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.
<p>I have been thinking </p><p>is their a cheap safe way to Current limit AC then to rectify it to DC thus keeping it current limited </p><p>I am wanting to drive 700~900ma LED's </p><p>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 </p><p>I would rather spend the money or more LED's then pay for expensive drivers :P </p><p>especially now that you can get 1000 3w LED's for $100 </p><p>the largest driver I can find is </p><p>$240 LED Power Supplies 200W 90-305VAC 143-285V CC DIMMING</p><p><a href="http://au.mouser.com/ProductDetail/Condor-SL-Power/LE200S70CD/?qs=sGAEpiMZZMt5PRBMPTWcaes6bEy7OfPPyZIV8IY2H4Y%3d" rel="nofollow">http://au.mouser.com/ProductDetail/Condor-SL-Power/LE200S70CD/?qs=sGAEpiMZZMt5PRBMPTWcaes6bEy7OfPPyZIV8IY2H4Y%3d</a></p><p>anything else I can find is low voltage or high voltage high current <br></p>
<p>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.</p><p>Take a look at this: &lt;<a href="http://www.dx.com/p/3-0a-100w-power-constant-current-source-led-driver-85-265v-47306" rel="nofollow">http://www.dx.com/p/3-0a-100w-power-constant-curre...</a>&gt; 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.</p><p>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.</p>
<p>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.</p>
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%.<br><br>The addition of control-ICs and other components can bring this up to almost 90%.
<p>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. </p><p>Thanks</p>
<p>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.</p><p>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.</p>
<p>Here is a neat little project to run LEDs from AC Power...</p><p>http://www.tinkerplayground.com/projects/ac-line-led-pilot-lamp-16</p>
<p>It's important to emphasize that the linked circuit is meant for 120vac (US, Canada, Japan) use ONLY.<br><br>In the build, the choice of &quot;any&quot; capacitor, even if in the proper range, is a serious concern. Unless it's a &quot;non-polarized&quot; 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.<br><br>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.</p>
<p>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.</p>
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. <br> <br>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? <br> <br>Anyhow I'm not happy with the results.
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. <br> <br>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. <br> <br>Other things to watch out for would be a supply mains voltage substantially over 120v, as well as mixing LEDs of different colors.
Thanks for your LED postings. All are very informative.
Thank you.
I know it's a little bit dated to ask. I plan to build one of your instructable. i have 2 questions. <br>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).<br>can i use capacitor after bridge rectifier to minimize the flickering effect? <br>thank you.
Hello qs.<br> 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. <br>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, ?).<br>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.
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. <br> <br>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.
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.
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!
Let me know if I am wrong.........<br><br>1.4 x 220v = 308<br>308 divided by 3.5 = 88<br>So 88 LED's on both sides without using a Bridge Rectifier.<br><br>positive cycle and negative cycle,<br>That means we have a mirror circuit with 88 + 88 LED's <br>Am I right?<br>
Yes, your calculations are right. <br> <br>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.
I think the Phillips SMT (Surface mount) LEDs will be costing hell of a lot of money?
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.
$150 of electricity over their life time is peanuts?<br>I would rather stick to the present cheap ones, cause if they give me 5 years my money is worth it.<br>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 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.<br /> <br /> Here's what I measured: V AC is about 540V and the current is about 2.5 A.<br /> <br /> 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.<br /> <br /> Running through the calculations, I get X=38184 ohms and C=1.4E-10 F when using the 30 kHz frequency.<br />
I know this reply is a bit dated, but in case anyone goes rummaging through the archives....<br> The answer is NO.<br> DO NOT attempt to run led's from the inverter output.<br> 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.<br> <br> If you MUST use that control board, I'd suggest tapping into the circuit somewhere safe, and before the inverter.<br> 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.<br>
I should note that I did these calculations with the measured VAC*sqrt(2). Thinking about that now, it was probably not correct.<br /> <br /> R = 540 / 0.02 = 27000<br /> <br /> C = 1/(2*PI*30000*27000) = 1.96E-10<br /> <br /> Those are most likely the values I need to use if this is something that might work.<br />
pak kalau diindonesia tegangan umumnya 220 volt bagaimana sama saja tidak skemanya atau beda terima kasih<br />
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
For anything other than 20mA LEDs, it is not advisable to use this method to reduce the voltage: use a transformer (See <a rel="nofollow" href="http://www.instructables.com/id/Using_AC_with_LEDs_Part_1/">part 1</a>) or a direct-connect switchmode system as in <a rel="nofollow" href="http://www.instructables.com/id/Using_AC_with_LEDs_Part_4_The_New_Technologies/">Part 4</a> of this series.<br/>
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.
Are you saying that by adding a 100uF capacitor the <em>measured</em> 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.<br/><br/>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.<br/><br/>For spotlight use, you may find the 25mA 10mm LEDs, with its tight +/- 6-degree beam-spread more usable.<br/>
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!
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.
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?<br/><br/>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*&#8730;2= 163V ?<br/>
The problem with calling what comes out of a full-wave bridge &quot;DC&quot; is the implication that it is the same current as what you might get from batteries.<br/><br/>The real waveform from your bridge, as this image <a rel="nofollow" href="http://en.wikipedia.org/wiki/Rectifier">Wikipedia</a> shows is still very much recognisably &quot;AC&quot;, except all the sine-waves appear in the same direction (polarity).<br/><br/>That is why you must add a capacitor to + and - to smooth out the &quot;ripples&quot; before your multimeter can recognize it as true &quot;DC&quot;.<br/>
thank you very much for your response. Does it mean that the reading of 105Vdc is the RMS?
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.
in d magic formula 230 * 1.4 / 3.3 , , , <sub> , , , 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</sub><br/>
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
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.
I added step (6) above with the calculation - hope it helps!
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 :)
For 230v, the 'magic' number of white LEDs is 230 * 1.4 / 3.3 or 98 LEDs.<br/><br/>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.<br/>
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
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.
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.
Take a look at the instructions posted <a rel="nofollow" href="http://www.instructables.com/id/Using_AC_with_LEDs_Part_2_and_make_this_handy_/">here</a> for the correct values.<br/>
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.
X is the resistance (reactance) of capacitor, C.<br/><br/>If we know we need a resistance of 7500-ohms at 50Hz, then the correct capacitor value is:<br/><br/>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.<br/>
Ok, thanks so much. Makes sense now.

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