LED from mains

[andrew2022]

i need an LED to display wether the mains power (230V) is on or off. whats the best way to do this without a transformer? can it be connected to a rectifier directly to the mains (obviously with a resistor in series)

 

[JimB]

A neon lamp would be much more appropriate that an LED if you want to run it directly from the mains.

JimB

 

[TheOne]

It can be done with a capacitor, resistor and diode, but it is bulky. The capacitor is chosen to drop most of the voltage due to the capacitive reactance. Make sure that the capacitor (rated for AC purposes) can sustain the required ripple current and have a high enough voltage spec.

As from the attached picture the current through the LED will be pulsating DC

 

[Dean Huster]

As Jim suggested, using a neon lamp such as an NE-2H or NE-51 is much more desirable and less bulky, even though it may be an "antique" device and not a modern solid state light emitter.

The problem with the LED is that it's a current-operated device. If you used nothing but a resistor to keep the current at 20ma, the resistor would dissipate nearly 5 watts and that's nearly as much power as a common nightlight. Using the capacitor as a reactive divider to knock down the bulk of the voltage helps that power dissipation problem, but you do have to trade a LOT of bulk for that advantage.

The neon lamp must also have current limiting, but it's current requirements are so much less than those of an LED that you can use a common 1/2-watt resistor (or maybe smaller) to do the job. The neon lamp won't be quite as bright as the LED, but is available in any color you'd like so long as it's orange. You could find an argon lamp if you'd like the color to be a less-visible near-UV blue. Argon lamps are harder to find, of course. Even though argon is a LOT more plentiful in the atmosphere, it's color just doesn't generate a lot of sales!

Dean

 

[panic mode]

...and if you do use caspacitor to knock down voltage, please do yourself
a big favour and put a resistor IN PARALLEL with C1 capacitor
(something like 100k 1W for example).
circuit as shown above would work but when you unplug it,
there is nothing to drain down the capacitor charge. (are you ready
for some nasty zapps?) also i would not use only half period.
small bridge rectifiers are dirt cheap and small (or just put 4 diodes).
if you want low power device, consider using small transformer from
wallwart. they are not any bigger than high voltage capacitor,
you get isolation and lowest power consumption...

 

[jrz126]

...and if you do use caspacitor to knock down voltage, please do yourself
a big favour and put a resistor IN PARALLEL with C1 capacitor
(something like 100k 1W for example).
circuit as shown above would work but when you unplug it,
there is nothing to drain down the capacitor charge. (are you ready
for some nasty zapps?)

Had this happen to me at work, I was doing some tests on a power filter involving some rather heafty caps. there were 4 in parrallel which had a resistor also in parrallel, and 2 that did not. I was taking the wires off, and I figured what the heck, I'll short them to see if any charge is still in there. Sure enough they were full, practially welded my screwdriver to the terminals. So I also recommend a bleeding resistor accross your cap.
Although wouldnt the LED drain the cap after switching off?

 

[panic mode]

Although wouldnt the LED drain the cap after switching off?

how?

if you turn the circuit off with a switch (inserter in series with the circuit)
or if you simply unplug the circuit from the receptacle, C1 would be
part of open circuit.

 

[TheOne]

What you guys/gals are saying is partly true. Remember in this circuit AC current is flowing because of the protection diode over the LED (the led has pulsating DC current through it). So the cap will have AC volts over it. The amount of voltage over the cap will depend where in the cycle you pull the plug. If you pull it on a zero crossing you will of course have no voltage over the cap. If you pull it on one of the peaks, you will have a max -ve or +ve voltage over the cap.

If we do some sums, the peak -ve or +ve voltage over this cap will be about 169v. Looking at the energy stored in the cap, will be about 7mJ

To put this figure in perspective, look at the extract I found on the web for the Taser Police unit.
Taser says the device operates at a fraction of the electricity used to resuscitate heart attack victims -- 1.6 joules is well below the 10 to 50 joule threshold above which cardiac ventricular fibrillation can occur.
. A joule is a unit of energy.

"That amount of energy applied to the chest, probably only a fraction ends up going to the heart," said Dr. Peter Kudenchuk, professor of medicine at the University of Washington.

"Simply the pain created by a shock of 1.6 joules might make the heart rate faster, but I'd be surprised if much of the energy reaches the heart itself," he said. "In terms of causing a cardiac arrest, the risk is probably low."

So 7 milli Joule of energy seems very small (230 times less than a taser).

It is obvious that for large capacitors there's a definite danger and bleeder resistors should be fitted.

 

[JimB]

If we do some sums, the peak -ve or +ve voltage over this cap will be about 169v. Looking at the energy stored in the cap, will be about 7mJ

That is true for a 120 volt supply, the original post stated 230 volts, so the peak will be 325 volts and the energy will be about 24 mJ.

JimB

 

[TheOne]

If we do some sums, the peak -ve or +ve voltage over this cap will be about 169v. Looking at the energy stored in the cap, will be about 7mJ

That is true for a 120 volt supply, the original post stated 230 volts, so the peak will be 325 volts and the energy will be about 24 mJ.

JimB

No, that is not correct, the AC voltage dropped over the cap in this circuit is 120v.

120 * SQRT(2) = 169v peak

But your figure does show the fact that the Joules will increase proportional to the square of the voltage over the cap.

 

[jrz126]

Although wouldnt the LED drain the cap after switching off?

how?

if you turn the circuit off with a switch (inserter in series with the circuit)
or if you simply unplug the circuit from the receptacle, C1 would be
part of open circuit.

I was thinking that the cap is in parrallel with the LED, so when you switch it off at the AC side, the cap and LED would still be connected, but that wouldnt drop the voltage down like you are looking to do.

 

[mstechca]

Why not use a lightbulb?

or if you really want to use an LED, make or buy a DC adapter rated 9V or less. connect a 60 (or more) ohm resistor in series with the LED and hook it up to 9V.

 

[JimB]

[quote="TheOne

No, that is not correct, the AC voltage dropped over the cap in this circuit is 120v.

120 * SQRT(2) = 169v peak

[/quote]

I do not agree.

The peak voltage of the 240 volt mains (used in your sim.) is 339 volts.
In the first sim the peak voltage at the C1/R1 junction is 25 volts.
339 - 25 = 314 volts across the capacitor.

Can I suggest that you check what your "voltmeter" is reading, is it true RMS, or RMS calculated from peak or average. It will make a big difference in this circuit.

JimB

 

[TheOne]

JimB

I based my calculation on the reading obtained by the voltmeter of the simulator (2nd simulation), without taking notice of the previous scope display value. Also the scope display in the first simulation was pasted into the diagram afterwards for the sake of illustration of the waveform-shape over the LED, as the program does not allow you to capture both at the same time. So it is possible that I saved another scope display later, while evaluating other cap and resistance values.

Based on your observation of the scope display as seen, the conclusion you are making is then 100% correct.

Also the only other info available on the voltmeter or adjustable parameters are, time constant and parallel resistance. I will have to hunt for the documentation on the program to see if there's more info regarding RMS etc.

 

[Oznog]

Why not use a lightbulb?

or if you really want to use an LED, make or buy a DC adapter rated 9V or less. connect a 60 (or more) ohm resistor in series with the LED and hook it up to 9V.

An electroluminescent panel is also possible, they use these for nightlights. They do have a limited life, though. A neon light is fine. Also a transformer could be used to make a low voltage for an LED.

 

[audioguru]

Hi Ms,
You like using a 60 ohm current-limiting resistor for a LED with a 9V supply, dontcha?
A common red LED has about a 2V drop and a 30mA to 50mA absolute maximum continuous current rating. With a 60 ohm current-limiting resistor and a 9V supply, it would draw 117mA and will melt or surely burn-out!
Even an "ultra-bright" blue or white LED with a 3.5V drop would draw 92mA! For a few milliseconds it would look like a searchlight before it burned-out. But you would be blinded and won't see the smoke.
For an indicator light only about 10mA is needed which would use a 700 ohm (use a standard 680 ohm) current-limiting resistor (for a common red LED) with a 9V supply.