9 led flashlight power supply help needed

[Abner]

I would like to build a 120vac power supply for these 1" dia x 3" 4.5 v. 9 led flashlights.

They are nice and bright, and could lend themselves to a wide variety of handy applications.

I have looked some on the net and my eyes glaze over because the information is more technical than I need.

I have tried working the problem from the led supply needs. I'm thinking of looking at what batteries can supply.

Anyone else tried to do this?

[torchroadster]

I haven't tried to build one but it should be pretty straightforward. What do you know about the LEDs? The key parameters that will affect the power supply are:

1. Are the LEDs in series or parallel connection?
2. What is the forward voltage drop of each LED? Most are about 2V
3. What is forward current rating of the LED?

Knowing this we can figure out the power supply. If you already have a battery powered flashlight and a multimeter you could make a couple measurements under battery power and estimate what you need.

An easier way, but less accurate is just make a guesstimate based on the number and size of recommended batteries.

[Abner]

I don't think I can easily find specs on the leds. I have a good meter.
measure voltage and amperage under load?
They use 3- 1.5 v AAA batteries.

[torchroadster]

I don't think I can easily find specs on the leds. I have a good meter.
measure voltage and amperage under load?

Yes. If possible use new batteries.

[lwalker]

The LEDs in those flashlights are arranged in a parallel array and for the most part the lights are very poorly designed, so do NOT just use a 4.5V DC supply generated from AC!! That will result in lots of exciting smoke and perhaps flames and exploding LED lenses.

The reason is that the flashlights depend on the internal resistance of the batteries limiting the current to the LEDs. If you use a supply with more current capacity than those batteries, the LEDs will be overdriven.

There are multiple ways to solve the problem. The ideal way to drive LEDs is with a constant current supply and they are easy to find these days. But it's not the simplest.
The simplest is using a resistor inline with your supply. You can purchase a 5.0V supply very cheaply. Also, cast off cell phone chargers are often 5 or 4.5V. This will be marked on the case.

So what resistor value to use? The average white LED has a forward operating voltage of 3.7V. This voltage varies from LED to LED and with temperature, but 3.7V is a good average. Those LEDs in the flashlight are likely safe up to about 20 milliamps (0.02A) so with a 5 volt power supply, an appropriate resistor to limit the current to 20mA would be (5 - 3.7)/0.02 = 65 ohms. Closest standard resistor value is 68 ohms. Using 20mA at 5V is .02 * 5 = 0.1W so a 1/4 watt resistor will handle the power easily. You can buy 200 of these 68ohm resistor from Digikey for about $2. Also available from many other suppliers. I have been buying from these guys for probably over 20 years and they are nice to deal with. They're a surplus dealer and it's a lot of fun browsing the site.

I have a basic LED tutorial here that I should really update. I just noticed that due to website restructure, some of my pictures are not linking.

Happy to answer any questions :-)

Lyndon

[Paul Alciatore]

You really need to know more. Since your "LEDs" are rated at 4.5 Volts, it is unlikely that they are simple LEDs. There is a good chance that there is a built in resistor. Or perhaps not. You really need to know more about these "LEDs". Can you get a spec sheet on them?

If you can not, then I would think that they are designed to work on three flashlight batteries. What size/type of batteries were used in the original flashlights? We could work from the specs of those batteries. Perhaps just a regulated (Voltage regulated) supply would work. But I would not use a 5V supply as it may make smoke, you would need a 4.5 Volt supply or a 5 Volt one that could be adjusted down to that level.

I haven't tried to build one but it should be pretty straightforward. What do you know about the LEDs? The key parameters that will affect the power supply are:

1. Are the LEDs in series or parallel connection?
2. What is the forward voltage drop of each LED? Most are about 2V
3. What is forward current rating of the LED?

Knowing this we can figure out the power supply. If you already have a battery powered flashlight and a multimeter you could make a couple measurements under battery power and estimate what you need.

An easier way, but less accurate is just make a guesstimate based on the number and size of recommended batteries.

[J Tiers]

This has been designed already.

A company called Power Integrations makes a bunch of chips suited for it, and in fact is supplying the power supplies for the CREE LED bulbs, as I recall.

Start here: The parts are available through Digikey and Mouser in the US. (you do not give your location)

http://led-driver.power.com/products/lytswitch-family/lytswitch-4/

They have plenty of pre-designed examples, etc. etc. Their parts are typically very easy to use, and they make non-LED supply parts as well.

If that is too much information, too technical, then you may want to do something much simpler, or not do the project at all. A transformer, rectifier and a resistor to set current would do the job, but is bulky, heavy, and not very elegant.

[lakeside53]

Scan ebay... there are zillions of led drivers available, packaged and bare. Unless you really want to "roll your own", they are very cheap to buy... including ac or optimized for battery use.

Your leds are not "rated at 4.5v" - they just happen to get connected to 3 x 1.5v (nominal) batteries). As other have suggested, yours likely has some form of current limiting device, probably a resistor already either per led or one resistor for all. If they were designed to run off AA or whatever batteries there a very good chance they will run at 5v without any adjustment, and if you are concerned, use an adjustable supply or add a suitable series resistor. If it was mine.... I'd use a constant current supply and "within reason" ignore the voltage. You could probably use a small USB supply (5v) - even those that are part of a duplex outlet (even 3 amp if you like). The body of your flashlights (if you are keeping that) could house any additional series resistor if you need it.


You might want to measure the current flowing at 4.5v, just get a constant current supply about the same current. If (example only) it happened to be 600ma (it won't!), then this supply would be perfect. http://www.ebay.com/itm/3W-LED-Drive...item3ce4a3b33b Example only... sooo many to choose from.

Also.. unless you have the "9 led" device in hand, look on ebay. There are so many similar devices to choose from even those with just one led of almost any power you want.

[mattthemuppet]

Personally I don't think it's worth the effort as you can get more and better light with discrete components, but if you want to use what you have just do what lwalker wrote above. To add to what he wrote, I'd suggest measuring the voltage across the LEDs first and then calculating the resistor value in case the voltage is closer to 3V

[jdunmyer]

Although I don't claim to be an expert on LEDs, I did exactly what the O.P. wants. I have a motor/generator that I belt up to one of my engines with the output voltage varying between about 7 and 11 volts. It goes to a 7805 voltage regulator I.C., through a resistor, to the light. Before doing this, I connected the light to an Astron variable voltage supply and played around a bit. Found that the light was "about right" with 3 1/2 volts feeding it, more or less. It definitely looked like it was going to fry at 4.5 volts. It's my understanding that these lights depend on the current draw being enough to drag the battery voltage down to limit the current to a safe level. Basically, lwalker summed it up well.

If it was me, I'd get a wall wart supply that outputs about 5 volts, tie in my resistor sub box, attach a meter, spin the knob to get 3.5 volts or so, read the number on the box, and go from there. You'd want to also measure the current draw so as to determine the power rating required of the resistor.

Ohms law is your friend.

[darryl]

Those 9 led flashlight using 3 tripleA cells are indeed relying on the resistance of the cells to keep the voltage across the leds in check. The leds are all in parallel and there is no series resistor in there. If you are going to adapt a power supply to run them properly, you have two options- one is to supply a regulated voltage and the other a regulated current. In this case the better option is a regulated voltage, since current has to divide amongst several leds, and that means some will take more than others. When that one burns out, which it will 'cause they are cheap crap, the others will have to divide the current. Another will burn out sooner, leaving the rest to divide the current, which will now be even higher per led. Soon the thing is garbage.

I've found that most white leds in the standard package, which these are, like to have about 3.2 volts across them. The best way to run the set then is to supply that voltage. Forget about series resistors or any possibility of a series/parallel combination of the leds.

For the most part, that type of led will be ok with about 20 ma or so each. That means that each 'led head' will want about 180 ma. Use this figure to determine how much current your power supply will have to be capable of. If for instance you want to power four heads, that means you'll need 3/4 of an amp.

The typical regulator IC is in a TO-220 case, rated at 1 amp or 1.5 amp. They are not commonly made with an output voltage of 3.2, so you would use the variable output type, which is what they call a 317. That and a single resistor of the right value sets the output voltage. There is a minimum voltage that's required to run the chip, which needs to be in excess of what it's delivering- probably about 2.5 volts or so, so the power supply you use needs to be capable of 3.2 + 2.5, or 5.7 volts at the current level you need. A typical 6 volt wall wart will be fine to run two heads through this regulator, and if the wart is capable of 1 amp output, then it can run four heads.

Today I've been toying with the idea of getting one of those led strips and cutting it into sections of 3 or 6 leds, then making a 'stripline' light from it. Two of these would flank the jaw area on my sheet metal bender so I can see what I'm doing from both sides of the jaw. Being a stripline thing, they would be narrow enough to not get in the way of vision. Those have a series resistor built in to each section, and are rated at 12 v, so my option is to run them through a 12 v regulator and feed that with a 15 v or higher power supply. If the power supply is already well-enough regulated that it can supply all the current required without the unloaded voltage being too high to begin with, then I would forget the regulator. The regulator does give a protection for the leds though, and keeps the brightness constant while the power supply output voltage might rise and fall with changes in the ac input voltage.

[Seastar]

I think all of you are overthinking this thing.
I'll bet a wall wort with a 4.5/5.0 Volt 300/500 mA DC rating will work without anything else.
The current limiting is probably built in.
I would buy one and try it.
What can you lose? One $3 flashlight.
Bill

By the way you can get a lot more than 180 mA out of AA batteries.

[Abner]

I think all of you are overthinking this thing.
I'll bet a wall wort with a 4.5/5.0 Volt 300/500 mA DC rating will work without anything else.
The current limiting is probably built in.
I would buy one and try it.
What can you lose? One $3 flashlight.
Bill

By the way you can get a lot more than 180 mA out of AA batteries.

I believe most of mine are free ones from HF coupons. Like collecting stray cats -red, blue, black.
Letting the smoke out isn't a big deal but I would like to avoid it.
I like the small size and I will look at my collection of chargers and see what I have.

Thank you,

[danlb]

There is a lot of good advice so far. Unfortunately, there are a lot of assumptions that are being treated as fact.

It has 9 leds, and people are assuming that they are the cheap 5mm style rated at 20ma. That's a good assumption, but other sizes and ratings exist. The fact that it runs from 3 AAA cells tells us that it is low power since AAA cells have pretty low capacity. I've seen these lights designed in many different ways. Some with no electronics at all. Some have just a resistor and some actually have a circuit board that regulates the power. Most of the "3 mode" flashlights have a PWM circuit provide high, medium and low beams.

5mm LEDs are often run with more current than the 20ma that they are designed for. That makes them brighter and (as a side effect of their design) bluer. 50ma is not uncommon.

If you want instructions on how to accurately determine the power supply, send me a PM. In the mean time you can safely buy a USB charger for a few bucks on amazon to power the light. The voltage is within 10% of the batteries that you normally use. They are current limited to 500ma. 500 ma divided by 9 LEDs is 55ma each.

To use a USB cable and the USB charger below as a power supply, cut off one end of the USB cable. The red wire is + and the black is -. Very simple.

http://www.amazon.com/lovable-miniat.../dp/B00R17UE2Y ( $2 )

Did I remember to say that I collect (and build) flashlights? Probably should have mentioned that.


Dan

[lakeside53]

Just put a series resistor inside the housing where the batteries went. Then pretty much anything will work.