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wcalvert Newbie
Joined: 05 Nov 2006 Posts: 4 Location: Austin, TX

Posted: Sun Nov 05, 2006 4:14 pm Post subject: LED lit Bevo (UT mascot) 


Ok, I'm a EE student at UT, and for a fun project, I'm going to make a Bevo to hang in my pool room.
I'm going to do 60 orange LEDs with a 2.2V 20ma rating, and to power it, I was going to pick up an AC to DC adapter from RatShack (probably this model: http://www.radioshack.com/product/index.jsp?productId=2049691 )
If I ran that at 12V, then I could do 12 series of 5 LEDs in parallel. According to the LED array calculator, I would need one 56 ohm resistor for each series.
By my own calculations, this would put the net resistance at 4 2/3 ohms and the current too high at about 2.6 amps with each set of 5 resistors getting 214 ma which is of course, way high.
By my own calculations, I would need 12 600ohm resistors. 1/(1/600 X 12) = a net resistance of 50 ohms. 12V = I * 50. I = .240 / 12 series = .020amps or 20ma each. I surely am wrong, I just can't figure out my error.
And PS, this is the Bevo shape I'm talking about:
Thanks! 

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wcalvert Newbie
Joined: 05 Nov 2006 Posts: 4 Location: Austin, TX

Posted: Sun Nov 05, 2006 4:44 pm Post subject: 


This is the approximate location of the LEDs.
To get a scale, each square is a square foot:


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wcalvert Newbie
Joined: 05 Nov 2006 Posts: 4 Location: Austin, TX

Posted: Sun Nov 05, 2006 4:49 pm Post subject: 


Wow, another thing I forgot to add. I plan on adding a photoresistor/transistor and have it automatically come on when it becomes dark 

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Alan Site Admin
Joined: 29 Jan 2006 Posts: 1399 Location: Winnipeg, MB

Posted: Sun Nov 05, 2006 8:21 pm Post subject: 


Hey wcalvert,
You must be a serious fan! I am sure the project will look great. I couldn't follow all of your calculations but you started out correct. You could light it by using 12 parallel sets of 5 LEDs and 1 resistor in series. Your calculation for the resistor is also correct ( a single 56 ohm resistor in series with 5 LEDs will provide the LEDs with 20mA given a 12 Volt supply). But I think you are going wrong when you total the current.
You will have 12 circuits that are each pulling 20mA. All you need to do is add them up. 12 * 0.02A = 0.24A or 240mA
So the transformer you are thinking about will work just fine.
Send in the picture when you are done, it sounds like it will be cool looking.
ps. when doing resistor calculations it is easiest to use an online calculator such as this one:
http://alanparekh.com/led_resistor_calculator.html 

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wcalvert Newbie
Joined: 05 Nov 2006 Posts: 4 Location: Austin, TX

Posted: Sun Nov 05, 2006 9:57 pm Post subject: 


What I was trying to get across was that another online calculator gave me the 56 ohm number, while I independently calculated the 600 ohm number. 

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Alan Site Admin
Joined: 29 Jan 2006 Posts: 1399 Location: Winnipeg, MB

Posted: Sun Nov 05, 2006 10:08 pm Post subject: 


wcalvert wrote:  What I was trying to get across was that another online calculator gave me the 56 ohm number, while I independently calculated the 600 ohm number. 
OK, I missed when you said online calculator in your initial post
Just incase you want to hand calculate it, the way to do it is add up all of the voltage drops of the LEDs in the circuit. In this case there are 5 LEDs with a voltage of 2.2 Volts. That is a total of 5 * 2.2 Volts = 11 Volts.
This means that is a 12 Volt supply is used there will be 1 Volt across the resistor, if you want 20mA of current flow you can use ohms law to calculate the resistor size. 1 Volt / 0.02A = 50 Ohms. 56 Ohms is a good common value to use.
The other thing you should do is calculate the wattage needed. Watts is Voltage * Current.
1 Volt * 0.02 A = 0.02 Watts. This means that a typical 1/4 Watt resistor will do just fine. 

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