Joule Thief - still not working....

[Jon Kirwan]

On Mon, 27 Jul 2009 00:20:15 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 27, 3:10 am, Jon Kirwan <j...@infinitefactors.org> wrote:

Mouser, Arrow, Newark, and Digikey all come up, for example.

Try:
 http://www.findchips.com/
 http://octopart.com/

The ones I looked at were all "US only".

Oh. That's me being provincial. Sorry about that. So Newark doesn't
sell to you?

Jon

 

[Jon Kirwan]

On Mon, 27 Jul 2009 01:00:39 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 27, 9:30 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Sun, 26 Jul 2009 23:33:48 -0700 (PDT), fungus

openglMYSO...@artlum.com> wrote:
On Jul 27, 6:28 am, greg <g...@cosc.canterbury.ac.nz> wrote:

I'm confused. Does "b_to_e" mean base-to-emitter?
I want to see *collector*-to-emitter.

This is collector->emitter:

http://www.artlum.com/jt/coll_to_em.gif

Can you provide some voltages for the y-axis and times for the x-axis?

It's about 20V on y and about 5 microsecond per wave (ie. the
jt is running at 200mHz).

Egads. I take it you mean 200kHz. That could be some of why the BJT
heats up a bit. Can you wind more wire?

 I'm assuming that the bottom span there is the BJT ON time

(thinks... ) Yes.

    Vout = 1.3*Vin

That doesn't look right to me if I assume you have a stack of LEDs
there.

Can you see if that's about right?

No...Vin is 3.8V and Vout is about 20 so the ratio is 5.3

But that isn't consistent. The OFF time seems way, way too long --
unless this photograph is taken very very early during startup time
and that isn't possible because you said the top part was near 20V. It
shouldn't be that long with that kind of voltage difference once it
reaches steady state. It should be shorter than the BJT's ON time.
Not longer. Makes no sense. Are you on the upside-down part of the
Earth?

There's something wrong in the communications here. Do you have two
diodes? One to protect the base and the other leading from the
collector over to the capacitor? Is that latter diode wired in the
right direction? On second thought, are they both wired right?

Yikes. Maybe someone else has more clues than I do here.

Jon

 

[fungus]

On Jul 27, 6:28 am, greg <g...@cosc.canterbury.ac.nz> wrote:

I'm confused. Does "b_to_e" mean base-to-emitter?
I want to see *collector*-to-emitter.

This is collector->emitter:

http://www.artlum.com/jt/coll_to_em.gif

 

[fungus]

On Jul 27, 6:31 am, David Eather <eat...@tpg.com.au> wrote:

http://www.national.com/mpf/LM/LM3914.html

LM3914 would seem to work a treat - it needs about .9 of a volt more
than the LED uses (page 5 "led current regulation drop out")

I hadn't understood that part (in fact I'm still not sure I do)
but does it mean I need an extra volt from somewhere?

Will I have to use a joule thief to drive the LM3914? :-(

 

[fungus]

On Jul 27, 3:10 am, Jon Kirwan <j...@infinitefactors.org> wrote:

Mouser, Arrow, Newark, and Digikey all come up, for example.

Try:
 http://www.findchips.com/
 http://octopart.com/

The ones I looked at were all "US only".

 

[fungus]

On Jul 27, 6:39 am, greg <g...@cosc.canterbury.ac.nz> wrote:

fungus wrote:
The problem seems to me to be that the transistor receives the
exact same current as the LEDs. As LED current rises, so does
transistor current.

With the output capacitor, it actually carries *more*
current than the LEDs.

(Thinks...) Yes, because once the capacitor is fully charged
the LED side of the circuit can't suck up all the available
electrons and the excess ends up going through the transistor.

 

[fungus]

On Jul 27, 9:12 am, David Eather <eat...@tpg.com.au> wrote:

fungus wrote:

Will I have to use a joule thief to drive the LM3914?  :-(

No. You will need a supply greater than the LED voltage + .9v for near
perfect current regulation. When the .9 volt is not available the
current regulation becomes a bit sloppy and the current drops a bit. If
you wanted to suck the batteries *really* dry then you might need a way
to get a bit more voltage.

Ok .... I think I'm going to order a couple of them off eBay to play
around with.

As an aside have you thought about using 6 volts (should be
easier/longer lasting to use with white LED) or a small 9v battery
(would last a bit more than an hour)?

Yes, I was just kidding about the joule thief. An extra battery
would be a lot more sensible/less work...

 

[fungus]

On Jul 27, 9:30 am, Jon Kirwan <j...@infinitefactors.org> wrote:

On Sun, 26 Jul 2009 23:33:48 -0700 (PDT), fungus

openglMYSO...@artlum.com> wrote:
On Jul 27, 6:28 am, greg <g...@cosc.canterbury.ac.nz> wrote:

I'm confused. Does "b_to_e" mean base-to-emitter?
I want to see *collector*-to-emitter.

This is collector->emitter:

http://www.artlum.com/jt/coll_to_em.gif

Can you provide some voltages for the y-axis and times for the x-axis?

It's about 20V on y and about 5 microsecond per wave (ie. the
jt is running at 200mHz).

 I'm assuming that the bottom span there is the BJT ON time

(thinks... ) Yes.

    Vout = 1.3*Vin

That doesn't look right to me if I assume you have a stack of LEDs
there.

Can you see if that's about right?

No...Vin is 3.8V and Vout is about 20 so the ratio is 5.3

 

[fungus]

On Jul 27, 10:18 am, Jon Kirwan <j...@infinitefactors.org> wrote:

Oh.  That's me being provincial.  Sorry about that.  So Newark doesn't
sell to you?

Nope.

If you click on their "Freight Estimator" it says 'US only'.

 

[fungus]

On Jul 27, 10:30 am, Jon Kirwan <j...@infinitefactors.org> wrote:

It's about 20V on y and about 5 microsecond per wave (ie. the
jt is running at 200mHz).

Egads.  I take it you mean 200kHz.

Ooops! Yes, you're right.

 That could be some of why the BJT
heats up a bit.  Can you wind more wire?

Can't fit any more turns on that one...

I used that one so you could see "worst case" but the transistor
doesn't really heat up any more than the 50kHz ones do (for the
same output current).

No...Vin is 3.8V and Vout is about 20 so the ratio is 5.3

But that isn't consistent.  ... Are you on the upside-down part of the
Earth?  

No wait, you're right it's upside down. Sorry.

 

[greg]

fungus wrote:

This is collector->emitter:

http://www.artlum.com/jt/coll_to_em.gif

Where is 0 volts on that picture?

--
Greg

 

[fungus]

On Jul 27, 1:32 pm, greg <g...@cosc.canterbury.ac.nz> wrote:

fungus wrote:
This is collector->emitter:

http://www.artlum.com/jt/coll_to_em.gif

Where is 0 volts on that picture?

At the top...

 

[fungus]

On Jul 27, 1:32 pm, greg <g...@cosc.canterbury.ac.nz> wrote:

fungus wrote:
This is collector->emitter:

http://www.artlum.com/jt/coll_to_em.gif

Where is 0 volts on that picture?

If you reload the image I drew a line where ground is....

 

[Jon Kirwan]

okay. Looking more at your scope display... With a new understanding
that it is inverted....

It looks 'about' right. There is a negative excursion that I'm
guessing is due to the collector winding (the slope is the same, but
inverted in sense in the display but still moving towards zero volts)
still having more energy to dump.

I think what happens here is that the base winding's voltage droops as
the collector winding's voltage also droops and at some point the
battery-opposing voltage in the base winding isn't enough to counter
the battery anymore and the BJT begins to turn on.

However, in my imagination here, the collector winding isn't done and
has significant energy left in it. The BJT turns ON prematurely,
supposedly bringing it's collector voltage down towards zero. But not
quite yet. There is that reverse transit time in action and,
unfortunately, the BJT can't yet cause the collector-emitter to be low
impedance. So the inductor current needs somewhere to either rise or
fall from its current level of 'I' at this time and to do that, it
starts driving current through the collector-base junction. To do
that, the voltage goes negative on the collector to forward bias that
diode junction.

Meanwhile, the BJT finishes getting through the reverse transit time
and is able to pull the collector into a low impedence with respect to
the emitter, the Vce jumps back up to 0V (about), and the current in
the collector winding rises again.

Maybe someone else can think about it and look at the curve and post
their own thoughts about all this.

Jon

 

[David Eather]

fungus wrote:

On Jul 27, 10:18 am, Jon Kirwan <j...@infinitefactors.org> wrote:
Oh. That's me being provincial. Sorry about that. So Newark doesn't
sell to you?


Nope.

If you click on their "Freight Estimator" it says 'US only'.

Try google for RS components, Farnell, Jaycar, Altronics and Radio Shack

 

[David Eather]

fungus wrote:

On Jul 27, 9:12 am, David Eather <eat...@tpg.com.au> wrote:
fungus wrote:
Will I have to use a joule thief to drive the LM3914? :-(
No. You will need a supply greater than the LED voltage + .9v for near
perfect current regulation. When the .9 volt is not available the
current regulation becomes a bit sloppy and the current drops a bit. If
you wanted to suck the batteries *really* dry then you might need a way
to get a bit more voltage.


Ok .... I think I'm going to order a couple of them off eBay to play
around with.

As an aside have you thought about using 6 volts (should be
easier/longer lasting to use with white LED) or a small 9v battery
(would last a bit more than an hour)?

Yes, I was just kidding about the joule thief. An extra battery
would be a lot more sensible/less work...

OK.

http://www.national.com/ds/LM/LM3914.pdf

on page 7 is the minimum circuit (you don't need all 10 LED's if you
don't want). R should be 680 ohms (18ma current) and the IC will run
very warm - but it should not be "burning" hot.

connect pin 7, pin 6, pin 5 and pin 9 to pin 3. Also connect a small
capacitor between pin 3 and 0v (any size bigger that about 2.2uF will do)

 

[fungus]

On Jul 28, 1:13 am, David Eather <eat...@tpg.com.au> wrote:

Try google for RS components, Farnell, Jaycar, Altronics and Radio Shack

Looks like RS and Farnell can send stuff here.

Shipping prices would kill me for small quantities but at least it's
an option....

 

[Jon Kirwan]

On Mon, 27 Jul 2009 22:00:47 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 27, 8:55 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

However, in my imagination here, the collector winding isn't done and
has significant energy left in it.  The BJT turns ON prematurely,
supposedly bringing it's collector voltage down towards zero.  But not
quite yet.  There is that reverse transit time in action and,
unfortunately, the BJT can't yet cause the collector-emitter to be low
impedance.  So the inductor current needs somewhere to either rise or
fall from its current level of 'I' at this time and to do that, it
starts driving current through the collector-base junction.  To do
that, the voltage goes negative on the collector to forward bias that
diode junction.

If I take out the protection diode that little hump turns
into a big sharp spike.

Hmm. I gotta find some time soon. In the interim, I may be liking
John's suggested thevenin arrangement, here. And/or a lower frequency
of operation. And/or better coupling (I'm worried about leakage
inductance, now.) Care to wind more windings? Tighter and better?
Etc?

Let's see:

: ,------+--------------,
: | | |
: | | o |
: | | |(
: | | L2 |(
: | | 120uH |(
: | \ |(
: | / R1 4:1 |
: | \ 1k winding | D1
: --- / ratio +----|>|---+-------, ,--,
: - V1 | | | | | |
: --- 4V | L1 o | | | | |
: - | ---- |/c Q1 | | | |
: | +-----^^^^---| 2N2222 | --- | ---
: | | 8uH |>e --- C1 \ / D6| \ / D3
: | | | --- 10u --- | --- LED
: | | | | | | |
: | \ | | | | |
: | / R2 | | | | |
: | \ 1.2k | | --- | ---
: | / | | \ / D5| \ / D2
: | | | | --- | --- LED
: | | | | | | |
: gnd gnd gnd gnd | | |
: | | |
: --- | ---
: \ / D4| \ / D1
: --- | --- LED
: | | |
: '----' |
: gnd

On this one (if you want to try it for now), wind L2 with 4 times as
many turns as l1. I'm not sure how many turns you already wound for
your 200kHz thing, but that wasn't enough! Put more on L2, less on
L1. See if you can balance it out at 4:1, or so. Note that there is
no protection diode. So it is kind of important that you keep to that
4:1 ratio. 2:1 and your BJT is at risk.

Jon

 

[fungus]

On Jul 27, 8:55 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

However, in my imagination here, the collector winding isn't done and
has significant energy left in it.  The BJT turns ON prematurely,
supposedly bringing it's collector voltage down towards zero.  But not
quite yet.  There is that reverse transit time in action and,
unfortunately, the BJT can't yet cause the collector-emitter to be low
impedance.  So the inductor current needs somewhere to either rise or
fall from its current level of 'I' at this time and to do that, it
starts driving current through the collector-base junction.  To do
that, the voltage goes negative on the collector to forward bias that
diode junction.

If I take out the protection diode that little hump turns
into a big sharp spike.

 

[fungus]

On Jul 28, 4:43 am, David Eather <eat...@tpg.com.au> wrote:

http://www.national.com/ds/LM/LM3914.pdf

on page 7 is the minimum circuit (you don't need all 10 LED's if you
don't want). R should be 680 ohms (18ma current) and the IC will run
very warm - but it should not be "burning" hot.

Ok.

If use less LEDs I'll connect them to alternate inputs to
toast the chip more evenly...

connect pin 7, pin 6, pin 5 and pin 9 to pin 3. Also connect a small
capacitor between pin 3 and 0v (any size bigger that about 2.2uF will do)

Is that because we're a bit low on volts and things are going
to get noisy when all those current regulators start switching?