110V powered LEDs

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I saw a LED assembly some time ago that consisted of two red LEDs in
parallel, a 3/8" diameter "bump" inline with the red & black wire with
a black & white wire out of that. The bump was about 1" long and
covered in heat shrink. The point was to run the two LEDs off 110VAC. I
don't know who made it, or even what country the assembly come from.
The point to all this, is to fire two red LEDs at an absolute minimum
cost from 110VAC. Wall-warts cost a lot.

Any ideas?

Thanks in advance!

 

[Ancient_Hacker]

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.

 

[Rich Grise]

I saw a LED assembly some time ago that consisted of two red LEDs in
parallel, a 3/8" diameter "bump" inline with the red & black wire with
a black & white wire out of that. The bump was about 1" long and
covered in heat shrink. The point was to run the two LEDs off 110VAC. I
don't know who made it, or even what country the assembly come from.
The point to all this, is to fire two red LEDs at an absolute minimum
cost from 110VAC. Wall-warts cost a lot.

Any ideas?

Thanks in advance!

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

 

[petrus bitbyter]

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.

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

 

[Jonathan Kirwan]

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.

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

 

[John Fields]

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.

---
"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?

 

[Jonathan Kirwan]

Mmm... Capacitors tend to be big and expensive.

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.)

Besides, the inrush current may be high enough to blow the LEDs.
<snip>

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

 

[Clark]

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.

 

[Spehro Pefhany]

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

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

 

[Jonathan Kirwan]

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.

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

 

[Zak]

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.

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