Using AC with LEDs (Part 3) - The BIG light

Step 3: Build time!

So, we can start our build of a simple all-LED + a bridge circuit to run off 110v mains.

You will need:

Lots of white LEDs - naturally! And TEST them all!

AC line cord

Perfboard

1N4003 diode or 200volt bridge rectifier

The first picture is what my circuit looks like when finished. Quick eyes will note that there are only 42 LEDs on board. Because of the need to accomodate the bridge on the board, and because of the relatively stable nature of our mains, we can run our lights a tad over 20mA.

The Bridge has 4 leads: 2 marked (~), a (+) positive and a (-) negative. The (~) ones go to AC Mains.

Start by connecting the Bridge (+) to the longer (+) lead of the first LED, then take the short lead to the long lead of the next LED. Do 1 row, double and triple check before soldering! Work your way down, ALWAYS connecting shorter to longer.

I have additional pictures below showing the various stages of completion. Print them out to help you do the wiring.

Comments

[elementarywatts]

If I use a bridge rectifier the dc rms output voltage is lower than the input AC voltage. Would I still use the rule to increase the DC V rms measurement by 1.4 for the purpose of determining the amount of leds per string? If I install a capacitor 100uF 200v on the DC side the voltage goes up. Do I need a capacitor? Would that increase overhead at the line source? Using the same method as you I am making a spotlight with some 400 leds + or - with serial / parallel. If I do this correctly would I expect the life of the leds to be long as they are rated 100,000 hrs.? Thanks 3.5v 20ma 5mm I would not like a lot of overhead only to drive these leds.

[qs]

Are you saying that by adding a 100uF capacitor the measured voltage goes up? That is probably because you are using a DC voltmeter to measure the rectified AC. Rectified AC is still sine-wave, so a DC voltmeter will not give you an accurate reading.

The addition of the capacitor forces the LEDs to work continuously and, in circuits involving large number of LEDs, heat becomes a problem. In cases where the LEDs are of good quality and operated within the rated current, the biggest factor affecting the life of the LED is heat. Ideally, they should never run over 80oC, but it they are placed close together without ventilation, they could reach over 100oC.

For spotlight use, you may find the 25mA 10mm LEDs, with its tight +/- 6-degree beam-spread more usable.

[opliko]

So if I were to make this for Higher powered leds, say 1 watt, would it work? Considering they use only slightly higher voltage, in what way could I figure out the current? I know it wouldn't work out to say 40 instead of 47 LEDs. Just for example lets say it is a 3.7V 350mA LED.

[qs]

Higher power LEDs can work in this circut as well, as long as the maximum currents are calculated and observed. Of course the sheer number of LEDs required, at $3-$8 apiece will make the project a bit more expensive. Then we have to worry about the heat generated by such an arrangement - the 'usual' star-type LEDs may have a 'hot' base, and so care must be taken not to short things out - at 150v it would be quite dangerous!

[acmefixer]

"..a bit more expensive." - I like that understatement. Forty-seven 5mm LEDs for $20 something, versus up to $200 for 1 watters..

[opliko]

Yeah, I realized it would be quite a few, haha. My other question would be, if the rectifier says "200V 1.5A" that means I need to "use up" all 1.5 amps or I would be over driving them? Last question, since there is only 154(6) volts to work with, would it really matter what the amp rating is?

[qs]

Voltage and current ratings are given as "do-not-exceed" values. In other words, running your rectifier at 10-volts and 0.1-amp is perfectly fine; but NEVER go over the limits. Amp x Volt equals power, so, in the rectifier, more amps mean more power-loss, and subsequently, more heat.

[opliko]

Thank You, you have been very helpful with lots of insight into AC led applications. So let me digest this really quick and spew out a circuit and figures to see if I am correct. If I used the bridge and the switch as in step 5, with 20 leds using 3.3V and 25ma each, would draw 500mA and 66V. Now with the switch only allowing one side of the AC current through, Would that stress the bridge as much as having the switch in the on position with both cycles going through? Kind of a stupid question but I like to be clear about things when working with AC especially high levels of voltage and current.
As for the high powered leds, say you have 4 that draw 350mA, that uses 1400mA total, with only 3.6V x 4 = 14.4V. Leaving 142V x 100ma = 14.2V wasted or dissipated as heat in the rectifier? Thank you again for all your patience.

[qs]

OK, I see we're treading in dangerous grounds here. It really helps to visualize an electric circuit like water going through a pipe.

Voltage is the pressure of the water, while the Amperage is the amount of water going through the pipe. The two are not interchangeable.

The current throughout the pipe (and circuit) STAYS THE SAME, whether it's gals/minute or electrons, this analogy holds. Meaning the current flowing through the first LED is identical to the current through the last LED.

What we are doing is reducing the voltage (pressure) by putting LEDs in between. If we know that the LEDs can 'use up' 3.3-volts each, then we need 154 / 3.3 = 47 of them to make the circuit work. Regardless of whether they are 25mA or 350mA ones, the magic number of LEDs for this circuit, is 47.

And, because everything is in series (one after another), the 'total' current through them stays at 25mA or 350mA - this is not added.

So, unless there are other 'restricting' elements in the circuit, putting fewer than 47 LEDs will force each LED to work beyond what they are designed to do. And this will likely burn them up, maybe even literally.

[opliko]

KK that is why I asked all these questions :). Wasn't sure on AC and bridges and such. Looks like I'll probably be using the DX.com constant current thing, or the previous step with the capacitors/resistors. Thanks

[japanjot]

in d magic formula 230 * 1.4 / 3.3 , , , _{, , , 230 is my ac volt, 3.3 is led volt but wot is 1.4 here????? wud b glad if u clear dis to me ...thnx again}

[qs]

The value reported for AC power is known as the RMS value, which is lower than the peak value. To find the peak value, you have to multiply by sqrt(2), or about 1.4

[japanjot]

ohhkkkzzz now i understand,wel u guided me so well, thank u frnd............this means on 230v rms the peak value ud be of 322v................nd one more thing if i connect 49 led's to the 230v ac,and i include a rectifie of max 400v, wt else wud be required more??? a capacitor??? if yes then plz tell of wt value.

[qs]

I added step (6) above with the calculation - hope it helps!

[japanjot]

plz help me, m new to this but want to learn alot from you people...... i ant to connect 44 leds in series and want them to work on 230 volts please tell me how can i without burning these leds nd myself 2 :)

[qs]

For 230v, the 'magic' number of white LEDs is 230 * 1.4 / 3.3 or 98 LEDs.

This is the number of white LEDs (the same type) that you can connect together in series and operate on 230V-AC without needing resistors or transformers. Just add a suitable bridge rectifier in front.

[japanjot]

well thanks alot bro for ur help, but if u don't mind can u also tell me abot bridge rectifier nd its construction nd values.... thank you

[qs]

A bridge (full wave) rectifier is simply a package of 4 diodes connected so that regardless of which direction (polarity) an AC voltage is, the + and - always go to separate outputs. Bridge rectifiers have 2 input connections marked (~) and 2 outputs. You have to make sure the specified voltage is higher than the peak voltage and the amperage is greater than what the LEDs draw. You can also make your own bridge by connecting 4 1N4007-type diodes (400V 1A) like in the diagram below.

[wkumtrider]

If I want to run 6 leds off AC, I would need a capacitor to reduce the voltage, correct? What is the best way to calculate the correct type of capacitor? Thanks for your help.

[qs]

Take a look at the instructions posted here for the correct values.

[wkumtrider]

Thanks qs, but I don't quite understand the formual. It has two variables C and X. I'm not sure how to solve the formula with two variables. I must be missing some data.

[qs]

X is the resistance (reactance) of capacitor, C.

If we know we need a resistance of 7500-ohms at 50Hz, then the correct capacitor value is:

C= 1/(2 x pi x F x X) = 1 / (2 x 3.1415... x 50 x 7500) = 0.424 uF, with 0.47uF as the nearest standard value.

[wkumtrider]

Ok, thanks so much. Makes sense now.

[ferociousllama]

Can I use the capacitor and resistor from part 3 with the bridge rectifier to make the leds brighter while still running 2-14 leds? I would put the cap and resistor with the leads that connect to ac on the rectifier. Would this work?

[qs]

If I understand your question, you are asking if you can add the rectifier to the circuit in part 3 and make the LEDs brighter. The answer is no - the rectifier will force the current to go through one set of LEDs, which will be brighter, but since the other set never turns on, the TOTAL light output is actually lower. Remember that the LEDs are diodes in themselves, and by connecting them back-to-back like in section 3, they act like a rectifier bridge for themselves.