more led questions

[new guy]

Does an LED work the same with AC voltage as it does with DC voltage? Is the formula the same for figuring out the resistor to use when dropping the supply voltage? Today I measured 49 volts AC going into an orange LED. This was after the resistor that dropped it down from 124 volts AC. There was an 8ma (.008) current flow and the resistor measured 231.7k ohms. The LED was not very brite. It was the kind built into an AC relay that lets you know it is energized. Why was the voltage so high? I thought all LED'S worked off low (2-5) volts?

[JPKNHTP]

-JPKNHTP

[Dave Dixon]

Hi, are you sure it isn't a neon bulb? It sure sounds like it may be. Dave

[HighFrequency]

Voltage across a forward biased LED is always the same. What matters is the current through it. You will never measure 49 Volts across an LED. Normal LED's have a forward biasing voltage on the order of 1.5V (if I remember my Devices 1 course). I suspect you measured the voltage incorrectly. KPKNHTP is correct, AC shouldn't be used. The reverse current flow could (and probably will) damage the LED. You may have been lucky and were only turning off the LED when the current flowed backwards (relative to the LED). You should rectify, and smooth AC to power an LED.

[Chris Smith]

LEDS are diodes, and thus AC works fine with them as long as the proper voltages and current are met. The only thing you have to worry about like any diode is don’t exceed any of the specs.

I make trickle chargers using four LEDs instead of four diodes to form my diode bridge which rectifies the AC into DC, with the added benefit of illumination showing the charge is happening.

[Robert Reed]

LEds work on average forward current within their limits. These can be pulsed over a variety of frequencys. Again you have to observe peak current limitations. There is nothing wrong with using them in an 120VAC circuit as long as you follow certain design rules. First you will need a diode of sufficient PIV (200V) in series with the LED polarized the same as the LED. This prevents LED breakdown from reverse polarity (line voltage reversing polarity). Second Calculate Series resistance for desired LED current. Remember you will have only 60 volts RMS to work with due to half wave cycles. So -- if you desired 10 milliamps of current thru your LED and by ohms law 60V/0.01 A =6000 ohms series resistor of a 1/2 watt minimum rating. More current-less resistance-higher wattage.This is a close calculation as I have eliminated diode Vf drops ,etc. I have no idea what the 270K resistor is doing in your circuit as that wouldn't even allow 1 milliamp throgh the circuit. Also 8 milliamps of current is enough to light up most LEDs decently. Go back and check your circuit again.

[HighFrequency]

For anyone not aware... PIV is Peak Inverse Voltage. This is the voltage that the diode can tolerate in the reverse biased direction without breaking down (smoke).

[Chris Smith]

PIV and Forward voltage will be the same going through a resistor under load.

A simple shunting or loading resistor to ground forms a voltage divider to ensure that both directions of current and voltage remain the same in both directions.

This simple two resistor network keeps the specs within the limits.

[MrAl]

Hi there,

If you want an LED to run off of 120vac you will
find that the required dropping resistor will
burn up power just like a regular night light.
Instead, here's a circuit that uses something
like 1/2 watt and the LED is bright as it can
be without exceeding any specs:
http://hometown.aol.com/xaxo/index.html

Normally, an LED by itself and a single
dropping resistor can not be run off of ac
unless the ac voltage is very low (5vac or so)
because the reverse voltage of the LED will
be exceeded when the ac switches phase.
Tricks around this are to use a diode in
parallel to the LED so that when the ac reverses
the diode picks up the current. Drawback is
diode conducts 50% of the time so the efficiency
is lower than optimum.
Next trick is to use a full wave bridge, which
makes sure the LED always gets dc voltage
(although pulsing). Drawback is if a dropping
resistor is used the dropping resistor uses more
power than the dang LED
Another trick is to use a diode in series with
the LED, but the drawback here is that if there
is appreciable leakage in the diode the LED
still gets banged with a high reverse voltage when
the ac reverses.

The basic idea in the circuit link above is to
use a capacitor in series with the LED (and a
full wave bridge) to convert the voltage down
to a lower value more suitable for the LED.
The capacitor uses very little power because
it can store energy. Thus, the LED lights the
same as with a resistor only no power is wasted.

You'll note that there are a few resistor in
the circuit too, to limit other specs for the
LED like peak current especially when the unit
is plugged into the wall when the ac is at its
peak.


Take care,
Al

[ecerfoglio]

Mr AL posted:

..... Instead, here's a circuit that uses something like 1/2 watt and the LED is bright as it can be without exceeding any specs:
http://hometown.aol.com/xaxo/index.html
....

** SAFETY NOTE: **

Be shure to use a capacitor with a voltage rating of at least 200 V (for a 110/115 Vac mains, the peak voltage is more than 150 V).

"Foreign note": If you have 220 Vac (like we have here in Argentina), you could use the same circuit but the capacitor should be a .3 (or .33) uF (that is, one half of the original .6 uF) ** BUT RATED AT 400 V **

[philba]

non-isolated circuits make me verrrry nervous. I'd use a polarized plug and make sure that the fuse is on the hot ("L") side. Note also that insurance companies may be able to weasel out of a claim relating to this circuit.

<small>[ December 07, 2005, 09:07 AM: Message edited by: philba ]</small>

[Robert Reed]

Mr Al

If you want an LED to run off of 120vac you will
find that the required dropping resistor will
burn up power just like a regular night light.
Instead, here's a circuit that uses something
like 1/2 watt and the LED is bright as it can
be without exceeding any specs:

Not meaning to step on your toes, but aren't 8 components a little overkill to perform a single funtion as lighting an LED ?

[Chris Smith]

A voltage divider consisting of a 9900 ohm resistor and a 100 ohm resistor will deliver 1.2 volts at the center tap, with a 12 milli amp draw.

A slight tweaking of these figures will deliver the exact current and voltage needed for any led to center tap the two resistors depending on your forward voltage and current requirements.

The over all power consumed is limited to 1.4 Watts from these two resistors values.

But Keep in mind all of the voltage values...
{www.ultracad.com/articles/rms.pdf}

........Our household voltage is 120 V. RMS, then 120 Volts is.707*V, where V is the peak voltage.

This means the peak voltage is 120/.707 or 169.7 Volts.

And the peak-to peak voltage at our wall outlets is 2 times this or 339.5 Volts!

Expressed another way, the formula for the voltage at our wall outlets
is169.7*sin(Q) where (Q) represents the angular
position within one cycle.

<small>[ December 07, 2005, 03:52 PM: Message edited by: Chris Smith ]</small>

[MrAl]

Originally posted by ROBERT REED:
Not meaning to step on your toes, but aren't 8 components a little overkill to perform a single funtion as lighting an LED ?

Hi there Robert,

Well, it depends on what you are using it for
and what you want to get out of it.

For example, if you use a dropping resistor to
light a high brightness white LED you'll want
at least 20ma to flow which requires something
like a 2500 ohm resistor in series, which would
eat up more than two watts and the resistor
gets almost hot, and you need a rather large
size resistor.

If you only need a *tiny* amount of light out of
the LED you might get away with a larger value
resistor in series, but keep in mind you wont get
the full brightness of the LED that way unless
you also are willing to eat the power and get
rid of the heat.
The heat is a big issue when you try to package
the circuit into a very small housing to make an
LED night light.

Also, commercial LED night lights are made using
this same technique with additional components
like cds light sensor to turn it on/off depending
on ambient light conditions. The nice thing
about them is you get the full brightness of the
LED but the package doesnt get hot.


Take care,
Al