Boost converter, inductor size?

On Jul 16, 9:38 am, Tim Wescott <t...@seemywebsite.com> wrote:
On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor.  Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot.  There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

Works fairly well at 84% efficiency.

I lowered the frequency to 12KHz to reduce the diode
switching losses. The current ramps from about 1 amp
minimum to 2.25 peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden
wrongaddress_at_att.net> wrote:

On Jul 16, 9:38 am, Tim Wescott<t...@seemywebsite.com> wrote:
On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor. Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot. There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1". .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is: r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries. That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.) Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51". From that,
I get about 107uH. Not 250uH. Let's call that 110uH and
talk about this later.

In any case, Tim's point about radiation may be important. An
air inductor has a very widely dispersed magnetic field (it's
not constrained much.) I don't know how to calculate the
loss to far field radiation here. But maybe someone can
discuss that a little bit.

Works fairly well at 84% efficiency.

What are you loading the output with? What is the output
voltage?

I lowered the frequency to 12KHz to reduce the diode
switching losses. The current ramps from about 1 amp
minimum to 2.25 peak.

Given the earlier discussion and what I think I learned from
Tim's comments, let's say your output voltage is 14V and your
diode circa 1A is running about 0.8V. From this, I gather
that the inductor must handle a percentage of the total power
by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%. (The
0.1V is your transistor switch drop and the 0.8V is the diode
drop estimate during conduction.)

With assumed usable input power near 9.5 watts, and 37.5% of
that in the inductor, about 3.56 Joules must be divided out
by 12000 pulses. I get close to 300uJ per pulse, here.
Assuming you actually had built 250uH into your air core,
this would be about 1.56A at the peak. However, using that
110uH I earlier estimated from that multilayer formula and
your other figures, I get about 2.3A at 110uH. Which is much
closer to your own comments about the observed peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe? It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

On Jul 21, 2:52 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden

snip

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6
layers. It resonates at 20KHz with a 0.22uF cap, so L=288uH.

snip

What are you loading the output with?  What is the output
voltage?

In normal operation, the load is the battery, which holds the output
voltage constant, regardless of input or output current.

The duty
cycle is fine tuned for a peak in output current, which seems the best
match.

But as the battery voltage rises, the duty cycle needs a small
adjustment. There is no feedback, so I just leave it set optimum at 13
volts. It's just a 555 timer driving a mosfet.

But using a 16 ohm load in place of the battery gives me 1.14A in at
9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe
83.5% efficiency.

But it was hard to get the same numbers twice since the sunlight kept
changing slightly. I'll run the test again using a power supply instead
of the panel.

snip
Bill:
I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe?  It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

I'm not sure of the current ramps, I was using a 0.1 ohm resistor in
series with the coil (for scope monitor) which upsets things a bit.

The peak to peak ramp appears to be about 1.25 amps offset by 1 amp,
or 2.25 peak to 1 minimum.

But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) =
14 volts.

I think it should be 9.5.

Probably inaccurate measurements, and
uncalibrated scope. Just ball park figures.

On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden



wrongaddr...@att.net> wrote:
On Jul 16, 9:38 am, Tim Wescott <t...@seemywebsite.com> wrote:
On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator.....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor.  Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot.  There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is:  r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries.  That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.)  Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51".  From that,
I get about 107uH.  Not 250uH.  Let's call that 110uH and
talk about this later.

In any case, Tim's point about radiation may be important. An
air inductor has a very widely dispersed magnetic field (it's
not constrained much.)  I don't know how to calculate the
loss to far field radiation here.  But maybe someone can
discuss that a little bit.

Works fairly well at 84% efficiency.

What are you loading the output with?  What is the output
voltage?

In normal operation, the load is the battery, which holds the output

I lowered the frequency to 12KHz to reduce the diode
switching losses.  The current ramps from about 1 amp
minimum to 2.25 peak.

Given the earlier discussion and what I think I learned from
Tim's comments, let's say your output voltage is 14V and your
diode circa 1A is running about 0.8V.  From this, I gather
that the inductor must handle a percentage of the total power
by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%.  (The
0.1V is your transistor switch drop and the 0.8V is the diode
drop estimate during conduction.)

With assumed usable input power near 9.5 watts, and 37.5% of
that in the inductor, about 3.56 Joules must be divided out
by 12000 pulses.  I get close to 300uJ per pulse, here.
Assuming you actually had built 250uH into your air core,
this would be about 1.56A at the peak.  However, using that
110uH I earlier estimated from that multilayer formula and
your other figures, I get about 2.3A at 110uH.  Which is much
closer to your own comments about the observed peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe?  It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

Try lowering the frequency a little?

On Wed, 21 Jul 2010 10:44:22 -0700 (PDT), Bill Bowden

wrongaddr...@att.net> wrote:
On Jul 21, 2:52 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden

snip

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6
layers.  It resonates at 20KHz with a 0.22uF cap, so L=288uH.

Okay.  I took things the other way.  (Thought that the 1" was
the length.)  I'll go with your measured values.

snip

What are you loading the output with?  What is the output
voltage?

In normal operation, the load is the battery, which holds the output
voltage constant, regardless of input or output current.

Yes, but what is that voltage it holds?

The duty
cycle is fine tuned for a peak in output current, which seems the best
match.

Okay.  Which implies a higher output voltage under load, I
think, as a higher voltage across your battery will mean more
current into it.  And all this means, as you already know,
that there is more power delivered, which is the goal.

But as the battery voltage rises, the duty cycle needs a small
adjustment. There is no feedback, so I just leave it set optimum at 13
volts. It's just a 555 timer driving a mosfet.

Okay.  Got it.

But using a 16 ohm load in place of the battery gives me 1.14A in at
9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe
83.5% efficiency.

Thanks.  I'm enjoying thinking about the problem and reading
about your observations.  It's teaching me to apply theory
and think better, generally.

Tim's knock in the head about the inductor's role in the work
involved helped a lot and now I'd like to see if paper and
pencil match experimental result, as it should!

But it was hard to get the same numbers twice since the sunlight kept
changing slightly. I'll run the test again using a power supply instead
of the panel.

Makes sense!!



snip
Bill:
I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe?  It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

I'm not sure of the current ramps, I was using a 0.1 ohm resistor in
series with the coil (for scope monitor) which upsets things a bit.

Given .2 ohms for the inductor itself, it's a 50% increase in
the DC resistance.  But you should loose an eighth watt or so
there, so I don't think it's all that terrible.  I might have
tried the same thing, since scoping makes observation of the
details clear.

The peak to peak ramp appears to be about 1.25 amps offset by 1 amp,
or 2.25 peak to 1 minimum.

Doesn't that bother you?  It just about cannot be.

But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) > >14 volts.

From this, I gather you are setting your 555 timer's pulse
width so that the mosfet is on for 25us, yes?

I think it should be 9.5.

I agree.

Assume you are right that L=288uH.  (I'm convinced you've
tested this.)  Assume you are right about 25us.  (I'm
convinced you can easily see this on the scope, by simply
looking.)  You and I both know the voltage should be 9.5V --
it's a given, considering the source you have.  So from that
the dI=V*dt/L, or .825A.  Since the others are solid givens,
this MUST be true.  It cannot be false.  So it is clear the
1.25A delta is off in some way.

Probably inaccurate measurements, and
uncalibrated scope. Just ball park figures.

Okay.

It still bothers me about the 1A minimum.  I know you can
easily establish where "zero" is at on your scope by simply
connecting the probe tip and ground and adjusting that line
to a graticule line for reference, so I don't think you can
mistake the fact that the pulse voltage bottom is above it.

This bothers me.  But I believe you are seeing a baseline
there, too.  Not that you have an exact figure for it.  But
the fact that it is present.

What is missing in my mind is WHY.  What would cause it?  Oh!
Obvious.  dt=(L/V)*dI, with V selected for the relaxation
time, which will be output+diode-9.5V, or so.  At 13V output,
this is maybe 13.7-9.65 or close to 4V.  This suggests a dt
of about 60us.  Added to 25us, this is 85us and exceeds what
is happening at 12kHz.  There just isn't enough time.

Try lowering the frequency a little?

Jon

snip
I got some better figures using a 9.5 volt PS and 16.3 ohm load.

Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts.
efficiency is around 90% which is closer to what I added up
considering 1/2 watt loss in the diode and another 1/2 watt in the
coil. The mosfet resistance is only 28 milliohms, so not much lost
there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33%

Assuming the inductor is 280uH, then di/dt is 9.5 / .00028 = 33929
amps per second, so the current should ramp up 1.36 amps in 40uS. But
the problem is on the down ramp, the voltage should be 12.7 + 0.8 for
the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should
be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But
this doesn't agree with the ramp up value of 1.36, so something is
missing.

How would you determine the minimum current point?

I think the idea is to drop the inductance from infinity where the
current is constant to a smaller value where the minimum current gets
close to zero.
So, I might take a few turns off the inductor and save some
resistance, and let the current get closer to zero, but I don't think
the efficiency will go much past 90% that I already have.

On 20 Jul 2010 21:34:07 (PDT), Bill Bowden <wrongaddress_at_att.net> wrote:

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1". .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is: r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries. That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.) Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51". From that,
I get about 107uH. Not 250uH. Let's call that 110uH and
talk about this later.

In article <l88d46h5bl21v35s21t3lvqh02r4n0j...@4ax.com>, Jon Kirwan wrote:
On 20 Jul 2010 21:34:07 (PDT), Bill Bowden <wrongaddr...@att.net> wrote:

SNIP what leads to this to edit for space



Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is:  r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries.  That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.)  Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51".  From that,
I get about 107uH.  Not 250uH.  Let's call that 110uH and
talk about this later.

  If Bill's spool is 1 inch long by 5/8 inch diameter and his
18 AWG wire is .065 inch in overall diameter, such as hookup
wire, then this inductor as described would have inductance
around 110 uH.

  I doubt it will have even .01 microfarad of interlayer
capacitance.

  Looks like I would want to hear more details about this coil.

  For example, if the spool was 1 inch in diameter and 5/8 inch long
and the wire was AWG 18 magnet wire more like .047 inch in diameter,
Bill could have wound 6 layers of 13 turns and a 7th layer of 12 turns.

  The radius difference c becomes .33  inch
  The average radius    a becomes .665 inch
  The length            b is      .625 inch

  The inductance in microhenries at this rate would be 222 microhenries..

 - Don Klipstein (d...@misty.com)

On Wed, 21 Jul 2010 17:03:24 -0700 (PDT), Bill Bowden wrote:
snip
I got some better figures using a 9.5 volt PS and 16.3 ohm load.

Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts.
efficiency is around 90% which is closer to what I added up
considering 1/2 watt loss in the diode and another 1/2 watt in the
coil. The mosfet resistance is only 28 milliohms, so not much lost
there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33%

Okay.  So about 120us period, 40us ON and 80us OFF.  Got it.

Assuming the inductor is 280uH, then di/dt is 9.5 / .00028  = 33929
amps per second, so the current should ramp up 1.36 amps in 40uS. But
the problem is on the down ramp, the voltage should be 12.7 + 0.8 for
the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should
be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But
this doesn't agree with the ramp up value of 1.36, so something is
missing.

This is exactly the point.  And I think you've got your
numbers right.

How would you determine the minimum current point?

That's a little trickier.  I had to think a moment to get a
clue.  I might still be wrong, but here is the argument.  If
I'm lucky, I will derive something quantitative below.

You are driving (forcing) the situation using a clock and
duty cycle.  The inductor doesn't control any of that.  It
is, instead, driven by it.  However, it does respond.

Imagine you set your duty cycle to some value that causes a
finite dI to be (t_on*V_on/L).  The V_on can't be controlled
-- it just is.  The t_on is set by your clock source.  It
also just is.  The same with L.  So dI is fixed.  It doesn't
matter what happens on the output.  The inductor has no
choice in the matter.  The dI will be some value.  What isn't
said here is what the baseline I happens to be.  Let's call
this baseline current through the inductor as I_0.  We hope
and expect I_0 = 0A.  But this isn't necessarily so.  What we
can compute is dI.  So the inductor current will have to rise
from I_0 to I_1 and it _will_ be the case that I_1=I_0+dI.
There is no choice on that matter.

In the steady state case, which we still aren't sure of, I_0
will be stable and so will I_1.  We hope I_0 is zero and I_1
is simply dI.  But hold that thought for a moment.

Now, your clock source also sets t_off.  In the steady state
case, we know what the dI must be for the t_off period, too.
It has to be the same.  If not, then it's not steady state.
And it turns out that we should think about that V_off value,
which will be L*dI/t_off.  Since dI=(t_on*V_on/L), this works
out to V_off=[V_on*(t_on/t_off)] -- the voltage that _must_
be across L when the switch is off if a steady state exists.

Now, this V_off might not actually be what you imagine.  In
other words, you might think it is simply the output voltage
plus a diode drop and less your input voltage because that is
what is needed.  But it isn't what happens.

Suppose t_off is shortened a bit.  Look at the equation for
V_off.  Note that t_off is in the denominator?  If t_off gets
smaller, V_off gets larger!!  This means the output voltage
also __increases__.  It has to (assume for a moment that the
diode voltage drop remains close to the same value as
before.)

But how can that be?  Doesn't that mean that the load
(resistive, for now) will use more power?  Yes, it does.  You
might think that is a good thing.  You've upped the output
voltage purely by reducing t_off!!  You could reduce it even
more and get a still higher output voltage.  But more output
power means more input power.  And you still have a fixed
number of clocks per second.  So more power must mean more
energy per pulse!

How do you get more energy per pulse??  You've already
determined that there is a fixed dI, completely determined by
L, t_on, and your input source voltage.  So that can't
change.  What can happen to increase the energy per pulse?

Well, take note that a fixed change in the magnitude of the
inductor current is NOT the same as representing a fixed
change in the magnitude of the energy stored in that magnetic
field.

You have these two circumstances:

   E_0 = (1/2) * L * I_0^2
   E_1 = (1/2) * L * I_1^2

The difference is then:

   dE = E_1 - E_0 = (1/2) * L * (I_1^2 - I_0^2)

But I_1 = I_0+dI.  So,

   dE = (1/2) * L * ((I_0+dI)^2 - I_0^2)
      = (1/2) * L * (I_0^2 + 2*dI*I_0 + dI^2 - I_0^2)
      = (1/2) * L * (2*dI*I_0 + dI^2)
      = (1/2) * L * dI * (2*I_0 + dI)
      = (1/2) * L * dI^2 + [L * dI * I_0]

Note the last term enclosed in brackets?  When I_0 is non-
zero and the same sign as dI, there is an additional amount
of energy in each pulse.

Another way of noting this fact is assume dI=1A and then to
compare when I_0=0A and I_0=1A.  dE when I_0=0A is some
number we can call X.  dE when I_0=1A will be 4X-1X, or 3X.
In other words, more energy is taken up and released in each
pulse when I_0=1A, even though there is the same dI in both
cases.

The point I'm getting to is that the inductor _must_ yield a
larger V_off if you shorten t_off.  To do that, the output
voltage increases.  That causes more power to be required.
Which, because the frequency is constant, requires more
energy per pulse.  But to do that, I_0 must rise upwards so
that each pulse _can_ deliver more energy to meet the power
requirements.  So it rises from zero.  But this then causes a
fixed DC current in the inductor, which means more loss in
the resistive parts and the diode, as well.

Which is not desired.

The computation of exactly what I_0 must level out as depends
upon the output load's response to a change in the voltage.  

for example, let's do some computations using some of your
numbers.  R_out=16.3, V_Lon=9.5V, t_on=40us, t_off=80us,
f=8333Hz, L=288uH, etc.  We don't know V_out, but we can
compute it from V_Lon*t_on/t_off.  This is 4.75V, in short.
Assuming your diode voltage is about 0.7V, this means 4.05V
added to the 9.6V input.  Call it 13.6V.  However, there is
some resistance you mentioned which at a couple of amps
average (grossly assuming a lot) is maybe .5V?  So call it
about 13.1V on the output.  The PCT of power in L should be
about 4.75V/13.1V or, say, about 36%.  And the power is
13.1V^2/16.3Ohms, times that 36%, or in other words about
3.79W is delivered by the inductor itself.  With f=8333, this
means almost 455uJ per pulse.

We know from t_on=40us that the dI=1.32A.  Energy from that
part, using (1/2)*L*dI^2, is about 251uJ.  Since we need a
total of about 455uJ, this leaves about 204uJ that needs to
be made up in the L*dI*I_0 term.  This suggests I_0=.534A.

Which may or may not be about where you are.  But that seems
to be where theory takes me, right now.

I think the idea is to drop the inductance from infinity where the
current is constant to a smaller value where the minimum current gets
close to zero.
So, I might take a few turns off the inductor and save some
resistance, and let the current get closer to zero, but I don't think
the efficiency will go much past 90% that I already have.

I think you are losing some efficiency you could get back,
simply because you have a DC current flowing all the time,
increasing the mean power dissipation in resistances.  Not to
mention that this would be a problem if you weren't using an
air core inductor.

I hope the above isn't too terrible to follow.  It took me
some moments of thought to realize _why_ the minimum inductor
current would rise above zero on its own.  But once I
realized that dI is determined by t_on and that the inductor
off-voltage is determined by dI and t_off and that shortening
t_off only means that the off-voltage goes UP, the rest
seemed to fall into place for me and make some sense.

Jon