Joule Thief - still not working....

[Jon Kirwan]

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

 

[David Eather]

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.

 

[David Eather]

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)

 

[fungus]

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?) .

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.

 

[fungus]

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.

 

[fungus]

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).

 

[David Eather]

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

 

[fungus]

On Jul 24, 1:22 pm, 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).

 

[fungus]

On Jul 25, 1:09 am, 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.  The 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]

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?

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.

 

[John Larkin]

On Fri, 24 Jul 2009 22:42:50 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Fri, 24 Jul 2009 14:47:58 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 24 Jul 2009 20:15:54 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

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

On Fri, 24 Jul 2009 07:50:41 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Thu, 23 Jul 2009 20:25:03 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 23 Jul 2009 23:34:04 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Thu, 23 Jul 2009 16:06:41 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 23 Jul 2009 20:32:20 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Thu, 23 Jul 2009 13:23:15 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 23 Jul 2009 19:04:43 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

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

On Thu, 23 Jul 2009 04:20:21 -0700 (PDT), fungus
openglMYSOCKS@artlum.com> wrote:

I just got some proper parts to start making joule thieves but I'm
still
having problems.

The circuit is this: http://www.artlum.com/jt/joulethief.gif

Except I have R1 and L1 one the other way around (as in the original
web page at http://www.emanator.demon.co.uk/bigclive/joule.htm )

The problem is that my transistors keep on overheating and dying.
Why should this be? I'm using a 2N2222 in metal can (as shown here
http://en.wikipedia.org/wiki/2N2222 ). These can switch at hundreds
of megahertz so I don't think it's because of slow switching.

I measured the current at point X and it seems high - over 100mA.
Could this be the cause of the overheating? Even if it isn't the
problem
it seems wasteful. I tried putting in a resistor there but the circuit
shuts down.
.
I also tried a honking big "high speed switching" transistor pulled
out of a PSU but it made the LEDs go very dim.

Any ideas?

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

Would you care to provide some sample values and analyze that circuit
for us?

No, too much work.

Hmm.

Just to goose things along, for the joule thief circuit I get
something like this for the frequency:

(Vbattery - Vsat) * (Vout + Vfreewheeldiode - Vbattery)
f = -------------------------------------------------------
Ic_peak * L_collector * (Vout + Vfreewheeldiode - Vsat)

Ic_peak may require an iteration or two with a datasheet to
approximate. I just go in with an assumed Ic, look up a beta estimate
for that on one curve and then grab the Vbe estimate from another
curve, and apply them into:

Ic_peak = beta*(Nratio*(Vbattery - Vsat) + Vbattery - Vbe))/Rbase

That Ic_peak is then used to repeat the process. When it settles,
I've usually got a reasonable figure that I can use to compute 'f'.
(Nratio is the turns ratio, usually just 1.) I tend to use Vsat=0.2V.

If your suggestion is so nicely designable, can't you at least provide
an approximate equation?

I see the RC node moving towards a bias point, but not really
setting the frequency at which the BJT goes on and off. But I haven't
sat down more than to glance over it, yet.

In general, "on" pulse width is set by the volt-second saturation of
the inductor (although a small value of C can make it shorter.)

So in your circuit case, it does depend on saturation of the core.
What would happen in an air core case?

The classic tube "blocking oscillator" had its ON time determined by
inductor saturation. If it can't saturate, the ON interval ends when
the transistor runs out of beta (or the tube out of plate current), or
when C runs out of charge to drive the base/grid. The "blocking" part
was the negative swing on the tube grid from grid current charging the
cap; it fired again when R1 charged the grid the other way, back to
the turnon threshold.


Base
current is limited by R2 (the one connected to the base.) While the
transistor's on, the base current charges up the cap, and that charge
will back-bias the transistor until R1 recharges the cap back up to
+0.7 volts, at which it fires again.

Something like that.

Try R1=1K, R2=100 C=100nF as very rough starting points. A lot depends
on the inductor. It won't Spice unless the model includes inductor
saturation.

Yes. I gather.

Unless L can't saturate, of course. Then it's not an official
"blocking oscillator."

Are the saturation of cores more predictable than BJT beta -- keeping
in mind that we are talking about the same part number AND
manufacturer in both cases?

It's probebly easier to use a Tiny Logic schmitt-trigger oscillator to
drive the transistor, and just use a single-winding inductor. Blocking
oscillators are tricky.

Single BJTs are cheap and, if you saw one of the web sites mentioned
some time back in the related thread, you'd have seen that the whole
thing is tiny enough to place inside a small flashlight bulb base.

If you don't mind the 2-winding coil, and the additional futzing, the
blocking oscillator is potentially cheap.


...

Since you write, "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," shouldn't it be the case that you can tell
me how to compute the frequency with ease? Isn't that the entire
point of saying all that? Or did I miss your point, here?

As I said, a blocking oscillator is complex. I can't define the
frequency "with ease." But having separate control over base drive and
rep-rate helps orthogonalize things. Having one part control two
circuit parameters can get awkward. Three is a nightmare.

But the existing schematic (the joule thief thing) does that, within
bounds. Assuming fixed battery voltage and fixed winding ratio of the
transformer, the base resistor sets the Ib. The beta then establishes
the peak Ic. I'm not sure of any advantages in the new arrangement
you suggest, yet. (And I suspect it's behavior is harder to analyze,
besides.)

In the OP's ascii-art circuit, R1 determines ON base current (and
perhaps ON time, if the inductor doesn't saturate first)

Yes. R1 determines the ON base drive, together of course with the
battery voltage, less something for the Vbe. The base drive
determines and BJT beta determines the peak sustainable Ic (leaving
out core satuation considerations), which then determines the ON
_time_.

and also sets
OFF time, as part of the L/R decay.

This, I do NOT see. Off time is determined by the required voltage
across the collector winding when the BJT goes OFF and the field
collapses, reversing the polarity. I really do NOT see how R1 plays
into that calculation, at all. The base winding really isn't tapping
much field energy -- most of which is being either driving out through
the LEDs or else via the freewheeling diode (1N5819?) and cap on the
output I suggested elsewhere. (I suggested also a diode to protect
the BJT base, but that's a separate issue.)

Can you explain how R1 plays into an L/R decay time? I'd like to hear
about it.

No need to get snippy about things.

I didn't think I was being. I was directly telling you what I see and
asking for an explanation why I am wrong. And yes, I really would
like to hear about it. I meant it. What else did you read into this?

When the transistor turns off, the flyback voltage gets dumped into
the LEDs, guaranteeing part of the OFF interval.

I'm with you, here. My first assumption would be that it determines
fully the OFF interval, without additional consideration (which I'm
not brushing off, as it is important to do.)

When the transformer
runs out of stored energy, Il=0 and the secondary essentially
disappears from the circuit.

Agreed. That's always as I saw it, too.

Now we have the feedback winding in
series with R into the base.

The secondary (base) winding which, initially, will have zero voltage
across it and thus there will only be one battery voltage to work
with, at first.

Base current will build up through that L/R until the transistor
turns back on and we get another positive
feedback cycle. It's complicated by the ringing in the transformer
(Millman and Taub caution about the complications of ringing!) which
can restart a cycle, so the turn-on situation is very messy. At any
rate, the value of R influences both the ON time and the OFF time.
snip

This is where I put my feet down and say I cannot follow you, John.

Yes. I agree that there is an L/R time constant here. Completely
agree. But it simply isn't important in the larger picture. In most
cases (those along the lines of discussion here, anyway), we are
talking about perhaps a hundred nanoseconds for tau. And well before
even that first tau is exhausted, the BJT is already ON. It happens
so fast it just 'doesn't count.' That's how I see it.

What I see happen instead, brushing away that unimportant detail, is
that the BJT turns on from the push of a single battery voltage at
first and then, once it turns on, there is another battery voltage
added to it by the secondary (base) winding to goose up the base
current to about twice (not quite) what it starts out as.

The few nanoseconds part of L/R before the BJT goes back ON just
aren't something to worry over. As I see it. And they certainly
don't contribute meaningfully to the OFF time, __as I see it.__

So this is where I'm stuck and cannot find agreement with you, yet.
The OFF time is NOT determined by R1, so far as I can tell. And your
explanation doesn't in any way suggest that it should be.

Can you address yourself squarely here? I'm seriously trying to learn
this.

Jon

The average voltage across the secondary must be zero.

We are talking about volt-seconds, yes?

Since it's
positive when it's turning the base on, it must be negative for some
time so that it averages zero.

When the collector winding is dumping energy during the BJT OFF time,
the voltage reverses, of course. When it does, so does the base
winding voltage, as well. No question. So yes, it is negative.
That's what turns the BJT off, in fact. If not, the BJT would just
stay on forever.

So immediately after the transistor
turns off (if it DOES turn off)

It should. Holding a voltage across the collector winding (which is
what would happen should it be possible for the BJT to 'stay ON')
would __require__ an Ic moving towards infinity. And we know that
can't happen.

the negative transformer voltage must
exceed the battery voltage-Vbe.

This isn't hard to understand. The voltage on the collector winding,
when the BJT is OFF, will be whatever is required. The initial
conditions are set by whatever Ic was, when the BJT turned off, and
decline from there according to V/L. However, whatever V is present
there on the collector winding appears on the secondary, as well
(assuming Npri = Nsec, of course.) This voltage opposes the battery.
If this V is large (as it will be with a stack of six 3V LEDs), then
the induced voltage on the secondary (base) winding will be pretty
large. (Which is why I suggested earlier that the OP place a diode
between base and emitter of the BJT to protect it from zenering and
possibly wrecking its beta in the process.)

In the OP's transistor barbeque, it may be that the negative secondary
voltage *doesn't* exceed Vb-Vbe, in which case the transistor never
turns off and the thing becomes more like a class A oscillator.

But it __will__ exceed Vb-Vbe for any case the OP is talking about.
The main thing is that Vb is less than the required Vout by some
reasonable margin. And with a stack of six 3V LEDs, it will be even
in the case of three 1.5V batteries.

There are too many possibilities here, too many branches in the
probability tree.

I don't find it at all that complex and the equations I've worked for
it appear to predict well enough, in practice. What I haven't added
to it is the non-linear behavior of a saturating core. But so long as
the volt-seconds are watched well (this is what pushes for a higher
frequency of operation and traps the design into the 20-200kHz range
of operation, bounded on the high side by the BJT's reverse transit
time), it works close enough.

Sometimes it's much easier to design a circuit than it is to analyze
it. This is such a case.
snip

Well, design does seem easier in most cases I've encountered. You
know what you are thinking about and ignoring. And unless you
document carefully your own thinking so that others can follow
readily, they have to reverse engineer that to follow the circuit.

But none of this addresses itself to what I was saying about the OFF
time. I find myself disagreeing with your earlier point that there is
some conflation going on with the base resistor, here. The base
resistor, R1, does impact the ON time. We agree there. However, my
hobbyist view does NOT see how R1 influences the OFF time for all
reasonable cases under discussion. The frequency of operation needs
to be high enough to avoid volt-second saturation issues (which would
require adding in more complex thinking about non-linear behaviors
better to just avoid in the first place) and low enough to avoid
uselessly wasting energy due to reverse transit time in the BJT. And
at those frequencies, 20-200kHz or so [closer to 100kHz is good], the
resistor R1's impact, via Lsecondary/R1, on BJT OFF time is entirely
ignorable so far as I can see. So it doesn't complicate the picture.

At least, that's how I see it.

Can you speak to this narrow point? The BJT OFF time influence of R1
for practical cases here? A significant part of your earlier premise
seems to be that R1 influences both ON and OFF times.

R1 affects everything: ON time, OFF time, average transistor base
current, rise/fall times, overall duty cycle. That's a lot of
responsibility for one small resistor.

ON time obviously affects OFF time, since the average voltage across
the inductor windings (ignoring copper loss) must be zero. If ON time
increases, then more volt-seconds pile up in the inductor, and the
flyback into the LEDs (ie, OFF time) will increase. So if R1 affects
ON time, it affects OFF time. You can't just look at the OFF interval
in isolation.

Without a capacitor in the base circuit, it's hard to determine the
on/off duty cycle. If it runs in continuous mode and the duty cycle is
not exactly right, huge currents can accumulate in the inductor,
limited by beta, and the transistor will fry.

If there's independent control of ON and OFF times, the duty cycle can
be lowered to avoid the current catastrophe. The performance can be
*designed* and not just tolerated.

John

 

[John Larkin]

On Fri, 24 Jul 2009 20:45:28 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 25, 4:33 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

R1 affects everything: ON time, OFF time, average transistor base
current, rise/fall times, overall duty cycle. That's a lot of
responsibility for one small resistor.


It doesn't seem to affect it an awful lot...

(Though I didn't try changing it on my magic new iron powder circuit).

One question: Does it matter where R1 is, relative to L1? I've seen
circuits with R1 before and after L1 in the circuit.

Things in series can be swapped without effect, usually.

John

 

[ehsjr]

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

 

[fungus]

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?

=================
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....

 

[fungus]

On Jul 25, 4:33 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

R1 affects everything: ON time, OFF time, average transistor base
current, rise/fall times, overall duty cycle. That's a lot of
responsibility for one small resistor.

It doesn't seem to affect it an awful lot...

(Though I didn't try changing it on my magic new iron powder circuit).

One question: Does it matter where R1 is, relative to L1? I've seen
circuits with R1 before and after L1 in the circuit.

 

[ehsjr]

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.

 

[fungus]

On Jul 25, 6:08 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

One question: Does it matter where R1 is, relative to L1? I've seen
circuits with R1 before and after L1 in the circuit.

Things in series can be swapped without effect, usually.

I was wondering if there's any current flow between L1 and l2 when the
transistor switches on. If there is then a resistor in between them
would make a difference.

 

[Jon Kirwan]

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

 

[fungus]

On Jul 25, 5:16 am, 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)

 

[greg]

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