LED light question for you electronics wizards...

[ParkRNDL]

I bought this LED flashlight for a few bucks, expecting to take it apart and gut it to make a headlight. I figure, cut a couple wires, solder up a battery holder, good to go. Only problem is, there's all kinds of circuit boards (well, okay, 2 circuit boards) that I don't know how to work with. The 5 LEDs are attached to one board that has 3 wires coming off it--two red and one black, which makes sense, since the light has a "low-high beam" set up where it's either 3 on or 5 on. I figure one red wire goes to the hot side of 3 LEDs, the other red goes to the hot side of the other 2 LEDS, and the black is the common negative. However, these three go to this other circuit board that's all kinds of complicated and attached to the 5-way switch and the batteries. This whole mess runs off 4 AA batteries. Here's my question: Is it safe to assume that these LEDs run off 6vdc (4 AAs by 1.5 volts each) and I can cut those three wires and hook them directly to batteries and a switch? Or is there likely to be voltage-reducing stuff on the other board so that when I hook up the batteries and flip the switch, it'll let the smoke out of the LEDs? :mrgreen:

(maybe I ought to post pics... I can do that later...)

--rick

 

[socal_jack]

Very unlikely you can hookthem straight without a voltage dropping resistor in series, without the aforementioned smoke show. The white LEDs and blue LEDs bias at about 3.2-3.6V. Most LED bulbs(not the raw LED) have a built-in bias circuit or as in your case an external bias controller circuits which is part of what you have on the other board, partly for battery life but also LED life.

You're gonna need to guess about the LED maximum current unless you can find a munfacturer part number and look up the specs. Generally you're going to need to run a simple calculation like this below which is for a 20mA current diode, which is typical for lighting LEDs. If you can find a P/N try to find a spec online at DigiKey maybe.

20 mA = 0.02 A

V/I = R
(6V - 3.2V) / 0.02A = R
140ohm = R

The circuit below assumes ~10 mA.

One of the other purposes of the bias circuit is for brightness matching, running off of a battery is going to be problematic. If you run them in parallel there will probably be a big difference, If you try to string them in series you can run 2 tops but still need a small series resistor so they will probably be dim. The bias boards usually have a small DC-DC converter to bump up the power supply for driving multiple LEDs

Here's a typical datasheet for one of these bias chips, mightgive a better idea, it might be better to find the manufacturer number on the controller chip and figure out how to control the switching inputs the way you want to from your own switches.
http://www.maxim-ic.com/appnotes.cfm/appnote_number/1804/

 

[ParkRNDL]

oh man. your explanation made sense, as far as the need to limit and equalize voltage ... but I have nowhere near enough knowledge (read: I have NO knowledge) to calculate or figure out how to control switching inputs or choose resistors. Any chance that you can tell from these pictures what I need to do in simpler terms (i.e. get XXXX resistor and solder it in between this red wire and the battery holder)? I really appreciate the time you took in your explanation, I just don't have the technical knowledge to use it...

The first pic is the back of the board that the blue-white LEDs are attached to. Nothing on the front except the LEDs themselves and this dull silver tape.

The second pic is the back of the second board that ties it all together... the wires going out the top of the pic go to the battery holder, the wires out the left go to a small board of 3 red LEDs, and the wires out the right go to the board in the first pic. That black box on top is a push-click switch that cycles through 5 "positions": 3 white LEDs on, 5 white LEDs on, red LEDs on, red LEDs flashing, off.

The 3rd pic is the back of the little board with the 3 red LEDs. Again, nothing on the front of the board but the LEDs.

Any way you can tell from looking if some of the stuff I need is already on the LED board, or if there's a simple resistor or two I can add between the LEDs and the battery to get this to work without the intermediate board?

thanks in advance for any info/insight...

--rick

 

[socal_jack]

It would be much easier hands on, but it looks like there are 3 LEDs in parallel and 2 LEDs in parallel each with a 68 ohm resistor in series. With that low of a value, there is most likely a constant current source on the other board. The two transistors with the caps near them are probably the flasher circuit, while the bulk of the other looks like TO-92 packages may be a combination of transistors and voltage regulators.

It would probably be easier to figure out something with the switch, how many leads are coming off that?

 

[ParkRNDL]

ack. It occurred to me AFTER I posted that last post that it might help to show the other side of that board too...

this is the back of the board. there are 5 leads coming off that switch:

top of the switch:

bottom of the switch:

i'm getting the idea that it would be easier to find a simpler light to start with...

thanks again for picking through all this...

--rick

 

[socal_jack]

OK, the main plus terminal comes in and directly feeds the white LED board common side, Q9 drives the black wire going over there(3 leds) and Q8(not sure if that's the number but next to R19) drives the middle red wire so those are the constant current sources for those puppies. Q6 feeds is the current source for the red while Q5, Q7 , C1 and C2 are setup in a classic multivibrator for the flasher. They are involved with Q6 for flash mode. It looks like main plus power goes into the switch on the middle solder connection.

Looking at the bottom of the main board in the 04 pic, it appears that the far left solder connection biases the Q8 transistor and next to it powers the Q9. The far right connects the red wire to the rear LEDs thru a 100 ohm resistor R14, the contact next to it enables the flasher.

I'd be willing to bet that the main switch just feeds single or multiple pins directly from the positive supply, you can check with a voltmeter easy enough. If not, one could with the switch in power off, run a wire directly to each of the outer 4 contacts, from plus side battery. For high/low main led both for hi, Q9 powered for lo(3 leds), for flasher probably both the right will be powered and only the far right if just red. Or unsolder/solder and experiment. It should be easy to rig up a DPDT switch to give red+low and red+high.