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

6 Steps
In Using AC with LEDs, part 1 and part 2, we looked at ways to adapt AC power to LEDs without the usual conversion to pure DC first.

Here, in part 3, we combine what we learned before to design a LED light that operated directly off AC mains.

Warning: AC mains is hundreds of volts, and is potentially lethal. Please take all necessary precautions before you start working with it!
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Step 1: The no-transformer transformer.

When we connected LEDs to AC transformers, the calculation we used was:

Vac / 3.3

to give us the number of LEDs we need to be able to properly handle the power without additional resistors and other parts.

What if we bypass the transformer completely and consider AC mains? In some ways it is simpler - the voltage from transformers could vary greatly with the load we put on it, whereas AC mains are much more stable.

If we use the 110v standard of the US, we first calculate the peak voltage, 1.4 * 110 = 156 and we can find the number of LEDs it can support:

156 / 3.3 = 47 LEDs

So, does that mean that if we put 47 LEDs in series, we can run the whole string directly off a 110v AC socket?

The answer is Yes! As long as we maintain the voltage across each LED at 3.5v or less, it will operate within its limits.

But then, let's not forget that for each positive cycle, there is a negative cycle! That means we need a mirror circuit like in (1).

Wow, that's an awful lot of bulbs!

However, if we add a blocking diode like in circuit (2), then we can safely operate our circuit. The 1N4003 is capable of handling 200 volts so is fine for US power.

For EU countries, the magic number is 103 LEDs (double if you want to use both cycles) and the diode for ckt (2) should be a 1N4004 or better.
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WVvan says: Sep 11, 2011. 8:12 PM
Thanks for your LED postings. All are very informative.
qs (author) in reply to WVvanSep 11, 2011. 8:33 PM
Thank you.
wiraksana says: Apr 24, 2011. 6:56 AM
I know it's a little bit dated to ask. I plan to build one of your instructable. i have 2 questions.
If i used combination of white led and amber led (to soften the light), should i use your formula too? let say if i use 80% of the voltage for white led (3.3volts) and use the rest for amber led (2 volts).
can i use capacitor after bridge rectifier to minimize the flickering effect?
thank you.
maanhe3r says: Mar 4, 2011. 5:57 PM
Hello qs.
I commend and thank you on a very interesting and educational Instructable. I really like the way you try to teach the principles instead of just showing how to slap something together. Your style has inspired me to want to try building the circuit in sketch 3 / step 2.
My question: How many of the circuits as in sketch 3 step 2 can I connect safely to one 120V outlet? What is it I have to be careful of ( current, voltage, power, ?).
I'm looking to replace a tube fluorescent lamp in the kitchen and am thinking I will probable need 4 or 6 such circuits (?) to get an equivalent lighting effect.
qs (author) in reply to maanhe3rMar 4, 2011. 7:09 PM
A crude rule-of-thumb is to multiply the wattage of the LEDs by 5 to get the equivalent output of a fluorescent. However, if the light is in a single direction, the multiplier can be as high as 10, since the LED is naturally directional.

Here, if you are using 117v, then the 50 or so LEDs will dissipate about 4-watts, making them equivalent to a 20W tube; or a 40W one if aimed down from the ceiling. Running 240v in other countries will see about 10watts.
clchee says: Feb 28, 2011. 2:40 AM
For the circuit with AC Cap, does it matter if Live and Neutral was connected the other way round ? Just in case deliberately or accidentally someone swapped the L and N positions at the outlet.
qs (author) in reply to clcheeMar 4, 2011. 6:58 PM
It doesn't matter which side is used, AS LONG AS YOU ADHERE TO STRICT HIGH-VOLTAGE PRECAUTIONS: Do not touch or adjust ANY part of the circuit if it is plugged in!
Dipankar says: Dec 9, 2010. 4:17 PM
Let me know if I am wrong.........

1.4 x 220v = 308
308 divided by 3.5 = 88
So 88 LED's on both sides without using a Bridge Rectifier.

positive cycle and negative cycle,
That means we have a mirror circuit with 88 + 88 LED's
Am I right?
qs (author) in reply to DipankarDec 9, 2010. 8:06 PM

Coincidentally, Phillips has announce THEIR version of this 'big' light for 230v where they place 96 SMT (Surface mount) LEDs in series with a bridge rectifier. This allows the LEDs to run cooler and perhaps extend their life.
Dipankar in reply to qsDec 10, 2010. 2:57 PM
I think the Phillips SMT (Surface mount) LEDs will be costing hell of a lot of money?
qs (author) in reply to DipankarDec 11, 2010. 12:15 AM
And you'd be right! No firm prices yet but bhey are claiming life in 10's if not 100-thousand hours and that a panel will save over \$150 of electricity over their life.
Dipankar in reply to qsDec 11, 2010. 2:26 PM
\$150 of electricity over their life time is peanuts?
I would rather stick to the present cheap ones, cause if they give me 5 years my money is worth it.
Over 2 years has passed and my LED Chandelier is being used daily is still going strong without any LED's packing up. Isn't that something?
-shtoink- says: Mar 28, 2010. 9:47 AM
I am hoping that you might be able to verify that I am going about this the correct way. I'm not actually trying to drive LEDs right from the AC in the wall, but rather from an AC inverter meant to drive a CCFL for a small LCD picture frame. It's a 7 inch wide screen LCD. I would like to still be able to make use of the features for controlling the backlight-off timer built in to the device, but may not be able to if this isn't a good option. Before I go ahead with this, does it sound possible or should I look at using DC to drive the LEDs from another location on the board.

Here's what I measured: V AC is about 540V and the current is about 2.5 A.

I wasn't able to measure the frequency, but looked of the values of other DC-AD inverters and they ranged between 30 to 50 kHz.

Running through the calculations, I get X=38184 ohms and C=1.4E-10 F when using the 30 kHz frequency.
ironsmiter in reply to -shtoink-Oct 21, 2010. 3:27 AM
I know this reply is a bit dated, but in case anyone goes rummaging through the archives....
DO NOT attempt to run led's from the inverter output.
In actuality, you may have measured 540VAC, but due to the frequency put out, it may actually be producing over 1000volt, and your meter just not able to cope with the switching speed.

If you MUST use that control board, I'd suggest tapping into the circuit somewhere safe, and before the inverter.
To retain the most factory like operation, I'd personally remove the inverter, anbd tap into the former input locations to power my led circuit. but that's just me.
-shtoink- in reply to -shtoink-Mar 28, 2010. 10:30 AM
I should note that I did these calculations with the measured VAC*sqrt(2). Thinking about that now, it was probably not correct.

R = 540 / 0.02 = 27000

C = 1/(2*PI*30000*27000) = 1.96E-10

Those are most likely the values I need to use if this is something that might work.
budiyanto says: Jan 13, 2010. 11:15 PM
pak kalau diindonesia tegangan umumnya 220 volt bagaimana sama saja tidak skemanya atau beda terima kasih
elementarywatts says: Sep 17, 2009. 8:39 AM
If the leds are 100ma instead of 20ma would the value of R change? The purpose of R is to protect against spikes in the current? How id the value of R calculated? Thanks
qs (author) in reply to elementarywattsSep 17, 2009. 10:03 AM
For anything other than 20mA LEDs, it is not advisable to use this method to reduce the voltage: use a transformer (See part 1) or a direct-connect switchmode system as in Part 4 of this series.
elementarywatts says: Aug 27, 2009. 7:07 PM
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 (author) in reply to elementarywattsAug 27, 2009. 8:39 PM
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.
rob_bisnar says: Aug 9, 2009. 3:24 AM
Hello qs! its me again. How will you measure the power consumption of this LED series? and How much does each LED consumes in this set-up? thanks in advance!
qs (author) in reply to rob_bisnarAug 9, 2009. 1:30 PM
Since we know that the MAXIMUM current we allow for is 20mA, then the power consumption is under 2.4-watts. Dividing this by the number of LEDs will give us 57mW per LED for 42 LEDs, or 51mW per for 47.
rob_bisnar says: Jul 28, 2009. 8:07 AM
Hi qs! I am very thankful for this, it really helps me a lot. I have a question though, you said, the max voltage is about 156V, is this equivalent to the DC voltage level?

I have arranged a bridge rectifier consists of four 1n4003, I measured its output voltage and the meter reads 105Vdc with measured input of 115Vac. Why is it the output is not 115*√2= 163V ?
qs (author) in reply to rob_bisnarJul 28, 2009. 4:10 PM
The problem with calling what comes out of a full-wave bridge "DC" is the implication that it is the same current as what you might get from batteries.

The real waveform from your bridge, as this image Wikipedia shows is still very much recognisably "AC", except all the sine-waves appear in the same direction (polarity).

That is why you must add a capacitor to + and - to smooth out the "ripples" before your multimeter can recognize it as true "DC".
rob_bisnar in reply to qsJul 28, 2009. 11:19 PM
thank you very much for your response. Does it mean that the reading of 105Vdc is the RMS?
qs (author) in reply to rob_bisnarJul 29, 2009. 3:10 AM
The meter is expecting a steady input so the reading will depend on the time frame your meter checks to see if the input has changed - the 'Samples/sec' number in the spec sheet.
opliko says: May 10, 2009. 5:11 PM
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 (author) in reply to oplikoMay 13, 2009. 4:35 PM
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 in reply to qsJul 12, 2009. 6:54 PM
"..a bit more expensive." - I like that understatement. Forty-seven 5mm LEDs for \$20 something, versus up to \$200 for 1 watters..
opliko in reply to qsMay 13, 2009. 6:43 PM
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 (author) in reply to oplikoMay 14, 2009. 7:50 AM
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 in reply to qsMay 14, 2009. 10:24 AM
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 (author) in reply to oplikoMay 14, 2009. 1:00 PM
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 in reply to qsMay 14, 2009. 3:39 PM
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 says: Jun 8, 2009. 2:25 PM
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 (author) in reply to japanjotJun 8, 2009. 4:41 PM
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 in reply to qsJun 8, 2009. 10:34 PM
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 (author) in reply to japanjotJun 9, 2009. 1:31 PM
I added step (6) above with the calculation - hope it helps!
japanjot says: Jun 6, 2009. 1:09 PM
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 (author) in reply to japanjotJun 7, 2009. 11:46 AM
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.
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