fungus wrote:> On Jul 25, 3:14 am, David Eather <eat...@tpg.com.au> wrote: >>> Aside: Does wire thickness make any difference? >> It won't make much difference at these low power levels. The main >> difference in this case is how many turns you can wrap on the ferrite bead. >> > > So the size of the magnetic field is purely down to current? > > ================== > > I've just another experiment and I've got some more fuel for > the raging debate.... > > I just did a comparison between a ferrite bead and an iron > powder bead. With the ferrite bead the current through the > LEDs drops off as the number of turns of wire increases. > With seven turns I get 12mA ... with 30 turns I only get 1mA. > > For comparison I just tried one of my iron powder rings > and I got completely the *opposite* effect - more turns > gave more output current. At 15 turns I was getting 6mA, > at 30 turns I was getting 12mA. > > I was doing this with a thicker piece of wire I pulled from a > transformer in the PSU* so luckily for the transistor I couldn't > physically get more than 30 turns on the ring. The trend was > very clear though - every turn I added produced a measurable > increase in LED current. > > nb. The transistor was getting hotter with every extra turn > despite the oscillation frequency going down, also the opposite > of what happens with ferrite. It seems that transistor temperature > is more strongly related to output current than frequency. > > Assuming I get the same result with thin wire this seems like > a really easy way to get any desired output current - just keep > adding turns until you get there. > > This also assuming we can solve the transistor heating problem. > What would happen if I put two transistors in parallel? Would the > load be halved or would differences in manufacturing tolerance > mean one of them took most of the load?It shouldn't be necessary the 2n22222 is quite a robust transistor> > > 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.... >
Joule Thief - still not working....
Started by ●July 23, 2009
Reply by ●July 25, 20092009-07-25
Reply by ●July 25, 20092009-07-25
greg wrote:> David Eather wrote: > >> In your circuit you could also try removing C1 - the LED's will light >> just fine with the JT pulses and your eye will make the circuit appear >> brighter for the same current > > No, it won't. An LED run continuously at 20mA will always > look brighter than one pulsed at 20mA.There is no current limiting in the LED circuit, it was reasonable to expect current peaks higher than 20ma. In that case, while the average current was 20ma, the higher current peaks might mean the output would be brighter without C1. It was worth a try. snip
Reply by ●July 25, 20092009-07-25
On Sat, 25 Jul 2009 20:36:25 +1200, greg <greg@cosc.canterbury.ac.nz> wrote:>David Eather wrote: > >> In your circuit you could also try removing C1 - the LED's will light >> just fine with the JT pulses and your eye will make the circuit appear >> brighter for the same current > >No, it won't. An LED run continuously at 20mA will always >look brighter than one pulsed at 20mA. > >Pulsing LEDs only helps if the average current available >from your supply is much less than the rated current of >the LEDs. E.g. if you can only manage 2mA continuously, >then 20mA at 10% duty cycle will give slightly more >light overall, due to nonlinearity of the LED's light >vs. current curve. ><snip>HP put out an extensive treatise on LEDs and includes a page or two on this point, in particular. It's called "Optoelectronics: Fiber-Optics Applications Manual," 2nd edition, 1981. It was hard-bound and cost $27.50 when new or else was given away by HP, if they liked you. They provide a curve on page 5.20 (for high efficiency red LEDs) that shows a step climb from 1 (100%) at 5mA peak current (which I assume is their benchmark continuous current, since the curve doesn't go to lower peak currents) up towards about 1.6 where it flattens out and doesn't change much, at about 40mA peak current. They hold the time averaged current fixed for the curve, so at 40mA peak current this implies that the duty cycle is 12.5%. (They also take a moment to say that all this only works before the junction reaches saturation. After that, additional current will obviously reduce this efficiency ratio.) This suggests to me that there is a limit to the advantage in increasing pulsed currents. At some point, gains aren't to be had. But also, HP's curve of efficiency is focused on current. It doesn't take respect of the fact that the voltage also increases, as well. So although efficiency vs pulsed current _in terms of current_ does rise up to a point, _in terms of power_ in the device the curve would likely be different as the drive voltage increases with higher pulsed currents. I'm betting that this creates a kind of parabolic shape with a peak, once voltage is added back into the mix. Another aspect though is that (using this particular curve) the value for 20mA is 1.43 while the value for 50mA is 1.61. The ratio of these, 1.61/1.43 = 1.13, suggests the relative value gained by pulsing 2.5 times the current with a 40% duty cycle. Not a lot for the work involved. (All this is in candelas, so I'm assuming this is relative to human perception -- actual radiant intensity ratios may be somewhat different.) And if the voltage drive were also taken into account (let's say, 1/3rd of a volt more so that it is 3 1/3 V and 3 2/3 V in the two cases), then this works out to about 1.1 times more _power_, which is might darned close to the 1.13 relative efficiency calc. Or, in other words, from a battery's perspective... almost no gain at all. Does anyone else have this book, by the way? It's pretty thorough, if not entirely modern. And if there is another to replace it, with newer technologies and covering as much in one volume, I'd like to know about it. Jon
Reply by ●July 26, 20092009-07-26
Jon Kirwan wrote:> On Sat, 25 Jul 2009 05:03:52 GMT, ehsjr <ehsjr@NOSPAMverizon.net> > wrote: > > >>Jon Kirwan wrote: >> >>>On Fri, 24 Jul 2009 14:17:33 GMT, ehsjr <ehsjr@NOSPAMverizon.net> >>>wrote: >>> >>> >>>><snip> >>>>The joule thief will "chew up" batteries quickly. >>>><snip> >>> >>>It's actually pretty efficient. I didn't get this from doing basic >>>calculations from theory, but by simply using LTSpice to do the calcs >>>of efficiency for me. It can be around 80-85%, or so. (It can also >>>be very bad, too.) At least, it seems so if there isn't 'operator >>>error' involved. >>> >>>Jon >> >>You snipped the content, completely. Joule thief efficiency >>is not the factor. At 100% efficiency, which is of course >>impossible, the op would be replacing AAA cells every 26 hours. >>The math is in my post. >><snip> > > > I'll stop it here. Yes, I snipped a lot. Mostly because that line > was a lead-in towards another later on where you wrote, "If you _must_ > use battery power, there are one chip solutions better than the joule > thief." I should have included that, as well. I'd read it, just > failed to quote it. The existing chip solutions aren't a whole lot > better, frankly. That was my point in writing as I did.No, it is not a lead in to that at all. It was a lead in to "Solution: mains power." Read the paragraph. There are one chip solutions that are better in all three areas I mentioned in the post: limited run time, LED brightness will decrease over time, and cost of batteries. I put the sentence below the three paragraphs with underscored headings to refer to all three of them. I don't _know_ if it qualifies as "a whole lot" better, but available one chip solutions can meet the op's stated requirement of keeping the current at 15-20 mA, and the joule thief cannot. Ed> > Sorry I wasn't more clear about it. > > Jon
Reply by ●July 26, 20092009-07-26
fungus wrote:> On Jul 25, 7:48 am, ehsjr <eh...@NOSPAMverizon.net> wrote: > >>The site you reference below, shows the joule thief producing >>BELOW 15 mA for every datapoint. >> > > > That's because the transistor dies if I go much higher... :-( >We'll get that fixed, eventually. :-) Here's a consideration, in the meantime: Run 3 joule thieves, each one powering 2 series LEDs. Or, run one joule thief with three transistors in parallel, but with 1 ohm resistors between each emitter and ground. Ed
Reply by ●July 26, 20092009-07-26
On Jul 26, 8:57=A0am, ehsjr <eh...@NOSPAMverizon.net> wrote:> > I don't _know_ if it qualifies as "a whole lot" better, but > available one chip solutions can meet the op's stated requirement > of keeping the current at 15-20 mA, and the joule thief cannot. >Can you maybe recommend one...?
Reply by ●July 26, 20092009-07-26
fungus wrote:> On Jul 26, 8:57 am, ehsjr <eh...@NOSPAMverizon.net> wrote: >> I don't _know_ if it qualifies as "a whole lot" better, but >> available one chip solutions can meet the op's stated requirement >> of keeping the current at 15-20 mA, and the joule thief cannot. >> > > Can you maybe recommend one...?LM3909 - but they are hard to get hold of now
Reply by ●July 26, 20092009-07-26
On Sun, 26 Jul 2009 22:24:34 +1000, David Eather <eather@tpg.com.au> wrote:>fungus wrote: >> On Jul 26, 8:57 am, ehsjr <eh...@NOSPAMverizon.net> wrote: >>> I don't _know_ if it qualifies as "a whole lot" better, but >>> available one chip solutions can meet the op's stated requirement >>> of keeping the current at 15-20 mA, and the joule thief cannot. >>> >> >> Can you maybe recommend one...? > >LM3909 - but they are hard to get hold of nowDavid, is this your recommendation for "one chip solutions can meet the op's stated requirement of keeping the current at 15-20 mA" for six 3.3V LEDs? It stacks the battery voltage with only one capacitor, to double the voltage. That's about it. And it doesn't control current over voltage source variations. Here is an LM3909 equivalent that works. R1 and C1 set timing. R8 is a current limiter, such as it is:>: +1.5 +1.5 +1.5 +1.5 +1.5 >: | | | | | >: | | | | | >: | \ | | | >: | / R7 | Q2 e>| |<e Q1 >: | \ 410 | 2N5401 |---+---| 2N3906 >: \ / | c/| | |\c >: / R8 | | | | | >: \ 12 | | | | | >: / | |/c Q4 | | | >: | +-------+---------| 2N3904 | | | >: | | | |>e | | | >: | | | | '-----+ | >: | | \ | | | >: | | / R3 | | | >: | \ \ 22k | | | >: --- / R6 / | | | >: \ / D2 \ 410 | | | | >: --- LED / | | |/c Q3 | >: | | +------------------------| | >: | | | | |>e | >: | | | | 2N3904 | | >: | C1 | \ | | | >: | || 150uF | / R4 | | | >: +------||-------+ \ 10k | | | >: | || | / | | | >: | | | | | | >: | | | | | | >: | | +-----------' | | >: | | | \ | >: | | | / R2 | >: | | \ \ 100 | >: | | / R5 / | >: | | \ 22k | | >: | | / | | >: | | | | | >: | | | | | >: | | gnd | | >: +--------------------------------------------------' | >: | | | >: | | | >: | ,--------+ | >: \ | | | >: / R1 | | 2N3904 | >: \ 10k | Q5 c\| | >: / | |--------------------------------------' >: | _|_ D1 e<| >: | /_\ BAT54 | >: | | | >: | | | >: | | | >: gnd gnd gndLots of parts and it doesn't meet with Ed's comment. Jon
Reply by ●July 26, 20092009-07-26
On Sat, 25 Jul 2009 20:48:54 GMT, Jon Kirwan <jonk@infinitefactors.org> wrote:>On Thu, 23 Jul 2009 09:24:55 -0700, John Larkin ><jjlarkin@highNOTlandTHIStechnologyPART.com> wrote: > >>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 > >I still can't make sense of that one. It bugs me a lot because I just >can't see why it would work well. > >This one does make more sense to me, though: > >> ,-------------------+----------, >> | | | >> | | | >> | | | >> | \ )| >> | / R1 )| L2 >> | \ )| >> | / )|o >> | | | >> | ,----+------+ | D8 ,-------, ,--, >> --- | | | +--|>|---+ | | | >> - V1 | | | | | --- | --- >> --- | | )|o | | \ / D6 | \ / D3 >> - | | )| | | --- | --- >> | | | )| L1 | | | | | >> | | | )| | | | | | >> | | | | | --- C1 --- | --- >> | | | | |/c Q1 --- \ / D5 | \ / D2 >> | | | +--------| | --- | --- >> | \ | | |>e | | | | >> | R2 / --- C2 | | | | | | >> | \ --- _|_ D7 | | --- | --- >> | / | /_\ | | \ / D4 | \ / D1 >> | | | | | | --- | --- >> | | | | | | | | | >> | | | | | | '-----' | >> gnd gnd gnd gnd gnd gnd gnd > >Is that what you were thinking of, instead? > >JonWell, John, you didn't respond to this one yet. However, I did play around with a first shot at design equations for it. (Not much different than the other concept except, as you say, I get to change the winding ratio and take better control of the reverse Vbe on Q1 so that I don't need D7 shown above.) Here is the new schematic: ,-------+----------------------, | | | | | | | \ )| | / R1 )| L2 | \ )| | / )|o | | | | | | DF | | +--|>|---+-------, ,--, --- | | | | | | - V1 +----+------, | | --- | --- --- | | | | | \ / D6 | \ / D3 - | | )|o | | --- | --- | | | )| | C1 --- | | | | | | )| L1 | --- | | | | | | )| | | --- | --- | | --- C2 | |/c Q1 | \ / D5 | \ / D2 | \ --- '--------| | --- | --- | R2 / | |>e | | | | | \ | | | | | | | / | | | --- | --- | | | | | \ / D4 | \ / D1 | | | | | --- | --- | | | | | | | | | | | | | '-----' | gnd gnd gnd gnd gnd gnd In this arrangement, I get to pick a winding ratio and much of the rest falls out. If X is the winding ratio (L1/L2) and Vr is the maximum reverse voltage you want to apply to the base of Q1 (referenced to ground, so usually negative), then: N = (Vout + Vdf - Vcesat) / (Vin - Vcesat) Ic_peak = 2 * Iout * (N + 1) (Vdf is the forward voltage of diode DF.) Then look up Ic_peak on the datasheet for picking off beta. Now to compute the Thevenin values: Rth = [beta * (X * (Vout + Vdf) - Vbe + Vr)] / Ic_peak Vth = X * (Vout + Vdf - Vin) + Vr Obviously, X must be selected so that Vth is _less_ than Vin and yet more than what's required to start Q1 (call that 1V or more?) At this point, we can compute: R1 = Rth * Vin / Vth R2 = Rth * Vin / (Vin - Vth) Which gets one started. I tried out the idea of an LED current of Iout=30mA, Vout=20V, Vcesat=0.1V, Vin=4.5V, Vdf=0.35V: N = 4.6 Ic_peak = 336 mA The beta for a 2N2222 at that current is about 90. I selected X=(1/4) after thinking for a moment. I used Vbe=0.85V. Rth = 599.33 Ohms Vth = 1.9625 V Then, R1 = 1374.26 Ohms R2 = 1062.85 Ohms I then plugged all this into LTSpice and ran the simulation. I used C2=100pF just to put something in. (I have already set up the six LED stack so that the current works out close to 30mA at close to 20V.) I set up the inductances to establish the ratio X (L1=L2/X^2, which I assume is how spice programs estimate the winding ratio of linked inductors.) The output of LTSpice was the following: Ic_peak = 332 mA Vr = -2.15V Pretty darned good for a first shot at it! Jon
Reply by ●July 26, 20092009-07-26
In article <eeOdnSNC0--N1vHXnZ2dnUVZ_hVi4p2d@supernews.com>, David Eather <eather@tpg.com.au> wrote:> fungus wrote: > > On Jul 26, 8:57 am, ehsjr <eh...@NOSPAMverizon.net> wrote: > >> I don't _know_ if it qualifies as "a whole lot" better, but > >> available one chip solutions can meet the op's stated requirement > >> of keeping the current at 15-20 mA, and the joule thief cannot. > >> > > > > Can you maybe recommend one...? > > LM3909 - but they are hard to get hold of nowSupertex CL2 - very easy to come by, though not every supply house carries them. Enough do, and they are current production. Or with twice as many parts (2) an LM317 and a resistor, in constant current mode. Neither is the ideal in terms of batteries, as they are basically linear "smart" variable resistors (well, I know the 317 is, and I think the CL2 works similarly, though I could be wrong) but if supplied with enough voltage above the minimum needed, they will provide constant current until the supply voltage drops too low, and then fizzle out quickly. A PWM buck, boost, or buck/boost solution would be potentially the most efficient method, but requires a few more parts and a lot more smarts to make it actually constant current. The Joule Thief (IIUC) is essentially a boost converter without any PWM. If starting form multiple cells, and thus more voltage, one of the above methods (paralleled out sideways to as many LEDs as you want to drive) is a whole lot simpler in terms of getting the job (if the job is 20 mA drive of LEDs) done. OTOH, the JT provides a lot of opportunities for messing about on the bench and trying things out, which is good for keeping your brain from rotting, if you get beyond monkey-see monkey-do and diversions into monkey doo-doo. -- Cats, coffee, chocolate...vices to live by
