this position period, we'll go down, with
a slope of minus v over
l, v is a little more positive than it
used to be, so
this slope is a little more negative than
it was during the previous period.
And we end up with some net increase in
current but
it's not as much as we increased in the
first period.
And we, we repeat
this process.
Okay, with the output voltage
slowly charging and eventually, after many
switching
periods, we, we end up at some current.
Let's, let me just draw it out here.
[COUGH].
And, and some output voltage where, the
amount we go up during
the, the first interval with a slope of vg
minus v over l.
Is exactly equal to the amount we decrease
during the second, second interval
where the slope is minus d over l so that
we end up at the same current we started.
Capacitor voltage
does something also and it ends up at the
same place it started.
Okay.
And so at this point, this is maybe
at time nTS and here's n plus 1 Ts, so
after n switching periods where n is some
large number we end up in steady state.
Where the, the wave forms become periodic.
Each switching
period has wave forms that are the same as
the previous switching period.
And the current and voltage end up at the
same place they started.
So you can see that that will happen in
when the output voltage rises to a large
enough value.
So V is large enough so that
these slopes have adjusted, and we our
final value of n plus one T S equals I N
of N T S.
And at that point, we say that we are in
steady state.
So the wave forms from then on have
the correct DC values where the converter
is supposed
to operate.
We've gone through some trangent that
eventually
has settled out and we reached steady
state.
[NOISE]
Okay.
It is
possible to find
[COUGH]
a simple relationship that expresses that
the circuit works in steady state.
And it comes from this notion that we end
up
at the same current that we started at for
the inductor.
And we can derive it, in fact from
the defining relationship of the inductor,
V is LDIDT.
What we do is we integrate this equation
to find the net change in inductor current
over one switching period.
So what we can do is integrate both
sides of the equation so integrate VL of
TdT over one switching period
[SOUND]
So we'll integrate
say from zero to Ts over one
switching period.
And here
we'll integrate from il of zero to il of
Ts.
And when we evaluate these integrals, the
right hand side gives us i L of
TS minus i L of zero, which is the net
change in current.
If I divide both sides by L, I can write
the other side of
the equation as 1 over L times the
integral of the voltage over one period.
So, in steady state, then the net change
in inductor current must
be zero, and therefore this integral of
the voltage must be zero.
Okay, and this
is the basic relationship that, where we
say
this is volt second balance on the
inductor.
That the integral of the voltage has
dimensions of volt
seconds and if you integrate the inductor
voltage over one
complete period, that integral or area
under the curve must
be zero if the inductor current has no net
change
over what that period.
Another way to write this equation is
simply to
divide both sides by switching period TS
and in
that case, we recognize that 1 over TS
times
the integral of the voltage is the DC
component.
So another way to say this is that the DC
component of the
voltage applied to the inductor Is zero
when the circuit operates on steady state.
We're going to use this notation of angle
brackets around a variable to
mean the average value.
So, the average value of the inductor
voltage is zero in steady state.
Okay?
Let's try that.
Try finding the applied volt seconds to
the inductor in our buck converter.
We previously found this wave
form was the voltage waveform applied to
the inductor.
The inductor saw voltage of vg minus v
during the first interval.
And minus v during the second.
So we have to integrate the inductor
voltage and the
easy way to do that is simply to find the
area under the curve so during this first
interval we
have an area that is the base DTS times
the height.
The height is EG minus V.
During the second interval, we have an
area that is the base
here, D prime TS.
That's the length of the second
interval times the height which is minus
D.
So if we add these two areas, they have to
add up to zero.
Okay so here
is that area, and if we equate it to zero
like this.
We get an equation and can simplify this
equation, D plus D prime
is, in fact, 1.
We'll call d prime as 1 minus d.
So, sum of d plus d prime works, works out
to be 1.
And we get dvg minus v equals zero.
You can solve this for the output voltage,
v.
And we find that the output voltage is
dvg.
Okay, well we already knew that for the
buck convertor.
But in fact this was a way to find that
the output voltage without
using the arguments of fourier series and
low pass filtering.
Here we, we simply draw the inductor wave
form and set it's average value to zero.
And this is something we can do for any
converter.
So we could similarly
use this approach find the output voltage
of
say the boost converter or the Buckman's
converter.
[SOUND].
So volt second balance plus the small
ripple approximation then
gives us a way to solve for the output
voltage.
Okay.
capacitor charge balance, or capacitor
amp second balance, is the dual of
inductor volt
second balance.
This is a, we can apply a similar
principle to a capacitor in the circuit.
What we do is we start with a defining
equation of the capacitor, I is CDBDT in
the capacitor, we can integrate both sides
of this, so integral of ICDT is the, C
times
the integral of DVC.
And if you integrate
over one period, from say 1 0 to
TS, then this part here gives us the
change in capacitor voltage over
one period.
The net change is B C of TS plus BC of
zero
and that's equal to one over C times the
integral of the current.
If we operate on steady state, then there
is
also no net change in capacitor voltage
over one period.
So the left hand side of this equation is
zero and
this says the integral of the capacitor
current must be zero when
the converter operates in steady state.
Again, another way to write this, we can
divide by ts multiply by c
and find that the dc component of
capacitor current must be zero in steady
state.
Okay, so we can use the principle of
capacitor charge balance to get a similar
equation to find the conditions under
which
the capacitor voltage will operate in
steady state.
This might be more familiar, than the
inductor volt second balance.
Or perhaps, more intuitive, if you put a
DC component of current into a capacitor.
Then the
capacitor will continue to charge and if
you keep putting
charge on the plates of the capacitor,
just positive charge,
say, then the charge will build up and the
capacitor
voltage will increase, and it won't
operate in steady state.
It's for part of the time we put a
positive current, say part of
the period we put a positive current
on the capacitor, making its voltage
increase.
For the other part of the period we must
put a negative current
to bring the charge back down to have no
net change in charge.
And therefore have no net change in
capacitor voltage.
Specifically we will employ the small ripple approximation to simplify the equations, and we'll derive the principles of inductor volt second balance and capacitor charge balance that give us the DC equations of the converter circuit.
So if you recall from the previous lectures we discussed the buck converter having a switch and having a low-pass filter that removes the switch in harmonics but allows the DC components or the DC component of the waveform through to the output.
Now, in a practical converter, we can't build a, a perfect low-pass filter. The low-pass filter will have continuation but it won't completely remove the switching harmonics of the waveform. [COUGH]
And therefore, the output voltage of a practical converter might look like this where there's a DC component capital V that as we previously found, is equal to the duty cycle times
the input voltage vg. And then in addition to that, we have some small ripple that is at the switching frequency and its harmonics that, where a small amount of the switching ripple gets through the filter and appears in the output.
And as sketched here, this ripple isactually probably
a lot larger than it would be in practice.
So, in a practical circuit, we'll have
some kind of specification on how
large this switching ripple can be, and
generally it's a very small number.
So in an output voltage of say a volt
or two for a computer power supply,
this ripple might be ten millivolts
or, or some very small number.
so therefore we can write the equation of
the
output voltage VOT, as being equal to
capital V which
[COUGH]
here I'm using the capitals letters to
denote the DC components.
So capital V is the desired DC component
equal to DVG.
And then plus some AC variation, that's
called V ripple of T here, that
is the undesired, switching harmonics that
make
it through the filter to the output
voltage.
[COUGH]
So in a well designed converter, the cell
c filter will have lots of attenuation.
And the ripple will be very small compared
to the DC component.
So we call this the small ripple
approximation, where the,
the ripple is small compared to the
desired DC component.
And under certain circumstances we will
neglect the ripple and
simply approximate the output voltage by
its desired DC component.
Okay, this has the effect of decoupling
the
equations, the differential equations of
the circuit, and making
them very simple to solve.
Okay, so this is called the small ripple
approximation to ignore this component and
simply approximate V of T by capital V.
Okay, brief discussion here.
[COUGH]
as we're
going to
see in a
minute,
the small
ripple
approximation
formally
can be
applied
only to
continuous
wave forms
that have
small
ripple.
So, we don't apply the small ripple
approximation to switched wave forms in
the circuit such as the switch output VS
of T
[NOISE].
And with these being a second order
circuit, in
general, the kinds of wave forms that we
get are
sinusoidal or decaying exponential in
nature.
So, the actual IL of T, when we solve
this circuit might have some wave form
that looks like
this, for example, with some initial value
IL of zero
and then some solution, of the second
order differential equation.
[COUGH]
Now in a well-designed convertor, this low
pass filter has a cutoff frequency
that is very low in frequency relative to
the switching frequency.
What that means is that the time
constants, or the time that it takes
this response, this ringing response, to
happen,
is long compared to the switching period.
So one switching period might be just this
long, for example.
And in the small ripple approximation,
what we're actually doing is
approximating this solution for a short
time with a straight line.
And for that reason we also often will
call
the small ripple approximation instead,
the linear ripple approximation.
And any continuous wave form
for short enough time can be approximated
by a straight line.
And so that is what we're in fact doing.
Of course if it's a discontinuous wave
form such as a switched
wave form then we cannot in general
approximate it with a straight line.
[COUGH]
So we apply the small ripple approximation
as appropriate to inductor
currents and capacitor voltages that have
responses such as this one.
That we can approximate over the switching
period or
a fraction of the switching period for the
straight line.
Okay, we can do a similar thing with the
switch in the second
position as well were the inductor left
side of the inductor then is connected
to ground instead of VG We get another
circuit to solve, and we can
find the, the wave forms for that interval
as well, in a similar manner.
So, let's look at, for example, for the
first interval, finding the inductor
current wave form.
using this small
ripple approximation.
So from solving this circuit, we have a
loop right here that the inductor is
connected in.
And the loop equation for this circuit is
the inductor voltage BL
is equal to the input voltage VG minus the
capacitor voltage.
Here it's called V.
So we get this equation for the loop the
the inductor is connected in.
Okay.
Now, V of T is the output capacitor
voltage, and as we've discussed
so far, we want this capacitor voltage to
have small ripple.
And so V of T again can be expressed as
its DC component, capital
V, plus the small ripple, and here is a
case where we can replace
[INAUDIBLE]
by its DC component capital V to find the
inductor voltage for this interval.
And so this is a good approximation over
the
short time of when the switch is in this
position.
Okay, so this is a small ripple
approximation.
Knowing the inductor voltage we can now
find what the inductor current does.
Using the well known defining equation of
an
inductor, that V is LDIDT in the inductor.
So if we plug this expression for the
approximate inductor
voltage into there, and solve for DIDT, we
get this
equation Which says that the inductor
current changes with a
slope that is given by the voltage over
the inductance,
with the voltage being equal to a constant
value, Vg minus capital V.
Okay?
So, for this first interval, The
inductor current will start at some
initial value,
IL of zero.
And it will increase with the, this slope.
And it will increase with that constant
slope.
Until this first interval is over, and
the, the switch
is changed to position 2, which happens at
time DTS.
[COUGH]
' Kay.
Then, during the second interval, we get
this circuit,
with the left side of the inductor
connected to ground.
And the loop equation now around this loop
where the inductor is
connected, says that the inductor voltage
VL is equal to minus V.
Like this.
We can again use the small ripple
approximation to replace V of T
with its DC component capital V, and
ignore the ripple during this interval.
And we can plug that voltage into the
defining equation of the
inductor again to find the slope of IL
during the second interval.
And what we find then is that the inductor
current for
this interval changes with a slope equal
to minus V over L.
[SOUND]
So we started at some initial value, we
went up
during the first interval and ended at
some point.
[SOUND]
And then during the second interval we now
change with a different slope.
And the slope is negative, assuming the
output voltage is positive.
So we get a slope for this interval of
minus V over L, and the current
goes down, until the switching period is
over, and we end at, at time
TS we end with some final value.
[COUGH]
' Kay, so the small ripple approximation
makes it simple
to, to sketch what the inductor current
wave form looks like.
We have straight lines that have constant
slopes and are easy to solve.
And this greatly simplifies the equations
so that
we don't have to write decaying
exponential type functions.
And have pages of algebra.
So then here's a summary of what the
inductor
current wave voltage and current wave
forms look like.
We found the inductor voltage during each
interval, it was positive
and equal to an approximately constant
value during the first interval.
And during the second interval, it was
equal to an approximately constant
negative value.
what this did was made the inductor
current go up during the first
interval and then go down with a negative
slope during the second interval.
Okay, one thing we can do with this right
off the, right
away is, get an equation that helps us
choose the value of inductance.
You can see that the actual inductor
current wave form has switching ripple.
The current goes up and down with a
traingle, triangular wave form at the
switching frequency.
And on top of that it has an
average value or dc component, which is
the dc value of inductor current.
That we'll call capital I.
To design the inductor, choose an inductor
value, we generally will
try to limit the amount of switching
ripple in the inductor current.
How much ripple we allow is a design
choise, and at least for the
next several weeks we will.
Discuss converters where this ripple is
limited in value to maybe ten or twenty
percent of capital I at full power.
' Kay, so the ripple, we commonly define
the peak to average ripple as delta IL.
And so two delta il is the peak to peak
ripple, which is from here to here.
And from this waveform and knowing the
slope, we can
easily write an equation for how large the
ripple is.
So for example, during the first
sub-interval, with the switch
in position one, the inductor current
changes from here to here.
And the net change is two delta I.
So we can write two delta IL is the
change in the inductor current, and this
is simply equal
to the slope during the first interval.
Vg minus V capital over
L times the length of the first interval
with is DTS.
Okay?
So knowing the slope and the time, we can
find the change.
And we can solve then for delta IL.
Just divide both sides by two.
And we get this expression.
Okay?
it's, this is the equation that is
commonly used to calculate
the value of inductance, and choose an
inductor for the converter.
So we can solve this equation for l.
Push l onto that side and get this
equation.
And from this equation we can solve for
the value
of l that gives a given value of delta I.
Okay, there's another thing we can do
besides finding the value of L.
We would like to find the steady state
voltage and current in the converter.
So, let's consider now what the inductor
current
does during a turn on trangant of the
converter.
[SOUND].
So, for example,
[COUGH].
Let's suppose that we, start the circuit
with the inductor
current equal to zero.
And the output voltage is zero also.
So we'll start at zero with the circuit
turned off.
And at t equals zero, we'll start
switching with So we'll
switch between positions one and two with
some fixed duty cycle.
Okay?
We've already found what happens, the
inductor current
will change with some slope during each
interval.
And so during
the first interval, the very first
switching
period, the inductor current will go up.
It with a slope that we've already found,
is equal to vg minus v over l.
Okay, well initially what v is the
capacitor voltage, vc in fact let's
just call this v, instead of vc.
So the voltage is initially
zero at the output, and our slope during
the first interval is V G over L.
Okay?
So we go up until the end of the
first interval at time DTS, we've reached
some positive current.
[SOUND].
Okay.
What happens to the output voltage?
Well, its inductor current is small, but
it's a little more than zero.
That inductor current flow to the output.
And it will slightly charge the capacitor
voltage up a little.
So V is still small but its a tiny bit
more then zero.
Okay during the second switch interval
with a switching position two, we found
that the conductor current changed with a
slope of minus V over L.
' Kay, so we have a slope here, of
minus V over L.
Well V, the output voltage,
is nearly zero.
so minus V over L is pretty close to zero.
And there's hardly any change in inductor
current during
this second interval because the slope is
nearly zero.
After one switching
period, we end up with the inductor
current at what IL of TS.
A little bit positive.
It's a little greater than 0.
Now we repeat the process.
So, for the second switching period, we'll
go up with a slope of Vg minus v over L.
V is a little positive now, so the slope
during the second interval,
this slope, is a little bit less than that
slope was, last time.
So we don't quite increase as much.
Okay, and then during the, second, with
the switch in position two, for
this position period, we'll go down, with
a slope of minus v over
l, v is a little more positive than it
used to be, so
this slope is a little more negative than
it was during the previous period.
And we end up with some net increase in
current but
it's not as much as we increased in the
first period.
And we, we repeat
this process.
Okay, with the output voltage
slowly charging and eventually, after many
switching
periods, we, we end up at some current.
Let's, let me just draw it out here.
[COUGH].
And, and some output voltage where, the
amount we go up during
the, the first interval with a slope of vg
minus v over l.
Is exactly equal to the amount we decrease
during the second, second interval
where the slope is minus d over l so that
we end up at the same current we started.
Capacitor voltage
does something also and it ends up at the
same place it started.
Okay.
And so at this point, this is maybe
at time nTS and here's n plus 1 Ts, so
after n switching periods where n is some
large number we end up in steady state.
Where the, the wave forms become periodic.
Each switching
period has wave forms that are the same as
the previous switching period.
And the current and voltage end up at the
same place they started.
So you can see that that will happen in
when the output voltage rises to a large
enough value.
So V is large enough so that
these slopes have adjusted, and we our
final value of n plus one T S equals I N
of N T S.
And at that point, we say that we are in
steady state.
So the wave forms from then on have
the correct DC values where the converter
is supposed
to operate.
We've gone through some trangent that
eventually
has settled out and we reached steady
state.
[NOISE]
Okay.
It is
possible to find
[COUGH]
a simple relationship that expresses that
the circuit works in steady state.
And it comes from this notion that we end
up
at the same current that we started at for
the inductor.
And we can derive it, in fact from
the defining relationship of the inductor,
V is LDIDT.
What we do is we integrate this equation
to find the net change in inductor current
over one switching period.
So what we can do is integrate both
sides of the equation so integrate VL of
TdT over one switching period
[SOUND]
So we'll integrate
say from zero to Ts over one
switching period.
And here
we'll integrate from il of zero to il of
Ts.
And when we evaluate these integrals, the
right hand side gives us i L of
TS minus i L of zero, which is the net
change in current.
If I divide both sides by L, I can write
the other side of
the equation as 1 over L times the
integral of the voltage over one period.
So, in steady state, then the net change
in inductor current must
be zero, and therefore this integral of
the voltage must be zero.
Okay, and this
is the basic relationship that, where we
say
this is volt second balance on the
inductor.
That the integral of the voltage has
dimensions of volt
seconds and if you integrate the inductor
voltage over one
complete period, that integral or area
under the curve must
be zero if the inductor current has no net
change
over what that period.
Another way to write this equation is
simply to
divide both sides by switching period TS
and in
that case, we recognize that 1 over TS
times
the integral of the voltage is the DC
component.
So another way to say this is that the DC
component of the
voltage applied to the inductor Is zero
when the circuit operates on steady state.
We're going to use this notation of angle
brackets around a variable to
mean the average value.
So, the average value of the inductor
voltage is zero in steady state.
Okay?
Let's try that.
Try finding the applied volt seconds to
the inductor in our buck converter.
We previously found this wave
form was the voltage waveform applied to
the inductor.
The inductor saw voltage of vg minus v
during the first interval.
And minus v during the second.
So we have to integrate the inductor
voltage and the
easy way to do that is simply to find the
area under the curve so during this first
interval we
have an area that is the base DTS times
the height.
The height is EG minus V.
During the second interval, we have an
area that is the base
here, D prime TS.
That's the length of the second
interval times the height which is minus
D.
So if we add these two areas, they have to
add up to zero.
Okay so here
is that area, and if we equate it to zero
like this.
We get an equation and can simplify this
equation, D plus D prime
is, in fact, 1.
We'll call d prime as 1 minus d.
So, sum of d plus d prime works, works out
to be 1.
And we get dvg minus v equals zero.
You can solve this for the output voltage,
v.
And we find that the output voltage is
dvg.
Okay, well we already knew that for the
buck convertor.
But in fact this was a way to find that
the output voltage without
using the arguments of fourier series and
low pass filtering.
Here we, we simply draw the inductor wave
form and set it's average value to zero.
And this is something we can do for any
converter.
So we could similarly
use this approach find the output voltage
of
say the boost converter or the Buckman's
converter.
[SOUND].
So volt second balance plus the small
ripple approximation then
gives us a way to solve for the output
voltage.
Okay.
capacitor charge balance, or capacitor
amp second balance, is the dual of
inductor volt
second balance.
This is a, we can apply a similar
principle to a capacitor in the circuit.
What we do is we start with a defining
equation of the capacitor, I is CDBDT in
the capacitor, we can integrate both sides
of this, so integral of ICDT is the, C
times
the integral of DVC.
And if you integrate
over one period, from say 1 0 to
TS, then this part here gives us the
change in capacitor voltage over
one period.
The net change is B C of TS plus BC of
zero
and that's equal to one over C times the
integral of the current.
If we operate on steady state, then there
is
also no net change in capacitor voltage
over one period.
So the left hand side of this equation is
zero and
this says the integral of the capacitor
current must be zero when
the converter operates in steady state.
Again, another way to write this, we can
divide by ts multiply by c
and find that the dc component of
capacitor current must be zero in steady
state.
Okay, so we can use the principle of
capacitor charge balance to get a similar
equation to find the conditions under
which
the capacitor voltage will operate in
steady state.
This might be more familiar, than the
inductor volt second balance.
Or perhaps, more intuitive, if you put a
DC component of current into a capacitor.
Then the
capacitor will continue to charge and if
you keep putting
charge on the plates of the capacitor,
just positive charge,
say, then the charge will build up and the
capacitor
voltage will increase, and it won't
operate in steady state.
It's for part of the time we put a
positive current, say part of
the period we put a positive current
on the capacitor, making its voltage
increase.
For the other part of the period we must
put a negative current
to bring the charge back down to have no
net change in charge.
And therefore have no net change in
capacitor voltage.
0 التعليقات:
إرسال تعليق