Reduce 6 Vdc to 4.5 Vdc

You could simply use a TLV431 and two resistors and drop 1.5V -OR- go more exotic with an LDO -OR- even put in a switcher and be quite efficient.

What do you want? ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

You have asked to hit a voltage but LEDs require a *current* drive.

If all it is to do is power some LEDs then a resistor to drop 1.5v at their nominal operating current is as simple as it gets. The ball park would be about 200mA current so 8.2R or 10R @ 0.5W is about right.

You could do a switched mode constant current source which would be more efficient. Chips exist to drive LEDs from a single cell.

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Regards, 
Martin Brown

A AAA battery is good for roughly 1 amp-hour, so your load is very roughly 200 mA. The simplest thing would be a series resistor, around

7 ohms, 1/2 watt. Try 5, 7, 10 ohms and see how you like the brightness.

Fancier circuits could give more constant current and squeeze out more battery life.

--

John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

...[to] power some LED lights.

If you want more time, you can throttle the current to something lower (add a series resistor). Slightly dimmer, much greater light time.

If this is a flashlight, some (inexpensive) ones need the AAA batteries' internal resistance, so adding batteries (like, to get to 6V) will vastly change the operating point of the LEDs (and maybe burn them out). If you add batteries AND add a resistor, the four batteries will go flat as quickly as three did (the extra voltage goes to waste heat).

Simple way: add resistor in series, that lowers the brightness but increases battery longevity.

Complicated way: put a current-limiting drive circuit in (this has the advantage of allowing AA or D cells instead of AAA, and various chemistries instead of just 'alkaline'). It'll stay full brightness until the batteries fail, then go dim FAST.

Most effective way: use a switchmode converter (which allows lower battery count as well as higher battery count, and doesn't waste most of the energy). That's complicated, though, and these days mainly requires lotsa resources (print up a PC board lately?).

Less, actually. But the light level is more stable.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com

Hmmm...if you have a 3 AA cell LED light string, one of those stick on mounts. Instead of powering with 4.5 worth of batteries you want to power it with a 6 Vdc supply, I don't know you might get away with it because it would simply 'look' like a larger voltage, brand new battery.

Siimple answer is try it. If the LEDs get super bright, stop it. If they stay simply bright like new batteries and nothing explodes, go for it.

Need more information.

I want to eliminate the rapid battery usage of some bicycle lights.

More details on what I would be powering using a

6 Volt - 4.5 Ah - UB645 - AGM Battery

Each of these uses 4.5 volts.

One light has 8 leds.

2nd light has 12 leds.

Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps

Thanks for your help gentlemen.

Andy

Depends on how "fancy" you want to go. A "Joule Thief" will squeeze more light out of a battery than a resistor ballast will.

I was thinking about a boost/buck switcher.

--

John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

Two silicon diodes in series?

The battery has a resonably flat discharge curve.

Assuming your load is ~1.25 amps.

I was going to say LM2575, but the Vsat gives you little headroom. I think a small DC-DC converter will work instead..

14V in so you can charge while the lights are on.

For $6 its worth try. I dont know what the LED current is so you could power 1 LED off one and the other LED and Bulb off a second one.

Cheers

very good! unaffected by load. or use a 2N3055 wired as a 'diode multiplier' if gets too hot.

This:

should do the trick for all the above. It is switch mode so it is efficient. You can set it for 4.5 volts and use it that way but you will get better, longer lasting results if you use it in constant current mode.

To do this you need a multimeter, crank the output voltage up as hight as possible (this is okay because the current will be limited - now forget about the voltage control - you will not use it again). Use it in constant current mode. Now set the current limit to the minimum or if that is not marked set the current limit to about half way. Set your multimeter to the

10Amp current range and use it to short out the output of the module. Power the module and adjust the current limit control to .65 amps for the 3watt led and 240 ma (or less) for the 12 led unit and 160 ma for the 8 led. You may have to change current range to get a good reading - make sure the current limit is set to a minimum before you do this.

For best results (efficiency and longest run time) bypass and resistors and electronics in the lamps and drive the leds directly from the modules.

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

So many ideas gentlemen, but I am overwhelmed.

I am a "recovering perfectionist".

Andy

Depends on the voltage drop and discharge curve, of course. They'll drop off pretty much the same way, at least for a conventional (resistor biased) 'thief'.

Point was, a proper CCS will drain the battery much more sharply due to the increasing current draw.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com

The problem with your plan is these devices that you wish to power are already engineered to run from a specific voltage. So unless you are going to open these devices up and re-engineer them, your best bet is to buck the AGM down to 4.5VDC and just power these devices directly.

To elaborate, your lights are probably not just a bunch of LEDs, but LEDs with some control circuitry.

"Andy K"

** Last time you asked this question -

you had three 4.5V LED lights and a 12V gel battery of 5AH.

So wiring the lights in series was a perfect solution.

What happened ?

.... Phil

"John Fields"

** Really ?

What is the full charge voltage of a nominal 12V SLA battery ?

What is the end point voltage of same ?

What is the end point of 9 x alkaline cells ?

BTW:

I can see the wink....

.... Phil