LED circuit?

Hi i'm new to the group, and i'm not really to awesome when it comes to electronics, so I thought I would ask people who really know!

Heres the situation: I play bass in a thrash/rock-punk band and I want to customise my bass a little. And my idea is to place red LED's under the scratch plate which I will raise up with washers/nut(s) to give a glow eminateing from underneath on dark gig's. Ideally I suppose 4 or 5 LED's on a 9v battery would be about as much as I could fit under there.

I dont know how to make a circuit with a switch and LED's or where to buy them in the UK.

A freind mentioned I could get a sort of fibre optic wire, that he has on his PC for cosmetic effect but I guess that runs off 12v?

Cheers for any advice + sorry for being so clueless!

There's a lot of stuff here:

and here

Some of it runs on as little as 2 AAs. or 9V.

Google "Electroluminescent Wire" or El wire

Seems like it's everywhere.

Richard

Buying is the easy part, most of the larger cities have a Maplin store, and you may also find some independents if you check the yellow pages. There are also several mail order stores for electronic parts in the UK, if you get a copy of elector magazine you should find their adds. For small orders however local stores are cheaper.

LEDs work best with a current of 10..20 mA, the voltage required to drive that current depends on the colour of the LED (about 2 V for red, yellow and green and 3.5 V for blue and white).

Thus assuming you want ot use red LEDs, you can have 4 of them in series from a 9 V battery, requiring a total of 8 V. The voltage of a new

9V-batt is about 10 V, so the remaining 2 V needs to be fried in a resistor, the resistance is calculated from Ohm's law R = U/I = 2 V / 0.02 A = 100 Ohms. As the battery empties, its voltage will drop and the LEDs will give less light.

The power fried in that resistor is P = U * I = 2 V * 0.02 A = 40 mW, so a standard 125 mW resistor should handle the load easily without turning black.

So your circuit will look about like this:

/ /

+Batt o----o o-----/\/\/\--->|--->|--->|--->|-------o -Batt Switch 100 R 4 x LED red SPST 125 mW superbright

Connection of the resistor and the switch may be in any orientation, but LEDs and the battery need to be connected in the right polarity.

For blue LEDs, you can have only 2 in series, and the resistor needs to be increased to 150 Ohms. But you can put several such circuits in parallel:

/ / 150 R 2 x LED blue

+Batt o----o o-----/\/\/\--->|--->|------o -Batt | | | | ---/\/\/\--->|--->|---- | | Since blue LED appear not as bright as red it may become necessary to use several such branches.

This of course is only the easiest version. As your knowledge and confidence with such projects grow, you could have the LEDs go on and off in the rythm of the music, or switch between different colours. You also may consider mains supply to avoid wasting batteries, but this is something you should do only when you understand the safety implications.

Hi im planning to do the same thing but im using it in my car. I plan to use a total of 6 blue led's. I just wanna know how many resistors should i use and it's rating. Is it possible to do it all in parallel? thanks very much

You can probably use 3 LEDs in series with a 150 ohm 1/4 watt resistor and another 3 LEDs and resistor for 6 total. Try this calculator to work out the exact values.

-Bill

Thanks! At the store though I found some color changing led's. It works well with 9v batteries but when I got home and tried it with 12v and a 750ohm resistor it just lit red and didnt change color. Tried again with lower voltage starting at 1.5 gradualy moving up without resistor... it changed color at 6v - and then burned! What am I doing wrong? It worked well with the 9v battery so why did it burn at 6v? Any ideas on how to get it to work?

Thanks and good day

I have no idea what circuit is inside the color changing LED. Does it come with a data sheet or any specs?

When it was operating from the 9 volt battery, how do you know what voltage it was getting? 9 volt batteries sag under load quite easily, especially if they are not alkaline or fresh. Rechargeable 9 volt batteries put out something like 7 volts.

--
John Popelish

Car probably means a 12 V lead acid battery, supplying about 14 V when full, blue LEDs need 3.5 V. So 3 of the LEDs in series would require

10.5 V, the remaining 3.5 V would be fryed by a resistor of 3.5 V / 0.02 A = 175 Ohms (180 Ohms is the next higher standard value) at 3.5 V * 0.02 A = 70 mW.

Since you want 6 LEDs, you need two such strings in parallel.

All, Instead of guessing, trial and error, and smoking those leds which stink badly there is a formula for this: R = (Vcc-Vled)/Iled where R = resistor to put in series. Vcc = supply voltage Vled = Led voltage across the Led Iled = led current

Michael