# How to make LED plug into wall AC socket?

Discussion in 'Electronic Basics' started by Me, Jul 12, 2004.

1. ### MeGuest

Hello,
I have a number of blue LEDs, that work when connected to a resistor
and a 9V battery (don't know much else about them!)

How would I go about making a wall socket to I could plug in a group
of LEDs?

If I got an old 6V or 9V AC/DC adapter from say an old walkman or
something, Could I safely connect that somehow to the LEDs?

Should I wire the LEDs in parallel?

Where can I find info to do this?
THANKS

2. ### Don KlipsteinGuest

http://www.misty.com/~don/ledd.html

among other places.

I recommend against paraleling LEDs.

With a 6-volt DC "wall wart", I would assume 7 volts with a light load.

Blue LEDs usually have a voltage drop near 3.5 volts and usually want 20
mA typical, 30 mA max, and I would recommend somewhat less (15 mA) if you
want to get the usually-advertised "100,000 hours" and you do not know
that the thermal situation is better than that in the LED manufacturer's
test lab. But you can probably get away with a little over 30 mA.

Back to the 7-volt "6-volt DC wall wart": Put each LED in series
with a dropping resistor, and then put all LED-resistor "strings"
in parallel with each other. Subtract the 3.5V LED voltage from the 7V
supply voltage, and that leaves 3.5 volts across the dropping resistor.
3.5 volts (resistor voltage, that is) divided by .02 amp 20 milliamps)
yields 175 ohms, and the nearest common value is 180 ohms. I recommend
220 ohms to play safe if you want a good expectation of really long life.
You will probably get away with 150 ohms and maybe with 100 ohms.

Do not try matching supply voltage to LED voltage. The current through
the LED will be of unreliable magnitude. The current through an LED as
voltage across it varies greatly with small changes in voltage, and the
voltage required to push a given amount of current through an LED varies
enough with temperature and manufacturing tolerances to make current as a
function of voltage unreliable.

If you have a 12-volt DC "wall wart", assume 13.5 volts for a light
load. You can put two LEDs and a resistor in series, and a few of these
"strings" in parallel.

With two 3.5 volt LEDs subtracted from 13.5 volts, you get 6.5 volts
across the dropping resistor. Divide by .02 amp (20 milliamps), and this
gives 325 ohms for a dropping resistor. The nearest common value is 330
ohms. I would use 470 ohms if I needed to count on a few 10,000's of
hours of LED life expectancy, although I forsee probably no quick failure
with as low as 220 ohms.

As for resistor wattage: Multiply 6.5 volts resistor voltage by .02
amp, and that means .13 watt. That sounds like a 1/4 watt resistor is
enough, but I recommend a half watt one if you want good reliability.
Resistor reliability decreases enough at over half rated power for
military 1/4 watt resistors to resemble commercial grade 1/2 watt ones,
and that 12 volt DC wall wart might just provide 16 volts to a really

Although you can probably put 3 blue LEDs and a resistor in series with
each other and run this from a 12 volt DC "wall wart", the LED current can
easily be greatly different from what you expect and may vary excessively
with temperature and presence of other loads on the "wall wart".

- Don Klipstein ()

3. ### Tom BiasiGuest

Hi,
Put the LED's in series; add the voltage drop of each LED. Subtract the
Use the difference as the V in R=V/I. Put the rated value for the LED's in
for ' I '.
This is the series resistor value that you need.
Take that same difference voltage and multiply it times ' I ' and double it
for safety and that's the wattage rating you need.
Use the closest standard value that doesn't go under.
Good Luck,
Tom

4. ### andyGuest

The way i would do it is to set up a test rig for one of the LEDs to work
out what voltage/current gives a good brightness. most normal leds expect
about 20mA - more for high brightness ones. This could just be like this:

9vDC-----------------
|
|
LED
|
|
1 k ohm Variable resistor
|
50-100 ohm fixed resistor (so you don't short it)
|
0V--------------------

Set the resistor to 1kohm, then slowly reduce it until the LED is at a
good brightness. When it's right, measure the voltage across the LED, and
the current through it, with a multimeter. You'll have to connect the
meter in series to measure current.

Once you've found a good working voltage, see how many of them fit into 9V
with a bit left over. (probably 2 for blue leds)

Then subtract the working voltages for however many leds from 9V, divide
this by the working current, to give the value for a resistor to put in
each series string.

Then wire them in series/parallel with a resistor in each series bit.

e.g. if the working voltage/current was 4V/20mA, then you would say 4+4=8,
leaving 1 volt over. Divide this by 20mA to get 50 ohms.

Then you would wire them like this:

9VDC --------------------------------------------------
| | | | |
LED LED LED LED LED
| | | | |
LED LED LED LED LED
| | | | |
50ohm 50ohm 50ohm 50ohm 50ohm
| | | | |
0VDC --------------------------------------------------

make sure you don't go over the current rating of the power supply.

you might be able to get away with wiring them up without any resistors,
but this would make the current more sensitive to changes in the supply
voltage, and you could blow the LEDs.

You could also work out the combined parallel resistance of the resistors,
(R/number), and use a single high power resistor instead. this might not
compensate so well for variations in each led though.

5. ### MeGuest

Hello,

I finally got around to experimenting...

For a Blue LED, with a 1000 ohm resistor,
Off a 9V battery,
the voltage was 9.57 without anything connected,
9.48 with the LED/resistor on the battery.

How many LEDs can I hook up to the following adapters I dug up:

12VAC, 1A
3V DV 500mA
9VAC 650mA

Does it matter if I use AC or DC (since it's an LED?)

Thanks

6. ### John PopelishGuest

LEDs do not last long if more than a little reverse voltage is
applied, especially blue LEDs. So DC is simpler to use than AC. But
you need about 4 volts DC minimum to light blue LEDS and have enough
extra voltage to waste a little to control the current. So that 3
volt DC supply is probably not going to work very well, )though it may
put out 4 volts or so under light load). I would add a bridge
rectifier to the 12 volt unit to convert it to DC and then connect in
parallel several strings of 3 LEDs (anode of one to cathode of the
next, etc.), each string having one 1000 ohm resistor to limit the
current. If you have a multimeter, you can measure the current
passing through this string (by putting the millimeter in series with
the string) and adjust the value of the resistor till the current
rises to around 10 milliamps. Then you can connect lots of similar
strings across the supply (up to 100) before the total current exceeds
the 1 amp rating of the supply.  