Portable LED Work Light Prototype

[Particle]

Alright. Here's my dilemma. I'm trying to design a circuit to handle a 5 x 5 grid of 10k mcd white LEDs to be used as a work light.

Setup:
[Panel]
- 5 x 5 grid of 10,000mcd white 20deg LEDs (@ 20mA)
- each LED will be coupled with a 100 ohm resistor (120's in 1/2w I guess are rare)

[Power]
- 4 1.2v NiMH AA batts
- LM7805 +5VDC regulator IC

Problem:
The voltage regulator is in place to prevent over driving of the LEDs. The original prototype was designed with only 30mA in mind. However, since LEDs are so cheap, I decided that extra power to the LED would be fine. To prevent the LEDs from being driven by more than about 50mA, I decided to use a 7805 to prevent excess voltage. In the case of NiMHs, the panel would be just fine. However, if alkaline batteries are used, the panel would experience more current than I am willing to allow--for the safety of the lamps.

When I tested the circuit with the regulator yesterday, I decided to test it with a .75A load. Unfortunately, the load experienced a problem. As soon as the load was attached, the voltage out of the 7805 dropped to 2.5VDC. Directly off the batteries, it was fine.

I've since come to realize that the 7805 needs at least 7.5VDC to operate correctly. Am I correct in assuming this was my problem and that the addition of two more AA cells would be the best way to solve my problem?

Estimated electrical characteristics of the panel:
4 ohms
1.25A maximum draw
500,000mcd @ 1.25A

The six NiMHs would provide 7.2VDC. Would this be enough or should I upgrade to 8 cells and use 9.6VDC instead? Either is acceptable, but the fewer required cells the better. The regulators I have available are rated at 1 amp, but there is a custom aluminum heat sink that is designed to be used for this circuit. Active cooling is not to be used due to excess current draw from the NiMHs and decreased running time.

Also, one last question. Does the regulator waste much energy?

 

[gerty]

I believe you're right about the battery voltage being too low . I would add 2 batteries. That way you wont be dropping more voltage across the regulator than you have to. Since the regulator is rated at 1a and the load is 1.25a why don't you split the load up and use 2 regulators, that way they won't run above max capacity..You shouldn't need 1/2 watt resistors for the leds..

 

[jbeng]

... why don't you split the load up and use 2 regulators ...

LM340T-5.0 (7805) regulators are cheap (~$1US) ...

 

[Oznog]

Best bet might be a current regulator. A current shunt resistor on the ground leg, feed the voltage into an op amp driving a MOSFET. Then there's no regulator voltage dropout. Keep the current limiting resistors on each leg though. And make sure all the legs are working before powering it up all the way, or it could overdrive the other legs.

Consider replacing the multiple LEDs with a single Luxeon. It produces the light of around 100 std LEDs for a reasonable cost.
https://www.luxeonstar.com/

With only a single LED, there's only one leg so you don't need limiting resistors and a whole lot of minimum wasted voltage/power, but there will be a lot of heat in the MOSFET regulator. I like to use a buck or boost converter with a regulated current output for this job, which requires involving an inductor and an appropriate IC. It doesn't waste heat and the batteries will last longer. LuxeonStar.com stocks some converters to do things like that too.

 

[Particle]

All of these are good solutions. I considered using Luxeon LEDs, but this way would actually use less power and be brighter. A typical Luxeon star achieves about 68k mcd @ 350mA. I'd be hitting 80k @ 200mA for every 4 diodes.

At 1.4A, the Luxeons would provide 272,000mcd where as I'd be getting about 500,000mcd at 1.25A with the normal LEDs. In addition, the normal LEDs are extremely cheap in comparison. $0.16 for a normal white LED. Heh, I appreciate the suggestion, but I've researched this already =)

The current regulator sounds like it may be the way to go. Can this be done relatively cheap?

 

[Oznog]

Doesn't sound right. Luxeon has a significantly higher luminous power efficiency than any other diode I know.

The "warm white" with a collimating lens was the only one with a comparable candela specification, and that showed 200k mcd @ 350 mA, over 10 degrees. 1.2 watt. The optics are also only 85% efficient, the batwing is less focused but will have more power than that overall.

Luxeons are also very well specified- don't bet too hard that you'll get what's advertised for normal LEDs unless you've got a thorough spec sheet.

 

[Optikon]

Alright. Here's my dilemma. I'm trying to design a circuit to handle a 5 x 5 grid of 10k mcd white LEDs to be used as a work light.

Setup:
[Panel]
- 5 x 5 grid of 10,000mcd white 20deg LEDs (@ 20mA)
- each LED will be coupled with a 100 ohm resistor (120's in 1/2w I guess are rare)

[Power]
- 4 1.2v NiMH AA batts
- LM7805 +5VDC regulator IC

Problem:
The voltage regulator is in place to prevent over driving of the LEDs. The original prototype was designed with only 30mA in mind. However, since LEDs are so cheap, I decided that extra power to the LED would be fine. To prevent the LEDs from being driven by more than about 50mA, I decided to use a 7805 to prevent excess voltage. In the case of NiMHs, the panel would be just fine. However, if alkaline batteries are used, the panel would experience more current than I am willing to allow--for the safety of the lamps.

When I tested the circuit with the regulator yesterday, I decided to test it with a .75A load. Unfortunately, the load experienced a problem. As soon as the load was attached, the voltage out of the 7805 dropped to 2.5VDC. Directly off the batteries, it was fine.

I've since come to realize that the 7805 needs at least 7.5VDC to operate correctly. Am I correct in assuming this was my problem and that the addition of two more AA cells would be the best way to solve my problem?

Estimated electrical characteristics of the panel:
4 ohms
1.25A maximum draw
500,000mcd @ 1.25A

The six NiMHs would provide 7.2VDC. Would this be enough or should I upgrade to 8 cells and use 9.6VDC instead? Either is acceptable, but the fewer required cells the better. The regulators I have available are rated at 1 amp, but there is a custom aluminum heat sink that is designed to be used for this circuit. Active cooling is not to be used due to excess current draw from the NiMHs and decreased running time.

Also, one last question. Does the regulator waste much energy?

Just a thought.. this is battery powered with LED loads that are quite heavy. Linear regulators will burn a significant fraction of your available power. Did you consider a more power efficient design? Small buck/boost converter circuits can provide better than 80% efficiency and take up just about the same amount of space as your linear IC.

 

[Particle]

I've been looking, and I found a buck/boost 5v DC/DC converter for about $2.50. My only problem is that I'd need many of them to achieve the power output needed. Are there larger packages available?

 

[Oznog]

There are some specifically meant to drive LEDs. Check Digikey.

Buck/boost converters can be done effectively with discrete components. You need a MOSFET, inductor, a Schottkey diode, current sensing resistor, and something to create a square wave adjusted to get a desired current (that's the tough part).

LEDs in parallel still need a resistor on each branch or they may not share current evenly and get toasted. The high boost ratio of putting everything in series requires a pretty big inductor. That's what's so great about Luxeons- one or two emitters in series, no current sharing resistors needed, and you've got something similar to the voltage of batteries commonly used to power it so the buck/boost ratio is modest.