# Help with resistors / Amateur Project

1. Aug 7, 2008

### anakinjay

Ok, just a warning... I'm an amateur engineer... I do things as a hobby (computer programmer by trade)

Anyways,
Here's what I'm doing:
I'm trying to mod a guitar cabinet with LED lights that pulse with the music. I want to power it from the signal that comes into the speakers from the power amp, so I don't have to worry about plugging anything else in.

Also, I'd like to use a potentiometer of some sort so I can adjust how loud of a signal triggers the lights. The goal is to get the lights to fade as the guitar gets quieter.

The signal coming from the power amp is 70v 500 watts.

Using an LED calculator, it seems I can make a 1x21 array with 70v input using a 39 ohm 1/4 watt resistor.

My question is:
If I try to push 70v 500 watts at a 39ohm 1/4 watt resistor, am I going to melt it's face off? :P I've never worked with anything bigger than normal 12v. Also, how would I wire up the potentiometer?

Or if there's a better way of handling this, I'm all ears... I'm a total noob at this stuff.

Thanks!

2. Aug 7, 2008

### mgb_phys

edit - sorry it saved an early draft.

Last edited: Aug 7, 2008
3. Aug 7, 2008

### anakinjay

There's a good chance you're right... like I said, never worked with anything this high voltage / wattage before...

Here's the array I get from the LED cacluator:

Solution 0: 21 x 1 array uses 21 LEDs exactly
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 39 ohms

The wizard says: In solution 0:
each 39 ohm resistor dissipates 15.6 mW
the wizard says the color code for 39 is orange white black
the wizard thinks ¼W resistors are fine for your application
together, all resistors dissipate 15.6 mW
together, the diodes dissipate 1386 mW
total power dissipated by the array is 1401.6 mW
the array draws current of 20 mA from the source.

Does that sound wrong?

4. Aug 7, 2008

### anakinjay

again, total noob here...

but in p = v2/r wouldn't you add up the total resistance of all the LEDS as well as the resistor?

that would make r around 593.4, and would put total watts at around 8.25'sh?

5. Aug 7, 2008

### mgb_phys

LEDS don't have a resistance as such, they have a fixed voltage drop, typically around 2V.
So you are dropping 40V accross the LEDs and 30V accross the resistor, that's about 30Watts, quite a lot for a 1/4W resistor.

I'm also not sure it would give the effect you are looking for - I'm guessing it would just glow gently and slowly turn on at the louder parts. You might need something a little more involved with some sort of time constant to turn the LEDs on in beats (like a graphic equaliser).

6. Aug 7, 2008

### anakinjay

Well, I'd probably be happy with just a soft glow, and then on when the guitar is playing...

I'm not looking for anything complicated like an EQ :)

Here's an example:

http://www.instructables.com/id/Sound-reactive-led/

this guy is just wiring the led straight up to the speaker.

the only real difference between what he's doing and what I want to do is I have a muuuuch higher input and more leds.

Make sense?

7. Aug 7, 2008

### mgb_phys

I don't know what the calculater did but -
You need the voltage drop accross the LED (or the whole array) - should be on the data sheet.
Subtract this from 70, this gives the voltage across the resistor.
You need the current for the LED (data sheet again)
You can work out the resistance from R = V/I, where V is the difference above and I is the current through the resistor (make sure it is in Amps not mA)

There is no problem with having a resistor with too high a power rating!
I don't know much about high voltage speaker lines, but I assume the amp will have short-circuit protection? You might also want to include a fuse just in case.

Ussual disclaimer about being carefull, don't electrocute yourself, blow anythign up, start a fire etc....

8. Aug 8, 2008

### anakinjay

if so then...
Each LED has a voltage drop of 3.3

3.3 x 21 = 69.3

that would give me .7 volts going through the resistor...

The LED's use 20ma (0.02 amperes ) so:

R = .7/.02

So that's 35ohms?

and P = .7^2/35

So the watts would only need to be .014 watts?

So if I'm reading this right, using the 39 ohms, 1/4 watt resistor like the calculator said should work ok... right?

9. Aug 8, 2008

### mgb_phys

If they are dropping 3.3V each then they pretty much add up to 70V anyway, probably no need to worry about the resistor. The voltage drop has a certain tolerance and is temperature sensitive so could easily become 3.33V.

If you have a meter you could set it to the mA scale and connect it in series with the LEDs. For a real system that you are going to install somewhere I would fit an inline fuse (available from car parts stores) say 250mA, in series with the LEDs - just in case.

10. Aug 8, 2008

### anakinjay

fantastic!

I'll definitely use a fuse, and I'll be sure to post pics when I get it all finished.

Thanks for all your help guys! Much appreciated!