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View Full Version : [ELECTRONICS HO!!] Need LED project help.

Grind Hard
02-09-2012, 07:11 PM
Ok I got a kindle. Best thing ever for a readaholic like me. LOVE IT.

I need a portable source of light.

On paper I have a "kindle holder" that I intend to fabricate out of brass. I would like to add some LEDs arranged at the top to provide light for reading.

Ideally I would like to use white LEDs, have a simple click-switch, and run the whole ball of wax off a nine-volt battery. Say four LEDs.

Is it as simple as wiring four LEDs together, wiring in the switch and attaching the battery? Or do I need OTHER PIECES?

AiR_GuNNeR
02-09-2012, 07:21 PM
LED's will draw all the current you can give them. Think of them as a dead short. You will need a current limiting resistor wired in series with them or they will quickly blow out.

Evan
02-09-2012, 09:15 PM
LEDs are a little more complicated than a light bulb. There is a minimum voltage that it must have depending on colour. For a white that is about 3 volts. With 9 volts you can then run three in series. BUT, they also have a maximum current rating. That depends on the type of LED, not the colour. The ordinary little clear plastic 5mm LEDs run on .020 amps. To figure that out you must compute what size of resistor will allow .020 amps to flow with 9 volts. Per Ohm's Law R=E/I where R=resistance, E=voltage and I=current. 9/.020 = 450 ohms. It does not need to be exact so something close will do. 470 ohms is a standard value.

I have ignored the effective resistance of the LEDs but that will always result in a conservative value so that isn't an issue. The LEDs are also polarised, unlike light bulbs. With a 5mm LED the leads are different length. The longer lead is the positive lead so they have to be connected in series with pos to neg to pos to neg etc. All 4 parts go in series. The resistor can be on either end and in either direction, it doesn't care.

In this particular case since the LEDs add up to 9 volts and the battery is a max of about 9.5 volts the effective resistance of the LEDs makes a big difference. You can get maximum brightness at the expense of battery life by using a 27 ohm resistor. DO NOT assume that means you can run them from a 9 volt power brick. They aren't really 9 volts.

darryl
02-09-2012, 11:02 PM
A normal white led will give light with 2.7 volts, and will be ok with up to about 3.4 volts. At 2.7, it will draw very little current and not be very bright- at 3.4 volts is will draw about 20 ma, which is about as high as you'd want to go. It will be quite bright at that level.

3 leds in series therefore will light up on about 8 volts, and will tolerate about 10 volts. A fresh 9v alkaline battery will deliver about 9.6 or so volts, as Evan suggested. At 3.2 volts each, the leds will draw about 15ma, which is close to optimum.

This is without a series resistor or any other component except the leds themselves, and the current level is easily within the safe operating area.

If you want more brightness than three leds can deliver, use two or more strings in parallel. Each string you add will draw another 15 ma or whatever, so now you calculate the draw on the battery based on that.

A 9v alkaline has a capacity of about 200 ma/hr, or 12 hrs with a 15 ma load, 6 hrs with a 30 ma load, etc. The more current you draw, the less of the ma/hr capacity you get to use. You wouldn't expect to get 3 hrs at a 60 ma draw for instance, but you might actually get more than 12 hrs at 15 ma. This gives you an idea of how much light you can get for how long, given the 9v alkaline as the power source. It is not quite this simple, though. As the battery voltage drops, the current through the leds also drops, which lessens the load on the battery. It also drops the brightness. The end result of this process is that the leds can still be glowing, although dimly, and the battery sees very little loading so the voltage doesn't drop very quickly anymore. You can find that you still have some light long after the bulk of the battery's capacity is gone. But of course it may not be enough to read by.

Batteries are great in one sense- they have a fully charged voltage level, a lower level under load, and a very low level where they are basically spent. The high level for a fresh alkaline will still be safe for 3 white leds in series. But as Evan also suggested, you can't expect this 'safety' if you use an adapter. A high level for a 9v adapter might be as much as 14 volts.

This is interesting though. If you started with an adapter which has more than about 13 volts output under a small load like a string of leds, you can use a regulator instead of a resistor to control the current through the leds. With one 3 legged part and one resistor, you can design in how much current you want the leds to draw, and it will remain constant within a close degree. All that's needed is about 3 extra volts from the power source.

Which brings us back to the battery- 2 leds in series will take about 6.6 volts to give an optimal operating point. For a fully charged alkaline, that leaves 3 excess volts. You are now able to regulate the brightness of the leds such that the light output remains stable from fully charged until the point where the battery voltage drops so low that the regulator doesn't get its minimum voltage. Then the leds will drop in brightness.

You mentioned 4 leds- that can be 2 strings of 2 leds each. Parallel those 2 strings, add the simple regulator circuit, and get about 5 to 6 hours of steady brightness out of it.

So, you have a few choices from super simple to a little more complex, but very little so.

darryl
02-09-2012, 11:12 PM
Got a little windy on the last post, but I had one more thing to add- I have an led flashlight which I made, uses one led and one lithium cell. Fully charged, the cell is about 3.4 volts. I wired it up to power the led directly, no regulator or resistor. It worked fine for about a year, and was still working fairly brightly when I lost it a couple months ago. Just the other day I found it, hiding in some underwear in my overnight bag. Still works, still quite bright. When it quits, it will be because my homemade switch contacts will have oxidized.

There is nothing wrong with using this simplest of circuits.

RWO
02-10-2012, 02:19 PM
Why reinvent the wheel?

http://www.amazon.com/b?ie=UTF8&node=1289281011

RWO

dockrat
02-10-2012, 02:25 PM
What RWO said. At today's price of brass and whatever your time is worth I think I would be buying one of those 15\$ kindle lamps from amazon.

garagemark
02-10-2012, 02:31 PM
Because we are hobbyists.

Even though this isn't my project, I learned some things in replies that may come in handy someday.

Most of us could buy most any part that we make on our machines. But that just ain't the same as sculpting it out of a block of XX. There's just something about a hobby. It isn't even always cost effective, it's just cost infectious.

I think reinventing the wheel is sometimes a very cool thing.

Mark

Grind Hard
02-10-2012, 04:15 PM
Why reinvent the wheel?

http://www.amazon.com/b?ie=UTF8&node=1289281011

RWO

What RWO said. At today's price of brass and whatever your time is worth I think I would be buying one of those 15\$ kindle lamps from amazon.

So what is the workshop full of tools for? So you can log on here and say "yep I have a genuine 1957 Model OMG-1 DS&G lathe with the thingy, the other thingy and the SUPER RARE ULTRA COVETED thingy. Mint condition, all I had to do was take it apart, grind and scrape the ways, fabricate new gears and cast a tail-stock!" :D

I use my tools to make things. Things that are useful to me. Because I ENJOY MAKING THINGS. Especially out of sheetmetal.

The cost of of a sheet of brass... to me... is trivial. I can get it from work from the drop-bin behind the shear. Because of my value to the company they let me take a reasonable amount of scrap for projects.

Making it myself -- if one person stops and says "where did you get THAT from? That's cool!" and I can explain my trade to that one person... totally worth the blood sweat and tears invested in my workshop. :)

And in the end... I don't see a kindle-holder with light that I actually LIKE on Amazon.

TO THE WORKSHOP! WHERE I SHALL REMEDY THE SITUATION -- :cool: <-- safety glasses

Evan
02-10-2012, 08:02 PM
A fresh 9v alkaline battery will deliver about 9.6 or so volts, as Evan suggested. At 3.2 volts each, the leds will draw about 15ma, which is close to optimum.

Maybe, some of the time. There is a lot of difference in the battery internal resistance between cheap brands and the top of the line. They are not all the same. It is entirely possible to overdrive 3 5mm whites with a fresh high quality 9 volt battery. There are also differences in the LEDs of up to several tenths of a volt in operating voltage and depending on the die the current/voltage curve may be different too. The safest bet is to use a current limiting resistor.

For this sort of application it would be a significant waste of power to use a linear regulator as a current limiter. Driving two leds with a 9 volt battery and a regulator would waste over a third of the power. There are lots of nice little PWM boost drivers with very high efficiency but that is getting a bit more complicated.

ikdor
02-11-2012, 01:33 PM
I would take another approach for a quick solution, run it off two penlites using a white led driver for flashlights. The dealextreme ones come to mind, but I'm sure you can find them in other places as well.
Run the leds parellel, if they're from the same batch their internal resistance will even out the slight mismatch. If it doesn't, put a ten ohm resistor in front of each led.

Two penlites store much more energy than the 9V block.

If I'd do it the hard way, I'd use a "white led driver" IC that runs on 1 penlite. Those drivers are specialised step up converters so you can connect all leds in series without any series resistors to waste power.
But then again, I'm more comfortable with electronics.

Igor

danlb
02-11-2012, 02:17 PM
The popular booklites use a simple design.

3 AAA batteries in series provide from ~4.7 volts (when fresh) to 3.2 when they are about dead. A simple resistor is used to limit the current to the LED(s).

A single 5mm white LED is enough to light a kindle from a distance of 6 inches. A second LED in parallel is nice to have when the battery voltage starts to fall. Light is additive, so two dim LEDs will be brighter than just one using the same battery. Making them parallel keeps the voltage requirement low.

LED considerations: The "correct voltage" for LEDs vary from batch to batch even from the same company. Putting 3 in series MIGHT cause one to be very bright and the others to be less bright. There is a minimum voltage required just to get them to light up, usually around 2.7 volts.

Battery considerations: A 9 volt battery is essentially a bunch of little button cells in series. At 20 ma current, the battery voltage will quickly fall below 9 volts in an hour or two. At 40ma it will drop almost immediately below 8 volts. Once the voltage is less than 8.1 volts ( 2.7 volts per LED ) all three will go out. A 9 volt is a very poor choice for a book light. It is expensive too. A good tech sheet is at http://www.duracell.com/media/en-US/pdf/gtcl/Product_Data_Sheet/NA_DATASHEETS/MN1604_US_CT.pdf

My wife uses the Mighty Bright Blue Xtraflex 2 LED Book Light with her Kindle every night running off 3 NiMH Eneloop AAA batteries. Every two weeks or so I swap the batteries. It's plenty bright enough with only one LED on. Assuming rechargeable batteries, that's a good design to emulate.

Dan

Grind Hard
02-13-2012, 08:22 AM
Ok lots of good information here, thanks!