LED as power on indicator with AC Mains Supply

[peterlonz]

Hi,
I'm new here & hope this question is posted in the right place.
I am interested in a simple but reliable circuit that will allow an LED to indicate "power on" in a 230VAC mains circuit.
I have Googled & found two circuits but neither are well explained & they differ markedly.
Neither appears to deal with transients, spikes or surges that I understand are more common than is generally known.
I am unsure how to post the circuits here, or even if this is allowed.
Can anyone refer me to a simple execution which can be made from readily available components (unlike the 1 MegaOhm, 350V resistor of unspecified power rating said to pass 30mA in one of the circuits I have).
I calculate this resistor dissipates 900 watts, so either it's incorrectly drawn or I am sillier than, well you know ......... LOL
Yes I know neons can do this, but they are not bright enough & tend to have a relatively short life.
Rgds to all & thanks,
Peter O

[jwax]

Welcome to the forum Peter O!
Ideally, you would use a transformer to transform the 230 VAC mains to a lower voltage, and drive the LED with that- typically 2 volts.
A transformer-less solution (although not recommended, but possible) would require a 10,700 ohm resistor and a 10 watt power rating.
You must calculate the required voltage drop in the resistor by the peak voltage, which is 230 X 1.4 = 322 volts peak. To have 2 volts at the LED, you must drop 322-2 = 320 volts.
Allowing .030 amp (very bright LED), requires a resistor of 320/.03 = 10,700 ohms.
Power dropped in the resistor = 320 X .03 = 9.6 watts.
Maybe that value resistor is available, but using a transformer would make life simpler, and safer!

[Lenp]

What about a series capacitor?
At 60hZ a .25uf would be about 10.6K, but I'm not sure about dissapation!
Len

[jwax]

With 230 VAC mains, methinks Peter O is on a 50 Hz supply?

[Lenp]

Well a .3 Cap is 10.6 at 50 Hz!

[haklesup]

I think the transformer is the best and safest way to go.

jwax calculations only take into account the forward biased condition of the AC waveform on the LED. In the other half of the cycle the LED will be reverse biased to 322V and that's far higher than the zener or reverse breakdown voltage of most LEDs. To use an LED at this voltage I would also put another Si or Schottky diode with at least 400V PIV (and 1000V PIV would be better) and adequate wattage in series with the LED pointing in the same direction as the LED. Assuming that diode only drops 2V at 30mA (which is a high guess) a 1W would be sufficient.

Furthermore at 1.7V and 30mA the LED will put out 51mW thereby not needing any special heatsink for mounting. While that 10.7K resistor needs to be 10W minimum, it will be fairly hot. A 20W or larger is recommended to prevent scorching your case and PCB and for reliability. While one could argue that the current flows only 50% of the time, I prefer to design with a 2x to 4X margin of safety so doubling the wattage over DC calculations gets you there. for AC circuits P= to the integril of I*V over a period of time

On the safety side you need to consider what happens if the LED fails, will it expose a user to dangerous voltages.

[Robert Reed]

1N4006 diode - 3.9K/5W resistor - LED
Wired series line to line.

[jimmy101]

I think Jwax's calc is off a bit.

Assuming an LED in series with a single resistor, the duty cycle is 50% and you use the RMS voltage not the peak voltage. Dropping the current back to 20mA to ensure the LED has a long life;

So, 230V/20mA = 11.5K
(I ignored the insignificant voltage drop across the LED.)

230V * 20mA = 4.6W but you only have 50% duty so actual power dissipated in the resistor is 2.3W. The LED will blink at the mains frequency, which shouldn't be noticeable.

Probably kind of hard to find a 11.5K (ish) 3W resistor. You could use say four 3K 1W resistors in series. Keep'm well separated cause they get hot since each will be dissipating about 0.6W. (1 watt resistors used to be pretty common. :} )

Like haklesup said, you probably need a diode to block the reverse part of the AC waveform.

A couple uF, 500V or 1KV cap in paralleled with the LED will reduce the chances that a voltage spike will cook the LED.

Another possibility would be an LED with a built in blink circuit. IIRC, their duty cycle is pretty small, like 10% and that'll decrease the power dissipated in the dropping resistor to the same extent. You might even be able to use a single 10K (ish) 1/4 watt dropping resistor with a blinking LED.

This is not a very electrically efficient way to do things. A small transformer would probably be better if you insist on an LED. Transformer, diode, cap, LED should be enough.

Or, do it the way it is usually done and use a Neon bulb (i.e., NE-2) and a 1/4 watt series resistor. You wont be wasting nearly as much power and the NE-2 isn't bothered by voltage spikes. For 120VAC I believe it is a 100K 1/4 watt series resistor. For 230VAC the resistor would be 200K 1/4 watt.

[jwax]

jimmy101- I disagree about using the RMS value. The LED will indeed see the peak voltage.
And yes, I neglected to consider the reverse voltage breakdown of the LED, sometimes rated around 5 volts.
That being said, I connected a LED with a 56K resistor, applied 240 VAC, 60 Hz, and the LED works fine.
Open discussion!

[Externet]

http://www.turbokeu.com/myprojects/acled.htm

[peterlonz]

Thanks everyone for your suggestions.
Can I make a few observations that to me are highly relevant:
1) It's no longer "usual" to use neons in this application mainly because they are not nearly bright enough & they deteriorate.
2) The link posted by Externet http://www.turbokeu.com/myprojects/acled.htm (thanks) is one of the circuits I found & had difficulty with. As I said where do you get a 1 MegOhm resistor for 350 V capable of carrying 30mA, that's 900 W !!. I also note the diode is used in parallel with the LED here which is not what most of you have suggested.
3) In pretty much any electrical store, supermarket etc you can now buy multi-way power boards with from 4 to 8 mains sockets. Most of the newer designs feature one or more 10 amp on/off switches, a 10A mcb & a bright "power on" LED indicator. So is it not reasonable to say these transformer-less designs are relatively safe & economical. I have many such around the house, but note I am aware my house switchboard offers circuit breaker and RCD protection.
4) One such power board failed recently after only 2 months service, luckily the non-indicating LED warned us, so the food in the deep freeze was not lost. The inbuilt mcb did not activate as you might have expected.
What I found after dismantling was:
* Failed 100 Deg C fusible link (non-resetable), heavens knows what that component was meant to do, my experience of these devices is that they are failure prone.
* 3 x GNR 14D471K varistors, apparently intact.
* 10A mcb not tested but presumed OK.
* LED still operational when supplied power.
* Diode small, lost & untested.
* 0.1uF 275VAC rated capacitor (yellow block type), untested.
* Resistor 2.5 mm diam x 9 mm lng, discoloured & failed, 4.1 KOhm I think.
* purpose made inductor, 9 turns of 1.2mm diam wire on 6.5mm diam core.
* 10A on/off switch, working OK.
* Circuit board which seemed unaffected & sound, sorry I destroyed it getting all the bits out, before realizing this design may not be so simple. BTW provision was made for an on board fuse but this was not fitted.

I am guessing this is a more cost & energy efficient "solution" than using even a small transformer, safety issues I'm not sure about, & in this case reliability looks very suspect.
Anyway I'd greatly appreciate more comment if the above helps, & particularly comment on the design in Externets link.
Thanks again everyone,
Peter O
Queensland Australia
230 VAC, 50 Hz

[jwax]

Peter O- The 1 Meg resistor does not pass 30 mA. It is across the 240 volts, so less than a milliamp is going through it. (I=E/R)
Either series or parallel diode perform the function of clipping the voltage going to the LED in the reverse direction.
The "power boards" you describe are called "power strips", or "outlet strips" in the US, and some are rather elaborate in offering transient voltage protection. Sounds as if yours failed!
Externet- nice find!

[dyarker]

Peter O,

To calculate power:
P (in Watts) = E (in Volts) times I (in Amps)
To get 900W you are multiplying Volts by 30, BUT 30mA is 0.03 NOT 30! Power 9W.

The 1M resistor is there to bleed off the capacitor when the unit is unplugged.
240V / 1000000 Ohm = 0.00024A
240V * 0.00024A = 0.0576W

Using a capacitor and resistor as a current limiter causes the unit to consume less power than using just a higher value resistor. The capacitor shifts the phase of the current relative to the phase of the voltage. (Learned how to calculate that in college, but haven't done it for decades.)

When using a capacitor, the diode needs to be reverse parallel with the LED. A diode in series with the LED would be a half wave rectifier. The capacitor would charge up, there would be no current and LED would be off.

Cheers,

[CeaSaR]

Dale,

You beat me to the reply re: 30 mA = 0.03 A, resulting in MUCH less power being dissipated.

CeaSaR

[dyarker]

Yeah, and I was editing (adding) when you posted.