Hacking an LED flashlight... question

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 8:53 pm

Iâ€™m working on headlamps for The Contraption and want to use LEDs. Iâ€™ve found some 12 LED amber turnlamp bulbs from the auto place that work on 12 volts. I thought theyâ€™d be cool since amber would look closer to a flame light. When I hooked them up, however, they were closer to orange and barely illuminated anything more than a couple few feet away (due to the color frequency and how our eyes pick it up, I guess). So I am now trying to find some white replacement bulbs, but so far Iâ€™m not having any luckâ€¦ all â€œamberâ€

MozyBonz · Post by **MozyBonz** » Wed Jul 18, 2007 9:10 pm

http://www.powerstream.com/dc6.htm

MozyBonz · Post by **MozyBonz** » Wed Jul 18, 2007 9:12 pm

google 4.5 volts.= 12 volts

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 9:17 pm

Thanks, Mozy!!

That would certainly work if it comes to it. I'm hoping for something to solder inline and hide inside the metal "antique headlight" I'm having to fab up.

Plus I'm wanting to learn about resistors and sizing them, too...

Heh Heh... multiple motives, you know.

MozyBonz · Post by **MozyBonz** » Wed Jul 18, 2007 9:21 pm

MozyBonz wrote:google 4.5 volts.= 12 volts

MikeVDS · Post by **MikeVDS** » Wed Jul 18, 2007 9:33 pm

I'm not certain your approach is the best but we'll start there. I cannot think of how you'd size a resistor for that unless you know the current of your light. Watt rating would get us the current, if you know that.

I'll assume each light is 45w so that gives you 90w total.
45w at 4.5v is 10 amps.

Then you have V = I * R (for two theoretical flashlights in series)
9 = 10 * R1
R1 = resistance built into two flashlights in series = 0.9 ohm

You want your final voltage to be
12 = 10 * (R1 + R2)
12 = 10 * (0.9 + R2)
R2= 0.3 ohm

So if your lights were 45 watts a 0.3 ohm resistor in series along with two of those lights and 12 volts would work nicely. Keep in mind I have no idea the stats on your actual light, but hopefully you can follow the same idea or get the real numbers. You can also throw something in series that actually uses the power instead of just a resistor.

Ok on with the other ideas. It sounds like your first LED try was just to use LEDs without a reflector? Without a reflector it will seem like your light is not very bright. I can guarantee that your flashlights that you're going to dismantle have reflectors to focus the light. Also, what is the watt rating on your LEDs and how many did you use? Compare that to your flashlight. Also you mentioned flame effects? Ive been looking into something along those lines and found this guy. Pretty good idea.

http://www.spookyblue.com/halloween/fli ... pookyfire/

rich_d_rich · Post by **rich_d_rich** » Wed Jul 18, 2007 9:38 pm

I'm doing this very thing. Here's how I'm going. I haven't finished yet but will update:

- I pulled the flashlight (cheap $2 headlight) apart. It has a battery pack, a control board with a push switch and a board with 5 leds wired in parallel.

- the push switch selects flash / steady / off and has a small chip (flasher unit) wired to it.

- my plan is to recycle all but the led array, which will be wired to a 4.5v battery along with some other stuff.

- I needed to know two things - the LED's forward voltage and the current. I measured these with a multimeter at 3.5v and 60mA.

- To do what I want I need a series resistor wired between the led array and the battery. (MATH FOLLOWS). The formula for this is:
(Vbattery - Vforward) / Current = (4.5 - 3.5) / 0.06 = 17R

- So I need a 17R resistor or thereabouts. Too large a value will make the leds too dim. Too small a value will blow them!

- How to avoid the math and faffing with multimeters: get a selection of resistors from your friendly local electronics store (about 2c each). You want everything below about say 150 ohms for a 12v battery (they come in what are called preferred values rather than an obviously logical sequence - there is a good reason for this which I won't bore you with)

- try each in sequence starting at the largest first, until the leds are about as bright as they are in the flashlight.

- Even simpler method. Find a kid who's doing high school physics or a teacher who teaches it. They will do this stuff as part of their course. Also, unless nowadays they do everything in computer simulation, they'll have easy to use kit all set up.

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 9:45 pm

MikeVDS,

I was actually using a reflector from a larger flashlight to test them. I cut a hole in the back of it large enough to hold the bulb assemby. This is the same reflector that I'm using in the headlamps and is about 3.5" in diameter. I think it was the color spectrum... it just was getting picked up by the eye. A three white LED flashlight put out much more detectable light, with the same 3.5" reflector, than the 12 LED amber bulb.

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 9:49 pm

MikeVDS wrote: I'll assume each light is 45w so that gives you 90w total.
45w at 4.5v is 10 amps.

Then you have V = I * R (for two theoretical flashlights in series)
9 = 10 * R1
R1 = resistance built into two flashlights in series = 0.9 ohm

You want your final voltage to be
12 = 10 * (R1 + R2)
12 = 10 * (0.9 + R2)
R2= 0.3 ohm

So if your lights were 45 watts a 0.3 ohm resistor in series along with two of those lights and 12 volts would work nicely. Keep in mind I have no idea the stats on your actual light, but hopefully you can follow the same idea or get the real numbers. You can also throw something in series that actually uses the power instead of just a resistor.

Ah... that may come in handy. I'll make a note of it.

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 9:55 pm

rich_d_rich wrote:I'm doing this very thing. Here's how I'm going. I haven't finished yet but will update:

- I pulled the flashlight (cheap $2 headlight) apart. It has a battery pack, a control board with a push switch and a board with 5 leds wired in parallel.

- the push switch selects flash / steady / off and has a small chip (flasher unit) wired to it.

- my plan is to recycle all but the led array, which will be wired to a 4.5v battery along with some other stuff.

- I needed to know two things - the LED's forward voltage and the current. I measured these with a multimeter at 3.5v and 60mA.

- To do what I want I need a series resistor wired between the led array and the battery. (MATH FOLLOWS). The formula for this is:
(Vbattery - Vforward) / Current = (4.5 - 3.5) / 0.06 = 17R

- So I need a 17R resistor or thereabouts. Too large a value will make the leds too dim. Too small a value will blow them!

- How to avoid the math and faffing with multimeters: get a selection of resistors from your friendly local electronics store (about 2c each). You want everything below about say 150 ohms for a 12v battery (they come in what are called preferred values rather than an obviously logical sequence - there is a good reason for this which I won't bore you with)

- try each in sequence starting at the largest first, until the leds are about as bright as they are in the flashlight.

- Even simpler method. Find a kid who's doing high school physics or a teacher who teaches it. They will do this stuff as part of their course. Also, unless nowadays they do everything in computer simulation, they'll have easy to use kit all set up.

Yes, yes!!! I think that between all you guys, I might have what I'm needing. The formulas makes sense and I can always use the trial and error method if that doesn't pan out.

Ok... If I can't find white LED 12v auto bulbs tomorrow on my search mission, then I have a place to start with the flashlights.

Thank you!!! (And if anybody watns to add any further advice, it will be appreciated as well)

MikeVDS · Post by **MikeVDS** » Wed Jul 18, 2007 10:02 pm

One handy bit of knowledge which I figured out one day is that anything in a circuit that uses power is just a resistor and a resistor is just a device that makes heat, using up power. I don't know if everyone knows that, but I went through physics and electrical engineering classes and still did not realize physically, that is all they are. They are just little power sapping heaters. It wasn't until I actually started designing circuits that I realized this (not just math problems where you want X output with Y input).

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 10:08 pm

MikeVDS wrote:One handy bit of knowledge which I figured out one day is that anything in a circuit that uses power is just a resistor and a resistor is just a device that makes heat, using up power. I don't know if everyone knows that, but I went through physics and electrical engineering classes and still did not realize physically, that is all they are. They are just little power sapping heaters. It wasn't until I actually started designing circuits that I realized this (not just math problems where you want X output with Y input).

I sort of knew this, but you raise a logical idea of using something that will do a beneficial thing (more lights?) as a resistor, rather than just sucking it off as heat and getting nothing out of it.

rich_d_rich · Post by **rich_d_rich** » Wed Jul 18, 2007 10:10 pm

Yes, if you wired some of the leds in series then it's a lot more power efficient. More work though.

MikeVDS · Post by **MikeVDS** » Wed Jul 18, 2007 10:15 pm

More lights either for the head lamps or if you wanted to illuminate something else on there, just wire it in series. Maybe a small decorative lamp or something, if you don't want/need more powerful headlamps? A resistor is basically using up 1/3 of your total used power, so if you want to be efficient, it could be worth it.

rich_d_rich · Post by **rich_d_rich** » Wed Jul 18, 2007 10:16 pm

Of course if you want full efficiency there is this:
http://www.linear.com/pc/productDetail. ... 1094,P1929

It uses a switching supply to drive the LED with a constant current so (in theory) all the power turns to light...

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 10:16 pm

True... its a damn sight more work, and I've only got so much space to hide the wiring on the thing. but still something to consider.

rich_d_rich...

So please tell me if this is correct. Theoretically (Iâ€™ll check real values tomorrow with the meter), but assuming some values here to check my thinking, Iâ€™m looking at the following:

(12 V battery - 4.5 V forward for the LED array) / .06 current = 125 ohm resistor.

In this case â€œforward voltageâ€œ is the voltage the LEDs are wired to run on currently, right? I just want to be sure I'm understanding your terminology.

MikeVDS · Post by **MikeVDS** » Wed Jul 18, 2007 10:18 pm

You could also use a third headlamp to illuminate your controls or something. That'd put you to 13.5 volts, which is ok and typical to get close to that out of a car battery anyway. Add a switch that goes to a properly sized resistor if you want to be able to turn that lamp off and maintain proper voltage.

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 10:22 pm

MikeVDS wrote:You could also use a third headlamp to illuminate your controls or something. That'd put you to 13.5 volts, which is ok and typical to get close to that out of a car battery anyway. Add a switch that goes to a properly sized resistor if you want to be able to turn that lamp off and maintain proper voltage.

Yeah, I'm thinking about something like that.

I've already got eight purple CCFL's underneath the thing so I don't need any more under there... but I'll be thinking on it.

MikeVDS · Post by **MikeVDS** » Wed Jul 18, 2007 10:26 pm

(12 V battery - 4.5 V forward for the LED array) / .06 current = 125 ohm resistor.

In this case â€œforward voltageâ€œ is the voltage the LEDs are wired to run on currently, right? I just want to be sure I'm understanding your terminology.

Looks like you've got that, as long as you mean one lamp by the "LED array". Both lamps should give you 9 volts in series.

LeChatNoir · Post by **LeChatNoir** » Wed Jul 18, 2007 10:29 pm

Yes, by "LED array" I mean one lamp (consisting of several LEDS) from one flashlight.

Nice!!

Thanks again to everyone. You've helped me a bunch!!

Token · Post by **Token** » Thu Jul 19, 2007 1:43 am

Guys, if I did this at the office, my ass would get fired right-quick.

A resistor in series with a diode will limit the current ONLY. It will NOT reduce the voltage. The voltage on either side of the resistor is the same.

If you hook up a 12V source to a diode that is rated for 4.5 volts, via a 125 Ohm resistor, the voltage across the diode will be 12V. The current will be limited to 96mA, but the voltage will still have a good chance of frying the diode.

Either drop the voltage down to 4.5, or use 2 or 3 of those diodes in series.

Most LED can be driven 50% past spec and survive.

T

Edit: forgot to state what formula is wrong:

(12 V battery - 4.5 V forward for the LED array) / .06 current = 125 ohm resistor.

This is bad ju-ju

The LED can handle 4.5V at .06A which is 270 mW

With 12V feed and 125 Ohm restor in series, you wil feed 12/125= 96mA

That will result in 12V * 96mA = 1152 mW

This is more than 4 times the power than spec.

MikeVDS · Post by **MikeVDS** » Thu Jul 19, 2007 6:37 am

If you hook up a 12V source to a diode that is rated for 4.5 volts, via a 125 Ohm resistor, the voltage across the diode will be 12V.

Are you certain this applies to this situation? Normally a diode will have negligible resistance compared to a resistor. You don't want diodes sapping your circuit. When using LEDs it's not just a diode but also a resistor. I'm definitely not a LED pro, but I always thought that's how they worked.

No (big) voltage drop across a diode, but there is a voltage drop across a LED.

MikeVDS · Post by **MikeVDS** » Thu Jul 19, 2007 7:28 am

With 12V feed and 125 Ohm restor in series, you wil feed 12/125= 96mA

You had me doubting myself for a few minutes but now I realize you're completely neglecting that Light Emitting Diodes have internal resistance of their own and act differently than just a diode. You're figuring out how much power the resistor will consume and applying it to a diode, which would use no power if it just acted like a normal diode.

robotland · Post by **robotland** » Thu Jul 19, 2007 7:59 am

Regarding the hunt for 12V WHITE automotive headlight diodes- They're somewhat harder to find than taillights since fewer vehicles use 'em...But they'e out there. Unlike the red ones, those amber LEDs need all of 12 volts to put out anything useful- The red cluster modules will produce blinding light from a 9V battery though.

Isotopia · Post by **Isotopia** » Thu Jul 19, 2007 7:59 am

LeChat,

You might check out the web page which describes our playafly project. Not really sure what you're looking for in the way of schematics, wiring, etc. but you might be able to gelan a few ideas from the instruction/assembly portion of the page. Have a look at: http://www.playaflies.com/

unjonharley · Post by **unjonharley** » Thu Jul 19, 2007 8:10 am

I have a 32 LED 12 drop light from HF.. At about 7 feet in the air it lights an area good enough to set camp.. On sale 14$ reg 24$.. You could raid them, make your own configuration..

capjbadger · Post by **capjbadger** » Thu Jul 19, 2007 9:09 am

dealextreme.com is always selling white led units and flashligths if you are still looking for more. Granted the wait time is 1-2 weeks...

Badger

fluffernutter · Post by **fluffernutter** » Thu Jul 19, 2007 10:06 am

I don't know if these links will help...

http://led.linear1.org/led.wiz

http://led.linear1.org/1led.wiz

Toolmaker · Post by **Toolmaker** » Thu Jul 19, 2007 12:10 pm

http://www.solorb.com/elect/solarcirc/d ... index.html
http://www.solorb.com/elect/solarcirc/1 ... index.html

I can't find the place that sells the 12v multi luxeon LED lights but I recall that they were not cheap. I noticed that you mentioned the need for brighter light, cf and cold cathode provide more light and run alot cooler. Hopefully those links will help ya with wiring your own circuits using leds.

Token · Post by **Token** » Thu Jul 19, 2007 1:23 pm

MikeVDS wrote:
With 12V feed and 125 Ohm restor in series, you wil feed 12/125= 96mA
You had me doubting myself for a few minutes but now I realize you're completely neglecting that Light Emitting Diodes have internal resistance of their own and act differently than just a diode. You're figuring out how much power the resistor will consume and applying it to a diode, which would use no power if it just acted like a normal diode.

Yep, My bad. Sorry, ignore me on this thread.Serves me right trying to do circuits after hours. I ignored the resistance of the diode as if it were a switching diode with .3v bias.

Just connect three of these in series and forgo the resistor.