At night, the smallest of the inflatable Hello Kitties were really dim! And the lights inside them were --gasp-- not user replaceable!
|Before: Small kitties are dim!|
|All but three kitties have been upgraded|
These inflatables have zippers at the bottom to access the internals. There's no need to unclip the LED.
|Epoxy-encapsulated single SMT LED|
Make sure you use a resistor rated for the power through it. Power is given by current times voltage: P=IV. So for example, 10mA x 12V = 0.12 watts. That's nearly 1/8W so I used 1/4W and 1/2W resistors I had laying around.
I selected wide-angle, 20mA, high brightness LEDs from Digikey, (Part No: C535A-WJN-CU0W0231-ND)
ConstructionI found some great protoboards on ebay. Long and narrow, it was easy to install the LEDs and resistors.
Since my wife has mastered soldering while teaching the 5th graders, assembly was a team effort.
InstallationInstalling the LED array was a matter of unzipping the un-inflated Kitty, reaching up in and grabbing the LED module, pulling out to access it.
Then, clipping the wires, I soldered them onto the protoboard.
I placed a dollop of Goop to glue the new board onto the old module.
Then added a zip tie (pink of course), belt-and-suspenders style, and we have bright kitties again!
Merry Christmas and/or Happy Holidays!
Oh, Here's an LED Refresher
It also means that connecting four LEDs in parallel draws four times the current as four in series. That's why I put the LEDs in series.
In this case, if I want I=15mA, and each LED drops 2.8V, then using Kirchoff's Voltage law, I can solve for the voltage across the resistor:
Then using Ohm's Law I can solve for the resistance, R, that results in a 0.8V drop and 10mA of current.
Closest E24 (5% tolerance) resistor is 56 ohms, but it's a good idea to see what happens in the worst case.
Most voltage regulators are within 5% (12.6V). Now if the resistor is at the low end of the 5% tolerance range (53 ohm), you'd end up with 26mA which is too high for the LED.
If you recalculated R for the worst case voltage and worst case current (20mA), you'd compute R=70. With the next E24 resistor of 75 ohms, you'd have a reasonably safe margin.
You could also check the LED datasheet and see if there is a worst-case (lowest) voltage drop on the LEDs. If you really wanted to get fancy, you could design a current source circuit instead that would be a lot less sensitive to power supply voltage error. But that's another story for another day.