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

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.

## Step 2: Pushing the envelope

Remember that, because we're using the diode to block half our cycle, the LEDs in circuit (2) only works 1/2 the time. How can we make them light up for the other half as well?

With a simple part called a Bridge Rectifier this can happen. This device is actually 4 diodes connected in a criss-cross way to make both cycles go in the same direction. Electronic fans will know this as part of the 'Full-wave rectification' circuit (as opposed to Half-wave).

With this addition, our LEDs will be turning on twice as often and we WILL get twice as much light from them.

## 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.

## Step 4: Behold!

...

And there is light!

Because of the hazardous nature of the components when plugged in, I covered the circuit board with a triple layer of parchment paper, which has a good dielectric value, and can withstand over 400F of heat.

Then I mounted the board on the lid of a take out container, using a foam spacer from a DVD spindle, with a cutout for the power cord.

The light output is equivalent to a 40-watt frosted bulb, but the container is barely warm.

Remember: Always unplug the circuit before you touch any exposed parts.

Also, the LEDs will be running close to their rated current, which could mean temperatures as high as 85C on their surfaces.

## Step 5: Variations

Too bright?

You can combine circuits (2) and (3) to give our light a Hi/Lo switch. In Hi, the switch shorts the diode so that it operates in Full-wave mode as in (3). Opening the switch only allows current to flow half the time, just like (2).

Ozzies and Brits: You too can use the 42/47 LED circuits - just combine the US version (.4uF and 1K-ohm) circuit presented in part 2 and you too can make a AC-mains light with just 42 LEDs! Or check out the calculations in the following step.

Oh yeah, our 'big' light is super thrifty - running off 110-volt mains, it barely consumes 3-watts.

Find out about more ways to light your house with LEDs off A/C mains here!

## Step 6: Crunching Numbers

Here is a recap of the calculations used for this project:

To operate white LEDs (nominal voltage 3.3v) safely off AC Mains without using any regulation (other than the diode bridge), the magic number is: Vac * 1.4 / 3.3. Which is the minimum number of LEDs in series that will run off AC without exceeding its 'comfortable' operating range. The choice of LEDs can be 20mA or higher - AS LONG AS they are all the same type and attached in series.

If you are using the full number of LEDs calculated above, that is all you need, but for arrangements using fewer LEDs (but no fewer than 30), we need to add the voltage dropping RC combination. R is always a 1K, 1Watt resistor, while the value of C is calculate as:

Vpk= Vac * 1.4
Vdd= N * 3.3, where N is the number of white LEDs we wish to use in series.
Iled = 0.02, the current we want for our LEDs.
C = 1 / (2 * pi * f * (Vpk-Vdd) / Iled), where f is the mains frequency, but you can simplify it to: (58 / (Vpk-Vdd)) in micro-farads (uF), and should range between .1 and .5 uF. Make sure it is a non-polar capacitor.

IMPORTANT: Parts must be rated for at least Vpk, and enough current to handle Iled.
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ChristopherS175 days ago

I have been thinking

is their a cheap safe way to Current limit AC then to rectify it to DC thus keeping it current limited

I am wanting to drive 700~900ma LED's

and planing on running them in series with an backup diode between each link so if 1 LED fails the diode takes its place until replaced

I would rather spend the money or more LED's then pay for expensive drivers :P

especially now that you can get 1000 3w LED's for \$100

the largest driver I can find is

\$240 LED Power Supplies 200W 90-305VAC 143-285V CC DIMMING

http://au.mouser.com/ProductDetail/Condor-SL-Power/LE200S70CD/?qs=sGAEpiMZZMt5PRBMPTWcaes6bEy7OfPPyZIV8IY2H4Y%3d

anything else I can find is low voltage or high voltage high current

qs (author)  ChristopherS174 days ago

The short answer is no, there are no cheap, safe (and accurate) ways to limit current in a variable supply situation. More so with AC circuits, since now you have to worry about current in both directions as well as the sine-wave power considerations.

Take a look at this: <http://www.dx.com/p/3-0a-100w-power-constant-curre...> Admittedly it's for 100w, but, at Au\$30, you can afford two. Quite a few Australian readers have good comments for this site, and they have literally hundreds of LED drivers to choose from.

You can put LEDs in parallel to multiply current, and then in series to multiply the voltage until it matches the output of the driver.

diywannaber2 months ago

If I were to use capacitors to drop voltage and create usable LED bulbs for replacing all the lights in my house will it in any way be harmful to the electric wiring in my house. I'm really curious because when it is possible to run LEDs with a simple driver circuit that you have described why do all the brand name LED bulb manufacturers include ICs, inductors and capacitors in their driver circuits. Thanks.

qs (author)  diywannaber2 months ago
As mentioned in the 'ible and elsewhere in the comments, the efficiency of using resistors and/or capacitors to drop the voltage to run LEDs is very low, typically under 25%.

The addition of control-ICs and other components can bring this up to almost 90%.
26 days ago

Using only capacitors as the voltage dropping element should waste a very small amount of power, shouldn't it. Why do you say the efficiency with this method is close to just 25%. Voltage drop using capacitors is almost 100% efficient, isn't it.

Thanks

qs (author)  diywannaber25 days ago

Whether you use resistors or capacitors to 'drop' the voltage, the waste is the same. While resistors convert the 'extra' into heat, capacitors use subsequent cycles to cancel the power. Less heat, but no more efficient.

In simple terms, if we are using 117v a/c at 20mA to drive 15v LEDs at 20mA, then the efficiency is 15 / (117 * 1.4) or 11%. The efficiency starts to climb up as we increase the number of LED, up to (117 * 1.4) / 3.3, such as this project, which then allow us to use 100% of the power in running the LEDs.

steelcity742 months ago

Here is a neat little project to run LEDs from AC Power...

http://www.tinkerplayground.com/projects/ac-line-led-pilot-lamp-16

qs (author)  steelcity742 months ago

It's important to emphasize that the linked circuit is meant for 120vac (US, Canada, Japan) use ONLY.

In the build, the choice of "any" capacitor, even if in the proper range, is a serious concern. Unless it's a "non-polarized" version, and the schematic seems to indicate it's not (the curved bar implies negative), then you will have -170v applied to it every half cycle. The cap will pop or explode over time.

A better approach is to put the 1N4004 diode in series with the LED (and facing the same way), at least this blocks off much of that potentially destructive voltage.

zeek31776 months ago

can someone tell me where, or better yet how and what i would need to run my 110v ac house electricity down to roughly 3v dc led lights? i know im going to need a circuit board with resistors, thats about all i know, ive never really made anything like this from scratch before, i only know how to wire simple things and solder.

hagen21 year ago
I built this. I used 42 extra bright white leds with a 3.6 forward voltage. I built a 1N4003 diode bridge for rectification. Plugged it in. Worked great. At first. Then it started to smell. Then some of the lights started to flicker. 10 minutes later half of them were no longer lit. I unplugged it and plugged it back in and now it won't come on at all.

Isn't everyone going to have this same experience? Won't this just overheat the leds which are in such close proximity to each other. I wonder if building this in a bar instead of a square would help with heat dissipation?

Anyhow I'm not happy with the results.
qs (author)  hagen21 year ago
Built correctly, the circuit consumes only 4 watts, so power dissipation should not be a major issue. A thermal scanner in my testing indicated 75C operating temperature, well within the tolerance of the electrical components.

You mentioned that half the LEDs failed first, to be followed by the other half. This would instead suggest a wiring problem -- the LEDs have to be all connected IN SERIES, with the current flowing in the same direction as the rectified power.

Other things to watch out for would be a supply mains voltage substantially over 120v, as well as mixing LEDs of different colors.
WVvan3 years ago
Thanks for your LED postings. All are very informative.
qs (author)  WVvan3 years ago
Thank you.
wiraksana4 years ago
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.
maanhe3r4 years ago
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)  maanhe3r4 years ago
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.
clchee4 years ago
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)  clchee4 years ago
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!
Dipankar4 years ago
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)  Dipankar4 years ago

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.
4 years ago
I think the Phillips SMT (Surface mount) LEDs will be costing hell of a lot of money?
qs (author)  Dipankar4 years ago
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.
4 years ago
\$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-5 years ago
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.
4 years ago
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.
5 years ago
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.
budiyanto5 years ago
pak kalau diindonesia tegangan umumnya 220 volt bagaimana sama saja tidak skemanya atau beda terima kasih
elementarywatts5 years ago
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)  elementarywatts5 years ago
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.
elementarywatts5 years ago
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)  elementarywatts5 years ago
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_bisnar5 years ago
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)  rob_bisnar5 years ago
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_bisnar6 years ago
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)  rob_bisnar6 years ago
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".
6 years ago
thank you very much for your response. Does it mean that the reading of 105Vdc is the RMS?
qs (author)  rob_bisnar6 years ago
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.
japanjot6 years ago
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)  japanjot6 years ago
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
6 years ago
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.
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