Wanted: 115-VAC LED Indicator

No - I saw an unjustified rule of thumb, which is simply incorrect it the Voltage is RMS. Your statement that your practical system cannot be dissipating anything like the three Watts we would expect can only mean that either you are deceiving yourself about the power, or the Voltage is not RMS, or the resistor value is not as you believe, or...

I'm sorry to be so blunt, but I'm beginning to suspect a reason why you didn't respond to the question about the measurement of the Voltage.

Physicist?,

It all about pulse feeding the LED using ~117Vrms.

It could just as well be a dc pulse with the same effect: 60mA derived from a pulse generator with an equivalent amplitude, pulse width and rep rate and power draw that produces the desired brightness effect.

I think I mentioned before that the circuit was running while I was making entries here.

Here is what I have:

1N5393 rectifier in series with

2800 Ohm carbon composition resistor 1/4 inch diameter. (measured resistance) in series with

A Red 20mA LED mounted on an

Archer Universal Breadboard connected to an

AC line cord plugged in the wall 118 volts at the time.

Instantaneous Peak line voltage would produce ~60mA.

Works fine, last long time. Try it and see if it is real.

Jon

The resistor will be dissipating somewhat over 2-Watts (see my explanation of the handbook rule-of-thumb in post #123). It will probably last quite well*. Carbon comps are also relatively tolerant of impulse energy.

On the other hand, the high random failure rate of even modern carbon composition resistors means I wouldn't choose them for any high-reliability application.

*There are a number of reasons, including:
The construction of carbon composition resistors is more robust than it was when the size-versus-Wattage ratings were originally established; both the thermal dissipation through the leads and the assembly's tolerance of high temperatures have been improved.
You are probably operating the resistors in a relatively open environment. The ratings should be for tight assembly on a board, typically with somewhat restricted air-flow.

Physicist?,

"The resistor will be dissipating somewhat over 2-Watts"

The 1 Watt resistor sees nearly 3 watts during the very short time each half cycle. The whole time is spent dissipating the overall equivalent of steady state heating of about 1 Watt.

I like to use larger wattage carbon film, metalfilm or wirewound resistors for the thermal and physical situations in the final application.

"ratings should be for tight assembly on a board, typically with somewhat restricted air-flow"

A good thing about mounting on a board or chassis is that it allows the board to become part of the heat dissipator. Especially if there is ground plane or dissipative meterial in the board designed for that type of application.

Mooseman's thermal requirements are restrictive even though the heat inside the box becomes stable and is dissipated through the material of the box.

Jon

I meant "somewhat over 2-Watts average" (i.e. heating with a steady state average of over 2-Watts). The peak dissipation should be V_peak²/R. With V_peak=115*√2=162-Volts, and just the resistor and a couple of diode drops, the peak dissipation should be 9.5 Watts - much more than the 3-Watts you suggest (can you explain where your 3-Watts peak came from)

Yes, in practice the board is part of the heat-sink, but som standard tests use minimal tracking widths so the resistor on a QA test board is often less well-cooled than a "free" resistor.

Mooseman's system is indeed restrictive - hence my concern about your apparently theoretically impossible observations. (In my experience this is usually a sign that there is something in the system that is not as one thinks it is - it could be that the series resistor misbehaving at high current, or there might be an internal resistor* inside the LED package or ...)

*1100 Ohms, for example, would halve the dissipation.

Been there, done that. For about 200 lamps. The circuit was in continuous use for about 5 years, when they replaced the equipment with stuff 30 years newer.

This is very similar to a circuit that I use except that the 1N4003 is series-reversed across the LED. I use a 100K 1/2W resistor with no problems.

.5*Vrms^2/R with 115vac and 100k is .0661Watts ---- check!

Current is just over 1 mA, so I don't think it would light a super bright. (Or not very brightly.)

True, not very bright but bright enough for my home made voltage detector. My intent was to show a low cost solution. Practical application might dictate a change in values.

Ha-Waiian,

You've read through the postings here on this thread. You should know that if you don't satisfy any criteria stated or unstated it will be questioned. I'm surprised nobody has questioned if the insulation will withstand five years of jungle mold, frost heave or how salt spray will effect integrity.

PW Slack,

Best idea I have seen so far.

Jon

I use an Idec pilot light with an LED, to sense both 115 and 240vac. They have LEDs that are off the the shelf rated for those voltages. They have different colors and are a bayonet style.

Look at their LSTD series

PWSlack,

I spent some time in there and found 5Vdc to 28Vdc units.

There was a need for 115Vac units at work and the highest we could get were 28Vdc units (1300 Ohm resistors in them) So we put a 1200 ohm resistor and a diode in series and they have been running since 1988.

Jon

Way to go PW,

Mooseman, this is from the source PWSlack has offered. With watts in the range of 1/25 to 1/4, and the size you wanted. About $8 with reductions for quantity. CJM

Beware - that's back to neon lamps for the 120-volt supply.

Get a 120 VAC relay with an indicating LED and connect the coil from line to neutral. When there is 120VAC present, the coil will be energized and the LED will be on.

Once you've done that and your satisfied, then remove the Red LED that's there and replace it with your Ultrabright LED.

Once your satisfied with that, then remove the resistor and the LED from the relay, remove the relay and put the resistor and the LED in place of the relay.

I'm wondering if relays with LED indicators are only popular in my next of the woods. Surely there must be a cabinet that has a relay with an LED indicator you could look at. Take a look at that first and you'll see how simple it is.

See Alley-Bradley: http://literature.rockwellautomation.com/idc/groups/literature/documents/pp/800l-pp001_-en-p.pdf

What about using a 240-5v transformer between the Li & L2 phases with a LED and suitable rectification and resistors on the 5v output.

I have a similar system with two transformers across a 440V three phase supply to indicate the loss of one phase

L1 - T1 prim - L2, L2 - T2 prim - L3

This has been working satisfactorily since 1970, although there are two bulbs in parallel across each transformer secondary

No room-too expensive.

Mooseman, PWSlack provided a good suggestion. There are plug-in replacements for some incandescant lamps available and there are a host of mfr's.

Digikey.com has a large selection.

You didn't indicate (no pun intended) the lamp type or color.

What about Tranilamp (usual disclaimer)?

Looks like a very clever device. I've got to remember this one. Unfortunately, It is large and looks expensive. mounting would be a problem -- Pc mount with SMD??

I Googled the device and copied it to " cool things to use ".

Thanks!

I would use a neon; cheap and long lasting:

120-Volt Neon Green Lamp (2-Pack)

$3.49

Model: 2720708 Catalog #: 272-708

No need for a power supply.

Mike

Mikerho, you are much like me. I never seem to read the questions or follow the clarifications. No neon-- not bright enough. No incandescent—burn out rate. 2 Phase brought to one box. Indicate voltage present on both lines –independently. Must be sensed at the box entry point (just after fuses). Low power dissipation –The box is sealed –heat rise problematic.

Well, I didn't read through every post. I did see that incandescent was out. For me, I love to play with LEDs, but the added step of converting from AC to low-voltage DC seemed a lot to go through.

It was just something that just came to me - thanks for the reply.

Mike

When I was first approached with this problem, I thought it to be a no-brainer, but, It it is a significant challenge. I think I now have a reasonable handle on the situation and will try several of the suggested solutions.

This is a great discussion group and I thank all of you for your responses.

Mooseman.

My experience with neon has NOT been long-lasting. I have several surge protectors and night-lights whose neon indicators either no longer work at all, or only flicker in the presence of ambient light. They are good for 3-5 years. I have a 30+ year old calculator, and every one of its many LEDs still work. Of course it has not been continuously on for all those years... I do have several other devices using LEDs (my TV antenna booster is the first thing that comes to mind) that have been on continuously for 30+ years, and still have what appears to be the original brightness.

I have used a capacitor in series with the standard resistor, diode, and LED to operate an LED on 120 V. The capacitor reduces the heat generated in the resistor.

I am sure that is the reason my client wants an LED. These indicators are the most expensive maintenance item in their system - and dangerous. Non-tech types have to get into the box and punch down sensor wires. If they do not have the power off, we may have a problem of removing bodies from the area.

Is there some reason why you cannot use a 110/220VAC/DC pilot light and a fuse?

http://www.wolfautomation.com/ProductList.aspx?CategoryID=5354&gclid=COTNn_Hbh5wCFQE0xgodygQ9Sg

Should have been more descriptive - these are 110/220VAC, LED pilot lights.

Hard to read text from the jpg's, but price of either one is under $20.

I know that the original query indicated two LEDs. Is this essential, or would a single lamp that says "both-on" suffice.

If you want separate indication, would a single dual-colour lamp that is red for one phase, green for the other and yellow when both are illuminated suit your purpose? If the latter, are the relative phases of the two supplies related, and if so, how?

P.S. how much dissipation can you afford? If it's 0.6-watts (total for both indicators) you could use a 12-Volt LED with integrated resistor. If its less you'd need to use the capacitor and overcurrent protection with lower resistance.

Re my PS: apologies - it wasn't clear that any arrangement would include the series capacitor and a limiting resistor.

Cheapest route will be to mount a 15K ohm power resistor to the inside of your chassis, possibly a TO220 like package so that the chassis acts as a heat sink. This will limit the current to the LED. Either use a bidirectional LED (Red/Green) to handle conduction in both directions or add a reversed biased diode in parallel to the LED. (Don't worry about reverse break down for one diode will always be on.) If the resistor generates too much heat then tough. 120VAC drawing 10 ma of current to turn on the LED will be 1.2 watts dissipated somewhere. To be more efficient you must convert the power to a lower voltage, or use a 115VAC relay to turn on the LED driven by your 12VDC.

Redfred,

At only 10mA during the peak of each cycle the LED will appear very dim.

Since the peak voltage is only there for a short time for each cycle the power dissipation averages out to less than a quarter Watt.

Jon

All depends on the LED selected, and background lighting. A Chicago Miniature Lamp super bright P/N CMD333URC-2 is rated to produce 2000mcd with 20mA of current. Their M8 LED indicator package P/N 19510330 includes current limitation to operate at 130VAC with only 5mA of current, but at a listed cost of $6.03 each (Newark). The latter package does not indicate brightness in my catalog. Depending on the luminous environment, 50mcd maybe sufficient. The project might require wiring for a socket-ed lamp base that could accept a standard 115 volt bulb but an LED replacement bulb available.

Redfred,

More new stuff that keeps getting cooler.

Jon

There are a lot of different approaches to doing the same thing. A 20mA current on an Ultrabright LED is very bright, 10 mA is actually enough. The LED is being view on its side, when looking at it at the proper 30 deg viewing angle, it's plenty bright.

For 20 mA the wattage will be 2.3 Watts. A 3 or 5 watt resistor is all that is required. The images are with a 2.74K resistor. Did not have any larger wattage than 1/4 for this demo. Adding a diode reduces the current and is not required.

Any resistor value less than 3 watts will get very HOT. Kudukdwelder, a 1/4 watt resistor will start changing colors in minutes and get smokin hot, you need to check your calculations.

Using 2 LEDs SR, they will both be on at the same time. A 50Hz pulse on an LED, light, etc cannot be distinguished from being on constantly. At least not by my human eyes.

LED + 2.7K or 3K 3 watt resistor is all you need.

If safety or ensuring that the power is off when the box is open, put a limit switch on the cover to disconnect power if that's possible, or an audible alarm when the box is open and power is on. One ear piercing blast and they'll check the power first the next time.

I have used this solution for years and the dissipation is below 1/4W and will operate on any AC line from 50V to 264V (240 +10%) AC

50 or 60Hz flicker is only a problem if the device is moving in which case you need another diode in series with the LED and an electrolytic 100microfarad 6V across the LED

The dissipation is low, but so is the mean current - about 0.65-mA. This may be OK for some situations - but it's certainly not a "bright" LED.

You are right, the current is small, but if the capacitor is 470nF (or, as they usually say o.47uF) then the peak current would be 14,4mA, which is more than sufficient to make a bright diode be seen in day light.

The capacitor resistor combination is the best (minimum heat dissipation) Now, if we talk power factor for the system....

The capacitor and LED ideas are pretty useful, but we can halve the dissipation in the LED protection resistor if we change the topology slightly, as below. The 1N4004 diode should be safe because it can withstand as much as 30-Amps for 8-ms (which is far more than can be applied via the series capacitor). The resistor value will depend on the peak current capability of the LED - for safety the resistor should limit the current to the safe single-peak current when a 200-Volt peak is applied across the resistor (you may need to check this with the LED manufacturer, as they do not generally publish non-repetitive peak figures). The resistor power dissipation can be calculated once the value of the series capacitor is known (the value chosen is to match 20-mA mean LEDs).
The other component that may be sensitive is the current-controlling capacitor; you may need a second (small) resistor in series with the 1N4004 diode to limit inrush to the level that the capacitor can withstand - but at least the capacitor specification sheet should specify the acceptable peak inrush current.

There will probably be bits that are unclear - so please don't hesitate to ask.

Hi Physicist,

This is, indeed the way to go. You bypass the R-LED combination with a 1N4007. It needs a limiting resistor.

30A - Peak Forward Surge Current 8.3ms single half sine-wave superimposed on rated load

(JEDEC method)

I would use your schematic a bit modified:

With a combo: 0.47uF and 10K resistor, the dissipation on the resistor would be around 1Watt.

The purpose of what I did was to reduce the dissipation relative to your (and others') original suggestions. By returning to the earlier version you have nearly doubled the dissipation you would get with my modification (for a fixed level of LED protection).

The energy that can be dissipated in the diode when charging the 1-uF capacitor to 200-Volts is 20-mJoule; I'm not certain whether the 1N4004 can absorb this much single-pulse energy; if not, there are two possible solutions - to use a diode with larger thermal mass, or to incorporate a lower-value resistor (than the one used to protect the LED) in series with the capacitor (and reduce the LED protection resistor accordingly). That would somewhat increase the total dissipation, but not as much as sharing a single resistor.

By the way, safety would also mandate a high-value resistor between live and neutral to discharge the capacitor when the circuit is depowered.

Bypassing the LED and resistor with a diode will reduce the real power dissipation of the resistor but will increase the power drawn by the whole circuit. If the reverse bias diode was forward biased just once (the transient strategically timed switch closure) I agree that this diode would probably survive. I'm just a little apprehensive of doing it 60 times a second with only the current limitation from the capacitor. Particularly because now the wire resistance, or poor contact may become the resistive load. Not to mention what will happen when the cap finally fails by shorting, bang.

Physicist,

You now what? You are right. Your schematic is not that bad

I assumed that there will be a huge current through the 1N4004 (probably by analogy with a rectifying diode followed by a large capacitor) without calculating. Talking about the pair values (0.47uF and 10K), the energy on the positive alternance, the way you drew, will not exceed 7mJ. Reported to a semi-period of 8ms, this is a poor 0.8W. Indeed, will be necessary a small resistor or another 1N4004 diode, in parallel with the first (yes, they might have not the same characteristics and the currents would not be equal).

I did miss one important issue - a large overdrive on the 1N4004 could cause appreciable reverse Voltage on the LED. You could in principle filter this using a capacitor around the LED; however, I think the capacitors will need inrush protection in any case, so this could be filtered with either a modest inductor or a resistor in series with the capacitor itself. As a 30-Ohm resistor would be sufficient to protect the LED, and even 100-Ohms would not significantly affect dissipation, the main consideration will be the impulse-energy rating of the resistor (~20000*C).

I also realised that we could reduce the value needed for the LED series protection resistor by replacing the single capacitor with a capacitive potential divider. The top capacitor would control the 60-Hz rectified current, and have the same value as in the original circuit. If the lower capacitor were to be made 9x the value of the upper one (and 1/10 of the Voltage rating) the resistor dissipation could be reduced by a factor of 10. I'll draw it up when/if I have time.

Hi Mooseman

Is there a reason why you can't use an LED indicator that is already manufactured like the ones at this site ? This site was already submitted by EElectrician earlier (GA to him).

Looks like quantity prices are available as well. I've used similar indicators as replacements in the past because I grew tired of changing filament & neon bulbs. They are more-than bright enough to be seen at distances of 30-50 feet.

Take a look if you haven't already.

Wow, a nice collection of analysis data to go along with this, too. Some here might complain about the part count but you have to answer the large group of "What if?" questions somehow. I like that it's also a completely passive circuit design. If this doesn't deserve a GA, I don't know what does. (Well at least you got one from me)

GA from me too!

I just wish there were a way for us to post higher resolution images... At first I thought they were both 1.2µF capacitors, instead of 1.2 and 12µF. Decimal points are soooo important!

Looks to me like you've thought of everything pertinent. I will certainly try it out!

dkwarner,

Yep µF can look like pF.

Russians used tick marks , instead of dots. Their poor paper quality made dots already and pencils dont made good dots unless you drag them a little.

It always confused me at first because I though they were commas.

We could look for alternate dot marks for our drawings for posting here.

1.2µF could be 1,2µF so the extra pixels give it more visibility/

Jon

Physicist?,

I agree with Redfred.

I give you a GA on that too.

Jon

Holy. Mooseman. A LED is a diode which will regulate the AC voltage into half wave DC. Take apart a transformer or AC adaptor with LED indicator to see how its powered. Its LED and a resister. Its been done and proven. Its is as simple as it is until you try to reason why it won't work. Stop thinking. Get a LED and a resister to reduce current to 20mA and plug it in to see if it blow.

LED breakdown Voltages are far too low for this to work. You definitely need a series diode if you wish to half-wave rectify.

How much voltage is across the LED when its getting 20mA from a resister in series?

Stop thinking, start doing.

Pineapple,

167 Volts peak every half cycle.

The forward current of 20mA lights it ok Reverse voltage more than 5 volts is death for the LED.

Using a rectifier and resistor to provide current in one direction lights it up nicely without damaging it.

It doesn't produce reverse Voltage or current but produces a lot of heat in the resistor to limit the current. The heat is a big disadvantage to Mooseman's design criteria. Therefore using the integrating and differentiating characteristics of capacitors combined with bleed resistors and a clamping diode seems to be the best way to go for the LED circuit.

Jon

Can be purchased off the shelf from Radio Spares (RS)

Or Farnell or Newark - not expensive

I hope this helps

http://australia.rs-online.com/web/search/searchBrowseAction.html?method=retrieveTfg&Ne=4294934505&Ntt=LED+115+V+Panel+mounted&Nr=AND%28avl%3aau%2csearchDiscon_au%3aN%29&Ntk=I18NAll&Ntx=mode%2bmatchallpartial&N=4294856115+4294831868&Ns=stockPolicy_au%7c1%7c%7cnew_au%7c1&Nty=1&binCount=307&multiselectParam=4294856115&selectSubRange=19mm%20-%2030mm%20Mounting#breadCrumb

Nap,

That was suggested in post number 25 by PWSlack and post 28 by Rakesh Semwal.

The op said they were too expensive.

Jon

I would say $26.00 is a lot of mula for a light bulb!

Mooseman,

I guess you should have said inexpensive in your original post.

Jon

Mooseman,

Check post 79. Exactly what you have asked for and around $8.

You could possibly do better if there is volume. CJM

8 Dollars? I think even 1$ is too high for an LED lamp.I can buy Atleast 4CFL of 20watts each in $8,

Well said rakesh, $8.00 is not a simple solution.

You know, I never thought that my request for advice about an LED 115VAC indicator would create such a debate. But, This is a great world that people are so willing to put forth their views and thoughts and advice on this and many other approaches to technical challenges. I thank all of you, but, it's about time I get off my duff and do some bench work. The time of thinking is over and the time of doing is here. I just picked up a true RMS meter from a friend and I'll do a little wiring and measuring. and I'm sure I will come up with a workable compromise. Thank you all.

Mooseman

Mooseman,

Best idea so far.

Jon

moose,

agreed - lots of good advice. fun to see "the boys" hash it out... probably saved a couple of thousand in consulting"fees.

its been fun reading, makes me want to get my hands on some parts and start to play.... I remember when I was a kid and did this stuff just for fun!

I'm glad you're moving to the next step, good luck!

What are you waiting for? It's been well over 50 years since I was considered a kid (by most observers), but I still "get my hands on some parts and start to play...." It's still fun!

You get what you pay for. Selecting the appropriate part for a budget is a key part of any engineering task. Heck, why don't we all just grab the raw materials from the beach, quarries, and oil fields, add a little elbow grease and fabricate 80 micro dot sized LEDs with a 1.5V forward bias voltage in series. 80*1.5V = 120V. That should work fine right . Oh but wait that's not taking into account the peak voltage of a sine wave. So 120*√(2)/1.5≈113 diodes. We should fabricate 113 red light emitting diodes in series, that should be bright enough. He he, yeah that'll work. But wait, don't try to change my plans because I'll just come up with some hidden criteria that you didn't understand. This should cost just a few pennies a pound. Ha Ha yeah, of course it will. No I don't like that white suit, put it away. Yeah just a few pennies for a thousand lights, thats just 9 lamps. No I don't like that suit, it buckels in the back. He he....

As was implied in answer to your original posting: the customer would not perceive the 115-Volt devices as being reliable enough, because they are neons, not LEDs as the title of the sheet says (from that source, only the low voltage devices are LED-based)

How much time have you wasted looking on CR4 ??, $26 ???

If you are looking for an standard industrial application LED with long life (20 years plus), I would not waste any more time.

If it is for home use, you can keep looking

Nap,

Getting it right and satisfying a customer brings them back followed by their friends.

Therefore it is not wasted time.

Jon

Its gone a long thread, Some members have impression that I supported the views of few others to buy this from market. but thats not true.

This what I said in one of my comment-"Exactly try RS"

Probably no one clicked on RS.

Rakesh,

Probably no one clicked on RS.

I did and all I got was your home page.

RS was suggested by someone in Oz and I got a site that had the item for about $25. It didnt say what kind of dollars.

Even Oz dollars is a lot for 115Volt led pilot lights made in China.

What do you think the price of a Rectifier, resistor and LED costs at a local electronics supplier costs? Or the LED and handful of capacitors resistors and diodes.

Jon

We are talking 400 systems in 4 weeks. The system sales price is about $600 US. My client has a P.O. for 4200 systems. The cost of the few components are < $1.50 US ea. ($6300.00) vs. $11.00-qty. price $46,200 US.

How much is your time worth? $40k?

If I were producing these, I would go for the hand full of components and pocket the 40 grand.

Mooseman,

That is SIGNIFICANT.

Jon

I don't quite grasp these numbers. 18AWG wire in 1000 ft lengths is about 25¢ a foot from Newark. (Yes, you can get a better price elsewhere. This is just a touchstone price.) So six feet of wiring will alone bust your $1.50 budget. That's not including any connectors, fuse or a box. Probably you meant <$1.50 for the two lights. But which dollar amount means what doesn't really matter for me. I suspect your major problem maybe the 4 week fabrication time for 400 systems without an established engineered design for these two lights. Hopefully, whatever method you choose will have parts readily available for your fabricator.

But go for it my friend! Blaze on into your design.

I just hope you haven't committed to an unrealistic schedule.

Redfred,

Silver plated wire? Nawww, copper is way up.

I used to work in a factory short order shop and once everything was documented it got done fast.

Especially nice if it isn't wanted all in one lump.

Jon

YOU are going to have to take off from the 115 VAC, put it thru a transformer to drop the voltage, and then convert it to DC then y ou can use it to power the LEDs, I know of no LEDS which operate directly on 115V AC (or any AC for that matter ). Still check the catalogs, you may find something.

livagain1,

Mooseman has already rejected that idea.

Rectified and current limited ac works fine but produces a lot of heat in the resistor.

The heated resistor was also rejected.

Read the entries with good answer scores at least.

Jon

Livagain1, you missed some of the previous discussion and it is a lot of techno-babel to absorb.

Yes indeed, the transformer approach is rejected but the Cap and diode (or bridge) is a viable approach. Its inexpensive and looks as though it will do the job. I just voiced concern over the heat dissipation. I think there will be enough mass and thermal conductivity of the enclosure to take care of that. I will build that circuit and check it out, but I think the over-all approach is reasonable and should work out ok.

I don't think the following has yet been properly evaluated:

If you are controlling the useful current with a capacitor (or a network), and limiting impulses with resistors, the resistors have to dissipate throughout the cycle, even if the LED only uses current during half the cycle. So it would make sense to place the LED inside a bridge rectifier.

The breakdown Voltage required of the bridge can be really low, and it will be impulse-limited the same as the LED, so you can use really cheap diodes. You can use this with any of the capacitor arrangements (including the potential divider) - but don't forget to halve all the capacitor values. Impulse energies will therefore be halved. Better yet, as resistor values will not need to be changed, RMS resistor dissipation will be reduced a factor of four compared with an impulse-protected capacitor-driven arrangement.

(There are of course additional losses due to always driving three diodes [= the bridge and the LED] in series, but you should still be winning at currents above about 5-mA).

Use a diode with a resistor the diode will protect the led from the 115 v ac and the resistor will drop the v ac to a level that will allow the led to work.

Portugalphilip,

If you read through you will see that is what I did for exactly that purpose. The on-time effective pulsed current for the LED became a small portion of the sine wave and the power and Ohmic rating for the current limiting resistor was significantly smaller than what was expected by those who did not use calculations for pulsed optical devices.

The relatively higher pulsed current and the repetition rate makes the LED appear as bright and steady as if it was operating on dc at its optimum rating.

The requested design had to limit the heating effect so a more sophisticated design was introduced.

Kuduk

Dear Kuduk ,

I have used this circuit I described lots of times for non critical purposes.

I was unaware of the criticality of the led I just thought it was for indication of presence of 115 V ac.

Sorry for not spotting your posts on the subject.

If the led is producing noise maybe it could be run off a separate rectifier smoother and regulation but there would be lots of things to go wrong then and the cost would be higher.

Regards.

Philip.

The Op did not include the details about thermal requirements in the enclosure until later in the thread. It would have been simple if the electronics in the enclosure were compensated for temperature. Low budget did not allow for more sophistication.

Kuduk