LED torch light

[weng]

hi Im new here and I need you guys to help up.I had made a LED torchlight with 15
LEDs connected to it.The LED used is white in colour,5mm in diameter and all connected in parallel.I had recycle 3 of the lightbulb reflectors to fit to fit in the LEDs
at 5 each.now my problem is upon finishing my build I used a 9volts 250mA battery to connect it but found out the the light is not bright enough.can anyone of you guys help me up to configure what is the voltage and current required to power up the LEDs until to some level of brightness? I had been thinking of building a voltage inverter circuit from 4.5volts to 9 volts for my torchlight. welcome to share my project to anybody interested ......thanks   

[Rob]

Hi weng, and welcome to the boards.

I think the best approach is to design to a particular drive current, rather than voltage. Current relates to brightness better than voltage does for more situations.

I'll follow this up with some details, but right now I'm pressed for time--check back today though.

[Rob]

I think I understand you to say that you have 15 LEDs all in parallel. And your source is a 9V battery.

You didn't say, are you using a resistor to limit current?

I'd be tempted to do your LEDs in series groups of three with no resistor:

Code:

Solution 0: 3 x 5 array uses 15 LEDs exactly
    +----|>|----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 1 ohms

The wizard says: In solution 0:
  each 1 ohm resistor dissipates 0.4 mW
  the wizard thinks ¼W resistors are fine for your application
  together, all resistors dissipate 2 mW
  together, the diodes dissipate 900 mW             
  total power dissipated by the array is 902 mW     
  the array draws current of 100 mA from the source.

Two notes--I cheated here by telling the wizard the forward voltage is 3.0V. I only did that to get this layout. Second, ignore the 1 ohm resistor. The battery will have internal resistance of about that or greater.

Three big advantages of this connection plan:
1) the overall current drawn from the battery goes from (approximately 15* 20  A) 300mA down to about (5 * 14 mA) 70 mA, so your 250 mAh battery will go from lasting under an hour to lasting about 3 and a half.
2) no resistors required--you're underdriving the LEDs a little bit but they should light up
3) smaller laod on the battery so you'll hold up closer to the nominal voltage, and for longer

A couple small disadvantages too:
1) less than the brightest possible solution at that low drive current, and for torch lights you want brightness
2) battery voltage will sag, and that will dim them further.


These disadvantages are pretty much hugely outweighed by the disadvantages of your choice of battery. 

What's so bad about a 9V? Nothing, but you're loading it way beyond design capacity with 15 LEDs in parallel. Even my 5x3 array dissipates almost a watt, and that's not what these batteries were designed for. They were designed or power situations about two orders of magnitude less than this.

http://www.duracell.com/oem/Pdf/new/1604_US_Ultra.pdf
Check out the discharge curve on the first page. For all intents and purposes, you would be the blue line if you use my parallel/series plan, and higher with your plan (I can't do the calculations--you didn't give me enough detail). You could expect the 9V to become 7V after about an hour under a 1W load according to that graph.

You noted it was dim--I believe this is your problem. If you have a 9VDC power supply, you could compare brightness and prove me wrong (or right).
 

[Rob]

So you are far from being the only one with this problem. That's the good news. The fine folks at linear (no relation) have recently developed a selection of driver ICs for situations like yours.

This design note from linear is exactly aimed at your situation. The LT1618 is the thing you need. Look at the efficiency curve: it gets more efficient as load current increases. That's sweet.

Now that bad news--it's in a package that is really difficult for a hobbyist to deal with. But if you need a flashlight that hits a wowrking compromise between brightness and lifetime, I think you need to be looking at solutions like this IC (and there are many, this is just one).

Hope that helps.

[weng]

Rob,
many thanks for our great info.so much appreciate your effort and your knowledge.
Actually I had already  enquire about LT1618 before I join this forum.The problem is in Malaysia the price is far out from my budget .At this cost I can purchase 16 pieces of 16 AA batteries.So with this factor I had decided to contruct a simple circuit to drive my 15 pieces of LEDs which shown in my attachment.the current for these batteries are roughly 2 amps each and my total current load for the LEDs are 300mA.All I need is a 10K ohm variable resistor to limit the current output.Of course it should be trouble some to tune the resistor once a while to get the brightness but with the consideration of the cost involve I think I worth the price.Thanks for your effort again and very grateful for the knowledge given.

KW NG