# Creating LED Flashlight



## lildragon555 (Apr 1, 2008)

Hi, I'm trying to create an LED Flashlight, I've seen a lot of them, but I've always wanted to create my own. I know how to wire LEDs and the battery together, but how would I add a *switch*?

You know, to cut the power when off, and to give power when on?
Also, what kind of switch should I get? I've always seen DPDT, DPST, SPDT, and SPST. No idea what they mean.

EDIT: If I wanted to create 3 rows of LEDs in a parallel, would I need 3 resistors? or I could I just used 1?


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## Basementgeek (Feb 7, 2005)

Not sure why you want to do this as the are really cheap any more.

Really to many variables to answer any specifics questions here, but this sight can help you some:

http://metku.net/index.html?sect=view&n=1&path=mods/ledcalc/index_eng

A simple off/on switch is all you need, SPST.

BG


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## lildragon555 (Apr 1, 2008)

Yea, I know, but I like to build things, get more experience out of it

Where do you connect the switch to? The battery? Resistor? LEDs?


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## Basementgeek (Feb 7, 2005)

Really any place along the line.

BG


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## Done_Fishin (Oct 10, 2006)

in any simple system there is a power supply that drives a circuit or load.
A switch is also known as a circuit breaker. It breaks a circuit so that the Power supply cannot feed the current that passes through the LOAD

If you are building a battery operated system the switch can go anywhere in the line so that the circuit is broken and no current will pass through the load.

If it's a transformer driven device, the best place to put it is where you want to isolate the current/power from feeding the load. You may have circuitry that requires 24/7 power but other parts that you want to work when required ie you place a circuit breaker at that point.

switches come in many shapes and sizes including mechanical & electronic. 
Normally on Battery operated devices the switch is placed at the positive output from the battery, but can also be placed so that it disconnects both positive and negative connections to the battery. The exact physical location is left to the designer along with the switch "device" and method of circuit breakage.


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## lildragon555 (Apr 1, 2008)

would this work? I don't know how to make a wiring diagram or schematic so I'm just gonna do w/e XD

8 AA Batteries = 12 Volts ------> Switch ------> 47 Ohm Resistor
------> ^^^ ------> Back to negative side of battery
^^
^^^

^ = LED

I wanna make them in 3 series, but in parallel. Would I need more than 1 resistor?

Additional Info: This is what I'm getting...do you think it's ok?
LEDs: http://www.radioshack.com/product/index.jsp?productId=3060980&clickid=cart
Resistor: http://www.radioshack.com/product/index.jsp?productId=2062313&clickid=cart
Switch: http://www.radioshack.com/product/index.jsp?productId=2062544&clickid=cart


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## Basementgeek (Feb 7, 2005)

I think this is what you mean:

Three in series and those 3 strings tied in parallel

Yes just one resistor.

Here is my source of LED's:

http://www.ledshoppe.com/led5mm.htm

Here you can get 100 5mm 13000mcd for $8.00 USD free shipping. I have bought from them several times and never a problem.

Check out their flash light prices.

BG


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## lildragon555 (Apr 1, 2008)

how would you make them in series and parallel? just put the positive and negative ends together on the end LEDs?

also, thanks for showing that to me, now I'll make a bigger one =D

and if I use 9 LEDs now instead of 8 I would need an 82 Ohm Resistor right? would it be ok just to use 100 Ohm?


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## Basementgeek (Feb 7, 2005)

Yes, put 3 in one string (+) to (-) on all 3.
Then tie the 3 (+)'s sides to the plus side of the line. Do the same with 3 (-) side.

You get to do the math using the link I sent you.

BG


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## Done_Fishin (Oct 10, 2006)

FW current 25mA; 
FW supply 3.3 (typical), 3.6V (maximum)

3 in series typical 9.9 Volts @ 25mA, might want to go to max 10.8 Volts though.

12 Volt supply from 8 batteries.

Initially a new fully charged battery is more than 1.5Volts and whilst discharging through a load the voltage will slowly drop. It may well still be functioning when it reaches 1Volt but this would obviously mean that you are now only giving 8 Volts to your lights.

lets go with the idea that you have 12 Volts and that your LED's require 10 Volts @ 25mA per string of three. You must dissipate 2 Volts @ 25mA so that the LED's don't get what is known as overvoltage. 
Resistance = Volts divided by current 
R = 12 / (25/1000) or (12 * 1000)/25 which is the same as 4*(12*1000)/25*4
(multiplying top and bottom of the equation by the same amount is just a simple way to make the figures easier to use.)
25 * 4 = 100 
1000/100 = 10 
4*12*10 = 480 

so your calculation for 470 Ohms is good. we only need an approximate value and the resistor is highly unlikely to be accurate , in fact it's listed as 47 ohm carbon-film resistors. 1/2-watt,* 5% tolerance*. 
5% of 470 is (5*470)/100 = 4*4.7 or approx 18 Ohms

power is the next thing to check out. 18 ohms at 25mA 
P=V*I or I*I*R 
V*I is simplest way to go, so we have (2 * 25) mW so no way that resistor is going to get warm. In fact it will run cooler as the battery voltage drops. In fact as the battery voltage drops so will the current, although only marginally at first.

The switch looks great it should be able to handle the 75mA for 3 sets of LED's with ease. 

Have you thought about how you expect to keep your 8 batteries together in a small package so that it will be useful when it is in use?
ie comfortable to hold, 
easy to replace batteries, 
waterproof in case it rains, 
isolated in case something metallic falls on it and might possibly short out a cell or two?
Spring loading to accomodate drops, shocks, thermal affects as batteries get warm.


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