Boost Converter Duty Cycle: Equations vs. Reality
Figbash
I'm trying to understand the physics of a boost converter. I have a
cursory understanding of the equation (see wikipedia), and how the
duty cycle falls out as a ratio of input to output voltage (regardless
of the values of L, C, or T), so a duty cycle of 50% should produce
double the input voltage applied to the top of the coil.
The problem is, no matter what values of components I swap in, I'm
seeing HUGE voltage gains at 50% duty cycle, way beyond 10x,
approaching 50x.
What is happening here that I'm missing?
Here are the parameter's I've changed with an input voltage of 3.3V
and a duty cycle of 50% (tried almost all the permutations):
sweeping frequency from 2Hz, 10Hz, 100Hz, 1kHz to 10kHz.
sweeping the cap across 100pF, 1nF, 10nF, 100nF, 1uF, 3.3uF, 100uF.
sweeping the inductor across 10uH, 100uH, 220uH, 110mH.
The lowest voltage above Vcc (of 3.3V) I see is ~13V across the cap
with C = 100uF and L = 10uH.
The highest voltage I've seen is 250+V.
Clearly none of these represent the basic equation, so I'm missing
something basic about the physical properties of the components that
isn't expressed in the equation.
Thanks,
P
Note that I'm using the basic circuit: a square wave driving a high-
power FET, sourced to ground, and the drain connected to Vcc via a
high-current inductor and a diode to a high-voltage capacitor, to
ground. The simplest possible circuit. The FET is an IRF630 which
can handle 100V+, and an ultra fast diode that can handle up to 8A at
100V+.
Jon Kirwan
<peter.j...@gmail.com> wrote:
>I'm trying to understand the physics of a boost converter. I have a
>cursory understanding of the equation (see wikipedia), and how the
>duty cycle falls out as a ratio of input to output voltage (regardless
>of the values of L, C, or T), so a duty cycle of 50% should produce
>double the input voltage applied to the top of the coil.
Without taking into account your flyback diode drop, your
MOSFET drop when on, then roughly speaking, "yes."
Duty = (Vout + Vdiode - Vin) / (Vout + Vdiode - Vsw)
If you ignore Vdiode (flyback drop) and Vsw (mosfet or bjt
switch drop when on) then it does reduce to 1-Vin/Vout as you
suggest above.
>The problem is, no matter what values of components I swap in, I'm
>seeing HUGE voltage gains at 50% duty cycle, way beyond 10x,
>approaching 50x.
>
>What is happening here that I'm missing?
Do you actually have a LOAD connected up?? Because if you
don't have a load, then the energy has to go somewhere for
each pulse. Eventually, you start getting enough losses that
nothing more happens or something will break down from the
excess voltage. But the voltage will rise up beyond your
"design" value, if there is no load siphoning away some
energy over time.
>Here are the parameter's I've changed with an input voltage of 3.3V
>and a duty cycle of 50% (tried almost all the permutations):
>
>sweeping frequency from 2Hz, 10Hz, 100Hz, 1kHz to 10kHz.
>sweeping the cap across 100pF, 1nF, 10nF, 100nF, 1uF, 3.3uF, 100uF.
>sweeping the inductor across 10uH, 100uH, 220uH, 110mH.
And the load is??
>The lowest voltage above Vcc (of 3.3V) I see is ~13V across the cap
>with C = 100uF and L = 10uH.
>
>The highest voltage I've seen is 250+V.
You need a load, I think.
>Clearly none of these represent the basic equation, so I'm missing
>something basic about the physical properties of the components that
>isn't expressed in the equation.
The load?
>Thanks,
>P
>
>Note that I'm using the basic circuit: a square wave driving a high-
>power FET, sourced to ground, and the drain connected to Vcc via a
>high-current inductor and a diode to a high-voltage capacitor, to
>ground. The simplest possible circuit. The FET is an IRF630 which
>can handle 100V+, and an ultra fast diode that can handle up to 8A at
>100V+.
Okay. Let's say you want Vout to be twice Vin=3.3V, or
Vout=6.6V. Just for starters. Let's say your load on the
output should use 55mA. This works out to 6.6V/55mA or 120
ohms. So go find a 120 ohm resistor and make sure it is on
the output.
Note that I'm insisting you need to specify the load, here.
You need one specified because you don't have a closed loop
controlling the voltage, so far as I can tell. Without that,
you are running open-loop. And that means you need to design
for some known load, at least.
So let's assume 120 ohms for now and 55mA @ 6.6V.
Your Voff (voltage across L when mosfet switch is off) should
be Vout+Vdiode-Vin. This is because L's voltage is stacked
on top of Vin, but you need to account for the Vdiode drop.
So let's select a reasonable guess for Vdiode (assume a
schottky) of 400mV. This sets Voff=3.7V.
Your Von (voltage across L when mosfet switch is on) should
be Vin-Vsw. You don't know what the peak current will be
just yet, but roughly speaking it should be about 2 times
Iout (55mA) times 1/(1-Duty). But before knowing Duty, we
can assume roughly .5 for now. This means about 4*55mA or
220mA. Your high power mosfet might be, let's say, 50mOhms
at that current. Or a drop of about 11mV. Negligible.
This means Duty = (3.7 / 7) = 53%. Refiguring the mosfet
drop here won't change much -- makes it 234mA instead of the
guestimate of 220mA. So still negligible. We now have Ipeak
at something like 234mA.
Okay. Now for the inductor. Let's go with your 10kHz
frequency, for now. The product of f*L must be about
Von*Duty/Ipeak or Voff*(1-Duty)/Ipeak. I get about 750uH for
that. But it doesn't appear you have one to try. Keep that
number in mind, though. I'll talk about it, later.
What about that 110mH!! It's huge. So the frequency will be
very low, I suspect. But plug it in and get a frequency of
about 67-68Hz. You might try that out.
Okay. Not it is later.
You could just design it, though, for some frequency. I'll
assume the mosfet works at 10kHz. Assume you get yourself
one of those #61 ferrite core types (size as yet unspecified)
with u_e = 125 and Bmax of 0.15T (1500 Gauss.) The magnetic
volume needs to be larger than:
Ve >= u_0 * u_e * Ipeak * [Von * (Duty / f)] / Bmax^2
The term in brackets is the volt-seconds, which is kind of
important. Anyway, this works out to Ve >= 286e-9 m^3.
Look at Material 61 here:
https://www.amidoncorp.com/specs/2-05.pdf
They list the term Ve (volume) in cm^3, which is units of
10^-6 m^3. You need at least .286 there. So that FT-50
looks like the first one that has at least that much.
Also from the table, AL for the FT-50(61) is 68mH/1000 turns.
And inductance goes by the square of N. L=68mH*(N/1000)^2.
From the basic V=L*dI/dt equation and the duty cycle being
53% at 10kHz, we have dt=.53/10000 or 53us, dI=234mA, and
V=3.3V. So this means that:
N = 1000*SQRT((3.3V*.53/10000)/(.234A*68mH)) = about 105.
So you should wind about 105 turns on the toroid. Inductance
would then be 68mH*(105/1000)^2 or 750uH. Bingo.
Note that this matches what we'd already decided we needed at
the outset. What we've done is worked through picking out a
core and fitting the windings to it, is all.
I'm just a hobbyist, though, and have no training at all. Not
so much as one credit hour. So if someone trained up on this
stuff tells you I'm wrong about all this, listen. But this
is how it looks to me, right now, and I hope I didn't stray
too far from the real reason why you are seeing higher
voltages than you'd expect.
Jon
P.S. From that table for FT-50(61):
Le = 3.02 cm = 30.2e-3 m
Ae = 0.133 cm^2 = 13.3e-6 m^2
From some work I'm not showing here, the number of turns are
also bounded this way. (It's not merely the volume that is
determined, but also the magnetic cross section and the
magnetic path length, as well.)
N <= Le / (u_0 * u_e * Ipeak / Bmax)
<= 123
N >= Von * Duty / (Bmax * f * Ae)
>= 88
The actual figure of 105 is, in fact, between these. That is
a good thing. However, not all cores will have the right
cross section and path length even if they have the right
volume.
So that's another guidance to let you know if a particular
core will or won't work for you. The volume, the cross
section, and the path length all must be correct. Usually,
the core designers create useful cores. So it often does
work out just fine.
Jon Kirwan
need to fit 105 turns into an FT-50(61). The inner diameter
is ID=.187". The total area is then pi*(ID/2)^2. If you
fill half that area, roughly, then each wire cannot have a
diameter more than about 2*SQRT((1/2)*(ID/2)^2/105) or
0.0129". Magnet wire has a little bit of insulation, as
well. So if you look at a wire table, you'll see that 30
gauge magnet wire looks about right.
It probably can handle the .234A peak, though it's getting on
the ragged end of things with 30 gauge. Also, it may be more
than 70 inches of wire to do it and with copper it may work
out close to 3/4 ohm. You might lose almost two tenths of a
volt across that at peak current. Another source of power
loss, which suggests you might not quite get your 6.6V across
120 ohms.
You _may_ choose to use something with more mH/k-turn to get
your turn count down more so you can use thicker wire. The
FT-50B, for example, gives 150mH/k-turn and has a slightly
larger ID, as well.
Jon
Jasen Betts
>
> I'm trying to understand the physics of a boost converter. I have a
> cursory understanding of the equation (see wikipedia), and how the
> duty cycle falls out as a ratio of input to output voltage (regardless
> of the values of L, C, or T), so a duty cycle of 50% should produce
> double the input voltage applied to the top of the coil.
>
> The problem is, no matter what values of components I swap in, I'm
> seeing HUGE voltage gains at 50% duty cycle, way beyond 10x,
> approaching 50x.
to get gain matched to duty cycle you need to run it in continuous mode
in discontinous mode it gives a much higher voltasge gain.
http://en.wikipedia.org/wiki/Boost_converter#Discontinuous_mode
increase the frequency, load and/or the inductance until you reach
continuous mode.
--
⚂⚃ 100% natural
Bill Bowden
> On Wed, 12 Jan 2011 22:23:44 -0800 (PST), Figbash
>
Yes, the problem may be no load with excessive voltage. When I was
playing around with boost converters I discovered the output voltage
is the same as the input without any switching action, or (zero%
duty), so the switch only needed to supply the extra voltage above the
input, and not all the power, just some of it above the steady state.
So, if the input was say 9 volts and I wanted 12 volts out, I only
needed 1/3 more, or maybe 33% duty cycle. But if you wanted say 100
volts out from 3 volts in, the duty cycle would be 97%.
-Bill
Bill Bowden
wrote:
> On Jan 13, 1:15 am, Jon Kirwan <j...@infinitefactors.org> wrote:
>
>
>
> > On Wed, 12 Jan 2011 22:23:44 -0800 (PST), Figbash
>
> > <peter.j.tore...@gmail.com> wrote:
> > >I'm trying to understand the physics of a boost converter. I have a
> > >cursory understanding of the equation (see wikipedia), and how the
> > >duty cycle falls out as a ratio of input to output voltage (regardless
> > >of the values of L, C, or T), so a duty cycle of 50% should produce
> > >double the input voltage applied to the top of the coil.
>
>
> Yes, the problem may be no load with excessive voltage. When I was
> playing around with boost converters I discovered the output voltage
> is the same as the input without any switching action, or (zero%
> duty), so the switch only needed to supply the extra voltage above the
> input, and not all the power, just some of it above the steady state.
> So, if the input was say 9 volts and I wanted 12 volts out, I only
> needed 1/3 more, or maybe 33% duty cycle. But if you wanted say 100
> volts out from 3 volts in, the duty cycle would be 97%.
>
> -Bill
Correction to the above:
The 33% should be 25%. I think the duty cycle is Vout - Vin / Vout, so
in the 9 to 12 volt circuit, it would be 12-9/12 or 25%. But it's
still a little foggy why the voltage gets excessive when the load
resistance goes too high? Maybe something to do with dropping out of
continuous mode?
-Bill
Figbash
Thanks for the thought-out walk through of the design.
First off, as I debugged the circuit, I started thinking that I should
focus on the behavior with a low frequency. I didn't realize I had
switched into discontinuous mode, nor that I was running open loop
without a load, I was thinking impedence of the capacitor would be a
load. So here are two mistakes in my analysis. But I have a few
other questions:
> Your Von (voltage across L when mosfet switch is on) should
> be Vin-Vsw. You don't know what the peak current will be
> just yet, but roughly speaking it should be about 2 times
> Iout (55mA) times 1/(1-Duty). But before knowing Duty, we
> can assume roughly .5 for now. This means about 4*55mA or
> 220mA. Your high power mosfet might be, let's say, 50mOhms
> at that current. Or a drop of about 11mV. Negligible.
How did you arrive at Ipeak approx = 2 * Iout * 1/(1-D)? Is it
because we know the the current into the drain of the mosfet is iL -
iLoad and you know the resistance of a mostfet is small (on the order
of ~50~100 Ohms?). Or something else? It is key in determining the
relationship of freq*L later on.
> Okay. Now for the inductor. Let's go with your 10kHz
> frequency, for now. The product of f*L must be about
> Von*Duty/Ipeak or Voff*(1-Duty)/Ipeak. I get about 750uH for
> that. But it doesn't appear you have one to try. Keep that
> number in mind, though. I'll talk about it, later.
Here's the later on part. What is the reason for freq*L having to be
about one of those two values? I'm guessing it is frequency response
impedence matching: freq * L = Zo * pi, but I don't quite see the
steps to derive the equation freq*L =~ Von*D/iPeak =~ Voff*(1-D)/
iPeak.
I can tune the frequency to the inductor, rather than trying to roll
my own inductor. I'm trying to ultimately build the circuit with the
components I have so that varying duty cycle and frequency controls
Vout.
So with my on-hand parts:
750uH * 10kHz = 220uH * 34kHz which is well within in the range of the
IRF630 that I'm using.
What about the capacitor value? Should I be thinking of impedence
here? 34kHz would imply that I'd need something above 1/
(34kHz*(pi*R)) >30~50uF to prevent a short to ground?
Thanks for the insight, I appreciate the walkthrough. It makes more
sense, but I still don't quite grasp the tradeoffs of the different
values/components, which I'm also guessing is part of the reason why
my transistor is smoking hot even with a heatsink.
Peter
Jon Kirwan
<bpe...@bowdenshobbycircuits.info> wrote:
><snip>
>Yes, the problem may be no load with excessive voltage.
><snip>
Yes. I think the "no load" is the only way he gets that kind
of result. And even then, I wonder at the mentioned 250V.
I talked about the discontinuous case because there didn't
seem to be any load. (Jason brought that up.)
Jon
Jon Kirwan
The OP is driving a circuit without closed loop feedback, as
far as I can tell and without any kind of load, either. If
there is no load, the inductor energy simply dumps onto the
cap and keeps on dumping. The voltage just rises until
something gives (or until the energy in each pulse is enough
wasted somewhere else that it doesn't add to the cap,
anymore.)
Jon
Jon Kirwan
<peter.j...@gmail.com> wrote:
>Jon,
>
>Thanks for the thought-out walk through of the design.
I think there are much better people to do this. I've zero
experience at it, basically. Just enjoyed thinking about it
on a few occasions. A while back, a Mr. Wescott caught a
significant error in my thinking and I learned a little more.
He's professional, trained, and has experience. Plus, he is
a writer, as well. So he'd probably be a lot better, if he
had the time and inclination.
So I thought I'd give it my own shot. The worst that can
happen is that I get corrected, again. And that would only
be good, not bad; for me, anyway.
>First off, as I debugged the circuit, I started thinking that I should
>focus on the behavior with a low frequency.
Okay. Let's start right there.
One problem with low frequencies is that with any kind of
duty cycle greater than zero the inductor is going to be held
with a voltage across it for a long time. The peak flux at
the end of that ON time is likely huge -- a value
proportional to the volt-seconds (or the voltage applied
while ON times the number of seconds held there.)
The flux is packed densely into the interior of the inductor
windings, but spread out by the inductor's diameter
cross-section area. If you divide the accumulated flux (in
Webers, and which is a physical constant determined by the
inductor's geometry times the applied volt-seconds) by the
cross-section area, you get the number of Teslas ... which is
a kind of 'flux density' thing.
For air core (or vacuum) inductors, it doesn't matter much.
But for anything else with atoms and grain structures and so
on, it does matter. Low frequency can cause "saturation"
problems in inductors that are not air-core and then the
inductance changes rapidly. And that messes up all your
assumptions about its value being fixed.
>I didn't realize I had
>switched into discontinuous mode, nor that I was running open loop
>without a load, I was thinking impedence of the capacitor would be a
>load. So here are two mistakes in my analysis.
Although some will argue that it really is a load, it's best
to just think about all this as a transfer of energy. You
turn on the switch (BJT or MOSFET) and apply a fixed voltage
across the inductor. In doing so, add energy to its magnetic
field. Then you turn the switch off and take away the
applied voltage, collapsing the field at some rate and
subtracting energy in its magnetic field. This energy needs
to go somewhere and the capacitor is a convenient place.
However, to store energy on the capacitor, it needs to
increase the charge differential across it. (In other words,
add stored charge to it.) This implies a current and luckily
that is exactly what the inductor has to offer.
If the OFF time is long enough, the entire magnetic field
collapses and delivers all of its energy to the capacitor
(less a little wasted in the flyback diode, wiring, etc.) And
this is called 'discontinuous.' If the OFF time isn't long
enough, then only _some_ of the magnetic field collapses and
delivers energy to the capacitor before the next cycle
starts. This is called 'continuous.'
If you recall the flux density issue, one difference between
these two methods is that if you are in discontinuous mode
then all of the flux and the attendant flux density is
allowed to go to zero before rising up, again. In continuous
mode, there is also a periodic flux change... but it sits on
top of some fixed flux.... like some AC signal riding on DC.
The peak flux and peak flux density may be higher for
continuous mode. So continuous mode _may_ require a little
more attention to the Bmax of the core material.
The short of it is that continuous mode is strangled by DC
winding losses and core saturation. The core saturation
problem suggests larger cross-sections to spread out the flux
and therefore larger inductors. And larger inductors may
mean more DC resistance in the wiring and attending losses.
Discontinuous mode is instead hampered by larger flux swings
and, thus, higher core losses and the related larger ripple
currents and, thus, AC winding losses.
So core saturation/DC winding loss dogs continuous mode and
core loss/AC winding loss dogs discontinuous mode. Also,
when thinking about energy losses -- in discontinuous mode
the optimum balance is found where core losses and winding
losses are about equal; while in continuous mode the DC
winding loss dominates. Of course, in continuous mode you
still have to worry about Bmax/saturation.
>But I have a few other questions:
>
>> Your Von (voltage across L when mosfet switch is on) should
>> be Vin-Vsw. You don't know what the peak current will be
>> just yet, but roughly speaking it should be about 2 times
>> Iout (55mA) times 1/(1-Duty). But before knowing Duty, we
>> can assume roughly .5 for now. This means about 4*55mA or
>> 220mA. Your high power mosfet might be, let's say, 50mOhms
>> at that current. Or a drop of about 11mV. Negligible.
>
>How did you arrive at Ipeak approx = 2 * Iout * 1/(1-D)? Is it
>because we know the the current into the drain of the mosfet is iL -
>iLoad and you know the resistance of a mostfet is small (on the order
>of ~50~100 Ohms?). Or something else? It is key in determining the
>relationship of freq*L later on.
I could take the long way 'round the barn and get there. But
I'll take the short cut.
Draw a horizontal line on paper. Make it a few inches long.
Label the left end point 0 and the right end point 1. This
is your time axis (x axis) and 0 represents the start and 1
represents the end of a cycle before the next one begins.
Starting at the end marked 0, draw a vertically rising "y
axis" line. This is the peak current axis. You now have
quadrant 1, an upper right quadrant of a standard Cartesian
plot. The x axis is time, the y axis is peak current.
Now choose a point in quadrant 1, but don't put it on the
time axis. It can be towards the left side or towards the
right side, doesn't matter. But pick a point and mark that
down. The time axis position represents the moment with the
switch is turned off. Draw a solid, vertical line from the
point down to the time axis and mark that position D. Now
draw a dotted, horizontal line from the point back to the y
axis and mark that intersection as Ipeak.
Make sense so far?
Now. Draw a solid line from (0,0) up to your point, and then
down from the Ipeak point to (1,0). While looking at this,
you should see two solid line bounded triangles. From time=0
to time=D, the BJT/MOSFET is ON. During this time, no
current is flowing to the capacitor. It is only being used
to add energy to the inductor's magnetic field. But from
time=D to time=1, current is flowing to the capacitor via the
flyback diode. So now draw hatching over the area of this
right side triangle to highlight it.
The area of this right side triangle is the integrated charge
that was transferred to the capacitor. Right? (Current
times time.) In other words, if you knew the area of that
triangle, you'd know how much charge was moved.
This area is simply (1/2)*(1-D)*Ipeak. Your basic triangle
area computation. But this must also be the same amount of
charge that is consumed by the load during the entire period
from 0 to 1. Assuming the load draws a constant current
called Iout, then:
Iout*(1-0) = (1/2)*(1-D)*Ipeak
Iout = (1/2)*(1-D)*Ipeak
2*Iout = (1-D)*Ipeak
2*Iout/(1-D) = Ipeak
And there it is.
Lots of other ways to get there. But more tedious and it all
goes to this same idea in the end.
>> Okay. Now for the inductor. Let's go with your 10kHz
>> frequency, for now. The product of f*L must be about
>> Von*Duty/Ipeak or Voff*(1-Duty)/Ipeak. I get about 750uH for
>> that. But it doesn't appear you have one to try. Keep that
>> number in mind, though. I'll talk about it, later.
>
>Here's the later on part. What is the reason for freq*L having to be
>about one of those two values? I'm guessing it is frequency response
>impedence matching: freq * L = Zo * pi, but I don't quite see the
>steps to derive the equation freq*L =~ Von*D/iPeak =~ Voff*(1-D)/
>iPeak.
Well, let's see.
I'm sure you'll agree that I can define things so that:
f = 1/(ton+toff)
With obvious meanings for ton and toff.
Also,
Duty = ton/(ton+toff)
So I could write instead,
f = Duty/ton
f = (1-Duty)/toff
You see that okay?
Recall this?
V = L * dI/dt
Assuming discontinuous mode operation (current starts at 0
and rises to Ipeak) and replacing infinitesimals with finites
and looking at both cases -- one where the BJT/MOSFET is ON
and the other where the BJT/MOSFET is off:
Von = L*Ipeak/ton, so that ton = L*Ipeak/Von
Voff = L*Ipeak/toff, so that toff = L*Ipeak/Voff
(I'm not worrying about sign, right now, just magnitude.)
Hence,
f = Duty/ton = Duty/(L*Ipeak/Von)
= Duty*Von/(L*Ipeak)
and,
f = (1-Duty)/toff = (1-Duty)/(L*Ipeak/Voff)
= (1-Duty)*Voff/(L*Ipeak)
Moving L to the other side of each gives,
f*L = Duty*Von/Ipeak = (1-Duty)*Voff/Ipeak
Does that make sense?
>I can tune the frequency to the inductor, rather than trying to roll
>my own inductor. I'm trying to ultimately build the circuit with the
>components I have so that varying duty cycle and frequency controls
>Vout.
I figured you might want that.
>So with my on-hand parts:
>
>750uH * 10kHz = 220uH * 34kHz which is well within in the range of the
>IRF630 that I'm using.
Okay.
>What about the capacitor value? Should I be thinking of impedence
>here? 34kHz would imply that I'd need something above 1/
>(34kHz*(pi*R)) >30~50uF to prevent a short to ground?
I tend to think about this is voltage ripple seen by the
load, when sizing the cap. During the switch ON time, no
current is flowing to the cap, so the cap is supplying all
the current to the load.
I = C*dV/dt
or,
C = I*dt/dV
You know that dt is the ON time, which is D/f. (Current will
be flowing from battery and inductor during the OFF time, so
the cap won't droop but will rise, then.) dV is what ripple
you want to allow during that time. Say that is 5mV just for
kicks. And if the load is that 120 ohm thing I mentioned
before and the Vout voltage is supposed to be 6.6V (the 55mA
I spec'd before) then I=55mA. If your duty cycle is that 53%
I figured before and your new f is 34kHz, then:
C = 55mA * (0.53/34kHz) / 5mV = 171.47uF
Make it 220uF.
>Thanks for the insight, I appreciate the walkthrough. It makes more
>sense, but I still don't quite grasp the tradeoffs of the different
>values/components, which I'm also guessing is part of the reason why
>my transistor is smoking hot even with a heatsink.
Well, I hope some of this helps. Let us know what happens,
too.
You do need some kind of load there if you aren't using any
feedback. So get some resistor in place.
Jon
Bill Bowden
> On Fri, 14 Jan 2011 13:19:57 -0800 (PST), Bill Bowden
>
>
>
I don't see the need for feedback. The output voltage should be a
function of the input and duty cycle alone, regardless of the load
resistance. It should regulate across a wide spectrum with no
feedback, but seems to go bananas when the load gets close to open
circuit. Maybe it's only a completely open circuit that causes
problems?
-Bill
Jon Kirwan
All of the energy built up into the inductor during the ON
time is delivered to the capacitor during the OFF time. The
capacitor's voltage must rise in order to store it. And it
must continue to rise with each cycle. There is no stopping
it.
>The output voltage should be a
>function of the input and duty cycle alone, regardless of the load
>resistance.
Agreed. The only factors usually used in determining the
duty cycle are the source voltage, the switch voltage drop,
the flyback diode drop and the output voltage. As you note,
there is no output current involved.
But that is "by definition." In other words, there was an
underlying assumption made in developing that relationship
which isn't true if there is no load at all.
Can you identify it?
>It should regulate across a wide spectrum with no
>feedback, but seems to go bananas when the load gets close to open
>circuit. Maybe it's only a completely open circuit that causes
>problems?
No, the problem lies in the definitions used when developing
certain relationships; and then forgotten when considering no
load and no feedback. In short, it's changing assumptions
without noticing and then finding things don't work as
expected. Nothing mysterious, really. Just unsound logic.
There is the combination of two definitions involved. Think
about it for a moment. I'll write more, but it is better if
you figure it out for yourself or at least try, beforehand.
Jon
Jon Kirwan
............................................
On Sat, 15 Jan 2011 18:09:15 -0800 (PST), Bill Bowden
<bpe...@bowdenshobbycircuits.info> wrote:
><snip>
>I don't see the need for feedback. The output voltage should be a
>function of the input and duty cycle alone, regardless of the load
>resistance. It should regulate across a wide spectrum with no
>feedback, but seems to go bananas when the load gets close to open
>circuit. Maybe it's only a completely open circuit that causes
>problems?
Let me start out by arriving at _your_ conclusion, but
showing excessively tedious work to get there. (I could
short-hand, but might as well take each and every step.)
The duty cycle is:
D = t_on / (t_on + t_off)
No argument possible. It's "by definition." D is defined to
mean that. Now, explore it a little:
D = t_on / (t_on + t_off)
D = 1 / [(t_on + t_off) / t_on]
D = 1 / [1 + t_off / t_on]
This last one will come in handy, shortly. The ratio between
t_off and t_on, in particular, is so handy that some assign a
special name just to that.
An important equation regarding inductors will now be
introduced:
V = L * dI/dt
Keep in mind, reading the following, that discontinuous mode
operation has the ampere-turns sit at zero for at least one
instant of time.
During the ON phase, defining V_on to be the voltage across
the inductor, I_peak to be the peak current achieved at the
end of the ON phase, t_on to be some fixed switch-ON time,
and assuming L doesn't saturate and change value, we know:
V_on = L * I_peak / t_on
Since I is assumed to reach zero (ampere-turns sits at zero
prior to the start of the ON phase) then it follows that dI,
which equals the difference between the current at the end
and start of the ON phase, is simply equal to I_peak.
Similarly, during the OFF phase, again assuming that we allow
just barely enough time for the all of the I_peak to reach
zero (just for an instantaneous moment and no more):
V_off = L * I_peak / t_off
(I've focused on magnitudes, not signs, here. The actual
case is that V_off is of opposite sign to V_on. But it's not
important because I'm just speaking in terms of magnitude.)
Since L*I_peak is the same in both equations, we can replace
one with the solution for the other, so that:
V_off = V_on * t_on / t_off
Or,
V_off * t_off = V_on * t_on
Or,
t_off / t_on = V_on / V_off
Now it is a good time to remember the earlier development of
the duty cycle equation, repeated here:
D = 1 / [1 + t_off / t_on]
We can substitute:
D = 1 / [1 + V_on / V_off]
D = 1 / [(V_off + V_on) / V_off]
D = V_off / (V_off + V_on) = 1 + V_off / V_on
If you ignore the flyback diode drop and ignore the switch
voltage drop (BJT or MOSFET), then:
V_on = V_in
V_off = V_out - V_in
so that,
D = (V_out - V_in) / V_out = 1 - V_in / V_out
Which is why you said what you did. It looks an awful lot as
though the input and output voltage entirely determines the
duty cycle. (One can slightly complicate the above by
bringing back in the flyback diode drop or the switch drop.)
Or, put in your words, the input voltage and duty cycle
entirely determines the output voltage.
............
Pause for a moment.
............
The problem comes from the fact that the required I_peak
depends upon I_out. Let's see why.
Assuming you already know V_in and V_out:
P_in = P_out
V_in * I_in = V_out * I_out
V_in * [(1/2) * I_peak] = V_out * I_out
Do you follow why I_in = I_peak/2?
Anyway, this results in:
I_peak = 2 * I_out * (V_out / V_in)
Since I_peak is held constant by holding t_on constant and
since V_in is also constant, then it follows from reordering
the above that:
V_out = V_in * I_peak / (2 * I_out)
= k / I_out, where k = (V_in*I_peak/2)
Note that V_out is a reciprocal function of I_out. And that
vanishing I_out values imply impossibly high values for
V_out.
Jon
Jasen Betts
> On Sat, 15 Jan 2011 18:09:15 -0800 (PST), Bill Bowden
><bpe...@bowdenshobbycircuits.info> wrote:
>>I don't see the need for feedback.
>
> All of the energy built up into the inductor during the ON
> time is delivered to the capacitor during the OFF time. The
> capacitor's voltage must rise in order to store it. And it
> must continue to rise with each cycle. There is no stopping
> it.
Parallel the high-side diode with a mosfet (and the low side mosfet with
a diode) and you get a bi-directional converter that always runs in
Jon Kirwan
wrote:
Thanks for the clue. I've just been looking via google using
the terms and I think I see what you mean. But I'm going to
have to think more. Here is some of what I saw right away:
http://www.vishay.com/docs/71917/71917.pdf
http://sibsil.narod.ru/199.pdf
http://www.nachhaltigwirtschaften.at/pdf/epsec01_bidi.pdf
Thanks,
Jon
Bill Bowden
Without a lot of math, seems like If you consider the case of the
infinite inductance and infinite output capacitance, the inductor
current and output voltage will be constant, no peaks or valleys, just
(flat line). So, if you preset the inductor current and output voltage
for the desired load, the switch just supplies the lost energy while
the switch was open.
I have built a adjustable duty cycle converter to boost 9 volts to 14
and found the output voltage was 14 volts at 600 milliamps. And
changing the load to 2K ohms, or maybe 6 milliamps only increases the
output 1 volt to 15 volts. So, it appears to regulate quite well from
6 milliamps to 600 milliamps while the output only changes from 14 to
15. I was using a fairly large inductor of 25 millihenrys at may 6
KHz.
-Bill
Jon Kirwan
okay. Did the inductor fall to zero amps each cycle? Can
you check that? It's important to know.
Let's go through some calculations. I'm going to tentatively
assume, as you don't say, that you computed a duty cycle
based upon D=1-Vin/Vout. In this case, would be D=35.7%,
yes?
f = about 6kHz
D = 0.357
ton = D/f = 59.5us ... call it 60us
toff = 107us
Ipeak = 9V*60us/25mH = 21.6mA ... call it 22mA
Power = (1/2)*L*I^2*f = .5*25mH*22mA^2*6000 = 36.3mW
max: Vout*Iout = 14*600mA = 8.4W
min: Vout*Iout = 15*6mA = 90mW
So tested from about 90mW to over 8W. All of which exceeds
the 36.3mW figure. This pretty much tells me that you were
in continuous mode.
Different rules apply here.
For testing in discontinuous mode, you need to set your
Vout*Iout to be less than your (1/2)*L*Ipeak^2*f figure. Then
I think you will see the output voltage more "mobile" and
following my earlier discussion better.
Jon
Bill Bowden
Ok, I put the thing on the bench and added a 0.2 ohm resistor in
series with the inductor to monitor current on a scope. I get about
250mA minimum and about 5 amps peak using a 22 ohm load with 9 volt
input and 13.3 volt output, or maybe 8 watts. The inductor measures
about 12 millihenrys and frequency is 8.3Khz and duty cycle is 50
microseconds on and 70 microseconds off. Next test was with a 1K load
and 27 ohm series resistor which indicated minimum current was about
10mA and peak was about 45mA with a much more linear waveform. Output
voltage increased to 15.2 with the 27 ohm resistor removed. The
inductor current appears to approach zero with about a 2K load and the
voltage increases to around 24 volts with a 8.2K load. So, it starts
to go out of regulation near the 2K point, and the output voltage
rises as the load resistance is increased beyond that.
-Bill
Jon Kirwan
<bpe...@bowdenshobbycircuits.info> wrote:
>On Jan 17, 9:30嚙緘m, Jon Kirwan <j...@infinitefactors.org> wrote:
>> On Mon, 17 Jan 2011 20:59:37 -0800 (PST), Bill Bowden
>>
>> <bper...@bowdenshobbycircuits.info> wrote:
>> >On Jan 16, 7:42嚙窮m, Jon Kirwan <j...@infinitefactors.org> wrote:
>> >> On Sat, 15 Jan 2011 18:09:15 -0800 (PST), Bill Bowden
>>
>> >> <bper...@bowdenshobbycircuits.info> wrote:
>> >> >> far as I can tell and without any kind of load, either. 嚙瘢f
>> >> >> there is no load, the inductor energy simply dumps onto the
>> >> >> cap and keeps on dumping. 嚙確he voltage just rises until
>> >> >> something gives (or until the energy in each pulse is enough
>> >> >> wasted somewhere else that it doesn't add to the cap,
>> >> >> anymore.)
>>
>> >> >> Jon
>>
>> >> >I don't see the need for feedback.
>>
>> >> All of the energy built up into the inductor during the ON
>> >> time is delivered to the capacitor during the OFF time. 嚙確he
>> >> capacitor's voltage must rise in order to store it. 嚙璀nd it
>> >> must continue to rise with each cycle. 嚙確here is no stopping
>> >> it.
>>
>> >> >The output voltage should be a
>> >> >function of the input and duty cycle alone, regardless of the load
>> >> >resistance.
>>
>> >> Agreed. 嚙確he only factors usually used in determining the
>> >> duty cycle are the source voltage, the switch voltage drop,
>> >> the flyback diode drop and the output voltage. 嚙璀s you note,
>> >> there is no output current involved.
>>
>> >> But that is "by definition." 嚙瘢n other words, there was an
>> >> underlying assumption made in developing that relationship
>> >> which isn't true if there is no load at all.
>>
>> >> Can you identify it?
>>
>> >> >It should regulate across a wide spectrum with no
>> >> >feedback, but seems to go bananas when the load gets close to open
>> >> >circuit. Maybe it's only a completely open circuit that causes
>> >> >problems?
>>
>> >> No, the problem lies in the definitions used when developing
>> >> certain relationships; and then forgotten when considering no
>> >> load and no feedback. 嚙瘢n short, it's changing assumptions
>> >> without noticing and then finding things don't work as
>> >> expected. 嚙瞇othing mysterious, really. 嚙皚ust unsound logic.
>>
>> >> There is the combination of two definitions involved. 嚙確hink
>> >> about it for a moment. 嚙瘢'll write more, but it is better if
>> >> you figure it out for yourself or at least try, beforehand.
>>
>> >> Jon
>>
>> >Without a lot of math, seems like If you consider the case of the
>> >infinite inductance and infinite output capacitance, the inductor
>> >current and output voltage will be constant, no peaks or valleys, just
>> >(flat line). So, if you preset the inductor current and output voltage
>> >for the desired load, the switch just supplies the lost energy while
>> >the switch was open.
>>
>> >I have built a adjustable duty cycle converter to boost 嚙� volts to 14
>> >and found the output voltage was 14 volts at 600 milliamps. And
>> >changing the load to 2K ohms, or maybe 6 milliamps only increases the
>> >output 1 volt to 15 volts. So, it appears to regulate quite well from
>> >6 milliamps to 600 milliamps while the output only changes from 14 to
>> >15. I was using a fairly large inductor of 25 millihenrys at may 6
>> >KHz.
>>
>> >-Bill
>>
>> okay. 嚙瘩id the inductor fall to zero amps each cycle? 嚙瘠an
>> you check that? 嚙瘢t's important to know.
>>
>> Let's go through some calculations. 嚙瘢'm going to tentatively
>> assume, as you don't say, that you computed a duty cycle
>> based upon D=1-Vin/Vout. 嚙瘢n this case, would be D=35.7%,
>> yes?
>>
>> f = about 6kHz
>> D = 0.357
>> ton = D/f = 59.5us ... call it 60us
>> toff = 107us
>>
>> Ipeak = 9V*60us/25mH = 21.6mA ... call it 22mA
>>
>> Power = (1/2)*L*I^2*f = .5*25mH*22mA^2*6000 = 36.3mW
>>
>> max: Vout*Iout = 14*600mA = 8.4W
>> min: Vout*Iout = 15*6mA = 90mW
>>
>> So tested from about 90mW to over 8W. 嚙璀ll of which exceeds
>> the 36.3mW figure. 嚙確his pretty much tells me that you were
>> in continuous mode.
>>
>> Different rules apply here.
>>
>> For testing in discontinuous mode, you need to set your
>> Vout*Iout to be less than your (1/2)*L*Ipeak^2*f figure. Then
>> I think you will see the output voltage more "mobile" and
>> following my earlier discussion better.
>>
>> Jon
>
>Ok, I put the thing on the bench and added a 0.2 ohm resistor in
>series with the inductor to monitor current on a scope. I get about
>250mA minimum and about 5 amps peak using a 22 ohm load with 9 volt
>input and 13.3 volt output, or maybe 8 watts. The inductor measures
>about 12 millihenrys and frequency is 8.3Khz and duty cycle is 50
>microseconds on and 70 microseconds off.
So this is definitely continuous mode, as the ampere-turns
doesn't go to zero. In continuous mode, I believe V_out is
indeed determined by the input voltage and the duty cycle,
V_in/(1-Duty). So 9/(1-50us/120us)=15.4V. But you are
losing power in the diode, the switch (bjt?), and your
inserted resistor, as well. With 5A peak, I'm not at all
surprised that the output voltage is 2V less than 100%
efficiency might suggest.
>Next test was with a 1K load
>and 27 ohm series resistor which indicated minimum current was about
>10mA and peak was about 45mA with a much more linear waveform. Output
>voltage increased to 15.2 with the 27 ohm resistor removed.
Which is still continuous mode (10mA minimum tells you this)
and note that with much lower peak currents (and attending
losses) the output voltage is nearly what I calculated above
with perfect efficiency. Sounds correct. (Again, I'm
assuming here that you didn't change the 50us on/70us off
timing.)
>The
>inductor current appears to approach zero with about a 2K load and the
>voltage increases to around 24 volts with a 8.2K load. So, it starts
>to go out of regulation near the 2K point, and the output voltage
>rises as the load resistance is increased beyond that.
V_out = 24V
I_out = 24/8200 or about 2.9mA
P_out = 24 * 2.9mA = 70mW
The power stored in the inductor during the 50us period is
found as:
f = 8.3kHz
D = 50us/120us = .41666
ton = 50us
toff = 70us
L = 12mH
Ipeak = 9V*50us/12mH = 37.5mA
Power = (1/2)*L*I^2*f = .5*12mH*37.5mA^2*8300 = 70mW
The power delivered by the 9V supply during the 70us period
isn't quite as obvious as it doesn't conduct the entire
period (usually.) So what we do know is that the voltage
across L will be 24V-9V-Vdiode or say 14.4V? We go with
that. dt=dI*L/V, so 37.5mA*12mH/14.4 or about 30us. That's
the free wheeling conduction time, from what you've reported.
So the battery power is .5*9*37.5mA*30us/120us or 42mW. So
the total is (70+42)mW or 112mW. As a rough guess. There
are losses in the inductor (call it 1mW for now) and free
wheeling diode (call it 4mW for now) and switch (call it
25mW.) So lets call the result P_out = 82mW.
From 24V and 8200 ohms, we computed about 70mW. Probably
close enough. Also, V_out = SQRT(82mW * 8200) = 25.9V. Which
isn't inconsistent with your 24V measurement.
Jon
Bill Bowden
> On Tue, 18 Jan 2011 19:16:57 -0800 (PST), Bill Bowden
>
>
>
> >On Jan 17, 9:30 pm, Jon Kirwan <j...@infinitefactors.org> wrote:
> >> On Mon, 17 Jan 2011 20:59:37 -0800 (PST), Bill Bowden
>
> >> <bper...@bowdenshobbycircuits.info> wrote:
> >> >> On Sat, 15 Jan 2011 18:09:15 -0800 (PST), Bill Bowden
>
> >> >> <bper...@bowdenshobbycircuits.info> wrote:
> >> >> >> there is no load, the inductor energy simply dumps onto the
> >> >> >> something gives (or until the energy in each pulse is enough
> >> >> >> wasted somewhere else that it doesn't add to the cap,
> >> >> >> anymore.)
>
> >> >> >> Jon
>
> >> >> >I don't see the need for feedback.
>
> >> >> All of the energy built up into the inductor during the ON
> >> >> capacitor's voltage must rise in order to store it. And it
> >> >> must continue to rise with each cycle. There is no stopping
> >> >> it.
>
> >> >> >The output voltage should be a
> >> >> >function of the input and duty cycle alone, regardless of the load
> >> >> >resistance.
>
> >> >> duty cycle are the source voltage, the switch voltage drop,
> >> >> there is no output current involved.
>
> >> >> underlying assumption made in developing that relationship
> >> >> which isn't true if there is no load at all.
>
> >> >> Can you identify it?
>
> >> >> >It should regulate across a wide spectrum with no
> >> >> >feedback, but seems to go bananas when the load gets close to open
> >> >> >circuit. Maybe it's only a completely open circuit that causes
> >> >> >problems?
>
> >> >> No, the problem lies in the definitions used when developing
> >> >> certain relationships; and then forgotten when considering no
> >> >> without noticing and then finding things don't work as
>
> >> >> There is the combination of two definitions involved. Think
> >> >> about it for a moment. I'll write more, but it is better if
> >> >> you figure it out for yourself or at least try, beforehand.
>
> >> >> Jon
>
> >> >Without a lot of math, seems like If you consider the case of the
> >> >infinite inductance and infinite output capacitance, the inductor
> >> >current and output voltage will be constant, no peaks or valleys, just
> >> >(flat line). So, if you preset the inductor current and output voltage
> >> >for the desired load, the switch just supplies the lost energy while
> >> >the switch was open.
>
> >> >and found the output voltage was 14 volts at 600 milliamps. And
> >> >changing the load to 2K ohms, or maybe 6 milliamps only increases the
> >> >output 1 volt to 15 volts. So, it appears to regulate quite well from
> >> >6 milliamps to 600 milliamps while the output only changes from 14 to
> >> >15. I was using a fairly large inductor of 25 millihenrys at may 6
> >> >KHz.
>
> >> >-Bill
>
>
> >> Let's go through some calculations. I'm going to tentatively
> >> assume, as you don't say, that you computed a duty cycle
> >> yes?
>
> >> f = about 6kHz
> >> D = 0.357
> >> ton = D/f = 59.5us ... call it 60us
> >> toff = 107us
>
> >> Ipeak = 9V*60us/25mH = 21.6mA ... call it 22mA
>
> >> Power = (1/2)*L*I^2*f = .5*25mH*22mA^2*6000 = 36.3mW
>
> >> max: Vout*Iout = 14*600mA = 8.4W
> >> min: Vout*Iout = 15*6mA = 90mW
>
> >> the 36.3mW figure. This pretty much tells me that you were
This is what I don't understand. Your example below says the peak
current is determined from the inductor value and time and applied
voltage, or 37.5mA in this case. This is happening while the load is
completely switched out of the circuit and should have no effect.
However, in the case of the 22 ohm load and 8 watt situation, the peak
current will reach 5 amps, which doesn't agree with what you have
below. Can you clarify?
f = 8.3kHz
D = 50us/120us = .41666
ton = 50us
toff = 70us
L = 12mH
Ipeak = 9V*50us/12mH = 37.5mA
Power = (1/2)*L*I^2*f = .5*12mH*37.5mA^2*8300 = 70mW
> The power delivered by the 9V supply during the 70us period
> isn't quite as obvious as it doesn't conduct the entire
> period (usually.) So what we do know is that the voltage
> across L will be 24V-9V-Vdiode or say 14.4V? We go with
> that. dt=dI*L/V, so 37.5mA*12mH/14.4 or about 30us. That's
> the free wheeling conduction time, from what you've reported.
>
> So the battery power is .5*9*37.5mA*30us/120us or 42mW. So
> the total is (70+42)mW or 112mW. As a rough guess. There
> are losses in the inductor (call it 1mW for now) and free
> wheeling diode (call it 4mW for now) and switch (call it
> 25mW.) So lets call the result P_out = 82mW.
>
> From 24V and 8200 ohms, we computed about 70mW. Probably
> close enough. Also, V_out = SQRT(82mW * 8200) = 25.9V. Which
> isn't inconsistent with your 24V measurement.
>
> Jon
-Bill
Jon Kirwan
I can try to clarify. Of course, one must always keep in
mind I have had zero training, too. So I love it when you
press on, as it challenges me and I need that. Thanks. That
said, here's my shot at answering your question.
Imagine the case of discontinuous mode (tentatively, let's
think about it as applying when the load is very light.)
There are three periods of time in the fixed-frequency case
and I'll make my own choice about their order:
(1) switch-ON, inductor magnetic field rising from 0 to X
(2) switch-OFF, magnetic field collapsing back to 0
(3) switch-OFF, dead time from 0 ampere-turns till next
cycle starts again.
I'll call these SON, SOFF1, and SOFF2, respectively.
As the load is increased, more current is drawn. During
SOFF1, there is current flowing via the battery and inductor
to charge the cap. But during SOFF2 and the following SON,
the storage cap supplies the current. Which means the
voltage droops during this time.
When the next SOFF1 period starts after that droop, the
voltage across the inductor that is required in order to
freewheel the diode into conduction is lower. This lower
voltage implies, from dt=L*dI/V, that dt lengthens. (dI is
taken as I_peak in discontinuous mode and is entirely
determined by the duration of the SON period, given a fixed
inductor and supply voltage.) This means that more time is
required for the inductor to relax. Which means SOFF1
lengthens in duration, while SOFF2 shortens.
Eventually, as the load continues to increase and the droop
also increases, SOFF2 gets zero time and SOFF1 occupies the
entire remaining period. This is the moment of switching
between discontinuous and continuous mode. The load
increases and at some point you move from one into the other
domain.
Okay. So let's assume we have a basic feel for discontinuous
mode and move on to continuous mode as an extension. What
happens as the load continues to grow? ... ??
As this point is crossed, the droop goes still higher and dt
grows beyond the time you've allowed. Remember that we are
talking about a fixed period of time and fixed duty cycle,
for now. So when there is no longer enough time for the
inductor to relax to zero ampere-turns, the current doesn't
reach zero (obviously) and the following SON period will
start with a non-zero inductor current. Since the SON period
is fixed (by definition in this example), the dI=V*dt/L will
still be the same but will "build up" upon the baseline
current left by the prior cycle. So I_peak will not be
entirely determined now by the SON period of time and I_peak
will be higher than used in computations for discontinuous
mode.
Another way of writing this: I_peak is entirely computable,
regardless of load, by knowing the inductor, source voltage,
and SON time period in discontinuous mode. You can figure
that out without knowing the load, a priori. What changes
with varying load, in discontinuous mode, is the relative
apportioning of time between SOFF1 and SOFF2. However, in
continuous mode SOFF2 doesn't exist and SOFF1 time is
entirely fixed by your fixed frequency and duty cycle. (Let's
just call it SOFF in continuous mode.) Since SOFF time don't
change on load variations, obviously, something else has to
give. What gives is I_peak. In continuous mode, I_peak
rises with rising load. The SOFF period simply sets how much
of I_peak is allowed to relax.
Since current is flowing all of the time in continuous mode
and since that current is also on average higher and always
flowing during SOFF onto the cap, there is more charge
transfer to the cap during SOFF and therefore a higher load
is supportable.
In all cases, discontinuous and continuous mode alike, the
V_out*I_out must equal V_in*I_in (ignoring inefficiency for
simplicity's sake.)
In discontinuous mode, I_peak is known (determined.) Current
linearly rises from zero to I_peak during SON, falls linearly
from I_peak to zero during SOFF1, and does nothing at all
during SOFF1.
In continuous mode, I_peak isn't known (it depends upon the
load.) Current rises from some non-zero floor value to
I_peak during SON, falls linearly from there back down to the
non-zero floor during SOFF, where the difference between
I_peak and the floor value is the same as I_peak would have
been computed for discontinuous mode.
That's the qualitative angle. I've stayed clear of writing
equations and expressions in this post and instead focused on
'understanding' things with broad brush strokes. (I did
equations before and it left questions.)
....
So let's take the discontinuous case and develop some
quantitative descriptions from the general understanding for
a situation where the load resistor, input voltage, duty
cycle, frequency, and inductor values are known but where we
want to know the output voltage and current that results.
Let's use your last tentative situation which I said ran in
discontinuous mode:
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 8200 Ohms
t_son= 50us
Add to the above, the following guesses:
V_sw = 0.1 V (BJT Vce drop, when on)
V_di = 0.5 V (freewheeling diode drop)
R_L = 0.4 Ohms (inductor resistance)
So,
duty = 50us*8300 Hz = 41.5%
Since in discontinuous mode, the inductor starts out by
entering SON with the ampere-turns being zero. So during
SON, we can compute:
I_peak = (V_in-Vsw)*t_son/L = (9-.1)*50us/12mH = 37mA
We want to know average values for V_out and I_out and I_in.
First off, let's write out the power equation:
P_out = P_in - (P_di + P_sw + P_L) = P_in
This illustrates that there are some losses. P_di is the
freewheeling diode loss, P_sw is the BJT switch loss, and P_L
is the inductor loss. Before continuing, let's get a bead on
them. A maximum figure would bound the problem. Let's see
where that takes us.
P_di = (1/2) * V_di * I_peak * t_soff1 * f = 5.4mW
P_sw = (1/2) * V_sw * I_peak * t_son * f = 0.77mW
P_L = (1/3) * R_L * I_peak^2 * t_son * f +
(1/3) * R_L * I_peak^2 * t_soff1 * f = 1.83mW
I used the worst possible value for t_soff1, which is 120.5us
- 50us, or 70.5us. In no case can it take longer than that,
if we stay in discontinuous mode. So, roughly speaking, 8mW
is the worst case loss. And the diode appears to be the main
culprit. The diode drop also factors in computing t_soff1.
So we can't entirely ignore it there, either.
Anyway, now that we have a worst case idea about losses,
let's use the figure 5mW for computations and label it
P_loss:
P_loss = 5mW
P_out = P_in - 5mW
V_out^2 / R + 5mW = V_in * I_in
So what is I_in, on average?
I_in = (1/2) * I_peak * (t_son + t_soff1) * f
That accounts for the rise during SON and the fall during
SOFF1. (It is zero during SOFF2.)
The problem is in figuring out t_soff1. It depends upon the
output voltage. But without controlling the duty cycle, you
don't know what V_out is, as it will depend upon t_soff1.
You can get a bead on it, though. From the above, we can
combine and re-arrange:
V_out = SQRT(R*(V_in*I_peak*(t_son+t_soff1)*f-10mW)/2)
and include,
t_soff1 = I_peak * L / (V_out - V_di - V_in)
Iterate. Start out by taking t_soff1 = 1/f - t_son and
compute V_out. Then plug that V_out into the t_soff1
equation and compute it. Then use that value back into the
V_out equation. Etc. Until it gets close. I get:
V_out t_soff1
0. ... 70.5 us
1. 36.4 V 16.5 us
2. 26.7 V 25.8 us
3. 28.6 V 23.2 us
4. 28.1 V 23.9 us
5. 28.2 V 23.7 us
6. 28.2 V 23.8 us
7. 28.2 V 23.75 us ... stop
It's actually a cubic equation that can be solved. But the
iterative approach works just fine.
So we get V_out = 28.2V with t_soff1 = 23.75us. Now we can
compute the average I_in as:
I_in = (1/2) * 37mA * (50us + 23.75us) * 8300 = 11.3mA
So,
P_in = 9 V * 11.3 mA = 101.7 mW
Clearly, using V_out = 28.2V, we get:
P_out = 28.2^2/8200 = 97mW
Which leaves us a difference of 4.7mW, which is very close to
the 5mW we used going into this process.
To reach the 24V you experienced:
P_out = 24^2/8200 = 70.2mW
That means that there must have been losses amounting to
about (101.7mW-70.2mW) or 31.5mW. Maybe 70% efficiency?
....
Now for your continuous case, which I have to say I don't
completely understand, either. You will quickly see why.
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 22 Ohms
t_son= 50us
The dI value will be the same, namely 37mA. But it will ride
on top of some I_base value. But!!! You said you were
getting a low of about 250mA and a peak of 5A!! What??? How
can this be??
This equation is pretty solid:
V/L = dI/dt
It won't give an inch. This means:
dI = V/L * dt
We know V is almost 9V. We know L=12mH (or think we do.) And
we know that dt=50us because you set things that way. This
means there is no escaping:
dI = (V_in-Vsw)*t_son/L = (9-.1)*50us/12mH = 37mA
It just sits there. But you were seeing (5A - 250mA) or
4.75A!! How??? The equation says you cannot get there from
here.
Well, the assumption that L is 12mH has to be wrong. I
believe your 50us. And I cannot see that your 9V supply
suddenly jumps up to huge numbers on its own. The only other
way you can get 5A is if L significantly drops in value.
And it can do that. Through saturation.
It's going to be hard for me to go any further because of
that fact.
Perhaps someone else knows. But that case makes no sense to
me, for now, except deciding that you are experiencing a
saturated core.
In the remaining case:
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 1000 Ohms
t_son = 50us
t_soff= 70.5us
In this case, you reported approximately I_peak=45mA and a
base line of 10mA. So a difference of 35mA.
Note this is VERY CLOSE to the 37mA I've been estimating. So
very likely there is no core saturation here.
I'm going to stop for now and let things soak in a bit. If
you buy all the rest I've said, I'll be happy to take on this
middle case you presented. But I think you will see that it
comes out sensibly, too.
Until then, let me know what you think about the rest.
Jon
Jon Kirwan
<jo...@infinitefactors.org> wrote:
><snip>
>In the remaining case:
>
> V_in = 9 V
> f = 8300 Hz
> L = 12 mH
> R = 1000 Ohms
> t_son = 50us
> t_soff= 70.5us
>
>In this case, you reported approximately I_peak=45mA and a
>base line of 10mA. So a difference of 35mA.
>
>Note this is VERY CLOSE to the 37mA I've been estimating. So
>very likely there is no core saturation here.
><snip>
In the continuous case, V_out is determined:
V_out = V_in * (1 + t_on / t_off) - V_di
or,
V_out = V_in / (1 - Duty) - V_di
I compute something less than 15.2V from this, which is what
you said you measured. But I'm using more than 0.2V for the
diode drop and it may be possible that the duty cycle value
is a tad bit larger (slightly lower frequency than 8300 or
slightly longer t_on time?) Or maybe the diode really isn't
dropping that much, being a schottky?
Anyway, it's close and I'm not surprised.
You noted a minimum current in the inductor near 10mA and a
peak near 45mA. The difference is 35mA, which is remarkably
close to the 37mA I'd estimated from values you'd provided.
So that suggests that your measurements are fairly close and
that ignored unknowns (like inductor resistance) aren't
confusing things much.
The only remaining question might be about the 10mA and why
it goes only that low. Guessing that power in must be close
to power out, ignoring losses entirely, I get:
I_min = V_out^2/(R*V_in) - dI/2
Using your measured values, where V_out=15.2V, R=1000 ohms,
V_in=9V, and dI=35mA, I compute I_min=8.2mA. Which you may
consider close enough. I can't say.
I considered adding some thoughts about volt-seconds. It
really helps in considering both discontinuous and continuous
mode. But I'll hold short, for now.
Jon
Jon Kirwan
>Using your measured values, where V_out=15.2V, R=1000 ohms,
>V_in=9V, and dI=35mA, I compute I_min=8.2mA. Which you may
>consider close enough. I can't say.
With losses, I_min would have to be higher of course.
Jon
Bill Bowden
I'm pretty sure the inductor is 12 mH. I resonated it against a .33uF
cap and got 2500 Hz, or 12 mH.
As for saturation, the core may be near saturation but the efficiency
is 85% so it looks good.
But just to eliminate the saturation problem, if it exists, I changed
the load to 55 ohms and got the following results:
9 volt input at 480mA, or 4.32 watts
14.5 volt output at 55 ohms, or 255mA, 3.69 watts
Duty cycle = 54uS on / 70us off
Ipeak = 2 amps.
Imin= 200 mA
So, now all we have to figure out is how the current increases from
200mA to
2 amps in 54us using a 12mH inductor and 9 volt supply.
-Bill
Jon Kirwan
I'm not doubting that.
>I resonated it against a .33uF cap and got 2500 Hz, or 12 mH.
Fine.
>As for saturation, the core may be near saturation but the efficiency
>is 85% so it looks good.
Until saturation and the current starts peaking like crazy,
should be okay.
>But just to eliminate the saturation problem, if it exists, I changed
>the load to 55 ohms and got the following results:
>
>9 volt input at 480mA, or 4.32 watts
>14.5 volt output at 55 ohms, or 255mA, 3.69 watts
>Duty cycle = 54uS on / 70us off
>Ipeak = 2 amps.
>Imin= 200 mA
>
>So, now all we have to figure out is how the current increases from
>200mA to 2 amps in 54us using a 12mH inductor and 9 volt supply.
(What is the capacitor value you are using on the output???)
Duty cycle is 54/124 = 43.55%. Given the usual computation
for continuous mode, I'd get V_out=V_in/(1-Duty)-V_di which
would be something in the vacinity of 15.2V-15.5V. You are
getting 14.5V.
However, big problem as you already know. dI=(V dt)/L and
that works out to 40.5mA. This is physics here, so I think
what you have is called a "swinging choke." The L value
declines as L saturates from excessive volt-seconds.
Saturation itself doesn't by itself mean you lose energy. It
means your peak current goes wild, though. And you will lose
energy to increased resistive losses. So your output likely
isn't quite up where it would be because of those losses that
arrive from such high currents.
Okay. So now it is 54us on. There is no way to violate the
following: V dt = L dI. So far as I'm aware. Since V is
set by a low impedence 9V power supply (I assume that's solid
enough) and since dt is set by your driver (I assume for now
there isn't any ringing/oscillation taking place, though you
might check) then the left side of that equation is fixed.
You are supplying 9V * 54us or 486u V-s. The right side
__must__ match that. Plain and simple.
Now you show 2A-200mA or 1.8A for dI. You work it out for L.
I get way less than 1mH.
So now the L in the above equation is replaced by L(t). Which
is a different beast.
Let's talk more about the inductor, now. Let me set the
deck, first.
Start with these (U0=4e-7*pi, Ur is relative permeability),
1. mmf = N I (in ampere-turns)
2. H = mmf / l (in ampere-turns/meter)
3. B = U0 Ur H (the relationship between B and H)
4. L = Ac U0 Ur N^2 / l (inductance)
5. V = L dI/dt
Combining equations 1 and 2 and taking equation 3, I can put
them into a slightly better form,
6. dH = (N / l) dI
7. dB = U0 Ur dH
Combining equations 6 and 7,
8. dB = (N / l) U0 Ur dI
This might be enough, if we knew the permeability of your
core, its magnetic path length, and the number of windings.
But let me go further, too. Just for kicks. Introducing
time to both sides of equation 8,
9. dB/dt = [(N / l ) U0 Ur] dI/dt
From equation 5 above we can extract dI/dt and substitute
that into equation 9 giving,
10. dB/dt = [(N / l ) U0 Ur] V/L
Substituting equation 4 as L in equation 10 and doing the
usual simplifying magic gives,
11. dB/dt = [(N / l ) U0 Ur] [(V l) / (Ac U0 Ur N^2)]
or,
12. dB/dt = V / (Ac N)
Rearranging gives,
13. dB = V dt / (Ac N)
Note that we already know (V dt) as 486u V-s. What I don't
know from you is N and Ac and Bsat for your core material.
Don't bother looking for Bsat right now. What I'd like to
have is N and Ac. Can you tell me the number of turns and
the area of the magnetic core? Or provide the magnetic path
length and permeability?
Looking at equation 13 suggests a few thoughts, without
knowing anything more. We can assume that N and Ac are both
fixed by your inductor design... so, constant and not likely
to change. Only (V dt) can change. Since we know that is
486u V-s, then the best suggestion I can provide right now is
that you reduce dt. In other words, don't use 54us. Cut
that back. I'm betting that your inductor cannot handle that
many volt-seconds and saturates. Instead, maybe increase the
frequency but cut back on the ON time. Drop it to, say,
40us? Or less? In other words, let's first find an ON time
that works consistently.
Otherwise, the equations won't help much.
Have you also tried putting your circuit into LTspice to see
what it says?? I've plugged in the numbers and LTspice
suggests an output roughly 15.3V and with dI being roughly
39mA and a I_min of 475mA and an I_max of 514mA. It doesn't
come close to agreeing with your experience. But it doesn't
take into account saturation with the model for L that I
used, either.
Jon
Bill Bowden
Yes, but the circuit was tested using a 1K load resulting in a
inductor current change of 10mA to 45mA in 50uS at 9 volts, or 12
millihenrys which agrees with textbook figures, so I'm fairly
confident the inductance is 12 mH. This was mentioned in an earlier
post.
The inductor was hand wound using about 74 turns on a unknown ferrite
core measuring 1.5 inch outside diameter and 7/8 inside diameter. It's
a fairly large beast with lots of area suggesting many millihenrys of
inductance. And if the inductance is low, the saturation current will
be high, since it will be more like an air core inductor that doesn't
saturate.
So, all I know is the circuit works well under minimal load conditions
and meets text book requirements, and also maintains efficiency at
high loads, but doesn't meet text book requirements at high loads. It
just does it's own thing at high loads.
Maybe some use is the current waveform at high load which is not
linear.
At minimal load, the scope waveform is very linear with straight lines
indicating the current rise and fall to the predicted values, just
like what you would expect. However, with heavy loads, the waveform
resembles a V shape with slowly beginning and ending points, sort of
like an RC or LC time constant curve. I can adjust the duty cycle and
notice the peak of the "V" waveform increases or decreases but looks
about the same overall. You might think this a saturation indication,
but I don't think so since it maintains 85% efficiency and I get
similar results at twice or half the load, and the inductor is a very
large beast probabaly not saturating.
-Bill
Jon Kirwan
Yes, it seems to be saturating. Like I said, that doesn't by
itself mean you lose energy in the process. (You will,
because of ohmic resistances, the Vce or Vds of the switch at
peak current, and the Vdiode at these higher currents, but
that's not the main point.) It just means that the calcs
which assume L is constant need to be replaced with L(t).
Frankly, if you are fine with all this ("maintains efficiency
at high loads") then I'm fine, too. I was just trying to
help you see why other measurements you made were as they
were.
Jon
Bill Bowden
>
> Yes, it seems to be saturating. Like I said, that doesn't by
> itself mean you lose energy in the process. (You will,
> because of ohmic resistances, the Vce or Vds of the switch at
> peak current, and the Vdiode at these higher currents, but
> that's not the main point.) It just means that the calcs
> which assume L is constant need to be replaced with L(t).
>
> Frankly, if you are fine with all this ("maintains efficiency
> at high loads") then I'm fine, too. I was just trying to
> help you see why other measurements you made were as they
> were.
>
> Jon
Yes, I think I see it now. The inductor is partially saturating
causing the current to rise faster as time goes on. In other words di/
dt is not constant, and therefore the waveforms are not linear.
I took a look at the voltage drop across the mosfet which indicates
the current does not rise linearly, and is increasing much faster near
the end of the switch-on time. The on-resistance of the mosfet is 0.18
ohms and the drain voltage rises to about 1 volt at the end of the
waveform indicating about 5 amps peak which agrees with the other
measurements using the 8 watt load.
So, it appears to be acting as a variable inductor where the effective
inductance decreases as the DC current increases. It seems reasonable
to have less effective inductance when the core is entering the
saturation region but not completely saturated yet. I suppose I could
test that idea with some DC current present, and try resonating the
inductor to see if the inductance drops off with DC current. I suppose
it will.
-Bill
Figbash
identify the region between continuous and discontinuous
http://www.ieee.li/pdf/introduction_to_power_electronics/chapter_05.pdf
Page 20 in particular gives two simple equations to determine the
region of operation, as well as assess current (now I know why I keep
burning parts out!).
Ultimately ended up going with an IRC2155 self-oscillating regulator
and a transformer trimmable feedback.
Still, a great diversion from digital electronics.
On Jan 24, 4:38 pm, Bill Bowden <bper...@bowdenshobbycircuits.info>
wrote: