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!

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


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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    102 Discussions


    3 years ago

    I found these last year

    and have used every one in the sample kit they will send you for free. One set for 10, 20, 30, and 50mA plus a couple adjustable ones as well... Super flexible, will run off pulsating dc (no capacitor - but the LEDs get noticeably brighter with one) built in surge protection and they run about $0.52 each from Digikey in small quantities


    P.S. Supertex has a line of similar devices that are slightly both, more expensive and more flexible

    1 reply

    Reply 1 year ago

    Those look like they run on 45v max so how do they help?


    2 years ago

    Hey qs!

    So if I wanted to run more LEDs from a 110V socket from the US exceeding the magic number. How would I do that? Would I just get a transformer and up the voltage so that I can add more LEDs in that strip? I'm just trying to make my own christmas lights for fun and make them super bright. I realize that I can only have around 47 LEDs from a 110Vac. You know how people add strips of christmas lights in series with each other? Like a strip of 100ct Led Strip connected to the wall(110Vac) and then people to add another one just like it to the end of the first LEDs strip to elongate the christmas light to put up all around the house. Is there a limit to how many I can run from a single socket? If so, if I wanted to run about 1000 LEDs from one household socket, what should I do?


    9 years ago on Step 1

    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 ?

    7 replies

    Reply 9 years ago on Step 1

    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".

    Patrick JamesOqs

    Reply 3 years ago

    How can we compute for the value of the smoothing capacitor?

    p.s.: Great instructables you have, qs!

    qsPatrick JamesO

    Reply 3 years ago


    The value of the smoothing capacitor is dependent on frequency, voltage, current drain and the amount of ripple allowed. Here is a simple formula to get an approximate value:

    C(farad) = i / (2 * (freq * V * (1 - Smoothing-factor))), where

    V = input Voltage (peak); i = current (amps), freq = frequency (Hz, or cycles/sec), and

    Smoothing factor = 0 for none and 1 for 100%; note that this means that 100% smoothing is impossible (division by zero).

    So if V is 110v a/c (60Hz), we will use the rectified peak value of 160 volts, and we're driving a chain of LEDs at 20mA, then for 90% smoothing, we'll need:

    C = 0.02 / (2 * 60 * 160 * (1-0.9)) = 10 uF (or higher, at 200v or better)
    (The 90% filtering is acceptable for LED lighting, but not for electronic applications, especially ones involving sound. To get 99% filtering, C becomes 100uF!)

    WARNING: This MUST be used in conjunction with a full-wave (bridge) rectifier circuit, or it's boom time!


    Reply 3 years ago

    Hey, can you explain me about calculation of filtering C in power supply or direct me any web link that has explained about calculating about it?
    I'm really confused in f, Vrms, Vp-p for DC!
    Although this may be because of my poor knowledge, Help me plzzz!!!


    Reply 3 years ago

    The a/c-mains (wall) supplied electricity in all countries has 2 basic values, Vrms and f. So they are easily discoverable by checking with your local power company or Google.


    Reply 9 years ago on Step 1

    thank you very much for your response. Does it mean that the reading of 105Vdc is the RMS?


    Reply 9 years ago on Step 1

    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.


    8 years ago on Introduction

    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?

    5 replies

    Reply 8 years ago on Introduction

    Yes, your calculations are right.

    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.

    Philips230v LED2.jpg

    Reply 3 years ago

    Something is wrong. Your pictured light is not 96 LEDs but rather 128 LEDs, so they cannot all be in series to run directly off 230VAC rectified.


    Reply 8 years ago on Introduction

    I think the Phillips SMT (Surface mount) LEDs will be costing hell of a lot of money?


    Reply 8 years ago on Introduction

    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.


    Reply 8 years ago on Introduction

    $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?

    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

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

    1 reply

    Reply 3 years ago on Introduction

    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: <> 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.


    4 years ago on Introduction

    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.