Enough power ?

Excellent decision to replace AAA batteries with AGM batteries. A 4.5AH 6V battery will supply 4.5 A for 1 hour, or 2.25 A for 2 hours, etc., before recharge. One could build a simple charger, using the fact that for this case, the maximum charging current would be 0.45 A, or 450 mA. Note that VRLA/ sealed lead acid batteries are VERY rugged.

not quite. the 4.5ah rating is at the 20hr discharge rate. the ah rating will be lower at faster discharge rates. also hr needs a float charger that will provide a maximum of 450ma *and* limit the output voltage. it's not hard with an lm317, but for longest life and best performance (especially if you are likely to forget to remove the charger when finished) you should get a small 3 stage charger that is sized for this battery.

I did a test which validated a comment I saw that "LEDS will use all the current they can."

A nine LED light used .125 amps when powered by 3 AA batteries.

That same light pegged out the .25 amp scale when powered from a 1.5 amp 5.0 volt source.

Thus I have to limit the current to keep them from burning out. :-)

Andy

It is really best to current limit supplies to diodes. There are COTS chips for this. If you don't mind low efficiency, I think there is some hack of a shunt regulator to make it into a current limiting device.

ps

I am looking for something that will not allow the LEDS to draw more than 5

Would this work ?

Input voltage: 4-40V

Output Voltage: 1.5-35V (adjustable)

Output current: rated current 2A, maximum 3A (heat sink required)

Conversion efficiency: Up to 92% (the higher the voltage, the higher the ef ficiency)

Switching Frequency: 150KHz

Rectifier: Non-Synchronous Rectification

Module Properties: Non-isolated step-down module (buck)

Short circuit protection: current limiting

Operating temperature: Industrial grade (-40 ? to +85 ?) (o utput power 10W or less)

Full load temperature rise: 40 ?

Dynamic response speed: 5% 200uS

use a LM317 in current limiting mode.

Jamie

Or a transistor. LED goes from e to 0v, base has resistors from + and 0v rails

NT

The AAA batteries have enough internal resistance to drive (some) LEDs in parallel with the 'right' current. It's wasteful, though, because battery capacity must be derated for the high drain condition.

With 4.5V to play with, you could use a voltage reference, amplifier, and pass transistor to regulate the current. Or, you could use a series resistor (different resistor for each of the LED arrays, they match because they're thermally close). Or, you could use a two-transistor, three resistor current mirror (different mirror for each of the LED arrays). Or, you could go with a full switchmode regulated LED drive (overkill, IMHO).

If you want long battery life in emergency, consider a bright/dim switch, too.

What about two diodes in series with the supply line to the LED's, silicone diodes have 0.7volt voltage drop so if you have two on a 6volt battery that will give 4.6volts output,

voltage drop of 0.7 volts per diode

|\ | |\ | 6v in | \ | + | \ | + 4.6v out

I can not find any .7 volt zener diodes.

What about going to a 12.6 volt battery ?

There should be some solution to this. :-)

Andy

Current source. A buck would save a lot of energy.

You can buy simple switching power supply modules for LEDs that are 80% to 90% efficient. That'a bit better than throwing out the extra voltage (75%).

AGM? Ugh. They're heavy, fragile, and their efficiency is awful except at ultra low discharge rates. Given a choice between AGM and Alkaline for portable, I'd go with alkaline batteries. "D" cells are 10 AH for heavy loads and over 16 AH at lighter loads.

Lithium iron phosphate is usually used to directly replace AGM. NiMH is OK if you can charge indoors with no load.

What do you think of this in terms of efficiency ?