110V powered LEDs

If the LED's were wired in opposite directions, the bump could be a capacitor. A resistor would be much cheaper, but have to dissipate about a watt of heat.

The dead simplest way to do this is to simply use a dropping resistor, and put the LEDs in antiparallel. Size the resistor for 110V * .02A, and I'll leave the arithmetic as an exercise for the student. :-)

Good Luck! Rich

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Mmm... Capacitors tend to be big and expensive. Besides, the inrush current may be high enough to blow the LEDs. To stay on the safe side you will need a capacitor and a resistor. Assuming the LEDs will produce light enough at

1mA (and a lot of them will do) you only need a 100k or so resistor which dissipates only about 100mW. You can even go downto 68k or 56k for more light (and more heat) but still use a common 250mW resistor.

petrus bitbyter

I hope you meant "size" in the sense of wattage and not in the sense of resistance.

Also, might try a .47uF capacitor. I see a metallized polyprop film from EPCOS at Digikey for USD 0.67/ea. I believe they are designed for AC lamp applications. Part number is 495-1315-ND. (The rating on Digikey's site shows 400V, but that is for DC.)

Jon

--
"Absolute Minimum Cost", IMO, would include not only the cost of
ownership of the device, but also its operating costs.  Such being
the case, a device with a reactive ballast might cost more to buy
but less to run, in the long term, than a device with a resistive
ballast. 

Consider a system with two white 3.5V 20mA LEDs connected in
antiparallel, in series with resistive, capacitive, or inductive
ballasts across 120V mains.

Which would you recommend, and why?

By comparison with 1/4 watt resistors, very much so. But I did find a

67 cent cap at Digikey. Not horrible and almost the same price as a 2 watt resistor, bought at Digikey in 1s. (All this assumes you have enough order quantity on other parts, of course, and discounts any shipping.)

Is this true? The maximum slope at 60Hz is d/dt (160*sin(wt)), which is 2*PI*60*160. I = dV/dt * C, which means the actual worst case is

2*PI*60*160*.47e-6 or about 28mA peak current, roughly.

I don't see how this can be a problem for most common LEDs.

Jon

I agree with Jonathan, the last time I saw Leds powered from 120vAC was using a 100 ohm on one lead and a .47uF cap 250 Vac cap on the other. Hardly no heat was generated. This only problem with this is a spike on the line will pass right through to the leds.

How do you *guarantee* that it doesn't EVER, even once, get switched on right at the peak of the AC line? Might take more than one or two components to do that.

Best regards, Spehro Pefhany

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That would be a problem. We have the effective dV/dI resistance slope of the diode, itself. But that would not be enough. I could easily see a few amps for periods of a few microseconds. A smaller resistor value of about 1k would nip that to something reasonable, but it would still need to be about 1/2 watt in size, I think.

Thanks for pointing that out.

Jon

I think you can with an extra cap and two diodes:

o-----||---- +->|--+----o | | line +->|-+ ===== LED | ----- | | o-------+----------+----o

The output cap needs to be fairly large, but it is low voltage (voltage limited by the LED).

Thomas