LED's for dummies

The voltages look like the typical forward voltages at the specified currents. Those currents, incidentally, will make the LEDs quite bright.

Assuming you are using a 9V wall wart, you could:

Power + ------ resistor1 -------- two blue leds ----- power - ------ resistor2 -------- two blue leds ------ power -

and so on so that the typical forward drops add up to about 7.2V in each chain

I don't know wall wart you would use. Let's take 9V for argument's sake

so Resistor 1 = (9V - 7.2V) / 20mA = 1.8 / 2E-2 = 90 ohm. Power dissipation in the resistor = 36mW. A typical (and easily available) value would be 100 ohm.

For another string, take 4 orange LEDs, 1.7V @20ma. Vf total ~ 6.8V. So R = (9 - 6.8) / 0.02 = 110 ohm. Once more I would use 100 ohm.

The actual currents won't be exactly 20mA - it's going to depend on the

Hope that helps with the idea!

Cheers

PeteS

Hi. You're stuck because you haven't chosen a wall wart voltage yet. Just for grins&giggles, try a 12VDC wall wart -- they're fairly common.

Now figure you want to have about 1/2 of the voltage in LEDs, and about

Now it's fairly easy. Combine two or three LEDs of the same type to get something between 4.5 and 7.5V (just add the nominal voltages), and then choose a series resistor so that (12V - LED voltages) / R = about

Example: 3 red = 2.3V * 3 = 6.9V

Choose 220 ohms or 270 ohms (standard values).

The reason you should choose LEDs of the same type is that you'll probably have to do some tweaking to get the LEDs to look nearly the same brightness.

Now put these strings of two or 3 LEDs with a series resistor in parallel, and you're good to go.

Good luck Chris

On 16 Aug 2006 13:34:16 -0700, in message , snipped-for-privacy@gmail.com scribed:

These are the voltages and currents that you can expect the LEDs to use when they are operating.

Your second problem is that you are limited to a reasonable maximum of 20 LEDs, 0.020 X 20 = 0.400, or 80% of your supply's rated maximum. I wouldn't go over that. Which, since your plan is 18-24, I suppose isn't a problem after all.

The resistors that you choose will be based on a rather simple mathematical formula. You want the optimum current to be 0.020A. Your supply is rated at 9.0V. For this design, it appears that two LEDs in series is optimal. Put two diodes in series, calculate the expected voltage drop and subtract from 9.0. Take the remainder and use Ohm's Law to calculate what value resistor will give you that voltage drop at 20mA. Voila! You have two diodes and a resistor in series that give 9.0V at 20mA. Rinse, repeat.

The resistor can go anywhere in the series combination.

there won't be any bulky electronics if you can drill it from the base up the centre you may be able to do it all from the bottom, start with a 1" (or larger) spade bit and hollow out the bottom half then switch to a 0.5" and go up further

more would be better...

it's basically the same thing.

yup. if you can, get a 12v or 24v wall-wart,

efficiency isn't real important. there's not enough power there to start a fire or anything.

if the wall-wart is 9V regulated: two blues 7.2V use resistor 91 ohms (should be 90) a green and two oranges, 6.9V use resistor 110 ohms (105) a sea green and two ambers, 6.9V 110 ohms (105) three reds, 6.9V 110 ohms (105) two greens 6.6V 120 ohms (120)

all those will be close enough.

hook the leds together in series (head to tail) and at one end put

(+) ---/\\/\\/\\--->|--->|--->|---- (-)

use heatshrink tubing to insulate the connections.

the resistor. user oneresistor for each series group of leds

(+) ---/\\/\\/\\--->|--->|--->|---- (-) (+) ---/\\/\\/\\--->|--->|--->|---- (-) (+) ---/\\/\\/\\--->|--->|--->|---- (-) (+) ---/\\/\\/\\--->|--->|--->|---- (-) (+) ---/\\/\\/\\--->|--->|--->|---- (-)

if you thread the wires into the holes first and then attach the LEDs that may give the tidiest result. "wirewrap" wire is made for this sort of thing.

you can grind the flange off the LEDs to make them fit drilled recesses better a sanding disk is good for this.

you'll need to be sure to connect them the right way round.

Bye. Jasen

Thanks, everyone. This is pretty much what I thought I would need to do. Time to break our the breadboard and see if I can't make it work before I actually start making things light up!

Just to add to the others' good suggestions: it would be good to measure the Wall Wart with a voltmeter, to verify it runs at the stated voltage. It could be a couple of volts higher than the rating, which would figure into your resistor calculation.

Mark

Just to give you another option: Create 3 or 4 strings of LEDs in series, using

Option 2 is probably best:

.------. ----- | +|----+--Vin|LM317|Vout---+----+------+------+ | | | ----- | | | | | 24V | | Adj | [LEDs] [LEDs] [LEDs] | DC | | | | | | | | | [.1uF] +--[240R]--+ | | | | Wall | | | [56R] [56R] [56R] | Wart | | [3.6K] | | | | | | | | | | | -|----+--------+---------------+------+------+ .------.

Voltage at the Vout pin will be 20.0 Each string of LEDs has a nominal Vf of ~18.9 volts. The 56 ohm resistor will limit the current through the string to ~19.6 mA

Ed

I'd go down to the dollar store and see what they have. They often have all sorts of weird and wonderful LED flashers. Easy, quick and cheap AND you can buy some spares.