Joule Thief - still not working....

fungus wrote:
> On Jul 24, 4:17 pm, 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  Obama 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.  Imagine 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:  mains power.  Mains 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?) .

How much time do you have until this thing must be right?

> 
> 
>> Hopefully, you are in this more for the experimentation than
>> anything else.  In that case, the joule thief is a wonderful
>> circuit to play with, and learn from.
>>
> 
> It's a "fun" circuit, yes.

On Sat, 25 Jul 2009 11:22:23 +1200, greg <greg@cosc.canterbury.ac.nz>
wrote:

>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,

Set up variable N as being the ratio,
  N = t_on / t_off

Total period is,
  t_total = t_on + t_off

>so the charging period of the capacitor is
>1/(N+1) of the whole cycle,

Yup.  To encourage more to follow the logic...

The charging duty cycle occurs during t_off, so,

   duty = t_off / t_total
        = t_off / (t_on + t_off)
        = t_off / (t_off * N + t_off)
        = t_off / [t_off * (N + 1)]
        = [t_off / t_off] / (N + 1)
        = 1 / (N + 1)

>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!

Energy in equals energy out (I prefer to not look so much at current
or voltage, as they are two facets of an underlying whole.)

Assume for a moment that the capacitor is 'large' and that there is no
change in voltage during operation (in other words, as current is
drawn the voltage doesn't change except by a very tiny amount.)

For one cycle of the period, t_total, we have:

  W_out = (V_out * I_out * t_total)

Assume the diode is perfect (has no voltage across it) and that
everything is in an equilibrium state, the collector winding's energy
at Ic_peak must be:

  W_in = (1/2) * L * I^2

But by definition, it's in equilibrium, so,

  W_out = W_in

Therefore,

  V_out * I_out * t_total = (1/2) * L * Ic_peak^2

or,

  Ic_peak = SQRT(2 * V_out * I_out * t_total / L)

or,

  Ic_peak = SQRT(2 * V_out * I_out / [frequency * L])

This isn't what you came up with.

Now assume for a moment that the capacitor is 'somewhat smaller' and
that there is some change in voltage during operation (in other words,
as current is drawn the voltage does change by some small, but
measurable amount.)

Assuming the voltage drop is linear vs time and goes from V_max down
to V_min for one cycle of the period, t_total, we have:

  W_out = [(V_min+V_max)/2] * I_out * t_total

Again, assume the diode is perfect (has no voltage across it) and that
everything is in an equilibrium state, the collector winding's energy
at Ic_peak must be:

  W_in = (1/2) * L * I^2

Therefore,

  [(V_min+V_max)/2] * I_out * t_total = (1/2) * L * Ic_peak^2
  [V_min+V_max] * I_out * t_total = L * Ic_peak^2

or,

  Ic_peak = SQRT( [V_min+V_max] * I_out * t_total / L)

Again, not much different, really.

It's about accounting for energy, not accounting for current.  When
the capacitor is charged up to its equilibrium voltage, it doesn't
take much current at that voltage to add substantial energy.

Of course, I'm a hobbyist.  So... that's only how I see it.

Jon

On Jul 24, 1:22=A0pm, greg <g...@cosc.canterbury.ac.nz> wrote:
> To get a higher peak collector current, you need
> to increase the base current. You can do that
> either by lowering the base resistor...
>

With the circuit I've got at the moment the output seems
to peak with a resistor value around 1k.

Higher than that and current drops off, lower than that and
it also drops off until it reaches a point where the circuit
shuts down (stops oscillating).

David Eather wrote:
> fungus wrote:
>> On Jul 24, 4:17 pm, 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  Obama 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"
>>

If your serious about NiMH, then this is a very different kettle of 
fish. NiMH put out about 1.2 volts and it is stable almost until the 
battery is completely flat. The easiest way to use 3 of them is just to 
use a resistor with each LED. Since the battery voltage is nearly 
constant so is the current .  Try about 15 ohms in series with 1 LED and 
measure the current. You want to read about 15-18 ma. The current 
through the LED will be slightly higher when the multimeter is removed.

6 LED's and 6 resistors and your done.

On Jul 25, 1:09=A0am, Jon Kirwan <j...@infinitefactors.org> wrote:
>
> The output voltage is mainly determined by the behavior of the stack
> of LEDs, and R1 and the battery voltage. =A0The winding ratios of the
> transformer has almost NO impact at all on any of this.
>

R1 really has very little effect. I've tried putting a
variable resistor in there to see what happens.

fungus wrote:
> On Jul 24, 12:07 pm, 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?  (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  (=192mW)
> 
> Batteries are at 3.75V and total current is 92mA (=345mW)
> 
> That's only 55% efficient but there's only seven turns on the inductor
> at the moment so I expect it can be better.
> 

55% is pretty good for such a simple circuit. You mentioned to someone 
else you might use NiMH batteries.  If this is so then there is a much 
simpler solution which I posted elsewhere in the thread. If you still 
want the JT then you could get some improvement in brightness and 
efficiency by removing D8 and C1 (connect the +ve end of the LED "chain" 
to the junction of L2 and Q1 - the same spot where the +ve end of D8 was 
connected)

On Jul 25, 1:22=A0am, David Eather <eat...@tpg.com.au> wrote:
>
> You're going to have to rewind the transformer.

I've got about 2km of wire, so....

> 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.

I'll give it a try. My wire is pretty tin so I should be able
to get loads of turns on.

Aside: Does wire thickness make any difference? The stuff I
got from eBay is thicker than the wire on the joule thief web
page but looks very thin to me (it's 30AWG). For a normal
transformer thin wire and lots of turns is good but would
thicker wire be better for a joule thief?

> 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)

Nope, C1 gives a huge increase in brightness - maybe twice as bright.

fungus wrote:
> On Jul 25, 1:22 am, David Eather <eat...@tpg.com.au> wrote:
>> You're going to have to rewind the transformer.
> 
> I've got about 2km of wire, so....
> 
>> 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.
> 
> I'll give it a try. My wire is pretty tin so I should be able
> to get loads of turns on.
> 
> Aside: Does wire thickness make any difference? The stuff I
> got from eBay is thicker than the wire on the joule thief web
> page but looks very thin to me (it's 30AWG). For a normal
> transformer thin wire and lots of turns is good but would
> thicker wire be better for a joule thief?

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.

> 
>> 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)
> 
> Nope, C1 gives a huge increase in brightness - maybe twice as bright.

I am surprised :-O

On Jul 25, 3:14=A0am, 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 bea=
d.
>

So the size of the magnetic field is purely down to current?

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

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

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?


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....

Jon Kirwan wrote:
> 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

You snipped the content, completely. Joule thief efficiency
is not the factor.   At 100% efficiency, which is of course
impossible, the op would be replacing AAA cells every 26 hours.
The math is in my post.

The problem starts with the op's requirement:
"b) I want them to be as bright as possible - the full 20mA or as
close to it as I can get."

I used that 20 mA figure in the math, but he also said
"c) It's a battery ... so voltage is going to drop over time
(from 4.6V to 3.3V), this makes part (b) problematic. I accept
that current will drop a bit, but if it can stay in the range
15-20mA then that's Ok. "

At 15 mA, he needs (10.8 * .015) 162 mw, which means replacing
batteries every 34.7 hours.

These numbers are theoretical, of course, since they are
based on 100% efficiency, presume constant current, and do
not take into account the battery discharge curve.  Nevertheless,
they demonstrate the fact that he will chew up batteries
quickly.  If he changed the circuit to high efficiency LEDs
he'd get much longer run time, and no size penalty.  With bigger
cells he'd increase the run time, with a size penalty.  He
mentioned that he wanted small size, so I don't know if
D cells would be acceptable.

Ed