Jon Kirwan wrote:> On Sun, 26 Jul 2009 17:06:24 GMT, ehsjr <ehsjr@NOSPAMverizon.net> > 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...? >> >>Manufacturer chips posted below are just the first few found by a >>Google search with "led boost drivers" in the search box. >> >>National recommends their LM3410X for this. >>http://www.national.com/ds/LM/LM3410.pdf > > > $2.50-$3 each. Lots around. > > >>TI shows the TPS61160 meeting the requirements. >>http://focus.ti.com/lit/ds/symlink/tps61161a.pdf > > > Hmm. Cheaper. $2 each. Lots around. > > >>Onsemi has the CAT3606-D >>http://www.onsemi.com/pub_link/Collateral/CAT3606-D.PDF > > > Couldn't find the -D around anywhere. But did find CAT3606HV4-T2 at > Digikey for $1 (and at only two other places.)That is the part number given in the datasheet for the CAT3606-D. http://www.onsemi.com/pub_link/Collateral/CAT3606-D.PDF> This device cannot > handle more than 4.2VWrong. Typical Vin is 3 to 5.5, max Vin is 7. See the datasheet.> input and must have at least 3V. It's designed > for Li-ion sources and can run in either 1X or 1.5X mode. I'm not > hyped on this as a 'solution.' It's a charge pump with regulation on > the current, I think.Yes - you program the current with one resistor.> > >>Linear's LT3598 will do it: >>http://cds.linear.com/docs/Datasheet/3598fa.pdf > > > Mucho expensive. I found them for over $7 each! (Some at under $5, > too.) Only a few places carry them. > > .... > > TI seems to be the one out of the above I'd focus more on. Looks nice > and seems to do the right job for a reasonable price and is at various > stores, as well. > > Jon > > >>I'm not recommending any one of those over any other, >>and there are other chips from those manufactures and >>others that may suit your needs. >> >>Ed
Joule Thief - still not working....
Started by ●July 23, 2009
Reply by ●July 27, 20092009-07-27
Reply by ●July 27, 20092009-07-27
ehsjr wrote:> 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.Yes, if you were just to let the inductor saturate in the standard Joule Thief circuit, it would still turn off by Ic reaching the beta limit. The only difference would be a big current spike at the end of the on period, which wouldn't do any good. So some kind of modification would be needed. In the pulse transformer circuit I posted earlier, there is a small resistor in series with the emitter. There's also a capacitor connected to the base. The way I think it works is that when saturation occurs, the voltage across the resistor begins to rise sharply. The capacitor prevents the base voltage from being able to keep up, so Vbe drops and the transistor starts to turn off, with feedback from the base winding finishing the process. I expect something similar could be done with the Joule Thief to make it work in inductor-saturation mode. -- Greg
Reply by ●July 27, 20092009-07-27
On Jul 27, 6:28=A0am, greg <g...@cosc.canterbury.ac.nz> wrote:> > I'm confused. Does "b_to_e" mean base-to-emitter? > I want to see *collector*-to-emitter. > >This is collector->emitter: http://www.artlum.com/jt/coll_to_em.gif
Reply by ●July 27, 20092009-07-27
On Jul 27, 6:31=A0am, David Eather <eat...@tpg.com.au> wrote:> > >http://www.national.com/mpf/LM/LM3914.html > > LM3914 would seem to work a treat - it needs about .9 of a volt more > than the LED uses (page 5 "led current regulation drop out") >I hadn't understood that part (in fact I'm still not sure I do) but does it mean I need an extra volt from somewhere? Will I have to use a joule thief to drive the LM3914? :-(
Reply by ●July 27, 20092009-07-27
fungus wrote:> On Jul 27, 6:31 am, David Eather <eat...@tpg.com.au> wrote: >>> http://www.national.com/mpf/LM/LM3914.html >> LM3914 would seem to work a treat - it needs about .9 of a volt more >> than the LED uses (page 5 "led current regulation drop out") >> > > I hadn't understood that part (in fact I'm still not sure I do) > but does it mean I need an extra volt from somewhere? > > Will I have to use a joule thief to drive the LM3914? :-( >No. You will need a supply greater than the LED voltage + .9v for near perfect current regulation. When the .9 volt is not available the current regulation becomes a bit sloppy and the current drops a bit. If you wanted to suck the batteries *really* dry then you might need a way to get a bit more voltage. As an aside have you thought about using 6 volts (should be easier/longer lasting to use with white LED) or a small 9v battery (would last a bit more than an hour)?
Reply by ●July 27, 20092009-07-27
On Jul 27, 9:12=A0am, David Eather <eat...@tpg.com.au> wrote:> fungus wrote: > > > > Will I have to use a joule thief to drive the LM3914? =A0:-( > > No. You will need a supply greater than the LED voltage + .9v for near > perfect current regulation. When the .9 volt is not available the > current regulation becomes a bit sloppy and the current drops a bit. If > you wanted to suck the batteries *really* dry then you might need a way > to get a bit more voltage. >Ok .... I think I'm going to order a couple of them off eBay to play around with.> As an aside have you thought about using 6 volts (should be > easier/longer lasting to use with white LED) or a small 9v battery > (would last a bit more than an hour)?Yes, I was just kidding about the joule thief. An extra battery would be a lot more sensible/less work...
Reply by ●July 27, 20092009-07-27
fungus wrote:> This is collector->emitter: > > http://www.artlum.com/jt/coll_to_em.gifWhere is 0 volts on that picture? -- Greg
Reply by ●July 27, 20092009-07-27
On Jul 27, 1:32=A0pm, greg <g...@cosc.canterbury.ac.nz> wrote:> fungus wrote: > > This is collector->emitter: > > >http://www.artlum.com/jt/coll_to_em.gif > > Where is 0 volts on that picture? >At the top...
Reply by ●July 27, 20092009-07-27
On Jul 27, 1:32=A0pm, greg <g...@cosc.canterbury.ac.nz> wrote:> fungus wrote: > > This is collector->emitter: > > >http://www.artlum.com/jt/coll_to_em.gif > > Where is 0 volts on that picture? >If you reload the image I drew a line where ground is....
Reply by ●July 27, 20092009-07-27
fungus wrote:> On Jul 27, 9:12 am, David Eather <eat...@tpg.com.au> wrote: >> fungus wrote: >>> Will I have to use a joule thief to drive the LM3914? :-( >> No. You will need a supply greater than the LED voltage + .9v for near >> perfect current regulation. When the .9 volt is not available the >> current regulation becomes a bit sloppy and the current drops a bit. If >> you wanted to suck the batteries *really* dry then you might need a way >> to get a bit more voltage. >> > > Ok .... I think I'm going to order a couple of them off eBay to play > around with. > >> As an aside have you thought about using 6 volts (should be >> easier/longer lasting to use with white LED) or a small 9v battery >> (would last a bit more than an hour)? > > Yes, I was just kidding about the joule thief. An extra battery > would be a lot more sensible/less work... >OK. http://www.national.com/ds/LM/LM3914.pdf on page 7 is the minimum circuit (you don't need all 10 LED's if you don't want). R should be 680 ohms (18ma current) and the IC will run very warm - but it should not be "burning" hot. connect pin 7, pin 6, pin 5 and pin 9 to pin 3. Also connect a small capacitor between pin 3 and 0v (any size bigger that about 2.2uF will do)
