Led lit surround shelving - diyAudio
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Old 27th July 2013, 03:49 PM   #1
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Default Led lit surround shelving

I'm building surround speaker shelves out of glass or plexi (haven't decided yet). What I want to do with them is have them glow slightly with led light (blue prolly). What I was thinking of doing is getting either the glass or plexi thick enough so i'm able to drill (dremel) a hole big enough into the edge of the material to tuck an led light inside therefore lighting the shelf and hopefully looking cool.. I've read a little bit about the simple circuitry involved with lighting a single led and somewhat understand it. Am I stuck with building a simple circuit with a double a battery and hiding it away behind the speaker or would I be able to power the led by running a small led of wire coming from the surround speakers to power it?
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Old 27th July 2013, 06:13 PM   #2
DUG is offline DUG  Canada
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Find a glass shop (or plexi) and get a small sample.

Use krazy glue to glue the led to the edge of the glass. That way the light will go into the glass. Use flat faced LED's or cut the rounded part off.

Then if the edges of the glass are frosted the light will show up on the edges.

Is this what you were thinking about?
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Old 27th July 2013, 07:41 PM   #3
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Quote:
Originally Posted by DUG View Post
Find a glass shop (or plexi) and get a small sample.

Use krazy glue to glue the led to the edge of the glass. That way the light will go into the glass. Use flat faced LED's or cut the rounded part off.

Then if the edges of the glass are frosted the light will show up on the edges.

Is this what you were thinking about?
Yep, that's the idea. Do you know if I can run the led using the speaker input from the amp?
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Old 27th July 2013, 08:19 PM   #4
DUG is offline DUG  Canada
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You can.

At max expected power:

(Vpk-2V) / .03 = resistor in series with LED.

The LED won't blow up but may not light very much at low levels.

Have fun.
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Old 28th July 2013, 01:47 AM   #5
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Default led lit shelving

gotcha.....so I might be better off picking out an led that requires less voltage i take it? would i be better off not running a resistor inline? I know it may cause the led to burn out prematurely but i just don't want to increase the load on the amp too much....

pardon the out of context smiley....just had a Hoegaarden ale....
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Old 28th July 2013, 04:47 AM   #6
DUG is offline DUG  Canada
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Quote:
Originally Posted by Daem View Post
...would i be better off not running a resistor inline? I know it may cause the led to burn out prematurely but i just don't want to increase the load on the amp too much...

.
Not running a resistor inline with the LED would cause more of a problem...much more.

Use the formula I suggested and look at the value of the resistor.

I think you will find it will not load an 8 ohm system down too much.

example:

An 8W into 8R is 11.312Vp.

9.312/.03 = 311R

I don't think that will be an issue for an 8R output.

Or look at it this way:

8W is 1.414Ap

0.03Ap is a 2.1% change in loading.

I doubt if your speaker impedance is withing 2% of 8R at all frequencies.

Math is a wonderful thing.

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Old 28th July 2013, 12:56 PM   #7
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Default led lit shelving

Dug, if you wouldn't mind....

could you run by me what the letters mean in the formulas below? I hope I didn't mislead when i mentioned i understood a little about the wiring of led's ....I'm just not that familiar with the abbreviations as far as this stuff goes...
I gather W is watt and R is resistance? maybe? not too clear on AP...
and if you could would you mind running through an example for me? I'm running 6 ohm surrounds and the channels are good from 4 to 6 ohms. the amp is capable of 110 per channel....
thanks again ....i appreciate the help on this.....
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Old 28th July 2013, 02:17 PM   #8
DUG is offline DUG  Canada
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W watts power...is RMS or root mean square...this is the equivalent in heating power in a resistor.

A is amperes (actually a large quantity of electrons per unit of time passing through your circuit) and is also rms unless specifier as Ap = Amperes peak or Ap-p = Amperes peak to peak. I is also used for Amperes in equations.

Similar is V is Electromotive force...AKA Voltage and is RMS but can also be labeled as peak or peak to peak. Sometimes E is also used for voltage.

Normally in audio usage we are talking about a sine wave. That way simple formulae can be used.

Vp = V * 1.414 or V = Vp*.707

Vp-p = Vp * 2 = V * 2.828

You can substitute A in the above equations.

W = V * I
W = V^2 / R
W = I^2 * R

I = V / R
R = V / I
V = I * R

And then there is impedance...

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Old 28th July 2013, 03:05 PM   #9
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Default shelving

Dug

I think i'm getting it. As long as i have the values for voltage, watts, and ohms I should be able to figure this out. can we go over the ohms bit? It's what I'm concerned with at the most (just don't want to blow up my amp...)

Thanks!
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Old 28th July 2013, 03:15 PM   #10
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Sometimes it might be useful to know the max. voltage U that a resistor can be connected to without exceeding its heat dissipation limit. With a little bit of manipulation, the formula becomes

U = √(P * R)

where P is the rated power dissipation of the resistor in Watts and R is the resistance of the resistor in Ohms.
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