Joule Thief - still not working....

On Jul 25, 11:35=A0am, greg <g...@cosc.canterbury.ac.nz> wrote:
>
> What does the voltage waveform between the collector
> and emitter look like? That's the important thing with
> regard to transistor dissipation.
>

Looks like this: http://www.artlum.com/jt/b_to_e.gif

It's running at about 166kHz.

On Jul 25, 2:55=A0pm, fungus <openglMYSO...@artlum.com> wrote:
> On Jul 25, 11:35=A0am, greg <g...@cosc.canterbury.ac.nz> wrote:
>
>
>
> > What does the voltage waveform between the collector
> > and emitter look like? That's the important thing with
> > regard to transistor dissipation.
>
> Looks like this:http://www.artlum.com/jt/b_to_e.gif
>
> It's running at about 166kHz.

More info...

The total power being drawn is about 400mW (105mA at 3.8V)
and there's about 220mW going through the LEDs (13mA at 16.6V)
so the missing power, ie. 180mW must be going through the
transistor (it's the only other path to ground).

The wave in that pic has about a 20% duty cycle. Does that mean
that about 45mW is being converted to heat by the transistor?

On Jul 25, 1:38=A0pm, Jon Kirwan <j...@infinitefactors.org> wrote:
>
> I'd probably start with something like 3.3k and work up or down from
> there. =A0Split the difference. =A0Then tweak Rbase until Ic_peak is wher=
e
> I need it to be (or where the LED current is about right.)
>

I've always had doubts about this theory because I tried
a variable resistor and it didn't do much.

But ... I just tried it with my new iron powder bead and it
works! I can dial the LED current up and down beautifully.

On my circuit I got 20mA at around 400 Ohms, lower than
that and it went towards 30mA but then I chickened out
because I was scared about my transistor.

I think we can definitely assume that the ferrite was
saturating or something and that we haven't found the
limits of the iron powder yet. I need to try the thin wire
and put a lot of turns on one until I find where it hits
the maximum.

Jon Kirwan wrote:
> On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
> <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
> 

<snip>

>> At any
>>rate, the value of R influences both the ON time and the OFF time.

<snip>

> The OFF time is NOT determined by R1, so far as I can tell. 

John Larkin is right.

Go to the bench, put a variable R in the base circuit
and observe for yourself.

By the way, did you ever build the air core version?
It eliminates core saturation from consideration when
trying to understand how the circuit functions.

Ed

On Sat, 25 Jul 2009 07:48:45 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

>I think we can definitely assume that the ferrite was
>saturating or something

Probably "or something."  

Iron powder usually has way more permeability than ferrite - ferrites
claim to fame is it is more permeable than air and able to handle high
frequencies with lower loss.  Some iron powder alloys can go to 100's
of megahertz, but iron is usually used in the kilohertz range.

In any event, when you select a core you are selecting for the
characteristics you need - and it is usually more accurate to say
"magnetic material, or magnetic material mix" than iron or ferrite.
-- 

fungus wrote:

<snip>

> 
> This also assuming we can solve the transistor heating problem.
> What would happen if I put two transistors in parallel? Would the
> load be halved or would differences in manufacturing tolerance
> mean one of them took most of the load?

One could take the whole load.  You can overcome that problem by
putting a small resistance in the emitter circuit of each
transistor.
                   Vcc
                    |
                 [Rload]
                    |
              +-----+-----+
              |           |
               \c       c/
                |---+---|
               /e   |   e\
              |    Ib     |
            [R1]        [R2]
              |           |
   Gnd -------+-----------+

When one transistor tries to conduct more current than
the other, the current through its emitter resistor
causes the voltage at its emitter to rise. That reduces
Vbe, which causes the transistor to conduct less current.

If you use 1 ohm resistors for R1 and R2, it will give
you a convenient measurement point. The voltage across
the resistor will equal the current through the transistor.

Ed
the 1 ohm resistor will read in miliamps



> 
> 
> PS: FWIW I measured the efficiency of this new circuit and it
> was 61% - a bit better that the 55% I get with a ferrite circuit
> with has similar output
> 
> 
> [*] PC PSUs are a real goldmine of parts if you really get in
>     there....
> 

greg wrote:
> whit3rd wrote:
> 
>> That kind of oscillator (blocking oscillator) depends on saturation
>> of the core
> 
> 
> Actually, no. That's what I thought at first, but
> Jon pointed out that the Joule Thief most likely
> works by a different mechanism.
> 
> The collector current rises until it reaches the
> maximum supportable by the base current, which
> depends on the induced voltage in the base winding
> and the base resistor. Then the collector voltage
> begins to rise, whereupon positive feedback via
> the base winding causes the transistor to turn
> off sharply.
> 
> So the on-time depends on the feedback ratio, the
> base resistor and the beta of the transistor. The
> latter is rather unpredictable, so you have to
> adjust the base resistor by experiment to get the
> result you want.
> 
> I've speculated that a version of the Joule Thief
> circuit could be designed to work by core saturation,
> and that the results would be more predictable.
> But I don't know of anyone who's actually built one
> that way yet.
> 

It works by transistor saturation, not core saturation.
A saturated inductor will take more current, as long as
the transistor is capable of pumping more current into
the inductor. As long as the current continues to flow
in the same direction, there is no polarity reversal
and no pulse created.

Ed

On Jul 25, 6:23=A0pm, default <defa...@defaulter.net> wrote:
> On Sat, 25 Jul 2009 07:48:45 -0700 (PDT), fungus
>
> <openglMYSO...@artlum.com> wrote:
> >I think we can definitely assume that the ferrite was
> >saturating or something
>
> Probably "or something." =A0
>

:-)

> Iron powder usually has way more permeability than ferrite - ferrites
> claim to fame is it is more permeable than air and able to handle high
> frequencies with lower loss. =A0Some iron powder alloys can go to 100's
> of megahertz, but iron is usually used in the kilohertz range.
>

I just made another one with my thin wire and an iron powder ring.
I was hoping hoping to get a lot of turns on it to see what happened
but unfortunately it behaved like the ferrite - current started
dropping
off.after about ten turns and varying R1 makes no difference to the
output of the circuit.

> In any event, when you select a core you are selecting for the
> characteristics you need - and it is usually more accurate to say
> "magnetic material, or magnetic material mix" than iron or ferrite.
>

Trouble is, I don't know the specs of any of these because I pulled
them out of a PSU.

On Thu, 23 Jul 2009 09:24:55 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

>That's a horrible circuit. Too many conflicting parameters depend on
>the value of R1. A proper blocking oscillator uses an RC time constant
>to set the rep rate, and a separate resistor to limit the base
>current.
>
>ftp://jjlarkin.lmi.net/BlockOsc.JPG

I still can't make sense of that one.  It bugs me a lot because I just
can't see why it would work well.

This one does make more sense to me, though:

>  ,-------------------+----------,
>  |                   |          |
>  |                   |          |
>  |                   |          |
>  |                   \          )|
>  |                   / R1       )| L2
>  |                   \          )|
>  |                   /          )|o
>  |                   |          |
>  |       ,----+------+          |   D8   ,-------,     ,--,
> ---      |    |      |          +--|>|---+       |     |  |
>  -  V1   |    |      |          |        |      ---    | ---
> ---      |    |      )|o        |        |      \ / D6 | \ / D3
>  -       |    |      )|         |        |      ---    | ---
>  |       |    |      )| L1      |        |       |     |  |
>  |       |    |      )|         |        |       |     |  |
>  |       |    |      |          |       --- C1  ---    | ---
>  |       |    |      |        |/c Q1    ---     \ / D5 | \ / D2
>  |       |    |      +--------|          |      ---    | ---
>  |       \    |      |        |>e        |       |     |  |
>  |    R2 /   --- C2  |          |        |       |     |  |
>  |       \   ---    _|_ D7      |        |      ---    | ---
>  |       /    |     /_\         |        |      \ / D4 | \ / D1
>  |       |    |      |          |        |      ---    | ---
>  |       |    |      |          |        |       |     |  |
>  |       |    |      |          |        |       '-----'  |
> gnd     gnd  gnd    gnd        gnd      gnd              gnd

Is that what you were thinking of, instead?

Jon

fungus wrote:
> On Jul 25, 3:14 am, David Eather <eat...@tpg.com.au> wrote:
>>> Aside: Does wire thickness make any difference?
>> It won't make much difference at these low power levels. The main
>> difference in this case is how many turns you can wrap on the ferrite bead.
>>
> 
> So the size of the magnetic field is purely down to current?

Not only current, also the number of turns and the type of former the 
turns are on etc

But all else being equal more current equals a stronger magnetic field

> 
> ==================
> 
> I've just another experiment and I've got some more fuel for
> the raging debate....

Oh, no....

> 
> I just did a comparison between a ferrite bead and an iron
> powder bead. With the ferrite bead the current through the
> LEDs drops off as the number of turns of wire increases.
> With seven turns I get 12mA ... with 30 turns I only get 1mA.
> 
> For comparison I just tried one of my iron powder rings
> and I got completely the *opposite* effect - more turns
> gave more output current. At 15 turns I was getting 6mA,
> at 30 turns I was getting 12mA.
> 
> I was doing this with a thicker piece of wire I pulled from a
> transformer in the PSU* so luckily for the transistor I couldn't
> physically get more than 30 turns on the ring. The trend was
> very clear though - every turn I added produced a measurable
> increase in LED current.
> 
> nb. The transistor was getting hotter with every extra turn
> despite the oscillation frequency going down, also the opposite
> of what happens with ferrite. It seems that transistor temperature
> is more strongly related to output current than frequency.
> 
> Assuming I get the same result with thin wire this seems like
> a really easy way to get any desired output current - just keep
> adding turns until you get there.
> 
> This also assuming we can solve the transistor heating problem.
> What would happen if I put two transistors in parallel? Would the
> load be halved or would differences in manufacturing tolerance
> mean one of them took most of the load?

Exactly correct! manufacturing tolerance would mean one transistor would 
take most of the load (and as the transistor gets hotter it will take an 
even greater percentage)

You would need to use a couple of small resistors to  make sure the 
transistors shared the current from L2 and the current into the base 
equally.

> 
> 
> PS: FWIW I measured the efficiency of this new circuit and it
> was 61% - a bit better that the 55% I get with a ferrite circuit
> with has similar output
> 
> 
> [*] PC PSUs are a real goldmine of parts if you really get in
>     there....
>