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How to make one flashlight from an old 9V battery.

Materials:
1 old 9V battery
1 super bright white LED
1 resistor 330R
1 little switch
solder
plasticine

Step 1: Disassemble the Battery

The first step is to disassemble the battery, using a knife to open the bottom of the case, as shown in the picture.

After that, remove the inner parts of the battery, keeping only the little strip connected to the negative pole and the case.

Step 2: Solder the Parts

In this step solder all the components to each other, like shows the picture, making a series, starting from the positive pole, soldering 1 pin of the switch directly to the positive pole and the other pin to the resistor, which will be connected to the LED.

The negative pole is directly connected to the Cathode of the LED with the strip of the battery that was kept in the step before.

Step 3: Fill With Plasticine

With all the parts connected and after the confirmation that the circuit lights the LED on, connecting it to another (good) 9V battery, it's time to fill the empty space of the flashlight with plasticine (or sugru), like shows the picture, to stow the components inside the case.

The case, that was kept from the old battery, need to have one rip, in order to fit the cap of the switch, in the assembling process.

Step 4: Put the Case On

To finish, put the case on, using a pliers to round the bottom, closing it.

And that's it - IT's DONE!

Step 5: Testing

Now we have one little shiny flashlight, to use wherever it's useful.
<p>I think filling it with sugru would be too expensive. Perhaps one could find a block of something non-conductant to pop in there.</p><p>It might be good to add some mention of how 9v might be too much voltage for some LEDs?</p>
<p>Thank you for your comment. I think you're right about the cost of sugru - I don't know that material because I've never used it before. The plasticine is the material that I'm used to aply since my elementary portuguese school time and it's the material that I've used in this instructable.</p><p>About the voltage applied in the LED's, usually the manufacturers refer the starter voltage and, in those cases, the starter voltages of the LEDs are from 1V to 4,5V. But the output power of any LED is directly proportional to it's direct current, which is determined by the value of the resistor used. Thus, the resistor that I've used in this instructable was 330R =&gt; I = V/R = 9/330 = 27,27mA. If you can see in any characteristic curve of any LED, 27,27mA it's right in the middle of the Intensity values admitted by the device.</p>
<p>For the benefit of Americans, Plasticine is a popular brand outside the US of what you call &quot;Modelling clay&quot; and has taken on the role of a generic name for that substance.</p>
Thanks. I appreciate the info. I'm learning about this stuff, and am beginning to understand the formula for this particular LED/resistor circuit. I'm curious about the other designs that add either a capacitor or a transistor and make the LED brighter (and perhaps more efficient?) but those are way over my head.

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Bio: I am a portuguese Engineer of Electronics and Telecomunications who loves inovations...
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