Eggcellent Night Light!

Step 3 Meet my Soldering Iron of Justice

This project requires a simple circuit with one LED.

Remember, the positive lead of the LED should be the longer one. The positive lead also has a smaller head *in* the build.

Do not solder yet. Make sure everything fits first. The battery pack door for batteries should sit in the tin on the exposed side.

Solder the switch into the circuit first and leave the LED bulb last. String the wires through the center hole of the base from the bottom, and fit the switch in the switch hole.

Use a glue gun or clear glue cement to glue the battery pack to the in the base.

Comments

[mje]

Great idea, using the egg. But you can't connect a pair of AA cells to just any white (or other) LED without burning it out. Usually you need a current limiting resistor in series.

[TXTCLA55]

depends what voltage the battery('s) equal and what the required amount need for the LED to light. normally anything above 3-3.5Vdc needs a current limiting resistor.

[mje]

It depends on the internal resistance of the battery and the voltage drop and current limit of the LEDs. For most small LEDs, anything over 20ma will severely limit their life.

[LasVegas]

He used a white 3v LED. It should have no problem without the current limiting resistor. This is an okay instructable, especially for a beginner. I would suggest pouring some melted paraffin wax into the egg, swirl it around so it coats the interior and adds a bit of strength to the fragile egg.

[sehrgut]

Umm . . . no problem except torching the LED prematurely, you mean? Brush up on Ohm's Law. An LED has a VERY low forward resistance (that's kinda the purpose of diodes, doncha know), and will draw far more current than is healthy for the semiconductor on 3V. If that doesn't help you, try "Why use current limit resistors on White LED's?".

[iscatel]

The page you supply the link to says that supplied by batteries, without resistor, the LED will outlive the batteries. All diodes have a forward resistance and a resulting voltage drop. All batteries have internal resistance. Ohm's law is very helpful, but like punching two significant digits into a calculator and getting out 14, can lead you down the garden path. Same here--build it and poke around with a meter, calculated values and real values always differ.

[sehrgut]

Yes, I'm quite aware of the internal resistance of batteries, and the voltage drop across an LED is part of the calculations cited on that page. As far as being glad an LED can outlast its battery . . . huh? Since when is it a "good mark" for a gadget to simply be able to last long enough to have to change the battery? Properly-used, an LED (white or otherwise) will last for years of constant illumination before any detectable performance drop: outlasting a single change of batteries is hardly evidence of proper use.

[iscatel]

Case in point: keychain flashlight. $1.00 at the dollar store, $1.50 to replace the 2 lithium cells included with. If it were for a more durable app, I'd agree but in this case when the batteries die, I'm tossing the light & buying another. I may pull the LED.

[LasVegas]

I would agree normally, but in this case he is using the less than the voltage that the LED is rated for. In this case, the current will actually below the LED's specification. If he were using 5VDC, he would have needed a current limiting resistor to drop the voltage across the LED, limiting the current to spec.

[momo!]

I was actually thinking of swirling modpodge into the egg to add some elasticity. The beautiful thing with that is that if the egg does crack lightly, the light will be able to show through.