Help wiring multiple flashlight LEDs

[slimeyhimey]

I recently built a custom LED light strip for my aquarium. I used 3 cheap Chinese made flashlights. Each flashlight has 9 LEDs (bright white), which are wired onto a small circular circuit board. 27 total LED's. The flashlights run on 3 AAA batteries. I wired all three circuit boards in parallel. I used a 4.5 volt adapter that puts out 300mA. This was an adapter I already had, so I used it instead of buying a new one rated for more amperage. The light works well enough.

My friend wants a similar set-up, but bigger. His aquarium is larger than mine and needs more light. He bought 8 flashlights. Same size, same flashlight. So he has 72 total LED's, on 8 circuit boards he needs to power. I'm assuming he needs about 1.5 amps to power these?

I have a 5 volt, 1 amp adapter I'm not using. If I wire in a resistor, will this adapter work? Or is it more complicated than this? Do I need more voltage?

I have spent a couple days reading, and calculating...but still kind of lost. Any help, to shed some light, would be great...thanks.

 

[Resqueline]

5mm LED's tolerates around 20mA and has a normal voltage drop of some 3-3.5V. In the Chinese flashlights they're seriously overdriven by the direct 4.5V application.
Yes, 20mA * 72 = 1.44A (max), but running them at 1A won't reduce the brightness much and will ensure a longer life. Use one resistor for up to 125mA on each board.
Try with 15 Ohms on just one board first. Depending on the actual voltage drop you measure you might want to adjust the resistance somewhat.
Ideally there should be 72 resistors instead of just 8, but that would complicate matters more than it's worth.
One alternative is to wire 2 & 2 boards in series and run them from a 9 or 12V adapter.

 

[slimeyhimey]

Ok...well I found an 18 volt, 1.4 amp adapter in the basement. So essentially if I wire 2 sets of 4 lights, in series (4x4.5v=18v), and then wire those 2 sets parallel, it should work?

So these cheap chinese flashlight LED's typically are rated for 3 to 3.5 volts? And running them on 4.5 volts is pushing it?

 

[(*steve*)]

So these cheap chinese flashlight LED's typically are rated for 3 to 3.5 volts? And running them on 4.5 volts is pushing it?

Yes. They generally rely on the internal impedance of the batteries to limit the current.

If you use a stiffer power supply (i.e. one that won't sag in voltage at this current) then it is reasonably likely that the LEDs will have "A short, but brilliant career".

You are better off putting fewer of these modules in series and then use a current limiting resistor or (even better) a constant current source.

You can make a constant current source fairly easily with an LM317. See here.

 

[slimeyhimey]

Haha, well I tried...and failed.

I wired up a combination of series and parallel. Take a look at my picture. For some reason, the first set of four lights lit up nice and bright, the second set of four lights did not at all.

How do you wire this up? I looked around on the net and did not find any good examples.

Thanks guys

 

[slimeyhimey]

Never mind. Figured the wiring out.

Now my question is this...I have a 12 volt 3.7 amp power supply, and a 16 volt 1 amp power supply. The 12 volt is only going to send 3 volts to each circuit board, but plenty of amps. The 16 volt will send 4 volts to each circuit board, but not enough amps to hit the 20ma per bulb allowance.

Do you like all my bad-ass technical talk? I have no clue

Is it better/brighter to have more volts or more amps going to the bulbs. I want bright, if I can get it!

Thanks

 

[(*steve*)]

You have no clue

The device will attempt to draw the current it requires at the voltage supplied.

4 of these driven from the 12V supply would be safest because you wouldn't be over-driving them. But you also may be under-driving them so much that they would be very dim.

If the 16V supply cannot deliver the current demanded, then it may get hot and be damaged.

I still really want to see what these devices look like. If they have no resistors on them at all then bad things are eventually going to happen.

At a guess, I would put three in series, place a 2 ohm 5W resistor in series with that, and connect it to the 12V supply and measure the current. If it is close to 1A (say +/-100mA) then wire the rest the same and place them in parallel (you may have to buy another module). If the current is too high or low you may have to consider a different size resistor.

Can you see if there are any components other than LEDs on these modules? Perhaps you can take a picture (of both sides) and show us?

 

[EmperorJJ1]

im looking at doing something similar persay. i wanted to use a few of the flashlights but in a car. Im looking at wiring either 3 or 4 of these depending on what fits. But i just attached a link to the backside of them. no resistors at all just 9 leds wired in parallel

 

[Resqueline]

In a car you may wire 3 such modules in series, and in series with a 20-22 ohm 1W resistor.
Alternatively you may wire 2 modules in series, and in series with a ~36 ohm 2W resistor.

 

[Merlin3189]

Driving LED arrays

1 - The lamp I have has 24 LEDs all in parallel, with no resistors at all and is driven by 4AA cells. The current distribution is inevitably very uneven and some LEDs are much brighter than others. As others have said, a bad situation relying on the cells' internal resistance (and possibly, that with 24 LEDs, it won't matter if a few burn out and a few are too under-driven to light up!) We really should have individual resistors, but as you say your arrays of 9 seem happy enough to run from a 4.5V adapter, let's go on from there.
2 - If one array of 9 is running at 4.5V and drawing 20mA per LED, that is 4.5V x 20mA x9 = 810mW. (Personally I'd guess they are running at a lower Voltage, say 3.5 - 4V, and I'd be happier if the average current were a bit lower, say 15mA, to allow for the uneven split between LEDs.* That would work out nearer 500mW.) Therefore, if you want 8 of these arrays, you need 8 x 810mW of power = 6.48W (or for me 4W). This you can get from either of your power supplies, 12V x 3.7A = 44W or 16V x 1A = 16W. And in each case there is plenty of power to spare for dissipation in some resistors.
3 - Now, how should we wire them up? Each module needs 4.5V and 180mA.
So the 12V supply can have up to 2 in series (2 x 4.5 V < 12V, but 3x4.5V >12V)
and it can have up to 20 in parallel (20 x180mA < 3.7A)
The 16 V supply can have up to 3 in series (3 x 4.5V < 16V)
and up to 5 in parallel (5 x 180mA < 1A)
So 2 in series x 4 in parallel will work on both.
This will need 2 x 4.5V = 9V and 4 x 180mA = 720mA.
4 - So for the 12V supply we need to drop 3V. This is best done for each of the 4 parallel circuits separately. So the resistor for each chain is 3V / 0.18A = 14.4 or say 15 Ohm.
For the 16V supply we need to drop 7V and the resistor for each chain is 7V / 0.18A = 39 Ohm.
Of these two, I would prefer the latter because the greater resistance gives a little more current limiting stability. It does however require that the resistor dissipate 7V x 0.18A = 1.26W, so at least a 2W resistor is needed.
5 - If you use the 12V supply, the best option might be to run 8 modules in parallel with a resistor in series with each module. This would require dropping 12V - 4.5V = 7.5V in the resistors. Each of the 8 resistors would be 7.5V / 0.18A = 41.6 or say 43 Ohm. This would dissipate 7.5V x 0.18A = 1.35 W, again at least a 2W resistor.
6 - And as I said I'd like to reduce the average current to say 15mA per LED = 135mA per module. This gives a resistor around 60 Ohm, say 62 Ohm which sets the average current per LED to 13.5 to 15.5mA (Sorry, it's getting late so I've skipped the calcs!)

Apologies for no diagram: if wanted, I'll scan one in tomorrow and post it.

* If 9 LEDs in parallel draw an average of 20mA, some will probably get more of the current than others. Those that get less will just be dimmer, but those that get more are likely to be getting overheated and shortening their lives. Due to the characteristics of LEDs, most will draw a little less than average and a few will draw a lot more: so the average needs to be well below the safe maximum. All of the above answers fail to deal with this problem: the only complete solution requires individual resistors for each LED.