Need some help...

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WipeoutFTW

Limp Gawd
Joined
Jul 17, 2006
Messages
357
Hey guys, I am currently building a wireless sensor bar for a Nintendo Wii and I was wondering if you guys see any flaws in my schematic. I build it and for some reason it isn't working. All the LEDs are im place correctly, I just tested all of them and my battery source an d it all works. If someone could go over my schematic and tell me what you think that would be a great help.

wiisensorbarschematictm2.png
 
The very 1st thing that strikes me is that looks like the polarity is reversed on the battery pack... + feeding led - ?

Also, unless you've got some ir led's that require less than 1v each, you don't have enough supply voltage to run a series circuit like that. For led's in series, you require a minimum of the sum of led voltages. Guessing that std ir led's might be 1.2v, you need at least 7.2v, and your AAx4 supply is only providing 6v.

Additionally, led's require a current limiting resistor to prevent them from drawing too much current and burning out. I don't see one in that schematic. Check out this calculator to help determine the resistor you need.

Actually, you might be better off with a series-parallel matrix of 2x3 led', that would reduce your power supply requirement. Check out this led matrix wizard for other possibilites.

Lastly, I'm guessing that you realise you can't see ir with your eyes, and are using some type of ir specific tester to verify they are working or not.
 
Thanks for such a great response! What Im going to do is instead of 3 LEDs on each side im going to take 1 off each side making it 2 on each side.

Thanks a lot.
 
A regular digital camera will work to see if the IR LED's are working. They are getting more IR filters, but still should show the LED's slightly purple when viewed.
 
DirtyLude said:
A regular digital camera will work to see if the IR LED's are working. They are getting more IR filters, but still should show the LED's slightly purple when viewed.
Click to expand...
Excellent suggestion...
 
WipeoutFTW said:
Thanks for such a great response! What Im going to do is instead of 3 LEDs on each side im going to take 1 off each side making it 2 on each side.

Thanks a lot.
Click to expand...
Just to be clear, there's nothing wrong with 3 per side if you did something like this:

Solution 2: 3 x 2 array uses 6 LEDs exactly
+----|>|----|>|----|>|---/\/\/----+ R = 100 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 100 ohms

The wizard says: In solution 2:
each 100 ohm resistor dissipates 84.1 mW
the wizard thinks ¼W resistors are fine for your application
together, all resistors dissipate 168.2 mW
together, the diodes dissipate 208.8 mW
total power dissipated by the array is 377 mW
the array draws current of 58 mA from the source.
 
agent420 said:
The very 1st thing that strikes me is that looks like the polarity is reversed on the battery pack... + feeding led - ?
Click to expand...
... that is actually traditional circuit diagram standards, you see originally scientists thought the the electrons flowed from the positive pole to the negative and all diagrams were drawn that way. When they discovered the flow is actually form the negative to positive terminal, the diagram didn't change. Just common knowledge of circuit builders, so remember it.

Also, if each diode uses 1.2V and the source produces 6 volts, you don't need resistors because the voltage will drop across the six diodes. And the diodes don't require a minimum voltage, if the supply is less than the standard operational voltage of 1.2V per diode, they will just produce less IR spectrum, but it will still work just fine.
 
BarneyGumble said:
... that is actually traditional circuit diagram standards, you see originally scientists thought the the electrons flowed from the positive pole to the negative and all diagrams were drawn that way. When they discovered the flow is actually form the negative to positive terminal, the diagram didn't change. Just common knowledge of circuit builders, so remember it.
Click to expand...
Yeah, I'm aware of the old electron flow philosophy, though I'm not sure I would call it traditional. But how many modern schematics use that? Point is, if you hooked that up as drawn it won't work.


BarneyGumble said:
Also, if each diode uses 1.2V and the source produces 6 volts, you don't need resistors because the voltage will drop across the six diodes. And the diodes don't require a minimum voltage, if the supply is less than the standard operational voltage of 1.2V per diode, they will just produce less IR spectrum, but it will still work just fine.
Click to expand...
I'm going to have give more thought to the summed voltage drop > supply not requiring any resistors...

As for the forward voltage, it may be that current can flow through an led with less voltage (guessing that the std .7v silicon still applies), the result being under spec ir output, but it makes me wonder what the point of that would be when the goal is to provide a good source of ir illumination...
 
yeah for some reason power wont transfer through the IR LED's, I used a multimeter to test if the connection was right and I didn't get any signal. All of my LED's are connected properly from what I can see... :confused:
 
WipeoutFTW said:
yeah for some reason power wont transfer through the IR LED's, I used a multimeter to test if the connection was right and I didn't get any signal. All of my LED's are connected properly from what I can see... :confused:
Click to expand...
You might want to check them one at a time... It only takes one bad led in a series for the whole thing not to work. Be sure and use the appropriate resistor (use the calculators above). Or, ir you multimeter has a diode-check function, you might be able to use that. Some or all of them may have fried if you previously connected them without a limiting resistor.

Are you sure of the led polarity? The battery + should be connected to the led anode. The resistor can be on either side, but it's more common to put it between the battery pos and the led.

Also, the specs I used earlier were based on a common ir led found at RatShack, it would be best if you located the actual data for the leds you are using.

I know this is probably self-evident, but the illustration for the battery pack is wrong in several ways...
 
I'm sure it's presented simply as an illustration of where it would be in the circuit, but in real life your AAx4 pack will have the positive wire leading from the top of a battery on one side, and the negative terminal will be at the bottom of the last battery in series. All the batteries will connect head to toe. An actual battery pack will have black and red leads, so it really doesn't matter unless you are homebrewing your own pack.

In the illustration, both wires originate at the top of a battery.
 
WipeoutFTW said:
yeah for some reason power wont transfer through the IR LED's, I used a multimeter to test if the connection was right and I didn't get any signal. All of my LED's are connected properly from what I can see... :confused:
Click to expand...


from what you have drawn the LEDs are reversed biaes.. of course it's not going to work. switch the polarity of the battery.. or flip all the LEDs around..

again.. i strongly suggest a current limiting resistor..

also.. IR light comes in various wave lengths.. so those LEDs may not create wave length the wii mote is designed to detect..
 
RancidWAnnaRIot said:
also.. IR light comes in various wave lengths.. so those LEDs may not create wave length the wii mote is designed to detect..
Click to expand...
Although I think those led's should work, it is important to clarify 'not working'... as in led's not illuminating or wii not detecting it...
 
The only thing I can think of is one of your LEDs might be connected the wrong way. If thats the case the current will not flow through the circuit, current can only flow in one direction through a diode.
 
well, using a multimeter to test, I wasn't able to complete a circuit through any of them...
 
Are you sure you connected them the right way around? I'm not trying to insult your intelligence, but things like that happen. The side of the LED with the flat surface is the cathode (-) leg. Like mentioned earlier, it won't be visible to the naked eye, so you'll need something to pick it up like a digital camera/camcorder. Another option is to measure the voltage across the current-limiting resistor when you have the whole circuit hooked up. If current is flowing, there will be some voltage across the resistor. If there's no voltage across the resistor, there's no current, and your LED is either busted or backwards. If the voltage across the resistor is the same as the battery voltage, you've got a short.
 
All of the LEDs are facing the correct way and I did use a camera to test to see if they lit up.
 
WipeoutFTW said:
All of the LEDs are facing the correct way and I did use a camera to test to see if they lit up.
Click to expand...


LEDs are probably burnt out.. if you got them set up right.. i'm telling you ... you really need to add a current limiting resistor..

connecting diodes without resistors is not good practice
 
BarneyGumble said:
Also, if each diode uses 1.2V and the source produces 6 volts, you don't need resistors because the voltage will drop across the six diodes. And the diodes don't require a minimum voltage, if the supply is less than the standard operational voltage of 1.2V per diode, they will just produce less IR spectrum, but it will still work just fine.
Click to expand...
I'm wondering if someone can point me to some technical reference that supports this... It caught my curiosity so I looked into it a bit, and as far as I can determine the forward voltage is the minimum potential required to overcome the junction barrier, and is dependent on the chemical composition of the device. So unlike a conventional silicon diode that requires .7 volts for current to flow, an ir led spec'd at 1.2fv will not pass appreciable current until that level is reached. The level of current passed with a forward potential less than fv would be extremely small. As led's are current operating devices, an extremely small current means virtually no irradiated output. While some small amount of current might be able to pass through the first diode, the combined voltage drops reduce the potential feeding the last one in the series to a point where no current flows. So you do in fact require a voltage source greater than the combined voltage drop of the diodes in series (and the appropriate ballast resistor).

Even in the unique case of source voltage = summed voltage drop exactly, there still needs to be some control of the current flowing through the circuit.

Just for kicks, I modeled this in my Spice simulator, which indicated that only .03ma current would flow. This is orders of magnitude less than the 29ma required by the led. I also ran a dc sweep that indicates measurable current doesn't begin to flow until near the expected 7.2fv level. This clearly seems to demonstrate that a 6v power source will not operate the circuit in the first post...

sweep.gif


As for the OP's problem, you may well have burnt them up. Buy another so you have a known good led for testing, and be sure to use an appropriate ballast resistor when doing so. These are very simple, easy to build circuits... just use the array calculator provided earlier, as the original design you provided is not correct. I highly recommend the 2 x 3 design, which would work fine powered from 6v.

EDIT -

Here is a link to one of the references I used.

Also, I don't know who came up with this schematic, but you should note the original version in the doctabu thread used only 4 leds, and included ballast resistors. Whoever thought of just tacking on 2 extra leds and removing the ballast resistors designs circuits as well as they design schematics like the one in the OP :rolleyes:
 
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