HOW TO – Make your own wireless Wii sensor bar!

Fun & Games
HOW TO – Make your own wireless Wii sensor bar!

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One of our makers Brian “DoctaBu” Moore IM’ed and said he was going to make a wireless Wii bar, and wireless Wii bar is what was made, he writes – “With just a bit of perf board, a few cables, some skillful soldering, four AA batteries, and a few IR LEDs, one can easily make their own, homemade Wii Sensor Bar, completely absent of wires. Enjoy!”Link.

4 thoughts on “HOW TO – Make your own wireless Wii sensor bar!

  1. MisterHay says:

    I think I’ll have to build one of these for the top of my projections screen.

  2. chriswa says:

    First, you need a DC supply. You could use batteries, or an AC/DC converter. I used an old cell phone charger; usually the positive wire is embedded inside the ground wire, so that if your cat sinks its teeth into the cable, she’ll hit ground before she reaches anything shocking. My charger was labelled as supplying 5.2 volts, but when I stripped the wires and hooked it up to my voltmeter, it turned out to supply 7.66 volts.

    IR LEDs typically cause a voltage drop of 1.5 volts, and should operate at 20mA, although I chose to run mine at 25mA. Running in series, 5 IR LEDs cause a voltage drop of (1.5*5=) 7.5 volts, meaning that I can run all 5 in a single series circuit from my 7.66 volt supply, supplying a resistor to drop the remaining (7.66-7.5=) 0.16 volts. Using Ohm’s law, if I want a current of 25mA and I have a voltage drop of 0.16v, I need (.16/.025=) 6.4 ohms of resistance. I dug through my resistors and couldn’t find any 6.4s, so I used a 10 ohm and 22 ohm in parallel to get (1/(1/10+1/22)=) 6.875 ohms. That gives me a current of roughly (.16/6.875=) 23.27mA through my circuit.

    If my power supply was 5.2 volts as advertised, I wouldn’t have been able to run all the LEDs in one series. I’d need to split them up into parallel circuits: 3 LEDs in one and 2 LEDs in the other. 3 LEDs would cause a drop of (1.5*3=) 4.5 volts, leaving (5.2-4.5=) 0.7 volts, meaning I’d need a (.7/.025=) 28 ohm resistor on that side of the circuit. 2 IR LEDs would cause a drop of (1.5*2=) 3.0 volts, leaving (5.2-3.0=) 2.2 volts, meaning I’d need an (2.2/.025=) 88 ohm resistor on the other.

    Since I wanted two banks of 5 IR LEDs, I simply duplicated the circuit in parallel and ran some wire between the two. Although not the most elegant solution, I mounted each bank in a plastic tape cassette case after melting some holes for power cabling and the LEDs to poke through. Works great!

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