Joule Thief - still not working....

[greg]

fungus wrote:

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]

John Larkin wrote:

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]

fungus wrote:

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]

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.

--
Greg

 

[greg]

Jon Kirwan 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?

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]

fungus wrote:

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

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

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

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

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

Any ideas?

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]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Any ideas?

That's a horrible circuit. Too many conflicting parameters depend on
the value of R1. A proper blocking oscillator uses an RC time constant
to set the rep rate, and a separate resistor to limit the base
current.

ftp://jjlarkin.lmi.net/BlockOsc.JPG

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

No, too much work.

Hmm.

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

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

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

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

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

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

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

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

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

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


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

Something like that.

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

Yes. I gather.

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

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

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

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

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


...

Since you write, "That's a horrible circuit. Too many conflicting
parameters depend on the value of R1. A proper blocking oscillator
uses an RC time constant to set the rep rate, and a separate resistor
to limit the base current," shouldn't it be the case that you can tell
me how to compute the frequency with ease? Isn't that the entire
point of saying all that? Or did I miss your point, here?

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

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

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

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

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

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

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

No need to get snippy about things.

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" <ehsjr@NOSPAMverizon.net> wrote in message
news:1ojam.1040$MA3.967@nwrddc02.gnilink.net...

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

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

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

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

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

Any ideas?

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

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]

On Fri, 24 Jul 2009 23:05:56 +1200, greg <greg@cosc.canterbury.ac.nz>
wrote:

John Larkin wrote:

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.

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]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Any ideas?

That's a horrible circuit. Too many conflicting parameters depend on
the value of R1. A proper blocking oscillator uses an RC time constant
to set the rep rate, and a separate resistor to limit the base
current.

ftp://jjlarkin.lmi.net/BlockOsc.JPG

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

No, too much work.

Hmm.

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

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

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

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

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

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

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

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

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

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


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

Something like that.

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

Yes. I gather.

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

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

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

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

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


...

Since you write, "That's a horrible circuit. Too many conflicting
parameters depend on the value of R1. A proper blocking oscillator
uses an RC time constant to set the rep rate, and a separate resistor
to limit the base current," shouldn't it be the case that you can tell
me how to compute the frequency with ease? Isn't that the entire
point of saying all that? Or did I miss your point, here?

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

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

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

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

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

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

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

No need to get snippy about things.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Jon

 

[Jon Kirwan]

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

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

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

Jon

 

[John Larkin]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Any ideas?

That's a horrible circuit. Too many conflicting parameters depend on
the value of R1. A proper blocking oscillator uses an RC time constant
to set the rep rate, and a separate resistor to limit the base
current.

ftp://jjlarkin.lmi.net/BlockOsc.JPG

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

No, too much work.

Hmm.

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

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

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

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

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

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

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

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

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

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


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

Something like that.

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

Yes. I gather.

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

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

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

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

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


...

Since you write, "That's a horrible circuit. Too many conflicting
parameters depend on the value of R1. A proper blocking oscillator
uses an RC time constant to set the rep rate, and a separate resistor
to limit the base current," shouldn't it be the case that you can tell
me how to compute the frequency with ease? Isn't that the entire
point of saying all that? Or did I miss your point, here?

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

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

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

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

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

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

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

No need to get snippy about things.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Jon

The average voltage across the secondary must be zero. 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

 

[Jon Kirwan]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Any ideas?

That's a horrible circuit. Too many conflicting parameters depend on
the value of R1. A proper blocking oscillator uses an RC time constant
to set the rep rate, and a separate resistor to limit the base
current.

ftp://jjlarkin.lmi.net/BlockOsc.JPG

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

No, too much work.

Hmm.

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

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

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

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

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

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

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

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

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

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


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

Something like that.

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

Yes. I gather.

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

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

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

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

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


...

Since you write, "That's a horrible circuit. Too many conflicting
parameters depend on the value of R1. A proper blocking oscillator
uses an RC time constant to set the rep rate, and a separate resistor
to limit the base current," shouldn't it be the case that you can tell
me how to compute the frequency with ease? Isn't that the entire
point of saying all that? Or did I miss your point, here?

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

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

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

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

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

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

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

No need to get snippy about things.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Jon

The average voltage across the secondary must be zero.

We are talking about volt-seconds, yes?

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

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

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

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

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

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

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

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

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

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

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

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

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

At least, that's how I see it.

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

 

[Jon Kirwan]

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

fungus wrote:
On Jul 23, 10:18 pm, "bw" <bweg...@hotmail.com> wrote:
Start with the original circuit, and get it to work on one battery.

then go from there to what you want to do.

That's what I'm doing...

The original circuit lights up a LED but the current
is very low - about 5mA.

To drive six LEDs at 20mA with one battery you'd
have to get the frequency up into the mHz (which
isn't going to happen).

No, the frequency thing is not correct.

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

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

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

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

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

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

Jon

 

[Jon Kirwan]

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

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

.... by L1 ...

 

[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.

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)

 

[greg]

In a fit of temporary insanity, I wrote:

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

Scrub that, it's completely wrong.

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

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

I hope I got it right this time!

--
Greg

 

[David Eather]

David Eather wrote:

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


OK, this is the next thing to try.

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

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

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

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

 

[David Eather]

Jon Kirwan wrote:

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

fungus wrote:
On Jul 23, 10:18 pm, "bw" <bweg...@hotmail.com> wrote:
Start with the original circuit, and get it to work on one battery.

then go from there to what you want to do.
That's what I'm doing...

The original circuit lights up a LED but the current
is very low - about 5mA.

To drive six LEDs at 20mA with one battery you'd
have to get the frequency up into the mHz (which
isn't going to happen).
No, the frequency thing is not correct.

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

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

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

Oh, crap! You are exactly right of course!

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

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

Of course, I'm just a hobbyist.

I'm just wrong.

So that's my view and I'm sticking to

it.

Jon

 

[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"


_Cost_

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

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

Solution: mains power. Mains power solves the other issues,
as well.


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

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

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


It's a "fun" circuit, yes.