whitch resistor

[landries]

i want to conect 2 P7 led's in serial on 2 battery's in serial type 18650 whitch resistor chould i use?
i have no place to put a driver so it must be a resistor
i tried direct drive ,but there is a little bit to much heat
ps sorry for mi poor englich

 

[Benson]

Just for talking, let's say you go with a 0.5 Ohm resistor -- that'll dissipate 4.5W of power, so you'll need a 5W resistor. You're sure you have room for one of those, but not a driver board?

The value of resistance you use will depend on what current you want. Do you know the voltage and current they run at in direct-drive?

 

[Mr Happy]

that'll dissipate 4.5W of power, so you'll need a 5W resistor

Actually a 10 W resistor would be better. A resistor in an enclosed space will get hotter than a resistor with open air circulation for cooling. In principle it might be good to heat sink the resistor to the lamp body for additional cooling. A 5 W heater in a small space will make things very hot...

 

[landries]

can i put several smaler resistors in paralel to get 10 watt,and do i need 0,5 ohm ??

 

[Benson]

can i put several smaler resistors in paralel to get 10 watt,and do i need 0,5 ohm ??

Yes, you could connect multiple resistors, series or parallel. But several small resistors take up a lot of space, too... if you literally don't have room for a driver, I doubt you have room for a proper resistor drive, however you work it.

As to whether 0.5 ohm is right for this, you still haven't given any indication of what operating point you're trying for, so I can't possibly tell you the bias resistor to obtain it.

Without knowing the Vf of your LED and the internal resistance of the battery, it's still not possible to specify the exact value, but by using typical values for those, we can at least come up with a starting estimate, then you can adjust up or down depending on actual current draw.

 

[Mr Happy]

Some experimentation on the bench is usually a good idea.

Grab a variety of resistors in the range from 0.05 to 0.5 ohms (say), and work down from the highest value.

Make sure the emitters are well heat sinked.

Try each resistor out, measure the current, check the brightness. Try it with freshly charged batteries and almost empty batteries. Feel how hot the resistors get. Feel how hot the emitters get.

You can combine resistors to get different values. For example two 0.5 ohm resistors in parallel will give you 0.25 ohms.

 

[landries]

i would like 2800 milliamps output and the v 2x 3.6 in series is 7.2v
the input come from a 2s2p battery made off 18650 from dx http://www.dealextreme.com/details.dx/sku.5790

 

[VidPro]

the 3.6v per cell is called a "nominal" voltage, each cell can actually put out ~4.2v when charged.
so if it WAS 3.6 then you would be all set
its the ~4.2 that your resistering, or drivering for, and the ~3.3v it gets down to that your drivering for.

then depending on how hard you drive the led and its "bin" each led gate (for white) is about ~3.5 to a max of about ~3.9v . and because a led can vary greatly even within the same speced item, that is where MrHappy said you put it in a more real situation and test.

if you just use various numbers (even mine) you will make a bad realistic formula that wont work well.
just the resistance of other things will effect thing also, like the protection curcuit will have some resitance, the light will have some , switches, springs, wiring, even the battery ouput capability and resistance will have to be included in your reality.

 

[Benson]

i would like 2800 milliamps output and the v 2x 3.6 in series is 7.2v
the input come from a 2s2p battery made off 18650 from dx http://www.dealextreme.com/details.dx/sku.5790

Well, I just measured some AW 18650s I have, and with three of them in series (and some Nd magnets holding the pack together) found the pack resistance to be 0.3ohm, or 0.1ohm each. Assuming yours are the same (which is not necessarily true), we can put the pack at 0.1ohm total (each pair in parallel is 0.05ohm; a "square" arrangement like 2s2p or 3s3p always has the same resistance as a single cell). 2.8A across 0.1ohm is 0.28V drop, so your battery pack will deliver 8.4 - 0.28 = 8.12V when charged. (You'll need to add any resistance for the light body, switches, wiring etc. here, when you do this.) Let's knock that 0.12V off, as the battery pack will very quickly settle, and a little overdrive for a short while won't hurt anything -- call it 8V.

Now we'll take a typical Vf of 3.6V at 2.8A (this depends on your emitter's Vf-bin, and would be a J-bin), so with two emitters that'll be 7.2V. The batteries start at 8V, so we need a total drop of 0.8V. 0.8V / 2.8A = 0.286 ohm, so you could try 0.3 ohm as a good starting point. For the power dissipated by the resistor, 0.8V * 2.8A = 2.24W. With good heatsinking, you might get by with a 2W resistor, but a 5W (still heatsunk well) is the smart choice.

Now you really need to find out at least the Vf bin of your emitters, and better yet their actual Vf. I-bin emitters are available these days, and if you were using the best I-bin emitters, the LEDs would be 3.25V * 2 = 6.5V, and that 0.3ohm resistor, + the 0.1ohm from the battery pack, would take up most of the 0.7V difference. The current through 0.4ohm at (8.4V - 6.5V) = 1.9V would be 4.75A!

If you don't know the bin, then you should start with more resistance than you need, and compare your actual Vf/If operating point to the graph 1 on page 6. The curve should be the same, but shifted by some voltage. Once you figure this voltage at the current you wound up running, you can then add or subtract the same voltage at 2.8A to get your emitter's Vf. Do that for each LED (or test the two LEDs together, if you prefer), and you'll be able to figure out an appropriate resistor value. Unless you have good data on your battery, though, you'll still have to experiment to compensate for that.

If, in this same test, you not only measure the LED voltage and the current, but also the battery voltage under load, and after you remove the load, then you'll be able to get the internal resistance of the batteries, and should nail the 2.8A current in one shot.

 

[HarryN]

Just for talking, let's say you go with a 0.5 Ohm resistor -- that'll dissipate 4.5W of power, so you'll need a 5W resistor. You're sure you have room for one of those, but not a driver board?

The value of resistance you use will depend on what current you want. Do you know the voltage and current they run at in direct-drive?

In general you are right, but I have some 2 and 10 ohm resistors rated for 50 watts that are only 1206 size, so that fixes that problem.

 

[Mr Happy]

In general you are right, but I have some 2 and 10 ohm resistors rated for 50 watts that are only 1206 size, so that fixes that problem.

If you put 50 watts into a soldering iron it gets hot enough to melt solder. Since a 1206 resistor is much smaller than a soldering iron it will certainly desolder itself if you try to make it dissipate even 10 watts, let alone 50...? :thinking:

 

[VidPro]

when you drive these high power leds to the higher levels, the bin numbers and math isnt very usefull , because your sorta overdriving it , vrses some really lightweight 350ma per die, they can be speced for. the voltage at the higher currents is not exactally low. like it is on paper, and the li-ion droops a bit before you get the first few minutes.

it is Still good to overestimate and move up, but the Voltage differance in reality is not as high as it looks in the math.

i am still agreeing with benson, but you really gotta have the resister array ready to tack a bit more items on in parellel or something, to tweak resistance down a bit. it is always good to have more on-hands parts on hand.
and you must make SURE that any testing with resistances added, translates to reality, like with added meters and wires and such.

smoke 'em if you got 'em

 

[Benson]

If you put 50 watts into a soldering iron it gets hot enough to melt solder. Since a 1206 resistor is much smaller than a soldering iron it will certainly desolder itself if you try to make it dissipate even 10 watts, let alone 50...? :thinking:

Well, if you stick the tip of a 10W soldering iron against an aluminum flashlight body with AA epoxy, I doubt it'll get hot enough to melt solder. (At 50W, it might, but the batteries will run out first. ) Heat-sinking is key for those little guys. (I'd never heard of them, though; I wonder if you don't need an Al PCB for 50W... surely a regular fiberglass board couldn't dump the heat fast enough.) Maybe also use lead-free solder -- it melts hotter.

when you drive these high power leds to the higher levels, the bin numbers and math isnt very usefull , because your sorta overdriving it , vrses some really lightweight 350ma per die, they can be speced for. the voltage at the higher currents is not exactally low. like it is on paper, and the li-ion droops a bit before you get the first few minutes.

We luck out with the P7, though! It's actually binned at 2.8A; it's the Crees that are binned at 350mA. For those, there's the extra step of going from Vf @350mA to Vf @2.8A, but that can be read off the I-V plot in the datasheet. I'm not saying you can nail a resistor drive dead on, anyway (the Vf bins are way too wide for that, in any case), but you can get pretty close by the numbers if you read and understand the datasheet and binning guide, and have good enough data for the rest of your system.

Dead right about needing more components on hand, though -- if you run to Radio Shock for every new component value you try, you'd better set up your workbench in their parking lot.

This is just one reason I don't like resistor or direct drive -- amc7135s are just as simple to implement, just as efficient as resistor, and always come out first try, with just the current you want (or less if your batteries simply won't put out).

 

[VidPro]

Off Topic:
Benson, i can easily put a Dimm2Dimm PWM thing in right after a amc7135 does it's little trim job, cant I ?

i have had this dimm2 dimm parked here because it doesnt like my massive overdrives for some reason, mabey its the PUNY little mosfet (george said its good for 3amp)
if i lower the max current using the amc7135 glorified resister, can you think of any reason a dim2dim PWM thing wouldnt go in there right after the current control?
like any harmonic distortion, or crasy internal networking between the devices that will occur , or wild PWM fluxuations that will cause the poor amc7135 chip thing to freak out and go home?

also has anyone ever noticed any issues with this multiarray of lineral regulators having some of the same issues of a multi array of parellel resistance?
because of heat differentials and all. i see sandwitches of them, and stuff like that. i am thinking i want to POT the whole wad, so all the little parts dont change over time and heat. like one time i had a chip on a computer board FALL OFF in my hand, unbelieveable, one of them flow soldered chips just popped right off an older curcuit board, with very minmum pressure.

i was thinking if i affix all this stuff in a BLOB of nonconductive/noncapacitive meld, and it still functions afterwards, then i could rely on it to last a few years.

i was really hoping for a amc700135, some day that is one fat all encompassing single item, drivers are ALWAYS well behind the needs were having for them, by the time i apply a 20W led, they might have the drivers for the 3 W ones oh wait i already DID.

 

[Benson]

Off Topic:
Benson, i can easily put a Dimm2Dimm PWM thing in right after a amc7135 does it's little trim job, cant I ?

I'd certainly expect so, in general, but I don't really know anything about this dimm2dimm -- a quick google didn't turn up anything looking like it, so if you could find me a link to start from, I don't mind looking at it.

The 7135 gives a pure DC output, rather than PWMing a high voltage to get the right average current, so it tends to work well with PWM dimmers.

i was thinking if i affix all this stuff in a BLOB of nonconductive/noncapacitive meld, and it still functions afterwards, then i could rely on it to last a few years.

Yeah, some people do that -- the good side is, things are fairly protected from failure, and heat can get out reasonably well, but the bad side is if something does fail, the whole wad's toast. (Or even if you just change your mind about drivers...)

Me? I "potted" the 2 AMC boards in my Mag (with cardboard insulating spacers between them) using hot-glue. I think it probably softens somewhat during long runs, but it hasn't melted all over everything yet. If I decide to redo things, I can put it in the oven and melt the glue enough to get everything out. Not something I'd really recommend, though; I'd say either pot it with epoxy, or find a non-permanent (i.e., screw-threads) way of fastening everything down. I was just too lazy to run to the store for epoxy.

 

[VidPro]

it was a taskled.com special. very simple i think a pic controller and a fast PWM and continuous dimming up when your down and down when your up. got all of 3 chips, no induction or boost or buck or nothing.

i didnt know it is no longer sold, i dont even see a spec sheet for it, oops gee mabey i should have copied the wiring setup oh crud, now i have to Guess again.

Hot glue? oh that is funny, that stuff has some really odd properties, seems to hold a load of heat in itself, like a vat of slow moving stickey oil, let it cling to the sides of a battery tube with it cooler and harder there, and soft and hot inside, wonder what that would really do.

 

[Benson]

Looks like this may be it? Looks like it would work fine -- I'd connect the d2dim between the battery and the amc7135 board, but I think it would be fine either way.

 

[VidPro]

Looks like this may be it? Looks like it would work fine -- I'd connect the d2dim between the battery and the amc7135 board, but I think it would be fine either way.

well if your sure this amc7135 is all analog acting, instead of ever spiking in current, then it would work on either side. i was just thinking of the controlled side, if there was any shifting draw from the battery.

how DID you find that, thanks, looks like i am history again, obsolete. discontinued
you know what we say about that, if it was good it was discontinued , specially at the radioshack, or cell store or whatever you call it now.

 

[HarryN]

If you put 50 watts into a soldering iron it gets hot enough to melt solder. Since a 1206 resistor is much smaller than a soldering iron it will certainly desolder itself if you try to make it dissipate even 10 watts, let alone 50...? :thinking:

Hi - sorry for the delayed response. The limit of these resistors is the ability to remove the heat - just like some of the more powerful drivers and the LEDs we like to push.

It is non trivial to remove 50 watts from a 1206 size component, so I happily run them at less than 2 watts most of the time. That can be removed pretty easily.

 

[qwertyydude]

Also keep in mind that when direct driving 18650's voltage will droop so a 5w resistor will be just fine. My P7 is direct drive and has been running no problem, You'll probably be fine with a .1 ohm 3 watt resistor.