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.

The site you reference below, shows the joule thief producing
BELOW 15 mA for every datapoint.  But your statement can be true,
by defining "life of the battery" as the time that it can maintain
at least 15 mA into the LEDs, and using a joule thief that produces
that much.  Using that definition you can measure the amount
of time it takes for the battery to reach end of life.


> 
> 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"

Two words: different cost.  If you can use rechargeables, why not
use mains power?  A 15 volt DC wall wart, an LM317 and a resistor
is all you need.  It'll cost you less than the NiMh cells, and can
be connected to the LEDs with small gauge wire, if necessary. It
will provide constant current - no decrease in brightness.
Fewer parts than a joule thief!

Why different cost?  An NiMh AAA at 750 mAh has a bit less than
half the power of an alkaline AAA at 1250 mAh.  So you'll be
replacing cells twice as often, putting a freshly charged set
into the assembly twice a day, and recharging the drained
set.  You'll cut the drain on your wallet drastically.  But you
won't maintain the schedule of twice daily cell swapping for
long.  The kicker to the cost is that you're likely to destroy
the NiMh cells by discharging them too low.


> 
> 
> 
>>_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.

Yes - you can get light out of a joule thief for far longer
than a day.  But that is not what you said you want.  You
said you want full brightness, 15-20 mA.  You will NOT get
that from 3 AAA cells powering a joule thief driving 6 LEDs
in series.  The light you get from a joule thief with a fresh
AAA cell driving 1  20 mA rated LED  will becom very dim as
time passes.  _That_ is the "few days of light" people talk
about.




> 
> 
>>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?) .

If you did, I missed it.  If you need it for only a few
hours, then NiMh or NiCd will be fine.  My concern that
you would be disappointed goes away.  :-)

Ed

> 
> 
> 
>>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 Jul 25, 5:16=A0am, fungus <openglMYSO...@artlum.com> wrote:
> 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've been thinking about this and this is what I think
is going on...

As I add turns, the current rises until I reach a peak then
the current starts to go down again do to something
saturating.

With ferrite this peak occurs very early, around six or seven
turns with my bead (other sizes may vary).

With the iron powder bead this peak arrives much later.
If I'd been able to add more turns it would eventually
start going downwards just like the ferrite.

The important thing is I can get the current much higher
with the iron powder and more turns should make the
circuit more efficient.

I've also been thinking about the transistor cooking
problem. It's been mentioned that most heating occurs
when the transistor is in a half-switched state. This means
that if the voltage at the base ought to rise as rapidly as
possible to make switch-on as fast as possible.

At the moment the voltage at the base of the transistor
ramps up as the inductor fills up.

What if I add a zener diode before the base of the transistor
and a pull-down resistor between the base of the transistor
and ground. This would mean the voltage at the base is
zero until the zener kicks in, than it should go up quite
suddenly meaning the transistor switches faster.

(As I said earlier , I know next to nothing about electronics
so all this could be complete rubbish)

On Sat, 25 Jul 2009 05:03:52 GMT, ehsjr <ehsjr@NOSPAMverizon.net>
wrote:

>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.
><snip>

I'll stop it here.  Yes, I snipped a lot.  Mostly because that line
was a lead-in towards another later on where you wrote, "If you _must_
use battery power, there are one chip solutions better than the joule
thief."  I should have included that, as well.  I'd read it, just
failed to quote it.  The existing chip solutions aren't a whole lot
better, frankly.  That was my point in writing as I did.

Sorry I wasn't more clear about it.

Jon

David Eather wrote:

> 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

No, it won't. An LED run continuously at 20mA will always
look brighter than one pulsed at 20mA.

Pulsing LEDs only helps if the average current available
from your supply is much less than the rated current of
the LEDs. E.g. if you can only manage 2mA continuously,
then 20mA at 10% duty cycle will give slightly more
light overall, due to nonlinearity of the LED's light
vs. current curve.

However, that's not the case here, because we're willing
to draw as much current as needed from the battery to
run the LEDs continously at their max current.

The purpose of the capacitor is to smooth out the output
current, so that you can get a steady 20mA through
the LEDs. Otherwise you would have to pulse them at
several times their rated current to get the same
light output, which wouldn't be kind to them.

-- 
Greg

Jon Kirwan wrote:

>   Ic_peak = SQRT(2 * V_out * I_out / [frequency * L])
> 
> This isn't what you came up with.

What I came up with didn't involve frequency or L,
so it's hard to tell at first glance. Let's try to
find out whether they're equivalent...

Consideration of the charging and discharging times
of the inductor gives

   t_on = L * I / Vin

   t_off = L * I / Vout

for a total period

   t_total = t_on + t_off

           = L * I * (1/Vin + 1/Vout)

Defining

   N = Vout / Vin

we have

   t_total = L * I * (N/Vout + 1/Vout)

           = L * I * (N + 1) / Vout

Substituting into

   Ic_peak^2 = 2 * Vout * Iout * t_total / L

we get

   Ic_peak^2 = 2 * Vout * Iout * L * I * (N + 1) / (Vout * L)

which cancels to give

   Ic_peak = 2 * Iout * (N + 1)

which I'm relieved to find is not entirely dissimilar
to what I came up with before. :-)

> Now assume for a moment that the capacitor is 'somewhat smaller' and
> that there is some change in voltage during operation

Considering those sorts of effects will complicate things,
of course. But the above should be good enough to get
a first approximation of the Ic_peak you need to aim
for, I think.

> It's about accounting for energy, not accounting for current.

Sure, that's another way of approaching it. The way
I derived it short-circuits any consideration of
frequency or inductance, though, and gets there more
directly.

> Of course, I'm a hobbyist.

So am I, and I'm learning things from all this too.
Wouldn't still be following the thread if I wasn't!

-- 
Greg

fungus wrote:
> 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.

Are you altering the number of turns on both
windings together, or only on one winding?

Theoretically, if you change the number of turns
on both windings by the same amount, and leave
everything else the same, there should be no
change in the output current -- the only thing
that should happen is a change in frequency.

However, this assumes that the core doesn't
saturate, the transistor doesn't run into some
operating limit, etc.

Without looking at waveforms, it's going to be
hard to make any headway on understanding what's
really going on in your circuit.

> nb. The transistor was getting hotter with every extra turn

That doesn't sound good.

> despite the oscillation frequency going down

Are you certain that the frequency *is* actually
going down, or are you just assuming that? Can
you tell what the frequency actually is?

> It seems that transistor temperature
> is more strongly related to output current than frequency.

You can expect some increase in temperature with
current, because Vce won't be *exactly* zero even
with the transistor turned hard on, so it will
dissipate some power. But again, without seeing
waveforms, it's impossible to tell whether this is
the major contributor to your transistor heating
problem or not.

> 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?

When putting BJTs in parallel, you need to put a
small resistor in series with each emitter to make
sure they share the current equally.

However, before doing this, it would be better to
find out why the transistor is getting hot, otherwise
you risk treating the symptom rather than the
real cause.

> 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

Yeah, but it means 39% is still going somewhere other
than the LEDs, and it sounds like most of it is ending
up in your transistor. :-(

-- 
Greg

fungus wrote:

> Aside: Does wire thickness make any difference?

At the frequencies you're using, all that matters is
keeping the resistance as low as possible, to
minimise losses. This means using as thick a wire as
you can while still fitting in the required number of
turns.

-- 
Greg

fungus wrote:

> At the moment the voltage at the base of the transistor
> ramps up as the inductor fills up.

Are you looking at this with a scope?

What does the voltage waveform between the collector
and emitter look like? That's the important thing with
regard to transistor dissipation.

Some variation in Vbe with collector current is
probably to be expected. It doesn't necessarily
indicate any inefficiency.

Vce, on the other hand, should be a sharp square
wave, otherwise you're not switching cleanly.

> What if I add a zener diode before the base of the transistor
> and a pull-down resistor between the base of the transistor
> and ground.

No, that doesn't make much sense.

What *might* help is putting a small capacitor
across the base resistor, to give the base a small
"kick" at turn-on and turn-off. Not too big, though,
or it could cause problems of its own.

-- 
Greg

On Sat, 25 Jul 2009 20:54:51 +1200, greg <greg@cosc.canterbury.ac.nz>
wrote:

>Jon Kirwan wrote:
>
>>   Ic_peak = SQRT(2 * V_out * I_out / [frequency * L])
>> 
>> This isn't what you came up with.
>
>What I came up with didn't involve frequency or L,
>so it's hard to tell at first glance. Let's try to
>find out whether they're equivalent...
>
>Consideration of the charging and discharging times
>of the inductor gives
>
>   t_on = L * I / Vin
>
>   t_off = L * I / Vout
>
>for a total period
>
>   t_total = t_on + t_off
>
>           = L * I * (1/Vin + 1/Vout)
>
>Defining
>
>   N = Vout / Vin
>
>we have
>
>   t_total = L * I * (N/Vout + 1/Vout)
>
>           = L * I * (N + 1) / Vout
>
>Substituting into
>
>   Ic_peak^2 = 2 * Vout * Iout * t_total / L
>
>we get
>
>   Ic_peak^2 = 2 * Vout * Iout * L * I * (N + 1) / (Vout * L)
>
>which cancels to give
>
>   Ic_peak = 2 * Iout * (N + 1)
>
>which I'm relieved to find is not entirely dissimilar
>to what I came up with before. :-)
>
>> Now assume for a moment that the capacitor is 'somewhat smaller' and
>> that there is some change in voltage during operation
>
>Considering those sorts of effects will complicate things,
>of course. But the above should be good enough to get
>a first approximation of the Ic_peak you need to aim
>for, I think.
>
>> It's about accounting for energy, not accounting for current.
>
>Sure, that's another way of approaching it. The way
>I derived it short-circuits any consideration of
>frequency or inductance, though, and gets there more
>directly.
>
>> Of course, I'm a hobbyist.
>
>So am I, and I'm learning things from all this too.
>Wouldn't still be following the thread if I wasn't!

Well, this is indeed fun.  So I did some more thinking about it and I
agree with you.  Case closed!

A design can now start with the computation of Ic_peak.  I'd modify
your N in this way, when there is a freewheeling diode (Vd) present
and taking into account the average Vcesat value (say 0.1V):

  N = (Vout+Vd-Vcesat)/(Vin-Vcesat)

That's very close to your own definition, but I think this one covers
a little more ground.

An interesting thing about all this is that the peak Ic value doesn't
care about the transformer inductance -- which you can change around
all you like and still get the same energy delivered per unit time. So
the frequency can be selected independently, at will.  Very nice.

So here goes.  Inputs are Vout, Iout, Vd (for a schottky diode likely
around .35V), Vcesat (probably a midpoint value between 0V and 0.2V,
or 0.1V), and the frequency of operation, f.

  N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
  Ic_peak = 2 * Iout * (N + 1)
  Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)
  Lsecondary = Lprimary ... unless you want to change this, of course.

Now look up Ic_peak on your BJT datasheet to find estimated Vbe and
beta.  Assuming Lprimary=Lsecondary and a new value that is the lowest
possible Vce where the BJT beta is still close to the value picked off
the datasheet, which I'll call Vce_beta and where I usually use a
value that is equal to Vbeon.  (The reasoning here is that the beta
picked off the datasheet is usually for Vce=1V.  If you want, you can
plug in 1V instead of setting Vce_beta=Vbeon, though.)

  Rbase = beta*(2*Vin-Vbeon-Vce_beta)/Ic_peak
        = 2*beta*(Vin-Vbeon)/Ic_peak            [if Vce_beta=Vbeon]

This should get pretty close, I think.

...

So, let's try some values.  Assume that these 3.3V LEDs are stacked
six up, for 19.8V... call it 20V.  Let's assume for a moment that at
20V for the stack, there is 30mA (I'm calling it a bit high just for
fun) flowing through them (assume this is validated with a simple
test.)  And let's assume Vcesat(avg) = 0.1V, Vd=.35V (schottky used),
and Vin=4.5V (three fresh batteries.)

  N = (20+.35-.1)/(4.5-.1) = ~4.6
  Ic_peak = 2*30mA*(N+1)   = ~340mA

Figure a frequency of 100kHz:

  Lprimary = (4.5-.1)*(20+.35-4.5)/(340mA*(20+.35-.1)*1e5) = ~100uH

Now the resistor.  Looking at the 2N2222 datasheet for Ic=340mA, I see
that Vbeon=0.86V and beta reads out 90 at 25C and Vce=1V.

  Rbase = 2*90*(4.5-0.86)/340mA = ~2kOhm

The other bound on this is to look at where the beta is at, at lower
currents -- about 200, or so.  If that were the case,

  Rbase = 2*200*(4.5-0.86)/340mA = ~4.3kOhm

I'd probably start with something like 3.3k and work up or down from
there.  Split the difference.  Then tweak Rbase until Ic_peak is where
I need it to be (or where the LED current is about right.)

Jon

On Jul 25, 7:48=A0am, ehsjr <eh...@NOSPAMverizon.net> wrote:
>
> The site you reference below, shows the joule thief producing
> BELOW 15 mA for every datapoint.
>

That's because the transistor dies if I go much higher... :-(