Building battery powered LED Lamp

Calculator83

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Nov 26, 2007
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I need to drive 100 leds 3.3v 20ma

Source+ --- 100leds para --- resistor --- Source-

Some one help me...

The Source and the Resistor is my question.?

 

Cogman

Lifer
Sep 19, 2000
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So, what is your question? How strong should the source be? How much resistance is needed? What? I would love to help but I don't know how, give me a hint. (also, cross posting is a bad thing, get the topic moved rather then posting a new one)
 

PottedMeat

Lifer
Apr 17, 2002
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wat?

Lets assume 100 LEDs in series, 3.3Vforward @ 20mA each. 100*3.3V = 330VDC@20mA.

You mean parallel? Resistance = ( Vsource - VLED ) / (LED Current) for each LED in the parallel ladder.

 

Calculator83

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Nov 26, 2007
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Originally posted by: Cogman
So, what is your question? How strong should the source be? How much resistance is needed? What? I would love to help but I don't know how, give me a hint. (also, cross posting is a bad thing, get the topic moved rather then posting a new one)

Right Right Right...


Ok So i DON"T know if i CAN use 100leds in Paralell like the Gentleman said above..

Which is why I'm actually skeptical of the whole THING..

I'm just trying to get as many LEds onto like 4nimhs AA s

So could I just hook up one big resistor to all of them?
 

bobsmith1492

Diamond Member
Feb 21, 2004
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It won't work to put too many in parallel. They won't share the current evenly; some will light but others won't.

Running it from 4AAs = 4.8V; that's only enough for one LED at a time (in series that is). What you'd have to do would be to either put them all in parallel with a resistor in series with each one (not practical) or use a boost switcher configured in constant-current (most efficient but would take some circuitry).

At any rate, that's 6.6W; you'd never run that from 4 AAs... maybe half that many LEDs at most.

Your best bet would be some lower number of LEDs, say 20, with a resistor in series with each individual LED. 75 ohms should do the trick. It wouldn't be that great, though, without a dedicated LED driver.
 

Calculator83

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Nov 26, 2007
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Originally posted by: bobsmith1492
It won't work to put too many in parallel. They won't share the current evenly; some will light but others won't.

Running it from 4AAs = 4.8V; that's only enough for one LED at a time (in series that is). What you'd have to do would be to either put them all in parallel with a resistor in series with each one (not practical) or use a boost switcher configured in constant-current (most efficient but would take some circuitry).

At any rate, that's 6.6W; you'd never run that from 4 AAs... maybe half that many LEDs at most.

Your best bet would be some lower number of LEDs, say 20, with a resistor in series with each individual LED. 75 ohms should do the trick. It wouldn't be that great, though, without a dedicated LED driver.

what does a dedicated led driver consist of.

and HOW many Watt,,,, Amp can 4x 2500mah output?
 

alpineranger

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Feb 3, 2001
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NiMH has some pretty serious internal resistance. It's like a resistor in series with each battery that you can't get rid of. Draw too much current and you'll spend a lot of it heating (and destroying) the batteries.

Here's some more things to keep in mind. If you have a bunch of LEDs you will want to run them in series, not in parallel. There are several reasons why. First, you would otherwise have to have a current limiting resistor for each led - not practical. Secondly, even if that was practical, real resistors and diodes have slight manufacturing differences which means (because the V-I relationship for diodes is exponential) you will end up with wildly varying power (and thus light output, lifetime, efficiency, etc) to individual diodes.

The drivers mentioned are constant current sources. They ensure, especially with multiple LEDs, that the devices are kept in the proper operating range. Fortunately LED drivers come in chips that you can buy cheaply, follow the app notes, and have yourself a working circuit.
 

PottedMeat

Lifer
Apr 17, 2002
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A dedicated constant current LED driver usually consists of a buck/boost/buck-boost controller chip + low value current sense resistor + inductor + schottky rectifier + some electrolytic capacitors.

Sounds like this is just a one time thing you want to get working though, so in your case I would just do the parallel thing. 4.8V @ 2.5Ah - ballpark calculation gives ~125 LEDs with 75Ohm resistors in series with each LED. Fry's sells 100 packs of resistors for something like $1.99.

See how long it lasts ( should be < 1 hour ) and cut down the number of LEDs until you get a satisfactory lifetime.

 

Calculator83

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Nov 26, 2007
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Originally posted by: PottedMeat
A dedicated constant current LED driver usually consists of a buck/boost/buck-boost controller chip + low value current sense resistor + inductor + schottky rectifier + some electrolytic capacitors.

Sounds like this is just a one time thing you want to get working though, so in your case I would just do the parallel thing. 4.8V @ 2.5Ah - ballpark calculation gives ~125 LEDs with 75Ohm resistors in series with each LED. Fry's sells 100 packs of resistors for something like $1.99.

See how long it lasts ( should be < 1 hour ) and cut down the number of LEDs until you get a satisfactory lifetime.

Is there a way I could NOT use 100s of resistors?
 

PottedMeat

Lifer
Apr 17, 2002
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You could spend ~15 bucks and buy a Luxeon or Cree 1Watt or 3Watt LED and a little driver circuit from dealextreme.
 

bobsmith1492

Diamond Member
Feb 21, 2004
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With one of these you could run 14 of the LEDs and maintain constant brightness until your batteries died.

7 in series with one resistor as one string and another string in parallel with that one

Here's a better one: 7 LEDs in series with a 45ohm resistor as one string, then hook 4 of those strings in parallel to get 28 LEDs. You'd need 6 AA NICADs or 4 alkalines, though.

One more for consideration: With this, you could run 9 LEDs in series with a 15-ohm resistor. Put 10 of those strings in parallel and you have 90 LEDs - that's as high as I'll go because I think your batteries won't make it with more than 28 LEDs or so. You'd need 6 AA NICADs or 4 alkalines, too.

One of these driving 40 LEDs would be the optimum solution - ~90% efficiency and complete control over your LED current. Unfortunately Digi Key is out of stock and Newark and Mouser don't stock them. It would require 6 NICADs or 4 alkalines and a couple of extra, carefully-selected parts as well. This is a real LED driver.
 

Mark R

Diamond Member
Oct 9, 1999
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LEDs shouldn't really be connected in parallel because there is natural variation in the LED's forward voltage. This means that the ones with naturally low voltage hog all the current and light up extra bright and burn out, and those with higher voltage, are dim. As the LEDs blow, so the current becomes progressively more concentrated through the remaining LEDs until they all blow.

You could certainly connect then in strings of 5 or 6 in series - each string with it's own resistor.

However, personally I wouldn't bother. Just buy a Luxeon K2 Star LED, and either use a suitable resistor (or preferably an appropriate LED driver circuit). One luxeon K2 LED will be brighter and more reliable than 100 standard LEDs. Just don't forget to heatsink the luxeon - some RAMsinks, or a northbridge cooler with the fan removed would be excellent.

 

Calculator83

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Nov 26, 2007
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Originally posted by: bobsmith1492
With one of these you could run 14 of the LEDs and maintain constant brightness until your batteries died.

7 in series with one resistor as one string and another string in parallel with that one

Here's a better one: 7 LEDs in series with a 45ohm resistor as one string, then hook 4 of those strings in parallel to get 28 LEDs. You'd need 6 AA NICADs or 4 alkalines, though.

One more for consideration: With this, you could run 9 LEDs in series with a 15-ohm resistor. Put 10 of those strings in parallel and you have 90 LEDs - that's as high as I'll go because I think your batteries won't make it with more than 28 LEDs or so. You'd need 6 AA NICADs or 4 alkalines, too.

One of these driving 40 LEDs would be the optimum solution - ~90% efficiency and complete control over your LED current. Unfortunately Digi Key is out of stock and Newark and Mouser don't stock them. It would require 6 NICADs or 4 alkalines and a couple of extra, carefully-selected parts as well. This is a real LED driver.

Looking at these make me feel Soo incredibly N00b. I have no idea where to begin.


Originally posted by: Mark R
LEDs shouldn't really be connected in parallel because there is natural variation in the LED's forward voltage. This means that the ones with naturally low voltage hog all the current and light up extra bright and burn out, and those with higher voltage, are dim. As the LEDs blow, so the current becomes progressively more concentrated through the remaining LEDs until they all blow.

You could certainly connect then in strings of 5 or 6 in series - each string with it's own resistor.

However, personally I wouldn't bother. Just buy a Luxeon K2 Star LED, and either use a suitable resistor (or preferably an appropriate LED driver circuit). One luxeon K2 LED will be brighter and more reliable than 100 standard LEDs. Just don't forget to heatsink the luxeon - some RAMsinks, or a northbridge cooler with the fan removed would be excellent.

Really? just ONE. Does this thing have major power requirements? Will I be able to run it off NiMH batts + resistor ,, like,, well?

I'm willing to go with an LED Driver thingamabober, but they seem to just look like a Chip, with a bunch of Pins,, What do I solder what to? And which one would be best for the Lxeon K2
 

Mark R

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Oct 9, 1999
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Originally posted by: Calculator83

Really? just ONE. Does this thing have major power requirements? Will I be able to run it off NiMH batts + resistor ,, like,, well?

I'm willing to go with an LED Driver thingamabober, but they seem to just look like a Chip, with a bunch of Pins,, What do I solder what to? And which one would be best for the Lxeon K2

Yes. Really.

If you get the high-power White K2 star (the 1500 mA one), it really is somewhere between 80-100x as powerful as a cheap white LED. It needs 80-100x the power though, so it gets very hot, and will burn out in seconds if it isn't attached to a heatsink - it doesn't need to be massive, just big enough to keep the luxeon's heat spreader temperature below 70 C.

You can use a resistor - but this is wasteful of power, and the resistor will get very hot. A better option is to use a dedicated high-power LED driver module. Like this one.

This driver is only 1000 mA - whereas the big white luxeon K2s are 1500 mA. This means you won't get full brightness, but the LED will run cooler and more efficiently.

The other advantage of using a proper driver is that the driver will automatically compensate for voltage fluctuations in the battery, so that you always get constant brightness. You can also change the battery voltage for longer run time. E.g. if you have 6 NIMH cells in series for 7.2 V, then you could upgrade to 10 NIMH in series (12 V) for upgraded run time. The driver will automatically regulate the power to the LED and only draw as much power from the battery as required.

The driver will also automatically compensate for additional LEDs in series. If you find that one luxeon isn't bright enough. Just connect a second identical one in series with it. The driver will automatically adjust for this (as long as your battery voltage is high enough).
You'll need a minimum of 5 NIMH in series for 1 luxeon, and 8 in series for 2, and 10 in series for 3 LEDs.

The driver I linked is also dimmable. You can connect a 5k ohm potentiometer to the module to dim the LED(s), or disconnect it to run at the driver's max power.
 

Calculator83

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Nov 26, 2007
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Originally posted by: Mark R
Originally posted by: Calculator83

Really? just ONE. Does this thing have major power requirements? Will I be able to run it off NiMH batts + resistor ,, like,, well?

I'm willing to go with an LED Driver thingamabober, but they seem to just look like a Chip, with a bunch of Pins,, What do I solder what to? And which one would be best for the Lxeon K2

Yes. Really.

If you get the high-power White K2 star (the 1500 mA one), it really is somewhere between 80-100x as powerful as a cheap white LED. It needs 80-100x the power though, so it gets very hot, and will burn out in seconds if it isn't attached to a heatsink - it doesn't need to be massive, just big enough to keep the luxeon's heat spreader temperature below 70 C.

You can use a resistor - but this is wasteful of power, and the resistor will get very hot. A better option is to use a dedicated high-power LED driver module. Like this one.

This driver is only 1000 mA - whereas the big white luxeon K2s are 1500 mA. This means you won't get full brightness, but the LED will run cooler and more efficiently.

The other advantage of using a proper driver is that the driver will automatically compensate for voltage fluctuations in the battery, so that you always get constant brightness. You can also change the battery voltage for longer run time. E.g. if you have 6 NIMH cells in series for 7.2 V, then you could upgrade to 10 NIMH in series (12 V) for upgraded run time. The driver will automatically regulate the power to the LED and only draw as much power from the battery as required.

The driver will also automatically compensate for additional LEDs in series. If you find that one luxeon isn't bright enough. Just connect a second identical one in series with it. The driver will automatically adjust for this (as long as your battery voltage is high enough).
You'll need a minimum of 5 NIMH in series for 1 luxeon, and 8 in series for 2, and 10 in series for 3 LEDs.

The driver I linked is also dimmable. You can connect a 5k ohm potentiometer to the module to dim the LED(s), or disconnect it to run at the driver's max power.

Ok,, I think this is cool.. So I need to Order THAT driver.

Which LED do I order, they all seem the same to me. K2,, K2 Star,, III,,, VI,
I think I want my Light to be with One white and One blue..

How come the Blue ones have such lower Lumens.??
 

Mark R

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Oct 9, 1999
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Originally posted by: Calculator83

Ok,, I think this is cool.. So I need to Order THAT driver.

Which LED do I order, they all seem the same to me. K2,, K2 Star,, III,,, VI,
I think I want my Light to be with One white and One blue..

How come the Blue ones have such lower Lumens.??

That driver is 1000 mA, and will drive a regulated 1000 mA through the LEDs connected to it.

There are 2 types of K2 LED. 1500 mA ones and 700 mA ones. If you connect a single 700 mA LED to the that driver, then it will burn out the LED. There are ways around this - e.g. you could connect 2 blue LEDs in parallel. I know I said LEDs are best not used in parallel, but the luxeons have better quality control and don't vary as much as cheaper LEDs, plus if you connect 2 7000 mA LEDs to a 1000 mA source - you're underpowering them, so you've got a reasonable safety margin, in case one hogs the power.

There are a whole bunch of different types of luxeon LED.

The first distinction is the difference between the emitter LED and the star LEDs. The Star LEDs consist of a emitter soldered onto a metal star shaped heatspreader. The heat spreader also has solder pads where you can attach wires - this is better because you don't have to solder the fragile and heat sensisive LED directly. Don't get the emitters, you need a heatspreader - but the emitter die is only about 2 mm in size which means perfect contact is required. The emitter core has to be soldered to the heatspreader in order to prevent the core from overheating. The stars are ready to go from the factory.

Luxeon I is the original design - 1 W of power, about 15-25x as powerful as a conventional LED.

Luxeon III is a more modern, more powerful version - 3 W of power, about 50-80 x as powerful as a conventional LED.

Luxeon V was a variant of luxeon I - basically 4 luxeon I chips in a single package. Very expensive, but very powerful.

Luxeon K2 is the very latest in LED technology. Available in 2 power levels - roughly equivalent to luxeon III and luxeon V - but these are single core, not quad core like the luxeon V.

There is an important problem with the Luxeon III and V star LEDs. The heatspreader is soldered directly to the LED die. This means that the heatspreader is connected to the LED electrically. You can't put multiple III or V LEDs on the same heatsink, because the power will flow through the heat spreaders and heatsink.

In the Luxeon K2 stars, more modern thermal technology allows the heatspreader to work, while being electrically insulated from the LED. This means there is no problem putting multiple LEDs on a single heatsink.

As to the difference in brightness - it's several factors:
1. There are 2 white LEDs on that page with 2 different power - 700 mA and 1500 mA. Whereas there is only 1 blue (700 mA).
2. The human eye is less sensitive to blue light, so blue LEDs appear dimmer (the 'lumen' takes into account how bright something looks to the human eye).
3. The catalog page I linked have been careless in the specs. Luxeon provide two power specifications - light at 50% electrical power, and light output at 100% electrical power. They've put the 50% rating down for the blue, and 100% for the white.

The way I'd do it, is to use the 1000 mA driver. Then use a SPDT switch to connect the driver either to: 1 White 1500 mA LED, or 2 blue 700 mA LEDs in parallel. All the LEDs can be attached to the same heatsink if you use K2 stars.

One final point. Luxeon LEDs have a very wide beam - they are like flood lights. Normal LEDs have a very narrow beam (spot lights). A lot of this is marketing. Cheap LED makers like to put very high intensity numbers on their spec sheets (mCd). By making the beam narrower, they make the beam more intense (so more mCd) even though the LED is the same. If you want a narrow beam from luxeon LEDs, then you can get add-on lenses that will change the beam shape. There is a wide variety of lenses to suit different purposes.
 

blahblah99

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Oct 10, 2000
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If your LEDS are 3.3V as you say they are, then 100 of them in series would net you 330V... so get a low current high voltage dc dc converter > 330V to do the job and stick 1 resistor in there.

That is, if your application allows voltages greater 330V to be running around...

If not, then you'll have to parallel them with a series resistor in each led and have at least a supply capable of sourcing 2A of current.
 

Twinpeaksr

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Aug 9, 2000
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Blue LED are GaN based and proved more difficult due to the shorter wavelength requiring wider band gap. GaN are also used for White and Green LED (usually mixed to InGaN). White LED is usually a Blue LED with a Yellow Phosphor to produce While light.

The other thing to remember is that your eyes perceive color differently. Your eye is more tuned to pickup green than Red, so you will need a less powerful green LED to have the same perceived brightness as a Red one. This curve also shifts depending on day or night (Photopic vs scotopic). In this shift, the peak moves more towards the blue end of the spectrum, hence blue is easier for your eye to pickup at night then during the day.

To answer why blue LED are lower power output than white is easy: White are used for backlighting applications and hence have more light loss in there application than an indicator application (transmission through light pipe, LCD, etc...) so most LED of very high brightness are white. Don't be fooled though, you will not need an equivilent power colored LED to produce a very bright light.

Word of caution, be careful with high output LED, they are VERY bright (I have some K2 LED as well as some OSRAM Golden Dragon and Avago Moonstone) try to avoid looking directly at them, they can hurt your eyes, even in the 50mA range.

One last note, use a constant current source whenever possible. LED forward voltage is not fixed, it will vary with temperature and current. Also, the wavelength of light will shift (most noticeable with LED over temp). If you use a resistor and a fixed voltage (which a set of Batteries will not deliver) you will have brightness variations that can be noticeable. Use a constant current source and you will not see the effects (well you may if you look really hard and have a comparison, but I am guessing you are not looking to operate in the 85C temp range). If you want to run in a battery operation you can also use a PWM drive to reduce power consumption, just make sure your duty cycle is high enough, minimum 200Hz, I usually design for 1Khz.

Hope this helps.