Minty LED Tester

Stop fumbling around for a 3V battery, build your own LED tester, and freshen your breath while you are at it!

Let's get started.

Step 1 Materials

You will need:

• Altoids Smalls tin: ~$1.50 at a drug store.  Or buy them in bulk on Amazon.
• (1) Wire Housing: $0.13 at Mouser.com
• (2) Wire Header: $0.14 each at Mouser.com
• 3V Coin Cell Battery: Find them for cheap on ebay.com
• Coin Cell Battery Holder: $1.56 at Mouser.com
• (2) Small Pieces of Hookup Wire
• (1) Rubber Grommet: 1cm across, can be found at any hardware store
• Dremel Tool or Rotary Device

Comments

[J-Five]

That's very Ingenious!

[iamseer]

cool idea. But the problem is some red and IR led works on 1.5V or 1.2V. That means the 3V cell may given them a sudden death because of the large current. But a single resistor will solve the problem. I think hundreds ohms will do.

[zack247]

maybe, but most red and ir leds are 3.6v ones, so you should be safe.

[iamseer]

May be the leds you find is more than one in serial. The longer the wavelength, the lower voltage the led needs. Of course the leds from different manufacture may work at a slightly different voltage. But the error can hardly exceed 20%. http://en.wikipedia.org/wiki/Light-emitting_diode#Colors_and_materials

[Goodhart]

Since there is so much extra space in the tin, one could set up 2-3 sockets and have different resistor values on each to provide lower current to the other sockets....then you could "start" at the lower end and if it doesn't light move up to the next higher level until the LED lights or is proven defective.

[mxc1090]

Same current...lower voltage. 1.8 - 2.0 for Red/Orange 3.0 -3.2 for Blue/UV/Green

[Goodhart]

Hmm, then in stead of current limiting resistors, one could set up "voltage dividers".

[mxc1090]

Correct.... So lets say you have a perfect 12V source. If you would like to achieve 3V across one resistor and 1.8V across another @ .025A (25mA) you would do something like this: V=IR 12V=(.025A)(R) Total resistance of the circuit = R = 480 ohms To achieve 3 V across one of the resistor the value of the resistor must be: R1=3V / .025A R1= 120 ohms To achieve 1.8 V across one of the resistor the value of the resistor must be: R2=1.8V / .025A R2= 72 ohms Total R must equal 480 so.... 480 = 120 + 72 + r So the value of 'r' (the last of the 3 resistors in this circuit) = r = 288 ohms then you make it so when your plug in an LED the current will bypass the resistor and flow through the LED and it would have no affect on the rest of the circuit. (Make sure you know the Amperage of the LED, it is vital to the circuit)

[Goodhart]

Oooo, I see LOTS of potential with this one :-)

[zack247]

cool! i need one of these.

[Brennn10]

I have a few extra LED tester kits lying around in my workshop. Send me a PM and I can hook you up with one.

[zombiefire]

email me at [e-mail] thanks

[zack247]

thats ok, making stuff is a lot more fun

[meh301]

Great! A small project where I can finally recycle almost useless coin cell batery holders!

[AndyGadget]

Would there be room in there to add a 2nd battery and a resistor (270R or so)? If so, it would be a lot safer for the LEDs.  
The 'voltage' of a LED isn't the voltage it's designed to run on, it's the voltage 'lost' inside the LED.  A typical red LED has a drop of 1.8V so the current through it from the 3V battery could possibly kill it.  A bit more info on LED voltage drops HERE.

[Brennn10]

There is definitely room left inside of the tin. Most of the LEDs that I use are safe with the 3V, but you can mod this any way you wish.