LEDs in bass cabinet

Discussion in 'Pickups & Electronics [BG]' started by teamup182, Apr 14, 2012.

  1. teamup182

    teamup182 Guest

    Apr 1, 2012
    so I recently started refurbishing an old 6x10 i had. and i want to put an LED light inside. the terminal/jack plate (5"x7") was plastic and had a broken corner and i just threw it out. and i had this cool idea of mounting my inputs and L-pad to a piece of plexiglass and i though it would look pretty sweet if i had a LED in there to kinda emphasize the plexiglass and the electronic components inside. i dont want it to be very bright. just barely so you can kinda see the inside. im already rewiring the whole cabinet but im not sure where i should put my LED without having it blow up. i know i should have a resistor in there somewhere but im not sure what kind or where it should go.

  2. A resistor may or may not be needed, depending on the LED and the voltage of the power source. In some cases, two AA batteries can be connected directly to an LED, but that may not be desirable, and some cheaper LEDs will burn out without a current limiting resistor.

    You will need to know the forward voltage and current rating of the LED first. Typical values range from 2V to 3.6V at 10mA to 20mA. You may wish to run a lower voltage to keep the light dim, but in any case, be sure to choose an LED with a lower millicandella output. (No superbrights.) From there, you can choose just about any low voltage DC power supply, or perhaps a 9V battery.

    Determine the value of the resistor using the following formula.
    R= (V[SUB]Supply[/SUB]-V[SUB]Load[/SUB])/I[SUB]Load[/SUB]

    For example, an LED rated for 10mA@3VDC, with a 9V supply requires a 600 Ohm resistor, as (9-3)/0.01= 6/0.01= 600

    Once the resistance has been determined, check to be sure the wattage rating of your resistor is sufficient to dissipate the thermal energy. You can simply multiply the voltage drop by the current load. The above example dissipates 0.06W, as 6*0.01= 0.06. A 1/4W resistor would be sufficient for such an application.
  3. Mykk


    Aug 22, 2010
    Prescott, AZ
    ^^ this.

    Isn't the terminal jack plate on the rear of the cab?
  4. How are you planning on powering it? If you don't want to use batteries, you could do the following: Using a multimeter, measure the voltage at your speaker jack. Tap your signal in at the jack, which is AC, rectify it (an LED needs DC to operate) and connect the appropriately oriented leads of the LED to the rectifier. Wire in a trim pot to one lead and adjust the resistance so your LED won't burn out. I can't imagine you'd need bigger than a 50k trim pot. A very simple rectifier can be made from 4 diodes arranged in a diamond pattern, with a 100uF cap to smooth the ripple and provide a more stable source.

    It won't stay lit solid like a battery-powered LED will, but will flash depending on how much current the LED is getting. Might actually be cooler that way.
  5. hartke20g


    Apr 12, 2006
    miami, FL
    care to go a bit further into that? i've been trying to do exactly that for years now, and i've never seen a simpler solution.
    what components do i use based on the voltage readings? where in the rectifier do i connect the LED and input leads?

    i don't mean to hijack, but i can't believe i never came across this solution before
  6. If you'll excuse my poor drawing, he's referring to something like this. (We must assume that no voltage exists above the rating of the regulator IC, however.)

    Personally, I wouldn't screw around with something like this. The circuit is not isolated from the speaker. Use a battery or a small DC power supply. A wall wart would be sufficient.
  7. walterw

    walterw Supportive Fender Commercial User

    Feb 20, 2009
    yeah, why would you want an LED that nobody can see interfering with your signal?

    just get a little battery-powered thing in there, maybe with a little switch to turn it on and off.

    (the whole thing seems wildly pointless, like using a bedazzler on the cables hidden in the back of your rack, but whatever :))
  8. Smilodon


    Feb 18, 2012
    I wouldn't do that either.

    The LM78XX only support 35V input. Speaker outputs can potentially go much higher than that. Especially in a low load circuit.
  9. I wasn't sure about the input voltage. How high does the voltage tend to go in bass amps? The problem either way, however, is that now you're getting into a tremendous amount of thermal dissipation. A DC/DC converter could potentially work.

    Again, seems like a much better idea to not do anything with the signal and just use a battery or DC power supply.
  10. spaz21387


    Feb 25, 2008
    Portland oregon
    Notice the 2x15s. We took those police style lights that were the led ones and put them inside the bass cabs where the tweeter was then covered the hole with clear plexiglass...

    IMAG1026 by spaz21387, on Flickr
  11. What was your power source?
  12. Smilodon


    Feb 18, 2012
    Rough calculations say that a 200W amp driving 8ohm loads will be about 40 volts. That's RMS. Peak to peak is about 56 volts.

    Bridging a LM78 and a LM79 won't work since they will need a common ground.

    The easiest way of doing is is to simply add a small transformer. For driving a LED it can be very tiny. It also have the advantage of galvanic separation. :)
  13. Yes, you need a transformer to provide a tapped secondary as a ground for the rail supply. I removed the rail supply comment, however, because I figured it wouldn't be a good idea to have the primary winding putting an inductive load across the speakers.
  14. Rune Bivrin

    Rune Bivrin Supporting Member

    Oct 2, 2006
    Huddinge, Sweden
    This is the simplest way. Two LED:s and a resistor.

    You don't really need a variable resistor. A 4.7k resistor will be large enough for almost any amp. If you make a habit of cranking you will want to make sure the resistor is rated a 1 watt.

    And don't worry, it won't affect your sound.

    Attached Files:

  15. The brightness will vary with voltage.
  16. Rune Bivrin

    Rune Bivrin Supporting Member

    Oct 2, 2006
    Huddinge, Sweden
    Yep, it will. As we say in the software business: "That's not a bug, that's a feature!". When the LED:s light up you know you're loud...

    There are certainly ways to achieve a more even intensity without involving external power supplies, but the circuitry will be somewhat more complex. Full bridge rectifier, resistor, zener diode, another resistor and then the LED.