Joule Thief - still not working....

On Fri, 24 Jul 2009 14:17:33 GMT, ehsjr <ehsjr@NOSPAMverizon.net>
wrote:

><snip>
>The joule thief will "chew up" batteries quickly.
><snip>

It's actually pretty efficient.  I didn't get this from doing basic
calculations from theory, but by simply using LTSpice to do the calcs
of efficiency for me.  It can be around 80-85%, or so.  (It can also
be very bad, too.)  At least, it seems so if there isn't 'operator
error' involved.

Jon

On Jul 24, 4:17=A0pm, ehsjr <eh...@NOSPAMverizon.net> wrote:
>
> I can see it now - you finally get your joule thief running
> without killing transistors - and your next questions will be:
> "why does it last for only a day?"
> "how can I keeep the brightness up? It gets dim over time"
> "can I get an =A0Obama bailout for the cost of all these batteries?"
>

From my experiments so far I think I can get current
to stay between 15-20mA for most of the life of the battery.

See: http://www.artlum.com/jt/jt_vs_res.gif

The problem at the moment is getting it to run at 20mA
without the transistor dying.

> "can I get an  Obama bailout for the cost of all these batteries?"

One word: "NiMH"


> _Cost_
>
> The joule thief will "chew up" batteries quickly. =A0Imagine the
> cost of replacing 3 AAA's every day or 3 D's every three weeks.

All the joule thief circuits on the net are usually about getting
a few days of light out of "dead" batteries so it can't be *that*
inefficient or you'd only get half an hour.

> Solution: =A0mains power. =A0Mains power solves the other issues,
> as well.
>

Part of the spec is that I might be walking around with it in a
procession (did I mention that?) .


> Hopefully, you are in this more for the experimentation than
> anything else. =A0In that case, the joule thief is a wonderful
> circuit to play with, and learn from.
>

It's a "fun" circuit, yes.

On Jul 24, 12:07=A0pm, David Eather <eat...@tpg.com.au> wrote:
> A different subject - I am seeking information.
>
> How long does this thing have to run on one set of batteries? and if you
> can how much current is coming out of the batteries when the LED's are
> getting their 18ma? =A0(if the 2n2222 is getting hot then this is a
> missing piece of information.)

Voltage drop across the LEDs is 16.6V and current is 12mA  (=3D192mW)

Batteries are at 3.75V and total current is 92mA (=3D345mW)

That's only 55% efficient but there's only seven turns on the inductor
at the moment so I expect it can be better.

On Jul 24, 12:17=A0pm, David Eather <eat...@tpg.com.au> wrote:
>
> No, the frequency thing is not correct. You can increase the output
> voltage by putting more turns on L2. It is the ratio of turns between L1
> and L2 that mostly determines the output voltage. To go from 1 LED to 6
> and upping the voltage by 3 times try doubling the turns on L2 and
> increasing R1 to about 3k (2.7k or 3.3 would both be fine)
>

OK, this is the next thing to try.

(I'm not sure I understand transformers which have input/output
 tied together).

On Fri, 24 Jul 2009 20:17:39 +1000, David Eather <eather@tpg.com.au>
wrote:

>fungus wrote:
>> On Jul 23, 10:18 pm, "bw" <bweg...@hotmail.com> wrote:
>>> Start with the original circuit, and get it to work on one battery.
>>>
>>> then go from there to what you want to do.
>> 
>> That's what I'm doing...
>> 
>> The original circuit lights up a LED but the current
>> is very low - about 5mA.
>> 
>> To drive six LEDs at 20mA with one battery you'd
>> have to get the frequency up into the mHz (which
>> isn't going to happen).
>
>No, the frequency thing is not correct.

Agreed, the frequency thing doesn't relate much to the output voltage.

>You can increase the output voltage by putting more turns on L2.
>It is the ratio of turns between L1 and L2 that mostly determines
>the output voltage.
><snip>

I disagree, here.  This has almost nothing to do with the voltage on
the output.  The only thing that changing the ratio of windings, L1 to
L2, does is change the BJT's base current... which changes the peak Ic
at which the BJT turns off... which affects the frequency.  In effect,
L2 is the primary and L1 is the secondary, for perspective purposes of
understanding what is going on.

The voltage applied across L2 when the BJT is ON is presented to the
base circuit and adds to the battery voltage.  After subtracting the
Vbe of the BJT, what voltage remains drives a current through R1 and
becomes a fairly stable (flat) Ib for the BJT.  When the BJT goes OFF,
L2 reverses it's voltage to maintain the current but now with a
declining I, and this reversed voltage (which is roughly the required
Vout determined by the load minus the battery voltage) yields a
now-opposing voltage in the base circuit.  In practical cases (where
Vout > 2*Vbattery), it will be enough to block the battery voltage and
will therefore cause Ib to go to zero (or a little less, via leakage)
and shut off the BJT.

The output voltage is mainly determined by the behavior of the stack
of LEDs, and R1 and the battery voltage.  The winding ratios of the
transformer has almost NO impact at all on any of this.

Of course, I'm just a hobbyist.  So that's my view and I'm sticking to
it.  ;)

Jon

On Fri, 24 Jul 2009 23:09:30 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

>The voltage applied across L2 when the BJT is ON is presented

... by L1 ...

In a fit of temporary insanity, I wrote:

> However, the transistor will be on for only
> 1/N of the time, so the *peak* transistor current
> will be N *squared* times 20mA.

Scrub that, it's completely wrong.

The on period of the transistor is N times the off
period, so the charging period of the capacitor is
1/(N+1) of the whole cycle, and the average current
charging the capacitor during that time must be
(N+1) * 20mA.

Since the current is a linear ramp during both
periods, the peak output current, and therefore
also the peak collector current, will be twice
that, or 2 * (N+1) * 20mA.

I hope I got it right this time!

-- 
Greg

fungus wrote:
> On Jul 24, 12:17 pm, David Eather <eat...@tpg.com.au> wrote:
>> No, the frequency thing is not correct. You can increase the output
>> voltage by putting more turns on L2. It is the ratio of turns between L1
>> and L2 that mostly determines the output voltage. To go from 1 LED to 6
>> and upping the voltage by 3 times try doubling the turns on L2 and
>> increasing R1 to about 3k (2.7k or 3.3 would both be fine)
>>
> 
> OK, this is the next thing to try.
> 
> (I'm not sure I understand transformers which have input/output
>  tied together).

You're going to have to rewind the transformer. It will be the same type 
as before. It will have 2 windings both with a separate start and a 
separate end, but L2 will have twice as many turns on it as L1. So 20 
turns on L1 and 40 turns on L2 or if that won't fit, 13 turns on L1 and 
26 turns on L2. You can start the same way as you did previously, 
winding both wires at the same time, but when you get to 20 turns (or 13 
turns) separate the wires and leave one wire alone while you make the 
extra turns. It is important that you know the start and end of each 
wire (as before) but you must also make sure L2 is the one with the 
extra turns.

In your circuit you could also try removing C1 - the LED's will light 
just fine with the JT pulses and your eye will make the circuit appear 
brighter for the same current (effectively an improvement in efficiency)

David Eather wrote:
> fungus wrote:
>> On Jul 24, 12:17 pm, David Eather <eat...@tpg.com.au> wrote:
>>> No, the frequency thing is not correct. You can increase the output
>>> voltage by putting more turns on L2. It is the ratio of turns between L1
>>> and L2 that mostly determines the output voltage. To go from 1 LED to 6
>>> and upping the voltage by 3 times try doubling the turns on L2 and
>>> increasing R1 to about 3k (2.7k or 3.3 would both be fine)
>>>
>>
>> OK, this is the next thing to try.
>>
>> (I'm not sure I understand transformers which have input/output
>>  tied together).
> 
> You're going to have to rewind the transformer. It will be the same type 
> as before. It will have 2 windings both with a separate start and a 
> separate end, but L2 will have twice as many turns on it as L1. So 20 
> turns on L1 and 40 turns on L2 or if that won't fit, 13 turns on L1 and 
> 26 turns on L2. You can start the same way as you did previously, 
> winding both wires at the same time, but when you get to 20 turns (or 13 
> turns) separate the wires and leave one wire alone while you make the 
> extra turns. It is important that you know the start and end of each 
> wire (as before) but you must also make sure L2 is the one with the 
> extra turns.
> 

Scratch the previous comment. It is not correct - I was thinking about 
my own circuit. Changing R1 is still the right thing to do as is the 
experiment to remove C1.

> In your circuit you could also try removing C1 - the LED's will light 
> just fine with the JT pulses and your eye will make the circuit appear 
> brighter for the same current (effectively an improvement in efficiency)

Jon Kirwan wrote:
> On Fri, 24 Jul 2009 20:17:39 +1000, David Eather <eather@tpg.com.au>
> wrote:
> 
>> fungus wrote:
>>> On Jul 23, 10:18 pm, "bw" <bweg...@hotmail.com> wrote:
>>>> Start with the original circuit, and get it to work on one battery.
>>>>
>>>> then go from there to what you want to do.
>>> That's what I'm doing...
>>>
>>> The original circuit lights up a LED but the current
>>> is very low - about 5mA.
>>>
>>> To drive six LEDs at 20mA with one battery you'd
>>> have to get the frequency up into the mHz (which
>>> isn't going to happen).
>> No, the frequency thing is not correct.
> 
> Agreed, the frequency thing doesn't relate much to the output voltage.
> 
>> You can increase the output voltage by putting more turns on L2.
>> It is the ratio of turns between L1 and L2 that mostly determines
>> the output voltage.
>> <snip>
> 
> I disagree, here.  This has almost nothing to do with the voltage on
> the output.  The only thing that changing the ratio of windings, L1 to
> L2, does is change the BJT's base current... which changes the peak Ic
> at which the BJT turns off... which affects the frequency.  In effect,
> L2 is the primary and L1 is the secondary, for perspective purposes of
> understanding what is going on.

Oh, crap! You are exactly right of course!

> 
> The voltage applied across L2 when the BJT is ON is presented to the
> base circuit and adds to the battery voltage.  After subtracting the
> Vbe of the BJT, what voltage remains drives a current through R1 and
> becomes a fairly stable (flat) Ib for the BJT.  When the BJT goes OFF,
> L2 reverses it's voltage to maintain the current but now with a
> declining I, and this reversed voltage (which is roughly the required
> Vout determined by the load minus the battery voltage) yields a
> now-opposing voltage in the base circuit.  In practical cases (where
> Vout > 2*Vbattery), it will be enough to block the battery voltage and
> will therefore cause Ib to go to zero (or a little less, via leakage)
> and shut off the BJT.
> 
> The output voltage is mainly determined by the behavior of the stack
> of LEDs, and R1 and the battery voltage.  The winding ratios of the
> transformer has almost NO impact at all on any of this.
> 
> Of course, I'm just a hobbyist.  

I'm just wrong.

So that's my view and I'm sticking to
> it.  ;)



> 
> Jon