Driving LEDs with a battery pack

[fungus]

On Jul 12, 12:14 am, Jon Kirwan <j...@infinitefactors.org> wrote:

On Sat, 11 Jul 2009 14:59:50 -0700, I wrote:
snip of Q of powdered iron vs ferrite
So you really do raise a good point for me.  I feel unable to give any
useful thoughts on this point.  I'm still learning, too.

Here's a link:http://powerelectronics.com/mag/0803flyback-transformer-choke-design.pdf

It seems to pretty much say that powdered iron is fine, as I read it.

I read it and it meant nothing... :-S

But ... there's a line on the second page which seems to say
it doesn't make much difference between 25kHz and 75kHz.
We're aiming at 50kHz so I guess we're OK.

I've got three of them so I hope so....

 

[fungus]

On Jul 11, 10:14 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

If I measure the current going through the LEDs it says
102mA

It's probably right.  (Unless you are averaging.)  You should be
seeing high peak currents in there.  But only for a short part of a
total duty cycle.  Are you averaging?

Ummm, I'm using a digital multimeter.

Does it mean the voltage is staying rock steady but the current
is going crazy?

 

[ehsjr]

Jon Kirwan wrote:

If you read some of my longer postings here (it seems you have not),
you will see a VERY DETAILED description. Not to mention the
equations I've provided. So I have a pretty thorough understanding
I'm applying here.

Jon

I saw the posts. I posted to back you up on your mention
in the thread that air core might work, proving, at least
for that one, that core saturation was not the issue.

I also posted to point out what your equations show, but that
may not be apparent to all, that Vbatt is key. Your detailed
posts are too long for some newbies to get the point.

As to the # of turns you asked about, I don't recall for sure,
but I don't think it was many. I had some pictures of the thing.
I'll see if I can find them, and post if I do.

Ed

 

[fungus]

On Jul 12, 2:34 am, Jon Kirwan <j...@infinitefactors.org> wrote:

So I guess I don't know what circuit you were using.

The one with the capacitor... (of course!)

If you were using the diode/capacitor thing I added, then 100mA is
probably looking pretty bright!!

A little bit dimmer than one of the LEDs at 20mA.

 Since you have a scope, place a small 1 ohm resistor
in series with the six LEDs and scope out the voltage across it.
Current will be equal to the voltage you see and you will get to see
it in real-time, that way!

I don't have 1 ohm, the smallest I have is 4...

With that resistor in there I see a wave which goes flat for a
little while (about 20% of the cycle), ramps up to 1V then
drops back down to zero. 1V is 250mA peak.

Now, if you take the average of the ramp (ie. 125mA) and
take away 20% for the flat part you get 100mA. Coincidence?
I think not...! I guess the meter is averaging the value.

 

[Jon Kirwan]

On Sun, 12 Jul 2009 03:55:02 GMT, ehsjr <ehsjr@NOSPAMverizon.net>
wrote:

Jon Kirwan wrote:

If you read some of my longer postings here (it seems you have not),
you will see a VERY DETAILED description. Not to mention the
equations I've provided. So I have a pretty thorough understanding
I'm applying here.

I saw the posts. I posted to back you up on your mention
in the thread that air core might work, proving, at least
for that one, that core saturation was not the issue.

Ah. My mistake.

I also posted to point out what your equations show, but that
may not be apparent to all, that Vbatt is key. Your detailed
posts are too long for some newbies to get the point.

Well, it's good to get it documented, I suppose.

As to the # of turns you asked about, I don't recall for sure,
but I don't think it was many. I had some pictures of the thing.
I'll see if I can find them, and post if I do.

The coupling of the two windings is my worry, plus how many windings
may be needed to get the frequency in the right range.

Thanks,
Jon

 

[Jon Kirwan]

On Sat, 11 Jul 2009 20:27:39 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 12, 2:34 am, Jon Kirwan <j...@infinitefactors.org> wrote:

So I guess I don't know what circuit you were using.

The one with the capacitor... (of course!)

If you were using the diode/capacitor thing I added, then 100mA is
probably looking pretty bright!!

A little bit dimmer than one of the LEDs at 20mA.

That doesn't sound good.

 Since you have a scope, place a small 1 ohm resistor
in series with the six LEDs and scope out the voltage across it.
Current will be equal to the voltage you see and you will get to see
it in real-time, that way!

I don't have 1 ohm, the smallest I have is 4...

Okay.

With that resistor in there I see a wave which goes flat for a
little while (about 20% of the cycle), ramps up to 1V then
drops back down to zero. 1V is 250mA peak.

Okay. At what frequency (rate?) How much time between these things?
And what size is the capacitor??

Now, if you take the average of the ramp (ie. 125mA) and
take away 20% for the flat part you get 100mA. Coincidence?
I think not...! I guess the meter is averaging the value.

I guess so.

Jon

 

[ehsjr]

Jon Kirwan wrote:

On Sat, 11 Jul 2009 20:50:17 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:


snip
BTW - I made one using an air core (we can eliminate core
saturation from that one)
snip


Do you remember how many windings, roughly, you used then?

Jon

Found the photo I mentioned in my earlier reply.
It is not the joule thief circuit, but it uses an
air wound coil and demonstrates that core saturation
is not required for the boost circuit to work.

19t
+---+-------[L1]--------+-------+
| | | |
| | | [100R]
|+ | + /c |
[Batt] +---[100uF]---+---| BC337 |
| | | \e |
| +---[4.7K]----+ | [LED]
| | |
+-----------------------+-------+

The red LED is lit at Vbatt 1.232 volts in the photo.
The coil id is ~1/4" and it is 19 turns, about 2 1/2" long.

Ed

 

[greg]

Jon Kirwan wrote:

The base winding also goes to
zero volts and the base current takes a hit. But it goes to about 1/2
the earlier value.

I don't follow. If the voltage across the base winding is
zero, the only base current left is that from the DC bias
across the resistor, and presumably that is a lot less than
what's required to sustain the peak collector current. So the
transistor will rapidly turn off by the same mechanism as
in the non-saturating case.

In both modes there will be a period of time when the
transistor dissipates, but it's not clear to me which
one will be better.

/Ipeak
^^^^

This is a problem, because beta varies greatly from one
specimen of transistor to another. The data sheet will
give you a "typical" value, but it needs to be taken with
several large grains of salt.

I suppose it's okay for a hobby project where you can
play around with it until you get it right. But even in
my hobby projects, I like to be able to calculate things
so they're at least roughly right!

Another thought: For driving several LEDs in series,
maybe it would help to use a separate secondary winding
with more turns for the output. That would give a
current step-down relative to the transistor current
and allow the output duty cycle to be increased for
the same output current.

But with the freewheeling diode and a capacitor, there is no need. And
that arrangement provides a nice smooth current.

Only because you've got a *resistor* in there, which
wastes a bit of power. One of the nice things about the
plain Joule Thief is that it doesn't waste any power
that way.

Pulsing with duty cycles not unlike 1:10 means
10-fold current increases and that is way past their efficiency curve
peak.

That's the idea behind the current step-down -- it lets
you increase the duty cycle to something much closer
to 100% while driving the LEDs at around their optimum
current, whatever that is.

On the surface, it seems easier and better to just add a diode and cap
than to wind another winding in a tiny bead.

So use a bigger bead. Remember that the capacitor is
going to take up space, too -- probably a lot more than
an extra winding would.

--
Greg

 

[greg]

fungus wrote:

Now, if you take the average of the ramp (ie. 125mA) and
take away 20% for the flat part you get 100mA. Coincidence?
I think not...! I guess the meter is averaging the value.

It also means you're being rather hard on your LEDs!

As I said, the current you get is rather random...

--
Greg

 

[greg]

Jon Kirwan wrote:

The problem, if I got it right, is energy storage.
Ferrites have little tiny air gaps (okay, not air, but about the same
in effect) between granules with iron in them. The energy gets stored
there, in the gaps, for release when the BJT turns off. Straight iron
cores won't store anything much.

Do you have a reference for this? It's the first time I've
ever heard anything about energy being stored "in the gaps",
or that iron doesn't store energy as well as other core
materials.

I don't think it can be right. According to the calculation
I just did on the back of an envelope, the energy required
to charge an inductor up to a current I by applying a
steady voltage V is given by

E = 0.5*V*L*I^2

So assuming that an iron core gives you more inductance than
something else, such as ferrite, then E goes up. If the extra
energy isn't stored in the inductor, then where does it *go*?

I suspect the reason iron won't work so well in this application
is mainly due to frequency considerations. The magnetic domains
in solid iron can't flip around so quickly. It's like putting
a bunch of bar magnets together in a row. If you let them touch
each other, they cling together strongly. If you keep them
separated a bit, they don't attract each other so much and are
easier to separate.

--
Greg


--
Greg

 

[Jon Kirwan]

On Sun, 12 Jul 2009 20:06:51 +1200, greg <greg@cosc.canterbury.ac.nz>
wrote:

Jon Kirwan wrote:

The problem, if I got it right, is energy storage.
Ferrites have little tiny air gaps (okay, not air, but about the same
in effect) between granules with iron in them. The energy gets stored
there, in the gaps, for release when the BJT turns off. Straight iron
cores won't store anything much.

Do you have a reference for this?

Yes. Since I've only been studying this recently, I do remember.

I'm looking at the Unitrode Magnetics Design Handbook (I've a paper
copy here.) Aka, MAG 100A. Looking it up on the web, I see that it
is here:

http://focus.ti.com/docs/training/catalog/events/event.jhtml?sku=SEM401014

Take a look at chapter 2 (Magnetic Core Characteristics) there. By
the time you've read the first two pages, you'll see what I see.

You should quickly see "Energy storage in a transformer core is an
undesired parasitic element." And there will be several more items to
read, soon. I think this is what is termed "magnetizing energy."
Desirable in flybacks. Undesirable in transformers.

It's the first time I've
ever heard anything about energy being stored "in the gaps",
or that iron doesn't store energy as well as other core
materials.

Well, please read the text and see if you see what I see. I think it
makes sense to me. Quite a bit, actually. So I suppose that's why it
doesn't bother me to say it.

I don't think it can be right. According to the calculation
I just did on the back of an envelope, the energy required
to charge an inductor up to a current I by applying a
steady voltage V is given by

E = 0.5*V*L*I^2

It's always been 1/2*I^2*L, agreed.

I am just learning this stuff, so don't take anything I say for much
right now. But if what my mind says right now as making sense is
right, a counter potential is created so that little or no actual
current is able to develop if there is no place for magnetizing energy
to be stored. There is always some place for it, though. Tiny gaps
in the core, outside the iron core itself in the surrounding air, and
so on. In effect, if the core is gapless and near perfect in the
sense of not allowing magnetizing energy there, then what happens is
that what energy must be stored is stored in the surrounding air and
the whole thing acts more like an air core inductor without an iron
core, at all. Unless there is a secondary somewhere for the flux to
link up with and expend energy.

That's how it looks to me.

So assuming that an iron core gives you more inductance than
something else, such as ferrite, then E goes up. If the extra
energy isn't stored in the inductor, then where does it *go*?

Well, I hope my above explanation (an internal mind thing of mine, for
now) is about right. It is what I've extracted from the Unitrode book
I just recently read through. This is all very new to me, though. So
I'm still putting stuff in my head about it.

I suspect the reason iron won't work so well in this application
is mainly due to frequency considerations. The magnetic domains
in solid iron can't flip around so quickly. It's like putting
a bunch of bar magnets together in a row. If you let them touch
each other, they cling together strongly. If you keep them
separated a bit, they don't attract each other so much and are
easier to separate.

Well, that isn't why I think so (frequency limits.) I think it is
because there are far fewer places to tuck away energy. The secondary
(base winding) certainly counts as a place. So does the surrounding
air. But it really makes a mess of things, if I've got it right.

Jon

 

[Jon Kirwan]

On Sun, 12 Jul 2009 18:17:49 +1200, greg <greg@cosc.canterbury.ac.nz>
wrote:

Jon Kirwan wrote:

The base winding also goes to
zero volts and the base current takes a hit. But it goes to about 1/2
the earlier value.

I don't follow. If the voltage across the base winding is
zero, the only base current left is that from the DC bias
across the resistor, and presumably that is a lot less than
what's required to sustain the peak collector current. So the
transistor will rapidly turn off by the same mechanism as
in the non-saturating case.

What happens is the primary (which is the collector winding, for
talking purposes) must go through zero volts on the way over to
driving the LED (which requires a reversal, so you are certain that
this happens if there is no other reason you are certain about it --
and there are other reasons to know.) When it is zero volts across
the primary, it is also zero volts across the secondary (the base
winding.) That's because there are no flux changes to induce one,
anymore, right at the point where the current hits a peak value and
before it starts declining when the LED is hit.

When the base winding (secondary) is at zero volts across, it doesn't
add anything anymore to the battery voltage. But it doesn't subtract
yet, either. So the battery alone is "seen." Since it is the case
that earlier while the BJT is on and approximately one battery voltage
is across the primary (collector winding), the flux is changing and
linked to the secondary, inducing approximately one battery voltage
there (in aiding fashion.) So a little earlier, there was the battery
itself PLUS the aiding base winding, yielding approximately two
battery voltages which drive through the resistor. When the situation
takes place that there is zero volts across the base winding, then
there is still one battery voltage left. That's the "1/2" I
mentioned.

In both modes there will be a period of time when the
transistor dissipates, but it's not clear to me which
one will be better.

Rbase = beta*[N*(Vbattery-Vceoff)+Vbattery-Vbeon]/Ipeak
^^^^

This is a problem, because beta varies greatly from one
specimen of transistor to another. The data sheet will
give you a "typical" value, but it needs to be taken with
several large grains of salt.

Yes. The way I look at it is that the beta only applies at some point
during the cross-over. You can either choose to degrade the beta or
else degrade the effective base current. I chose to do the latter in
my calculations and keep the beta, because I can look it up in the
datasheet.

Look, I can walk you through the exact details (it's iterative in the
sense that it takes two or three passes through the datasheet to nail
the exact values by hand.) If you want, and go grab up a copy of the
On-Semi datasheet for the PN2222, I'll let you pick a situation and
we'll walk through it in detail so that you see what I see. Then, if
you have LTSpice handy, we can plug in the exact circuit there and you
and I will each see that our hand calculations nail down exactly what
we see in LTSpice, which is far, far more sophisticated.

The equations work pretty well. Been there, done that.

I suppose it's okay for a hobby project where you can
play around with it until you get it right. But even in
my hobby projects, I like to be able to calculate things
so they're at least roughly right!

Me, too. This isn't a case where I white-washed. I both understand
the theory AND why the practical details work out. Well, except that
I'm still learning about core materials.

Another thought: For driving several LEDs in series,
maybe it would help to use a separate secondary winding
with more turns for the output. That would give a
current step-down relative to the transistor current
and allow the output duty cycle to be increased for
the same output current.

But with the freewheeling diode and a capacitor, there is no need. And
that arrangement provides a nice smooth current.

Only because you've got a *resistor* in there, which
wastes a bit of power. One of the nice things about the
plain Joule Thief is that it doesn't waste any power
that way.
snip

No, you can remove the resistor. As I said, I was just adding it for
kicks. It is NOT at all necessary. What happens is that the load
pulls some current while the transistor is ON and the freewheeling
diode is OFF. As it does, the capacitor droops a little, (I/C)*dt.
But then the BJT goes OFF, the freewheeling diode goes ON, and the
energy is dumped onto the cap. You can see that it is easy to select
any C value you want to match the dV you want to limit to. The
resistor doesn't really matter, at all. I just put it in for playing
around.

Jon

 

[Jon Kirwan]

On Sun, 12 Jul 2009 08:28:55 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

I am just learning this stuff, so don't take anything I say for much
right now. But if what my mind says right now as making sense is
right, a counter potential is created so that little or no actual
current is able to develop if there is no place for magnetizing energy
to be stored. There is always some place for it, though. Tiny gaps
in the core, outside the iron core itself in the surrounding air, and
so on. In effect, if the core is gapless and near perfect in the
sense of not allowing magnetizing energy there, then what happens is
that what energy must be stored is stored in the surrounding air and
the whole thing acts more like an air core inductor without an iron
core, at all. Unless there is a secondary somewhere for the flux to
link up with and expend energy.

That's how it looks to me.

Sorry, I should have added something else. The effective inductance
goes very close to zero if there is no place to put energy. This can
allow an I that you'd imagine as worthy of some energy storage when,
in fact, the L just went effectively close to nil.

The key here is that 1/2*I^2*L is always true. But there must be a
place to put the energy, too. If there is no place to put it, either
there is no I or else there is no L. One of the two, or both, stay
low enough to match up with the magnetizing energy being stored. L
isn't quite the fixed value you may imagine it is.

Jon

 

[Jasen Betts]

On 2009-07-12, Jon Kirwan <jonk@infinitefactors.org> wrote:

On Sun, 12 Jul 2009 18:17:49 +1200, greg <greg@cosc.canterbury.ac.nz
wrote:

Jon Kirwan wrote:

The base winding also goes to
zero volts and the base current takes a hit. But it goes to about 1/2
the earlier value.

I don't follow. If the voltage across the base winding is
zero, the only base current left is that from the DC bias
across the resistor, and presumably that is a lot less than
what's required to sustain the peak collector current. So the
transistor will rapidly turn off by the same mechanism as
in the non-saturating case.

What happens is the primary (which is the collector winding, for
talking purposes) must go through zero volts on the way over to
driving the LED (which requires a reversal, so you are certain that
this happens if there is no other reason you are certain about it --
and there are other reasons to know.)

where are you getting this from? it sure doesn't sounde like the joule
thief blocking oscillator layout.

collector winding is always +ve (much of the time by only V_CEsat),
base winding goes negative at transistor turn-off





























When it is zero volts across

the primary, it is also zero volts across the secondary (the base
winding.) That's because there are no flux changes to induce one,
anymore, right at the point where the current hits a peak value and
before it starts declining when the LED is hit.

When the base winding (secondary) is at zero volts across, it doesn't
add anything anymore to the battery voltage. But it doesn't subtract
yet, either. So the battery alone is "seen." Since it is the case
that earlier while the BJT is on and approximately one battery voltage
is across the primary (collector winding), the flux is changing and
linked to the secondary, inducing approximately one battery voltage
there (in aiding fashion.) So a little earlier, there was the battery
itself PLUS the aiding base winding, yielding approximately two
battery voltages which drive through the resistor. When the situation
takes place that there is zero volts across the base winding, then
there is still one battery voltage left. That's the "1/2" I
mentioned.

In both modes there will be a period of time when the
transistor dissipates, but it's not clear to me which
one will be better.

Rbase = beta*[N*(Vbattery-Vceoff)+Vbattery-Vbeon]/Ipeak
^^^^

This is a problem, because beta varies greatly from one
specimen of transistor to another. The data sheet will
give you a "typical" value, but it needs to be taken with
several large grains of salt.

Yes. The way I look at it is that the beta only applies at some point
during the cross-over. You can either choose to degrade the beta or
else degrade the effective base current. I chose to do the latter in
my calculations and keep the beta, because I can look it up in the
datasheet.

Look, I can walk you through the exact details (it's iterative in the
sense that it takes two or three passes through the datasheet to nail
the exact values by hand.) If you want, and go grab up a copy of the
On-Semi datasheet for the PN2222, I'll let you pick a situation and
we'll walk through it in detail so that you see what I see. Then, if
you have LTSpice handy, we can plug in the exact circuit there and you
and I will each see that our hand calculations nail down exactly what
we see in LTSpice, which is far, far more sophisticated.

The equations work pretty well. Been there, done that.

I suppose it's okay for a hobby project where you can
play around with it until you get it right. But even in
my hobby projects, I like to be able to calculate things
so they're at least roughly right!

Me, too. This isn't a case where I white-washed. I both understand
the theory AND why the practical details work out. Well, except that
I'm still learning about core materials.

Another thought: For driving several LEDs in series,
maybe it would help to use a separate secondary winding
with more turns for the output. That would give a
current step-down relative to the transistor current
and allow the output duty cycle to be increased for
the same output current.

But with the freewheeling diode and a capacitor, there is no need. And
that arrangement provides a nice smooth current.

Only because you've got a *resistor* in there, which
wastes a bit of power. One of the nice things about the
plain Joule Thief is that it doesn't waste any power
that way.
snip

No, you can remove the resistor. As I said, I was just adding it for
kicks. It is NOT at all necessary. What happens is that the load
pulls some current while the transistor is ON and the freewheeling
diode is OFF. As it does, the capacitor droops a little, (I/C)*dt.
But then the BJT goes OFF, the freewheeling diode goes ON, and the
energy is dumped onto the cap. You can see that it is easy to select
any C value you want to match the dV you want to limit to. The
resistor doesn't really matter, at all. I just put it in for playing
around.

Jon

 

[fungus]

On Jul 12, 7:49 am, Jon Kirwan <j...@infinitefactors.org> wrote:

On Sat, 11 Jul 2009 20:27:39 -0700 (PDT), fungus

openglMYSO...@artlum.com> wrote:
On Jul 12, 2:34 am, Jon Kirwan <j...@infinitefactors.org> wrote:

So I guess I don't know what circuit you were using.

The one with the capacitor... (of course!)

If you were using the diode/capacitor thing I added, then 100mA is
probably looking pretty bright!!

A little bit dimmer than one of the LEDs at 20mA.

That doesn't sound good.

 Since you have a scope, place a small 1 ohm resistor
in series with the six LEDs and scope out the voltage across it.
Current will be equal to the voltage you see and you will get to see
it in real-time, that way!

I don't have 1 ohm, the smallest I have is 4...

Okay.

With that resistor in there I see a wave which goes flat for a
little while (about 20% of the cycle), ramps up to 1V then
drops back down to zero. 1V is 250mA peak.

Okay.  At what frequency (rate?)  How much time between these things?
And what size is the capacitor??

This is the wave without capacitor:

http://www.artlum.com/jt/without.jpg

This is the with capacitor:

http://www.artlum.com/jt/with.jpg

The frequency is 60kHz and it's about 1V amplitude
with a 4 ohm resistor (would be four volts with a 1 ohm).
The capacitor is 470uF (the smallest one I had with
decent legs on it)..

Weirdly it seems to have changed since yesterday.
Yesterday the "down" part of the wave with capacitor
was completely horizontal at zero volts.

What's changed? Well, I tried removing LEDs to
see what would happen (100mA should be ok for
five, and a little bit over for four). One of them died
and another one didn't look too healthy afterwards
so I changed it.

If the wave was different then maybe they were
already damaged. 100mA average certainly seems
to indicate we're really abusing the LEDs..

 

[fungus]

On Jul 12, 2:11 pm, fungus <openglMYSO...@artlum.com> wrote:

This is with capacitor:

http://www.artlum.com/jt/with.jpg

This makes no sense at all to me. Surely the
capacitor should be smoothing that out and
taking up the slack when there's no current
coming through from the rest of the circuit.

The voltage across the LED chain is a series
of narrow spikes without the capacitor. With
the capacitor it's a perfectly flat DC voltage
as you would expect. How can the current
be varying from zero to 250mA?

 

[ehsjr]

Jon Kirwan wrote:

On Sun, 12 Jul 2009 03:55:02 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:


Jon Kirwan wrote:

If you read some of my longer postings here (it seems you have not),
you will see a VERY DETAILED description. Not to mention the
equations I've provided. So I have a pretty thorough understanding
I'm applying here.

I saw the posts. I posted to back you up on your mention
in the thread that air core might work, proving, at least
for that one, that core saturation was not the issue.


Ah. My mistake.

I don't think so. My post did not make the point clear,
so the mistake is mine, not yours.

I also posted to point out what your equations show, but that
may not be apparent to all, that Vbatt is key. Your detailed
posts are too long for some newbies to get the point.


Well, it's good to get it documented, I suppose.

Yes! I like it, and I assume others do, too. But
I've talked with some newbies who glaze over when there is
too much detail for them. I think that is why the series
of books "xyz for Dummies" is successful. They allow
a newcomer to the subject to "gear up" slowly. Those books
are not for everyone, of course.

As to the # of turns you asked about, I don't recall for sure,
but I don't think it was many. I had some pictures of the thing.
I'll see if I can find them, and post if I do.


The coupling of the two windings is my worry, plus how many windings
may be needed to get the frequency in the right range.

Somebody else posted about winding an aircore with bifilar to
get the coupling. Way back when I was messing with these things,
I wound some ferrites with bifilar, noted the results, then
re-wound with single to compare. Of course, all I got was 1:1 with
the bifilar. And in any event, I don't have the notes and no
recollection of the results. :-( And I didn't do a bifilar aircore.
I did do an aircore transformer - 2 complete coils where the id of
one was just a little greater than the od of the other - but again
I don't recall the results. I know I was experimenting with varying
the coupling by moving the smaller coil farther into or out of
the larger coil's inside diameter. Again, whatever notes I had
are gone.

Ed

Thanks,
Jon

 

[ehsjr]

fungus wrote:

On Jul 12, 2:11 pm, fungus <openglMYSO...@artlum.com> wrote:

This is with capacitor:

http://www.artlum.com/jt/with.jpg



This makes no sense at all to me. Surely the
capacitor should be smoothing that out and
taking up the slack when there's no current
coming through from the rest of the circuit.

The voltage across the LED chain is a series
of narrow spikes without the capacitor. With
the capacitor it's a perfectly flat DC voltage
as you would expect. How can the current
be varying from zero to 250mA?

Can you post a schematic of your setup, and indicate where you
are attaching the scope probe and ground clip?

On another point you mentioned earlier, using a DMM to measure
the output voltage or current may give you meaningless readings.
It simply may be incapable of accurate readings, given the
waveshape it is "seeing". Your use of the scope for making
measurements on this circuit is the right approach.

Ed

 

[Jon Kirwan]

On Sun, 12 Jul 2009 17:56:14 GMT, ehsjr <ehsjr@NOSPAMverizon.net>
wrote:

Jon Kirwan wrote:
On Sun, 12 Jul 2009 03:55:02 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:

If you read some of my longer postings here (it seems you have not),
you will see a VERY DETAILED description. Not to mention the
equations I've provided. So I have a pretty thorough understanding
I'm applying here.

I saw the posts. I posted to back you up on your mention
in the thread that air core might work, proving, at least
for that one, that core saturation was not the issue.

Ah. My mistake.

I don't think so. My post did not make the point clear,
so the mistake is mine, not yours.

Hehe. No, it's my mistake! (We can go back and forth with this.

I also posted to point out what your equations show, but that
may not be apparent to all, that Vbatt is key. Your detailed
posts are too long for some newbies to get the point.

Well, it's good to get it documented, I suppose.

Yes! I like it, and I assume others do, too. But
I've talked with some newbies who glaze over when there is
too much detail for them. I think that is why the series
of books "xyz for Dummies" is successful. They allow
a newcomer to the subject to "gear up" slowly. Those books
are not for everyone, of course.

Cripes, I'm one who also sometimes just glazes over with too much
detail. I started reading the Unitrode book in earnest about a month
ago or so when I first posted in sci.electronics.design about a
transformer I wanted to use for a rocket launcher idea. I knew almost
nothing about magnetics in electronics design at that point because
the few times I'd tried to glance over detailed material it seemed
'difficult.' I mean, it actually is kind-of, for exactly the reasons
that John Larkin recently mentioned that inductors are pretty inferior
by comparison with other electronic parts -- there is a complex
reality to grasp of many different and yet also important factors,
none of which can really be ignored to get it right. With capacitors,
for example, yes there is reality there too but the unwanted ones are
in pretty good control for the most part and a more simplified view
works well with them.

As to the # of turns you asked about, I don't recall for sure,
but I don't think it was many. I had some pictures of the thing.
I'll see if I can find them, and post if I do.

The coupling of the two windings is my worry, plus how many windings
may be needed to get the frequency in the right range.

Somebody else posted about winding an aircore with bifilar to
get the coupling. Way back when I was messing with these things,
I wound some ferrites with bifilar, noted the results, then
re-wound with single to compare. Of course, all I got was 1:1 with
the bifilar. And in any event, I don't have the notes and no
recollection of the results. :-( And I didn't do a bifilar aircore.
I did do an aircore transformer - 2 complete coils where the id of
one was just a little greater than the od of the other - but again
I don't recall the results. I know I was experimenting with varying
the coupling by moving the smaller coil farther into or out of
the larger coil's inside diameter. Again, whatever notes I had
are gone.

Thanks, Ed. I wondered how air-core versions might be done well. I
haven't any bifilar wire here, that I'm aware of, so I need to go get
some samples and look at them and try them out, now, along with
winding air cores like you suggest above, too. Just to see what the
results actually _can_ be.

I just wound myself four separate transformers, late last night. So
I'm motivated to learn more and try my hand so that I can learn to be
better, later, at this stuff. It's an important nook that I've
skipped over all these years.

Jon

 

[Jon Kirwan]

On 12 Jul 2009 10:26:45 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

snip
where are you getting this from? it sure doesn't sounde like the joule
thief blocking oscillator layout.

It's the 'mental theory' I used to develop accurate equations to
estimate behavior I observe, drawn from my understanding of
theoretical books I've also read.

However, you need to the fuller context (which is in my other posts)
to understand that part. I wasn't saying this is what occurs for any
length of time. It is only for an instanteous moment during a
transitition. I was just trying to explain the "1/2" I'd earlier
written. It's not how I see the operation, generally, though.

collector winding is always +ve (much of the time by only V_CEsat),
base winding goes negative at transistor turn-off
snip

But I think I agree with you, Jason. The collector winding, for most
of the time, remains with the entire battery voltage across it (less
the gradually and slowly rising Vce of the BJT.) And indeed the base
winding goes negative at BJT turn-off. I said that in earlier posts.
You need to read those, as well, to see the fuller context.

Jon