{THE LED LIGHTBULB}

Step 2: [CALCULATIONS]

Now, like any project with electricity, certain calculations are necessary. Although the equations might look a bit threatening, I can assure you that it's quite simple. Before starting your calculations, make sure you have your LEDs' detailed specifications from your supplier.

FV (Forward Voltage) - This is the voltage used by each separate LED. It is expressed as a range, so a minimum and maximum value is present.
AC MAX/MIN - AC Mains are not always at a constant voltage and are not always the same across a whole house. There is actually a range present. In the US, the range is 110-125VAC. In other nations, the range is 220-250VAC.

EQUATIONS
[AC MAX] X 1.4 = A
A / [FV MAX] = [# LEDs]

CHECK
[AC MIN] X 1.4 = B
B / [# LEDs] = C
C represents the forward voltage and must be within the range.

Your final result represents the number of LEDs you can put in each series. Think of this as a basic unit. The total amount of LEDs on the light bulb must be a multiple of this number. In each "unit," the LEDs connect positive to negative in order to distribute the voltage. All the series may then be connected together, positive to positive and negative to negative. Below is a sample of my calculations.

EQUATIONS
125 X 1.4 = 175
175 / 3.8 = 46

CHECK
110 X 1.4 = 154
154 / 46 = 3.3478
C is in range. (exhale)

Comments

[LGProspects]

So I did the build I am in the US so based everything off here. 5MM leds, 3.0-3.2 FV. Did the math and got 55 LED's needed. During the check it came back as 2.8. So I figured it was UNDER voltage at 55 so went for it.


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

[meissler]

Could you just use any 12V power chord (like a charger for a computer or something) and hook the +/- ends up to the right areas? Of course use resistors too. Would that work or no, not really?

[downgrade]

 Yes, it would. You could also try to find a wallwart to some broken or lost electronic device and use that. A DC power supply would bypass the the rectifier as well. It would be more likely to last longer (read: fewer parts = fewer issues, generally but not always). Hopefully whatever power supply you find has some power regulation going on so it would be less likely to spike and such.

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.

[orthodoxmonster]

Questions: in this series crt, 110-125Vdc/1.5A is initially applied to 1st Led (rated at 3.2-3.8Vdc/30mA, doesn't this cause damage to the Led? Thanks.

[chrwei]

where does the 1.4 value come from in AC MAX X 1.4 = A ?

[metal_flowboard]

AC voltage cycles from zero to a peak voltage and back down to zero again. The voltage given is an RMS value, a kind of average, it is not the peak voltage value Multiplying by 1.414 gives a good approximation of what the peak voltage is. By working this out you find the maximum voltage that your led's will be subjected too rather than the average. That way your leds don't blow up on that first AC peak hope that helps, if not I'll try again

[chrwei]

so AC MAX is really a MAX average? I know AC voltage fluctuates, but 40% seems excessive

[metal_flowboard]

Well its a bit more involved than just a simple average,
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}

[chrwei]

I think I see now, well after looking up Root Mean Squared. the short answer is: 1.4 = sqrt(2)

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

[wirecutter]

GOOD Ideas. But the thing about LED's is that as the current through the diodes heats them up the voltage across them drops, then the current rises, the diodes gets warmer, and warmer, the voltage across drops till PHUT! its called thermal runaway As an aside if you connect each LED to an identical LED but the other way round you dont need to have a bridge rectifier to give a dc supply just ac.

[kscience]

NEVER run white, blue, green or uv "cmos" LED's on AC. Very little reverse voltage will blow them up short circuit....

[chrwei]

sure, but then you have a nice flicker effect that'll drive you mad

[Hardwyre]

This is an excellent instructable; and thank you for answering my other question, just another one for ya. I noticed you made three sizes of lights; looking again at your calculations I THINK I'm understanding how you did it. For the small one you did one series of 46 LEDs, the mid was what, 2 series? And the large was 3? As long as you're using series of the number of LEDs you calculated, there shouldn't be much worry about blowing them, correct? I'm curious how the issue of maximum milliamps is handled.
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.

[mattsouthgate]

Hardwyre,
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

[Al1]

That's a good idea about using LEDs for the bridge rectifier. The only slight downside (I could see) is that those four LEDs would only illuminate for half of the AC cycle as compared to the main bank of LEDs so they may appear to flicker or appear dimmer.

[charlieb000]

your rectifier can consist of leds too, in this case you have two arrays so you would need two led rectifiers