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LEDs - I used 5mm LEDs. You can change the type of LEDs as long as you augment the calculations accordingly.
Bridge Rectifier - Converts AC to DC.
Perfboard - The size of perfboard you buy will depend on the size of light bulb you wish to create.
Soldering Iron and Accessories - The cheapest of soldering irons will do.
Base Plug - This product has a normal bulb's base at one end and a normal household outlet on the other. Your local hardware store will most definitely have some.
Cable Ties - The question here is not if you have them, but, rather, how many hundred of them do you have.
Cardboard - This is the main support piece for all the component of the bulb.
Wax Paper - I used a silicon cookie sheet (that rolled up blue thing) instead. You can use practically anything that has a high temperature tolerance and doesn't conduct electricity, but I have found these to be the best materials for the job.
Drill and Small Bit - I used a 1/8" drill bit.
X-ACTO Knife and Whole Puncher - These items will be used to prep the insulation.
20AWG Wire and PVC Pipe - These are required to connect the bulb's base to the rest of the assembly.
Multimeter - I always keep my multimeter handy in order to check conductivity and make sure there are no shorts.













































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I did everything following your diagram and it was a nice glitter show for a microsecond. It appears about 10 LED's are dead.
What did I do wrong?
Lance
Long story short: yes, and for the most part I would recommend it, especially if you go with 12v you could easily convert it to an off grid solar project as well, since smaller projects like that stick with 12v.
Everything I found so far is for low voltage, don't know if it would be as simple as slapping in the Mosfet in series with the LEDs.
I used 5 mm Ultra Bright White LEDS. VF is in the range of 3.4 Volts to 3.6 Volts. The Hydro here, is in the range of 115 +/- 5 Volts AC 60 Hz. It's the Country, things go up and down. The Math given did not account for numbers on the other side of the decimal point. So 46.05263 was rounded off to 46 LEDS for a Voltage of 125 Volts AC or 175 Volts DC..
Using the Math given 120 Volts X 1.4 = 168 Volts DC.
168 Volts DC/3.6 VF Max = 46.6666 # LEDS Do we round off to 47 or leave it at 46?
Check
110 X 1.4 = 154 Volts DC
154 Volts DC/47 # LEDS = 3.27 Volts this is lower than 3.4 Volts VF Min.
The AC Voltage range it will work in is then:
47 # LEDS x 3.4 VF Min = 159.79 Volts DC/1.4 = 114.14 Volts AC Min
47 # LEDS x 3.6 VF MAX = 169.2 Volts DC/1.4 = 120.857 Volts AC Max
this is a 6.717 Volts AC Variation.
I used 46 LEDS
46 # LEDS x 3.4 VF Min = 156.4 Volts DC/1.4 = 111.7 Volts AC Min
46 # LEDS x 3.6 VF Max = 165.6 Volts DC/1.4 = 118.28 Volts AC Max
I made 9 all the same type with 46 LEDS in series, powered by a Bridge Rectifier. No Epoxy this time. 2 failed after 2 seconds of power on. The rest worked Ok so far.
I'll monitor the line voltage, for maximum and minimum spikes for a while, see what the AC voltage variation actually is. Might have to fix this with a resistor, or more LEDS or something??
Teleran
I'm not trying to be negative, and I hope with some explanation I won't be "kept in the dark" for any longer...
*cough* bad pun *cough*
I made those fireballs but I used metho and you couldn't hold it but you could throw it at people if you didn't grab it for more than about half a second.
One thing I can say for you, may your parents never, ever, ever, ever, ever find out what happened.
FYI - prefab LED-PCboards are available (at extra cost of ~$4)
see http://c-leds.com/shop/product_info.php?cPath=26&products_id=78
... Me too. High Five!
The LED's are DIODES. Yes only half of them will be on at any point in time if the strings are wired anode to cathode, but your eyes will not see the blinking .
Since the diodes (LED's) are specified in DC terms i.e. forward voltage drop of 1.8 to 3.8 VDC (IR to UV) it's easier to think in V=IR terms. However the "source" is AC (60 Hz), try to think in terms of reactance and frequency
Try a simple circuit of 2 LED (wired cathode to anode) in series with a .47uF 200V (for 110VAC input) capacitor is series with a 1K resistor (to limit the current into the capacitor if it's switched on near a peak AC cycle). Works great as a power-on indicator.
Buy a "cheap" set of LED Christmas lights and take apart the "blob" in the power line to see what's in there....also a cheap (60 of them for $1) source of LEDs, but with unplated (steel) wires.
How do you determine which bridge rectifier to use?
If I understand it correctly, if you have two strands of 46 LEDs in parallel, the positive lead of the first LED in each set connects to one of the rectifier leads, and the negative leads, and the negative leads of the last LED in each set connects to one of the rectifier leads?
thanks =]
Steve
RMS stands for Root Mean Squared
The square Root of the Mean value of the integral of the wave Squared
If you were to draw the sine wave on a graph,
you could then find the mean area between the line and the x axis,
finding this area will give you a set of limits which you calculate between,
you use the limits to find the area of the integral squared,
then square root the whole thing.
if you work through the proof for the general case you can skip the above calculations as it comes to a general formula.
RMS voltage = peak voltage x 1 / root 2
where 1/root 2 =0.707 (rounded)
and to go back
peak voltage (AC MAX) = RMS voltage x ( root 2 )
where root 2 = 1.414 (rounded)
So AC MAX is the actual peak voltage (maximum voltage) at the crest of the waveform.
The peak voltage is only there for the smallest fraction a second, but in the case of the instructable above would be long enough to pop your led's of you didn't take it into account.
The RMS value comes into play when calculating things like power usage, where you can drop it into equations like ohms law as if it were DC
Hopefully that makes it a bit clearer :) much further I'll have to get my maths book out don't think I'll be able to do it from memory anymore
the longer answer, the AC MAX of 120 is the RMS max, not the actual AC peak, which is more like 170V but only for a tiny fraction of 1/60th of a second. (since AC goes from 170 to 0 to -170 to 0 to 170 in 1/60th of a second). the formula being Vpeak = Vrms * sqrt(2) and the square root of 2 is about 1.4.
if anyone wants to know where that comes from, http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power
- In Japan, typical outlet voltage is 100V (sorry I don't have max/min).
- You might want to describe the RMS to peak conversion in the first equation. It took me a little while to figure out what you were doing.
Also, you might put a fuse in the schematic, just in case there's some construction problems. I could see a lamp going up in flames.... :-oOne LED in the string being able to conduct in reverse would not normally be a bad thing (despite losing its light), but its forward voltage and series resistance may drop when it fails short, so then your only method of limiting current is suddenly letting a lot more current through, which only increases the chances that the rest of the LEDs experience excessive heating and fail sooner.
That being said, I don't know how likely these failure events would be. Using the fuse at the connections to the mains would absolutely be the way to go, and I would recommend that for any project, not just for testing.
Newer AC powered LED lights are quite new and the little 1.5W=40W (?) bulbs from Lights of America (available at Costco and some Walmarts) are very reasonable (3 for $15.00).
I replaced four 9W CFLs in outdoor fixtures (those date back into the 70s) with the LED lights. I used dead CFLs for their base and connected a miniature bulb socket to them. They provide nearly as much light directly underneath (about 8 feet high) as the CFLs at considerable savings.
The ballasts in the fixtures provide an extra kick (that was a surprise; i thought they would reduce the voltage) and the bulbs are getting 140Volts. I have two other LED bulbs in non-ballasted applications. I will be interested in seeing if the balasted LEDs have shorter life spans.
I also put one of the 4.5W = 75W (?) in an outdoor spotlight and that thing is real bright.
The great thing about these is the directionality. Very little light out the side and all concentrated beneath.
Another forum describing these bulbs in more detail explains you can dis-assemble them to get some bulbs for other projects.
Regards
I was trying to use the LED calculator at http://led.linear1.org/led.wiz to give me an idea, using 120volts as the input voltage, and trying different configurations, and it keeps telling me I need giant resistors at multiple wattages.
I'm just a little wary of spending lots of time to make a light and then have it over-driven and die after a handful of hours. I greatly appreciate the answers though.
I've taken a look at the Linear calculator and it seems to work okay (except that won't calculate for UK mains voltages - they're too high apparantly and so I'll just scale the figures anyway). Try:
Source voltage 150
diode forward voltage 2
diode forward current (mA) 30
number of LEDs in your array 72
You should get:
- 220 ohm resistor dissipates 198 mW
- the wizard thinks 1/2W resistors are needed for your application
- together, all resistors dissipate 198 mW
- together, the diodes dissipate 4320 mW
- total power dissipated by the array is 4518 mW
- the array draws current of 30 mA from the source.
Is this similar to your calculations?The trick is to build a LED array that has a comparable forward voltage to the supply so the voltage differential is small so only small resistor is needed.
I hope this helps.
Matt
I was going to work out the power (to get an idea of how bright the bulbs are, I know how bright 5W of LEDs are etc.) but I don't see any current calculations in your numbercrunching. This makes me slightly twitchy, are you not at risk of overdriving your LEDs and not getting the bazillions hours of life you should? I may be confusing the theory of how LEDs work and how they work in the real world, however, so if this isn't really a concern then I'll ignore it and get on with life :)
Just for the sake of it, if we assume you are using the 13000mcd LEDs,
80mW per LED * 46 LEDs per string = 3.68W, equivalent to perhaps a 5W CFL or a 30-40W incandescent. However, living in the UK I'd be using more like 90 per string, which gives 90*0.08W = 7.2W, equivalent to maybe a 9W CFL or a 60W incandescent. Plenty to light up a small-medium room with white walls :)