Joule Thief - still not working....
fungus
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?
David Eather
Yes. Figure out what you want to do and state it explicitly and exactly.
Then work to that goal in steps you understand.
default
Check the phasing of the feedback coil versus the collector coil
-switch ONE only pair of leads around .
The coil can be wound with a tap or separately. If you use a tap and
keep winding in the same direction it will be phased right.
OR the "start" of one winding must be connected to the "finish" of the
other and wound in the same direction.
Phased wrong and it won't oscillate, but will pull current.
Once you get it working. Keep a load on the output. It is possible
to get voltage spikes over 80 volts with no load and that may be
enough to eat your transistor.
I haven't fried a 2N2222A due to over voltage, but some 2N4401's I had
died almost instantly.
BTW this circuit (and some variations like cap across the resister or
electrolytic cap in series with the resistor and variable resistors)
is rightly called a "blocking oscillator." It dates back from the
days of vacuum toobs.
I put one on my bicycle for a horn - it appeared in Radio Electronics
under the heading of "Build the Sonic Shake Table" - basically driving
a speaker voice coil with variable frequency (you could put mercury or
flour, or other fine particles - or liquid with a plastic liner - in
the upturned woofer speaker cone and watch the standing waves form in
the material as you tuned through its resonant frequency).
--
default
<opengl...@artlum.com> wrote:
PS
Another possibility is the choice of a poor core material. You may
have something that has too low permeability for this application. If
that's the case, a few more turns of wire may make a difference.
(like 30-40 turns instead of 10-20)
Check out this site, he's got some better drawings and circuit
variations:
http://cappels.org/dproj/ledpage/leddrv.htm#Rusty_Nail_Night_Light
--
fungus
>
> Check the phasing of the feedback coil versus the collector coil
> -switch ONE only pair of leads around .
>
> The coil can be wound with a tap or separately. If you use a tap and
> keep winding in the same direction it will be phased right.
>
If you get it wrong the LEDs don't light up at all.
> Once you get it working. Keep a load on the output. It is possible
> to get voltage spikes over 80 volts with no load and that may be
> enough to eat your transistor.
>
I'm doing that. The LEDs are lighting up nicely but the transistor
gets hotter and hotter until eventually it gives up. I thought it
was because the voltage at the transistor base was going
to negative 10V. Adding D7 this morning cured that but the
transistor is still getting hot.
fungus
> fungus wrote:
> >
> > Any ideas?
>
> Yes. Figure out what you want to do and state it explicitly and exactly.
> Then work to that goal in steps you understand.
a) I want to light up some LEDs (eg. six of them) using batteries, eg.
three AAAs. Circuit is decorative and has to be small because I want
to hide it.
b) I want them to be as bright as possible - the full 20mA or as close
to it as I can get.
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.
I've figured out that a Joule Thief is much closer to these
characteristics
than a simple resistor circuit doesn't. See the graph I plotted here:
http://www.artlum.com/jt/jt_vs_res.gif
But ... at the moment it's eating up transistors.
fungus
>
> Another possibility is the choice of a poor core material. You may
> have something that has too low permeability for this application. If
> that's the case, a few more turns of wire may make a difference.
> (like 30-40 turns instead of 10-20)
>
I've just got my reel of magnet wire in the post so the next step is
to play with the number of turns to get the right output current.
OTOH I'd be raising the current, not lowering it, and I don't see
how the number of turns would be related to the transistor
temperature.
fungus
>
>
> PS
>
> Another possibility is the choice of a poor core material. You may
> have something that has too low permeability for this application. If
> that's the case, a few more turns of wire may make a difference.
> (like 30-40 turns instead of 10-20)
>
> Check out this site, he's got some better drawings and circuit
> variations:
>
> http://cappels.org/dproj/ledpage/leddrv.htm#Rusty_Nail_Night_Light
>
There's an interesting bit at the bottom which says:
"I was intrigued with getting the BC107 to run and added a small
capacitor (22nF) across the base resistor to 'kick-start' the
oscillations.
It worked so well - with various transistors and coils that I was
further intrigued to see how much I could increase the resistor value
- hence the 20k trimpot. (The 22nF also got over the problem of the
oscillator failing when I tried to add an ammeter in the battery
circuit).
I found it would continue to oscillate right up to 20k ohms and this
also had the effect of reducing the supply current (osc only) from
90mA to 800microamps - very important if using batteries."
...
Jon Kirwan
BJT temperature is related (obviously) to its power dissipation. That
power dissipation comes from a variety of possible corners from my
hobby viewpoint: (1) The transistor has been damaged (diode put in
later, after it was already ruined perhaps?) and isn't operating well
anymore; or, (2) the base-emitter junction current; or, (3)
collector-emitter current times collector-emitter voltage; or, (4)
frequency of operation is too high for the reverse transit time of the
BJT.
Your comment about it being able to operate at 100's of MHz is mostly
wrong, by the way... what you saw in the Wiki article is the f_t,
which is NOT what you can operate it at in this circuit. That is the
frequency where the short circuit signal current gain drops to 1 and
it varies with Ic... quite a bit, if I recall. I think the spec you
are reading assumes Ic=20mA, or so. You don't operate BJTs anywhere
near their ft for this. Divide by a factor of 20, at least. Also,
capacitances can become increasingly important as your frequencies
rise. I think these are in the 10's of pF for the 2N2222, so I don't
think power dissipation due to capacitances will be the problem in
this case.
For higher frequencies, I think the reverse transit time will be a
problem.
Item (2) has Vbe which remains high during the BJT on cycle (0.9V?)
and the current remains fairly fixed. Item (3) has a linear rise in
current that gets pretty high near the end, but the Vce voltage
remains relatively low (under 0.2V) for most of the time. Hand
waving, I'd guess that power dissipation in item (3) should be about
4-5 times higher than in item (2), so I'd focus on Ic*Vce as more
likely the problem if your frequency of operation is under 100kHz or
so.
But as the frequency rises much above that (and few turns/low
inductances in your transformer will do that to you), the power
dissipation (for the same output current) goes way up as you start
encroaching on the reverse transit time. With the 2N2222, this figure
is about 1/10th of a microsecond. So by the time you get up to 1Mhz
or so (reverse transit is about 10% of your total time in this case),
you are in serious trouble with item (3). This means that there is a
very large Vce on your 'off' transistor __but__ there is still a
significant Ic while charges are swept out of the BJT -- which means
power dissipation.
So wind more turns and get the frequency near or under 100kHz, or so,
where the reverse transit will be only be a few percent or so and
won't be wasting a lot of power.
The
petrus bitbyter
"fungus" <opengl...@artlum.com> schreef in bericht
news:ddeb7f42-6adb-4135...@e27g2000yqm.googlegroups.com...
Well, you bluntly got a circuit designed to run on <=1.5V and powered it
with 4.5V. Suppose the coil worked fine in the original circuit (did you
try?) two important things will happen:
- The frequency will raise. It easyly raises beyond the maximum frequency
the coils core can handle. As a result, the transistor will stay long in the
lineair region and get hot. Too hot.
- The base-emitter voltage of the transistor will fall below -5V, which will
make the transistor malfunction and may even damage it. The diode you added
makes things worse as the off-time of the transistor will be shortened.
So you need to move the diode. Place it between R1 and the base of the
transistor. Alternatively you can place it between the emitter and gnd but
make sure it can handle the current. In both cases make in point the same
direction as the be-junction.
Next step is adding more turns to the coil as others stated already. Guess
you will need 1.5-2 times the original number of turns.
petrus bitbyter
Jon Kirwan
<opengl...@artlum.com> wrote:
>On Jul 23, 2:26�pm, default <defa...@defaulter.net> wrote:
>> On Thu, 23 Jul 2009 04:20:21 -0700 (PDT), fungus
>>
>> <openglMYSO...@artlum.com> wrote:
>>
>> PS
>>
>> Another possibility is the choice of a poor core material. �You may
>> have something that has too low permeability for this application. �If
>> that's the case, a few more turns of wire may make a difference.
>> (like 30-40 turns instead of 10-20)
>>
>> Check out this site, he's got some better drawings and circuit
>> variations:
>>
>> http://cappels.org/dproj/ledpage/leddrv.htm#Rusty_Nail_Night_Light
>
>There's an interesting bit at the bottom which says:
>
>"I was intrigued with getting the BC107 to run and added a small
>capacitor (22nF) across the base resistor to 'kick-start' the
>oscillations.
Called a 'speed-up' capacitor. For a short time, it acts as a short
circuit across the resistor. And in this case, it will dramatically
increase your BJT dissipation due to substantially increased base
current for a short time near the early on-time. If it is anything
near 1000's of pF, anyway. (22nF is way, way too much, I think.)
>It worked so well - with various transistors and coils that I was
>further intrigued to see how much I could increase the resistor value
>- hence the 20k trimpot. (The 22nF also got over the problem of the
>oscillator failing when I tried to add an ammeter in the battery
>circuit).
>
>I found it would continue to oscillate right up to 20k ohms and this
>also had the effect of reducing the supply current (osc only) from
>90mA to 800microamps - very important if using batteries."
The amount of delivered power to the LEDs goes down. In other words,
at some point it is a bad idea because of the transistor's increased
dissipation.
You made me curious enough to try a simulation with 200uH on each half
of the transformer, a 2N2222 BJT, a 1N5819 freewheeling diode, 10uF
output cap (across the 6 LEDs with Vfwd=3, Ron=12.6), and 3 fresh
batteries. The Spice results look like:
Battery C BJT LEDS
436mW 0pF 28mW 19.9mA
428mW 22pF 16mW 20.0mA
425mW 220pF 19mW 19.7mA
392mW 2.2nF 50mW 16.5mA
173mW 22nF 142mW 0.0mA
So battery power does go down. But BJT power goes up and they
eventually cross and the whole endevour is a waste. In other words,
the BJT starts eating up all the power and eventually eats up
everything there is.
If you look at the above closely, there does seem to be a small range
of values that work well. Say up to a few hundred pF in the example
case I let Spice run on. Everything depends, though. So mileage may
vary.
Jon
John Larkin
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
John
default
Core material and number of turns will affect the frequency of
oscillation and that can affect power dissipation. As frequency goes
up (fewer turns or less permeable core) the transistor may get warmer
due to core losses and storage time (staying on when it should be off
for a few nano to microseconds - doing it more often).
Then there's D8. That should be a high speed diode not a run of the
mill 60 Hz rectifier. A cheap 1N4148 or 1N914 switching diode will be
good enough for LEDs - they switch very fast and can tolerate 100
milliamps and up to 100 Volts (manufacture specs may vary, but that's
a general idea of what they are rated at).
The cap across the resistor is, in my opinion, a good idea as long as
you don't get too carried away. (and .022 uf sounds a little high -
but a lot depends on how fast it is running too) It serves to drive
the transistor harder when the initial switching occurs and that is a
good thing. The base emitter dissipation will rise - but the emitter
- collector voltage should go down (and that usually accounts for more
dissipation since the current is much higher there). You want it to
switch fast and not dilly-dally in the linear region. (all other
things being equal . . . )
--
Jon Kirwan
Would you care to provide some sample values and analyze that circuit
for us? 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.
Jon
bw
"fungus" <opengl...@artlum.com> wrote in message
news:ddeb7f42-6adb-4135...@e27g2000yqm.googlegroups.com...
>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
>
Start with the original circuit, and get it to work on one battery.
http://cappels.org/dproj/ledpage/leddrv.htm#Rusty_Nail_Night_Light
then go from there to what you want to do.
John Larkin
No, too much work.
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.) 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.
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.
John
Jon Kirwan
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
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?
>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.
>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.
...
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?
Jon
fungus
>
> ...a 'speed-up' capacitor.
> ...
> You made me curious enough to try a simulation with 200uH on each half
> of the transformer, a 2N2222 BJT, a 1N5819 freewheeling diode, 10uF
> output cap (across the 6 LEDs with Vfwd=3, Ron=12.6), and 3 fresh
> batteries. The Spice results look like:
>
> Battery C BJT LEDS
> 436mW 0pF 28mW 19.9mA
> 428mW 22pF 16mW 20.0mA
> 425mW 220pF 19mW 19.7mA
> 392mW 2.2nF 50mW 16.5mA
> 173mW 22nF 142mW 0.0mA
>
> So battery power does go down. But BJT power goes up and they
> eventually cross and the whole endevour is a waste. In other words,
> the BJT starts eating up all the power and eventually eats up
> everything there is.
>
Ok, that's scary. I'll stick to playing with the number
of turns of wire for a while. I'm going to try a ferrite rod
instead of a ring because they're much quicker to add
and remove wire (and you can even do it while the
circuit is running!)
fungus
<pieterkraltlaatdit...@enditookhccnet.nl> wrote:
> The diode you added makes things worse....
>
Wasn't me that added it...
> So you need to move the diode. Place it between R1 and the base of the
> transistor.
I'll try that, thanks.
> Next step is adding more turns to the coil as others stated already. Guess
> you will need 1.5-2 times the original number of turns.
>
Yes, this seems to be the place to experiment.
fungus
> keep winding in the same direction it will be phased right.
>
I assume the winding after the tap has to go over the top of the
previous winding, right?
default
No. They can be side by side. There may be some slight advantage to
covering the whole toroid (ferrite bead) core (if that's what you
have) with the collector winding to keep magnetic leakage low.
I'd put the collector winding closer to the core or just wind them
side by side.
I like to wind bifilar when winding two windings with the same turns
count on a core. It is slightly more hassle to phase correctly, but
if my turns count is off, it will be off proportionately on both
windings (which is important when the winding has to be balanced -
which is not the case here)
--
fungus
>
> Start with the original circuit, and get it to work on one battery.
>
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).
The solution seems to be to raise the input voltage
so that's what I'm trying to get working. Unfortunately
I never studied electronics beyond what they told me
in high-school physics class so I'm at a disadvantage.
fungus
>
> BJT temperature is related (obviously) to its power dissipation. That
> power dissipation comes from a variety of possible corners from my
> hobby viewpoint: (1) The transistor has been damaged (diode put in
> later, after it was already ruined perhaps?) and isn't operating well
> anymore;
Fresh transistors are quite frequent at the moment...
> or, (2) the base-emitter junction current;
I've moved the resistor in between the inductor and transistor
to try to limit this (ie. R1 is now exactly as shown in the
scematic). Doesn't seem to help much.
> (3) collector-emitter current times collector-emitter voltage;
I'm suspecting this at the moment.
> (4) frequency of operation is too high for the reverse transit
> time of the BJT.
>
Not sure I understand that.
> So wind more turns and get the frequency near or under 100kHz, or so,
> where the reverse transit will be only be a few percent or so and
> won't be wasting a lot of power.
>
ie. Lower frequencies mean the transistor will be switched on for less
percentage of the time and more electrons will go through the load
instead of being dumped to ground via the transistor.
(Yeah, I know - http://xkcd.com/567/ ).
fungus
> On Thu, 23 Jul 2009 14:30:02 -0700 (PDT), fungus
>
> >On Jul 23, 2:03 pm, default <defa...@defaulter.net> wrote:
>
> >> The coil can be wound with a tap or separately. If you use a tap and
> >> keep winding in the same direction it will be phased right.
>
> >I assume the winding after the tap has to go over the top of the
> >previous winding, right?
>
> No. They can be side by side.
>
I'm not sure I'm being clear (my bad), I mean do
they have to be like this:
ABABABABABABAB
=================
| BABABABABABABA
tap
or can they be:
AAAAAAA BBBBBBBB
==================
AAAAAAA | BBBBBBBB
tap
(view with monospace font...)
On the rusty nail he uses a twisted pair
John Larkin
<jo...@infinitefactors.org> wrote:
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."
>
>>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.
The MIT RadLab books are full of blocking oscillator theory and
circuits, especially vol 19. Tube radars were full of them, as
oscillators, comparators, pulse regenerators, and frequency dividers.
Some texts referred to rf squegging circuits as blocking oscillators.
John
David Eather
What voltage are you running the joule thief on?
default
<opengl...@artlum.com> wrote:
>On Jul 23, 11:54锟絧m, default <defa...@defaulter.net> wrote:
>> On Thu, 23 Jul 2009 14:30:02 -0700 (PDT), fungus
>>
>> <openglMYSO...@artlum.com> wrote:
>> >On Jul 23, 2:03锟絧m, default <defa...@defaulter.net> wrote:
>>
>> >> The coil can be wound with a tap or separately. 锟絀f you use a tap and
>> >> keep winding in the same direction it will be phased right.
>>
>> >I assume the winding after the tap has to go over the top of the
>> >previous winding, right?
>>
>> No. 锟絋hey can be side by side.
>>
>
>I'm not sure I'm being clear (my bad), I mean do
>they have to be like this:
>
> ABABABABABABAB
>=================
> | BABABABABABABA
>tap
>
>or can they be:
>
>AAAAAAA BBBBBBBB
>==================
>AAAAAAA | BBBBBBBB
> tap
>
>(view with monospace font...)
>
>On the rusty nail he uses a twisted pair
Either way. The top example might produce slightly better results
but I couldn't say for sure. As the winding spreads out, inductance
per turn goes down and resistance goes up (all other things being
equal again) so the Q might be slightly lower, for any benefit gained
by keeping the leakage lower.
No advantage, I can think of, to using a twisted pair.
You mentioned a ferrite rod or bar? Leave a little bit of ferrite rod
with no coil over the ends (about 1/2 -2 X the diameter of the rod)
don't wind right to the ends.
When I was a kid we'd use center tapped filament transformers for
blocking oscillators, and then use them to step up voltage to get
500-1000 volts from a battery supply. I tried that with a tapped
modem transformer recently and even though the primary to secondary
ratio was only 1:2 the output voltage was ~350 volts with 12V input
(because the wave form contains a large spike).
--
whit3rd
> 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
That kind of oscillator (blocking oscillator) depends on saturation
of the core, which is quite difficult to engineer (and somewhat
energy-lossy). Unless you really need a minimum-parts solution,
it's easier to run a '555 oscillator into an autotransformer to
boost voltage. Remember to use DC blocking capacitors,
and you can get nearly half a watt from such a circuit.
John Fields
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
---
Translation:
I don't have a clue.
JF
Jon Kirwan
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
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.)
>The MIT RadLab books are full of blocking oscillator theory and
>circuits, especially vol 19. Tube radars were full of them, as
>oscillators, comparators, pulse regenerators, and frequency dividers.
I think someone posted some of that, some time back, in
sci.electronics.design. I'll see if I can track any of that down.
sadly, other than that possibility, I don't have ready access.
>Some texts referred to rf squegging circuits as blocking oscillators.
I'll look to see if I can find a lucid description.
Thanks,
Jon
David Eather
Also what is the resistance value of R1?
In the interim try this:
1 - increase R1 by a factor of 10 and next by a factor of 100, does that
reduce the heating of the 2n2222? (if you are over-driving the base of
the transistor and the inductance of the transformer is too small this
would cause the problem you are seeing - I think JK is also thinking
along these lines)
2 - increase the number of turns in both l1 and l2 by the same ratio e.g
. double the number of turns on both or triple the number of turns on
both. This should make the transformer more efficient (and you probably
want to do that anyway)
fungus
>
> What voltage are you running the joule thief on?
Right now? I just measured 3.8V.
David Eather
> fungus wrote:
>> On Jul 23, 1:52 pm, David Eather <eat...@tpg.com.au> wrote:
>>> fungus wrote:
>>>> Any ideas?
>>> Yes. Figure out what you want to do and state it explicitly and exactly.
>>> Then work to that goal in steps you understand.
>>
>> a) I want to light up some LEDs (eg. six of them) using batteries, eg.
>> three AAAs. Circuit is decorative and has to be small because I want
>> to hide it.
>>
>> b) I want them to be as bright as possible - the full 20mA or as close
>> to it as I can get.
>>
>> 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.
>>
>> I've figured out that a Joule Thief is much closer to these
>> characteristics
>> than a simple resistor circuit doesn't. See the graph I plotted here:
>> http://www.artlum.com/jt/jt_vs_res.gif
>>
>> But ... at the moment it's eating up transistors.
>
> Also what is the resistance value of R1?
>
> In the interim try this:
>
forget about point 1 - I made a wrong assumption about your circuit. I
assumed you had a resistor connected from the base of the transistor to
the junction l1 and d7. If you want to try it put a 1k resistor in -you
may need to change this value upwards if the transistor still runs too
hot or downwards if the circuit stops working.
fungus
> fungus wrote:
> > On Jul 23, 1:52 pm, David Eather <eat...@tpg.com.au> wrote:
> >> fungus wrote:
> >>> Any ideas?
> >> Yes. Figure out what you want to do and state it explicitly and exactly.
> >> Then work to that goal in steps you understand.
>
> > a) I want to light up some LEDs (eg. six of them) using batteries, eg.
> > three AAAs. Circuit is decorative and has to be small because I want
> > to hide it.
>
> > b) I want them to be as bright as possible - the full 20mA or as close
> > to it as I can get.
>
> > 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.
>
> > I've figured out that a Joule Thief is much closer to these
> > characteristics
> > than a simple resistor circuit doesn't. See the graph I plotted here:
> >http://www.artlum.com/jt/jt_vs_res.gif
>
> > But ... at the moment it's eating up transistors.
>
> Also what is the resistance value of R1?
>
1k
> In the interim try this:
>
> 1 - increase R1 by a factor of 10 ... does that
> reduce the heating of the 2n2222?
>
Yes, but the current going through the LEDs
dropped from 18mA to 5mA.
John Larkin
<jo...@infinitefactors.org> wrote:
In the OP's ascii-art circuit, R1 determines ON base current (and
perhaps ON time, if the inductor doesn't saturate first) and also sets
OFF time, as part of the L/R decay. It may be hard to pick one value
that does both right, namely produces an efficient duty cycle, as
witnessed by the many blown up transistors.
The big advantage of the circuit I posted is that rep-rate can be set
independent of pulse width... two knobs to turn. That allows low duty
cycles which won't fry transistors. And brightness control, if you
want it.
>
>>The MIT RadLab books are full of blocking oscillator theory and
>>circuits, especially vol 19. Tube radars were full of them, as
>>oscillators, comparators, pulse regenerators, and frequency dividers.
>
>I think someone posted some of that, some time back, in
>sci.electronics.design. I'll see if I can track any of that down.
>sadly, other than that possibility, I don't have ready access.
>
>>Some texts referred to rf squegging circuits as blocking oscillators.
>
>I'll look to see if I can find a lucid description.
Terman's "Radio Engineering" texts refer to squeggers as blocking
oscillators. But Terman was a little weird.
Millman&Taub's classic "Pulse and Digital Circuits" has a whole
chapter "Pulse Transformers and Blocking Oscillators", with a bunch of
analysis. All tubes.
John
Jon Kirwan
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
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.
>It may be hard to pick one value
>that does both right, namely produces an efficient duty cycle, as
>witnessed by the many blown up transistors.
Well, until I gather your point about the OFF time's L/R, I have to
withhold further comment.
>The big advantage of the circuit I posted is that rep-rate can be set
>independent of pulse width... two knobs to turn. That allows low duty
>cycles which won't fry transistors. And brightness control, if you
>want it.
I need to first fathom your L/R point and then I need to spend more
time with the suggested circuit you gave before I can agree. I
apologize for my ignorance on this, but that's where I'm at right now.
>>>The MIT RadLab books are full of blocking oscillator theory and
>>>circuits, especially vol 19. Tube radars were full of them, as
>>>oscillators, comparators, pulse regenerators, and frequency dividers.
>>
>>I think someone posted some of that, some time back, in
>>sci.electronics.design. I'll see if I can track any of that down.
>>sadly, other than that possibility, I don't have ready access.
>>
>>>Some texts referred to rf squegging circuits as blocking oscillators.
>>
>>I'll look to see if I can find a lucid description.
>
>Terman's "Radio Engineering" texts refer to squeggers as blocking
>oscillators. But Terman was a little weird.
>
>Millman&Taub's classic "Pulse and Digital Circuits" has a whole
>chapter "Pulse Transformers and Blocking Oscillators", with a bunch of
>analysis. All tubes.
I'm modestly familiar with vacuum tubes, load lines, grid leak
resistors (what a pain I had with those), etc. So I may somehow hope
to be able to follow along.
Jon
David Eather
Iff (if and only if) you are in the mood for experimenting, more turns
on L2 might increase this current through the LED while R1 is 10k.
Also the range of currents 18ma to 5ma (4:1) vs the range of resistance
1k to 10k (1:10) suggests that some value between 1k and 10k will reduce
power consumption of the transistor with a much smaller decrease in
current to the LED. Perhaps there is a suitable compromise.
>
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.)
David Eather
> 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. 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)
If you really wanted to nail it, triple the turns on L1 and put 6 times
the number of turns you already have on L2 and change R1, but just L2
and R1 will get you closer to where you want to go.
greg
> The problem is that my transistors keep on overheating and dying.
> Why should this be?
My first suspicion would be switching losses in the
transistor.
What does the voltage waveform across the emitter and
collector of the transistor look like? Does it have
nice sharp vertical edges?
If not, then you need to lower the operating frequency
until the switching time is negligible compared to the
on and off times. Increasing the number of turns on
the inductor should do that.
> I'm using a 2N2222 in metal can (as shown here
> http://en.wikipedia.org/wiki/2N2222 ). These can switch at hundreds
> of megahertz
Like Jon said, f_T is NOT the relevant parameter here,
it's only about amplifying small signals. You can't
expect the transistor to switch between full-on and full-off
at anywhere near that speed.
> I measured the current at point X and it seems high - over 100mA.
From the Wikipedia article, the 2N2222 should be able to
handle that, so I don't think it's a concern, provided the
transistor is switching efficiently.
> I also tried a honking big "high speed switching" transistor pulled
> out of a PSU but it made the LEDs go very dim.
If it's designed to handle very high currents, it may not
work very efficiently at the modest power levels you're
concerned with here.
From what Jon said, it sounds like the 2N2222 should be
capable of the job as long as you don't try to run it
at more than 100kHz or so.
You really need to look at things with a scope and find
out what's going on. Just fiddling randomly with the
parameters is likely to lead to a lot of frustration and
dead transistors.
--
Greg
greg
> I don't see
> how the number of turns would be related to the transistor
> temperature.
Increasing the number of turns increases the inductance,
so the current rises at a lower rate when the transistor
is on. Assuming the base current remains the same (which
it will be if you increase the number of base winding
turns by the same amount and keep the base resistor the
same) then more time will be taken for the collector
current to reach its maximum value and trigger a turn-off.
This means that the operating frequency will be lower,
and the switching time of the transistor will be a
smaller fraction of the on-time, leading to less power
dissipation and a cooler transistor.
That's the theory, anyway.
--
Greg
greg
> ie. Lower frequencies mean the transistor will be switched on for less
> percentage of the time and more electrons will go through the load
> instead of being dumped to ground via the transistor.
No, the point is that the transistor should be either
fully on or fully off for as great a proportion of
time as possible.
The time that the transistor takes to switch on or
off (given a sharp driving waveform to the base) is
fairly constant and depends on the characteristics
of the transistor.
So increasing the duration of the on-time makes the
proportion of in-between time smaller.
--
Greg
greg
> In general, "on" pulse width is set by the volt-second saturation of
> the inductor
In the circuit you gave, it seems to me that if the
inductor saturates, rather than terminating the
on-pulse, the collector current is just going to
shoot sky-high.
--
Greg
greg
> 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).
>
> The solution seems to be to raise the input voltage
That's not the only way -- you can also raise
the peak collector current, so that you draw
power from the battery at a low voltage and
high current, and deliver it to the LEDs at a
higher voltage and lower current.
To get a higher peak collector current, you need
to increase the base current. You can do that
either by lowering the base resistor, or increasing
the number of base winding turns relative to the
collector winding.
Be careful, though -- keep an eye on the average
LED current and make sure it doesn't go over
20mA.
Also keep the maximum current rating of the
transistor in mind. If the voltage drop across
the LEDs is N times the battery voltage, and the
average LED current is 20mA, then the average
transistor current will be N times 20mA.
However, the transistor will be on for only
1/N of the time, so the *peak* transistor current
will be N *squared* times 20mA.
--
Greg
greg
> 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.
--
Greg
greg
> 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?
I can't say for sure, but I'd be surprised if it wasn't.
Transistor beta is a spectacularly crappy parameter to
rely on -- it would be quite hard to do any worse!
Also I know for a fact that pulse-generating circuits
using saturating inductors were quite common at one
time, so they must have been reasonably predictable.
BTW, I've had another thought about the use of saturation
in a Joule Thief type circuit. Using it for pulse
generation is one thing, but in that case you're only
interested in the output voltage. The inductor doesn't
have to store much energy, so you can use a high
permeability material that is easy to saturate and
has a fairly sharp saturation characteristic.
But in our case we need energy storage, so we need
lower permeability. That makes the core harder to
saturate, and may also make it saturate more "softly"
with a less sharply-defined saturation point. So
it might not work so predictably after all.
But that's pretty much guesswork on my part at the
moment.
--
Greg
ehsjr
> 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 )
>
>
> I measured the current at point X and it seems high - over 100mA.
> problem
> it seems wasteful. I tried putting in a resistor there but the circuit
> shuts down.
> .
> out of a PSU but it made the LEDs go very dim.
>
Yes. You are likely to be disappointed by the joule thief,
for what you are trying to do, unless you just want to experiment
with it. You are also likely to be confused by all the hand
waving and arguing going on in replies in this thread.
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?"
_Limited run time_
You want to run 6 LEDs at 20 mA. Let's assume that each LED has a
1.8 volt Vf. That means you need to boost your supply to 10.8 volts.
The power needed by the LEDs is 10.8 * .02 or 216 miliwatts. A
typical new AAA cell is rated at 1.5 volts, 1250 mAh. To produce
10.8 volts at 20 mA from that cell, you must draw 144 mA assuming a
perfect conversion circuit. (Your joule thief is far from perfect.)
If you use three cells, you can get (mathematically) only 26 hours
run time.
Solution: larger batteries and/or more of them. To overcome losses
in the joule thief, use a better circuit, but you are in all cases
limited by the power available from the batteries vs the power used
by the LEDs.
_LED brightness will decrease over time_
The joule thief will not deliver constant current to the LEDs, so
brightness will decrease as battery voltage drops.
Solution: a better (constant current or PWM) circuit.
_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.
Solution: mains power. Mains power solves the other issues,
as well.
If you _must_ use battery power, there are one chip solutions
better than the joule thief.
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.
Ed
John Larkin
<jo...@infinitefactors.org> wrote:
No need to get snippy about things.
When the transistor turns off, the flyback voltage gets dumped into
the LEDs, guaranteeing part of the OFF interval. When the transformer
runs out of stored energy, Il=0 and the secondary essentially
disappears from the circuit. Now we have the feedback winding in
series with R into the base. 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.
Adding a cap across R enforces a more predictable OFF time. Adding a
parallel RC in series with R is even better, and is a variant on the
circuit I posted. That also allows independent base drive and OFF time
tuning.
I wasn't kidding when I said blocking oscillators are complex.
Impressive for so simple a circuit.
>
>>It may be hard to pick one value
>>that does both right, namely produces an efficient duty cycle, as
>>witnessed by the many blown up transistors.
>
>Well, until I gather your point about the OFF time's L/R, I have to
>withhold further comment.
>
>>The big advantage of the circuit I posted is that rep-rate can be set
>>independent of pulse width... two knobs to turn. That allows low duty
>>cycles which won't fry transistors. And brightness control, if you
>>want it.
>
>I need to first fathom your L/R point and then I need to spend more
>time with the suggested circuit you gave before I can agree.
I don't see why. The circuit I posted enforces OFF time deliberately
and un-complicates things.
John
bw
"ehsjr" <eh...@NOSPAMverizon.net> wrote in message
news:1ojam.1040$MA3...@nwrddc02.gnilink.net...
My experiments show that the efficiency is dependent on minor adjustments to
the windings due to ugly waveform. I'll check my notes, about 50 percent was
the best I found, at 50 to 100 kHz, using a ferrite core, with 10 percent
duty.
John Larkin
wrote:
Yes, very briefly. The resulting rise in collector voltage is coupled
into the base circuit such as to turn the transistor off, in a
positive feedback loop (just as the turn-on was positive feedback.)
The turnoff snap is usually very fast, or at least should be if
everything is done right. A capacitor across the resistor helps here.
The cycle can also terminate if the transistor runs out of beta
(against the current ramping up in the inductor) or if the capacitor
runs out of charge driving the base. So the inductor need not
saturate, although that is the most efficient way to use the
magnetics.
I think the origial joule-thief (or whatever) worked well because the
1.5 volt battery was about half of Vbe_on, which made the duty cycle
work about right, ballpark 50%. Higher battery voltages push the ON
duty cycle up and toast transistors. Something like that.
John
Jon Kirwan
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
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
Jon Kirwan
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
John Larkin
<jo...@infinitefactors.org> wrote:
The average voltage across the secondary must be zero. Since it's
positive when it's turning the base on, it must be negative for some
time so that it averages zero. So immediately after the transistor
turns off (if it DOES turn off) the negative transformer voltage must
exceed the battery voltage-Vbe.
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.
There are too many possibilities here, too many branches in the
probability tree.
Sometimes it's much easier to design a circuit than it is to analyze
it. This is such a case.
What I would do is pick a candidate transformer and transistor and
connect it to the battery and the LEDs and drive the transistor base
from a pulse generator. Find a frequency and duty cycle and base
current that work well and efficiently. Then design the feedback
circuit to hit those values.
I always prefer to design circuits than to analyze them.
John
fungus
>
> 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
> 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.
Jon Kirwan
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
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. [You wrote,
"Too many conflicting parameters depend on the value of R1," following
it by, "R1 determines ON base current (and perhaps ON time, if the
inductor doesn't saturate first) and also sets OFF time, as part of
the L/R decay."] This is, as I try and understand your points, why
you felt strongly enough in your first posting on this topic to say
that there are "too many conflicting parameters depend on the value of
R1 (ed., to make this a 'good circuit.')"
However, it still appears to me that at least the BJT OFF time does
not depend on it -- it's ignorable.
Jon
fungus
>
> 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).
Jon Kirwan
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
Jon Kirwan
<jo...@infinitefactors.org> wrote:
>The voltage applied across L2 when the BJT is ON is presented
... by L1 ...
greg
> 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
David Eather
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
> 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.
David Eather
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.
David Eather
> 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?
Jon Kirwan
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
fungus
> To get a higher peak collector current, you need
> to increase the base current. You can do that
>
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
> 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.
fungus
>
> 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.
David Eather
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
>
> 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.
David Eather
> 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
John Larkin
<jo...@infinitefactors.org> 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.
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
fungus
> > Aside: Does wire thickness make any difference?
>
> 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
<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.
John Larkin
Things in series can be swapped without effect, usually.
John
fungus
<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.
ehsjr
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
ehsjr
> 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
fungus
> 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)
Jon Kirwan
wrote:
>Jon Kirwan wrote:
>> On Fri, 24 Jul 2009 14:17:33 GMT, ehsjr <eh...@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
greg
> 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
greg
> 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
greg
> 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.
I don't see how that can happen. The off-time
will be just what is needed to reduce the current
in the inductor back to zero.
So it doesn't matter that you can't control the
on and off times separately, because you don't
need to, or even want to. The duty cycle takes
care of itself.
--
Greg
greg
> 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
greg
> Does it matter where R1 is, relative to L1?
Shouldn't make any difference that I can think of.
The only way it could matter would be through
changes in stray inductances and capacitances,
and that's not likely to be a problem at these
frequencies.
--
Greg
greg
> 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
greg
> 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
Jon Kirwan
wrote:
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
fungus
>
> 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... :-(
fungus
>
> 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.
fungus
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?
fungus
>
> 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.)
>
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.
ehsjr
> On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
> <jjla...@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
John Larkin
wrote:
If you get the turns ratio right, it does. If the base winding doesn't
produce enough turnoff voltage (ie its negative peak doesn't exceed
Vbatt) the transistor doesn't totally turn off and efficiency will be
rotten. If the turns ratio is too high, the transistor will zener and
beta will degrade over time. If it's just right, R1 will see a lot of
voltage when it's pumping the base of the transistor (Vbatt +
Vtransformer - 0.7) and will waste a lot of power. Terrible circuit,
especially as the battery goes past a single cell. Leakage inductance
and ringing and variation in battery voltage and various boost ratios
will make it worse.
One improvement is to drive one end of the feedback winding from a
voltage divider. That way, you can use a much higher turns ratio,
reduce the feedback winding voltage, use a much lower R, and get most
of the base current from ground, not from V+. There's a diode trick
that's even better.
John
default
That looks way too "clean." What is the bandwidth of your scope?
Blocking oscillators are notorious for producing overshoot,
undershoot, and ringing.
If the transistor heats that's where most of the waste power is going.
You can't always count on a digital meter to give accurate readings in
circuits where there are non-sinusoidal wave forms. Even when
measuring what are assumed to be DC readings.
--
default
<opengl...@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.
--
ehsjr
<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
ehsjr
> 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
fungus
> On Sat, 25 Jul 2009 07:48:45 -0700 (PDT), fungus
>
> >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.
>
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.
Jon Kirwan
wrote:
>Jon Kirwan wrote:
>> On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
>> <jjla...@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.
In the sense that Ic_peak is higher. The off time equation is:
t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)
I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.
So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:
N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)
And then using Lprimary by selecting a desired 'f':
Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)
Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:
Rbase = 2*beta*(Vin-Vbeon)/Ic_peak
In other words, the ratio of OFF time to OFF+ON time is already cast
in concrete by the desired output voltage and the available input
voltage. That's a done deal. Ic_peak is then set by the desired
Iout, taking into account that required ratio. Lprimary is used to
then tweak in the desired frequency of operation (wind or unwind, as
needed.) And then Rbase is designed to compensate for the BJT's
operating beta and the design Ic_peak (tweak up or down, as needed.)
The frequency of operation, within practical bounds set by volt-secs
and reverse transit times to name two, has no apparent impact on power
out. Ic_peak, itself a function of both Vout and Iout, is established
by that. So the frequency (if you only use Lprimary windings to
adjust it) can be moved around independently without changing Ic_peak,
as Rbase has already fixed that in place.
I was so focused on that aspect of it that I didn't realize John was
talking at cross purposes to where my mind was mired.
>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.
I will try and get some time to do that, today. Actually, I just
wound up some various ferrite cores, as well (7 of them.) What I
don't have handy
Jon
Jon Kirwan
<jo...@infinitefactors.org> wrote:
>What I don't have handy
...is a way to measure inductance. (Never bought such a tool.) But
actually, I can approximate that by observing frequency of operation.
Jon