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I'm working on a project where I want to have Christmas lights (or LED strips if needed) react to music. In this case, the end goal is to have them pulse to the beats/bass of the music. There's a lot of LED strips that come in kits where there's a microphone that listens to the music but this doesn't work for me for a few reasons:

  1. seems cheesy
  2. doesn't give me just the bass
  3. I'd prefer a hard-wired solution

There's a lot of solutions on YouTube where people hack in to their speakers and simply power the lights from the signal being sent to the subwoofer in the speaker. This works, but I can only drive a small number of lights this way and this is also highly dependent on the volume of the music.

Another approach is to build a SSR (solid state relay) based circuit like this but the problem here is that I'm not looking to do any high voltage wiring myself (just an EE student here) due to the fact im not comfortable with it and I don't know anyone who really is.

I thought of the idea to purchase an audio amplifier (or I guess I could make one with a transistor and potentiometers) that would normally drive a set of speakers and instead have it drive my lights. I know it would put out a lot of power I wouldnt need, but I can adjust this using the gain controllers on the amp (power efficiency isn't the goal here- overkill is acceptable.) Would this even work? Fundamentally I think this would work, but am I missing something? -perhaps resistive load issues

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  • \$\begingroup\$ If overkill is acceptable, any resistive load issues can be buffered with a series-connected resistor (that'll waste some power). \$\endgroup\$
    – glen_geek
    Commented Sep 7, 2017 at 14:10
  • \$\begingroup\$ But would this work? Assuming I worked out any lack of resistance \$\endgroup\$
    – Alex
    Commented Sep 7, 2017 at 14:47
  • \$\begingroup\$ Consider arranging LEDs with bi-directional symmetry. All LEDs going one-way (driven by capacitance-coupled amplifiers) would be a problem. \$\endgroup\$
    – glen_geek
    Commented Sep 7, 2017 at 15:06
  • \$\begingroup\$ This sort of thing is often called a "light organ". \$\endgroup\$ Commented Sep 7, 2017 at 15:42

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The old term for what you describe is a "color organ", a very popular hobby project in the 60's and 70's. Search for color organ schematic to see examples. Old ones are built to drive 120 Vac light bulbs directly, and newer ones drive LEDs.

Your idea of using an additional power amp just for the lights also is valid. There are many low-cost amp modules on ebay. Since audio fidelity is not an issue, almost anything will work. The one you link to should work. Or, you can get an amp module and a used computer or laptop power supply, and wire ti up yourself.

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Google "music activated lights" but you'll find old schematics.

If you want only the bass, you'll need to run the signal through a lowpass filter followed by a peak detector(ie, a diode). Signal can come from a microphone or the amp's output, or RCA connector...

Then you'd use the output of that to turn the LEDs on or off with a MOSFET, for example.

If the original power supply has a voltage output, you can even reuse it.

DO NOT use a MOSFET to switch the AC mains! The switch goes on the low voltage side...

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  • \$\begingroup\$ But the goal here is not to turn the lights on/off, its to have them fasing in/out with the bass \$\endgroup\$
    – Alex
    Commented Sep 7, 2017 at 14:46
  • \$\begingroup\$ You can use the MOSFET in linear mode also, or a bipolar transistor, or PWM it... \$\endgroup\$
    – bobflux
    Commented Sep 7, 2017 at 15:34
  • \$\begingroup\$ I think often the timeconstant of the peak detector can be tuned to already act as a lowpass filter. \$\endgroup\$
    – Joren Vaes
    Commented Sep 7, 2017 at 15:40
  • \$\begingroup\$ Could be, yeah. You can try something like this: makezine.com/2010/10/18/circuit-skills-led-color-organ-spon only keep the bass section (the one on the bottom), remove the input preamp (opamp on the left) and replace the output transistor with a TO220 MOSFET. Should work. \$\endgroup\$
    – bobflux
    Commented Sep 7, 2017 at 20:22

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