This is a Compact LED Light on the lines of a COMPACT FLUORESCENT LIGHT. (CFL). It runs on 220 Volts AC or can also be made to run on 110 volt AC, if the value of the CONDENSER is changed to .47uF / 250Volts, AND the RESISTOR is changed to 1K / 1Watt.

Step 1: STEP-1

Items you will need.

30 Extra bright White LED's of 5mm size.
3 Strips of Sun-mica Lamination Board.
One Condenser- value of 0.22uF / 400 Volts.
One Resistor- value of 1K - 1/2 Watt.

Making of the base BOARD.
Three strips of Sun-mica Lamination Board, is cut to size as to make a Triangle to fit the inside of the PVC base of the light. The base is taken from the bottom portion of a fused CFL light, as shown in STEP-3.
The length of the three strips of Sun-mica Lamination Board is to be determined by you, but 4.5 inches will do.
Drill small holes for the legs of the LED to penetrate through the board.
Each 3 sections of the board is laid together and taped with cello tape at the back to reinforce it and hold it together.
I've read your detail,but I still don't understand.<br>I've read if peak of AC voltage for 220V is:<br>220V * (2)^0.5 about 311V.<br>Voltage for 15 LED = 15 * 3.3V about 50V.<br>Commonly current for LED is 20mA,so: (311 - 50) / 0.02 about 13K ohm.<br>13K ohm = 1 / (2 * pi * freq * C); where pi = 3.14, freq = 50Hz, so: 1 / C = 13000 * 314, so value of C is about 0.24uF.<br>Or maybe function of the resistor 1K so voltage on the capacitor is lower than (311 - 50)V or 261V?<br>Thanks...
<p>Your compact led light is really cool!Thanks for the design but if you would please, give me a technical explanation how it works. I made this and am curious how it works! Thank you in advance sir!</p>
<p><strong>What you should be interested in is the current through your <br>LED's, and that can be calculated this way:</strong></p><p><strong>For each cycle (+ or -), you have 15 of them. So that means <br>together they 'need' 15 * 3.3v or about 50v.</strong></p><p><strong>Now the 220 vac puts out 220 * 1.4 or 308v <br></strong></p><p><strong>So, the &quot;leftover&quot; volts the cap need to handle is <br>308-50, or 258volts peak.</strong></p><p><strong>Now the REAL maths: the resistance of the .22uF cap:</strong></p><p><strong>1 / (2 * pi * freq * C) which works out to 14.5K-ohms for <br>50c/s. (that's why we need AC)</strong></p><p><strong>Now current, i = V/R, so for we have 258/14500 = 18mA, which <br>is also what is allowed to go through the LED's in your circuit. Now remember <br>that this is only for half the time, so it's REALLY safe for the LED.</strong></p><p><strong>But fortunately, we have another set of LED's lighting in the <br>OTHER half of the time, plus our eyes have a thing called persistence of vision <br>which remembers the brightest light for about 1/20 of a second, and so your <br>light 'appears' to be almost as bright as a LED that is constantly lit at 17mA, <br>which is about 75% of maximum brightness.</strong></p>
Why is it 220v? Is that a european thing?
Americans use 110 volts and the English use 220 volts that is why 220 volts is an European thing.
The light is a bit flashy type of,what can i do to fix it.can i use a bit bigger capacitor like 0.23uF? Or anything else
Where do you get the Sun-mica Lamination Board? What store sells it? Thanks, Great Instructable!
In INDIA you get it from any Ply-Board store. It is a common item in a Ply-Board store. It is used to be pasted on ply-board Table top.
Thanks for the quick reply. Can you give an American equivalent? We would call Ply-Board, Ply-Wood and I think you might be talking about a plastic laminate that is glued on top. Dose that sound right? I am having a hard time finding a .22uf Cap @ 400 Volt. Would .22uf @ 250 Volt work? Thanks, Tom
(I think you might be talking about a plastic laminate that is glued on top. Dose that sound right?)<br/>You are <strong>RIGHT</strong><br/>Would .22uf @ 250 Volt work? NO NO NO NO <strong>NOT SAFE</strong><br/>For 220 volts 400<br/>For 110 volt 250.<br/>
Hello,<br><br>It is hard to find nonpolarized capacitor .22uF @ 400 for 220, can I use .22uF @630 volts for 220 ?<br><br>Thanks.
Yes you can use a .22uf @ 360volts
excuse me, sir. You might read my post incorrectly ...I asked 630 volts , can I use it? Many thanks in advanced
Try it, it should work.<br>If the voltage is more no problem but it should be .22uf.
Thanks for your quick replies...I would try and post the result soon, again, thanks
Thanks for your quick replies...I would try it and post the result soon, again thanks.
please...answer ....anyone ...help...thanks in advance
ello dipankar I want to make a Combination in your thread, I want to know the details sequence. therefore, could you send the circuit and specification of components in use?
I am unable to follow your question. Please write in detail.
I am unable to follow your question. Please write in detail.
sir your circuit works fine. but not able to download as pdf.
Tell me if and where I'm wrong !<br /> <br /> AC mains is a sinusoidal signal, so at each half period the current will reverse its flow, and you have during this half period a voltage varying like this : 0-110-0. If you neglect the voltage drop through the LEDs (say around 15 V for 3V@20 mA LEDs for casual 5mm LEDs), you'll get a voltage of max 110 V between the legs of the resistor, that is a 110/1000=110 mA. During a half period, only three out of six branches conduct the current, so you'll get a 110/3 mA=36 mA through the LEDs, which is too much to keep a decent lifetime to your LEDs. BUT this works because the 110 Volts value is a peak value, the real mean value is closer from (110-15)/sqrt(2)=67 V, ie 67/3=23 mA flowing through the LEDs, and this is true only half of the time, as 3/6 branches don't conduct while the 3 others do, so you're not overloading your LEDs. Okay so this works, this is a quite pretty design, intelligent at first sight. BUT HIGHLY DANGEROUS as you have not planned any failure protection on a 220 V feeded circuit. In the worst case, your LEDs may have to handle 70 mA, which they won't do and finally overheat and burn out. In addition, your resistor is crossed by a 67 mA current with a 100 V bias that make a 6.7 W dissipation power ! How come this works ?<br />
110V or 220V is not peak value, they are RMS value (root mean square), the peak voltage should be higher than the rms voltage. I don't remember how to calculate the peak voltage for sine wave alternating current.
You do not need to calculate anything.<br>Just use a non-polar Capacitor .33uF / 250 or 300 volts for 110 volt AC.<br>And .22uf / 400 volts Capacitor for 220 volt AC.
<style type="text/css"><![CDATA[p.MsoNormal, li.MsoNormal, div.MsoNormal { margin: 0.0in; font-size: 12.0pt; font-family: Times New Roman; } div.Section1 { page: Section1; } ]]></style> <p class="MsoNormal">This is from my friend QS who has explained it very well in his Instructable <br /> <br /> V is shorthand for the Voltage for your country,</p> <p class="MsoNormal">C is the value of the Capacitor,</p> <p class="MsoNormal">R is the resistor value and n is the number of LEDs per side.</p> <p class="MsoNormal">the * is used here for the multiply sign and / is to divide.</p> <p class="MsoNormal">So 22200 * V means 22200 multiplied by 220</p> <p class="MsoNormal">&nbsp;</p> <p class="MsoNormal">With a .47uF capacitor, you will be running the LED&rsquo;s quite hot - around 80-degrees, so make sure they are in open air to cool them a bit. LED&rsquo;s will slowly lose their brightness (in months) if they are run too hot. A .33uF should be chosen if you want the LED&rsquo;s to last a few years. Remember to get at least 400V, non-polarised ones.</p> <p class="MsoNormal">&nbsp;</p> <p class="MsoNormal"><b><span style="color: blue;">In your voltage and a .33uF (or .47uF), you can have 31 + 31 before the LED&rsquo;s start looking 'dim'. You can have up to 100+100 before they stop working.</span></b></p> <p class="MsoNormal">&nbsp;</p> <p class="MsoNormal"><b><span style="color: red;">The formulas that you should use are:</span></b></p> <p class="MsoNormal">&nbsp;</p> <p class="MsoNormal">C = 1 / (22200 * V) = 0.2uF<span style="">&nbsp;&nbsp; </span>for 220vAC (non-polarized, rated for 350V)</p> <p class="MsoNormal">R = 1.5K, 1/2W</p> <p class="MsoNormal">n = max number of LED&rsquo;s in each direction</p> <p class="MsoNormal"><span style="">&nbsp; </span>= .14 * V = 31 LED&rsquo;s before it starts to be noticeably dim.</p> <p class="MsoNormal">&nbsp;</p> <p class="MsoNormal">If you make C too large, you will be letting too much current through the LED&rsquo;s and that will cause it to heat up too much, which will drastically shorten the life of the LED&rsquo;s.</p>
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Hello Dipankar, I am in a similar type issue here: http://www.instructables.com/community/LED-Lamp-Parallel-Circuit/ I need a parallel circuit or led connection in a bridge connection. can you explain what do I do to make this.. I had posted a full detailed info on my thread. Please go through it ............ Thank You Marl&ouml;Pax
Send your E-Mail ID for me to send the CIRCUIT.<br>
ello dipankar I want to make a Combination in your thread, I want to know the details sequence. therefore, could you send the circuit and specification of components in use?
Send your E-Mail ID so that I can send the CIRCUIT.
THIS IS MY EMAIL : fizaroozora@yahoo.com thx be4
marlopax@gmail.com Plz mention the name of elements, coz I am novice and plz see the photographs which I posted on the above url. Thank You Marl&ouml;Pax
ello dipankar I want to make a Combination in your thread, I want to know the details sequence. therefore, could you send the circuit and specification of components in use?
&quot;This design uses 10 x 20mA LEDs in series, giving us a total of 200mA draw through each of the three circuits.&quot; FALSE !<br /> <br /> Currents are not additive along a branch ! I don't understand why the author didn't point out this absurd comment. It is obvious that most of the enthusiasts on this project just don't have a clue of the way this circuit works. I'm asking the author : why don't you explain clearly the electric principle IF YOU KNOW IT (which I don't believe) ? &quot;write a private message if you want explanations&quot;, this is not a good way to do, particularly when a project involves mains supply in such an unsual way, your LEDs being practically unprotected. BTW, this is a highly dangerous circuit where each of your LEDs should better not fail, or this would turn into a fire in one second.<br /> <br /> PLEASE DON'T BUILD THIS unless you FULLY UNDERSTAND what you are doing (in which case you wouldn't build this...).<br />
This page is taken from Bowden's Hobby Circuits. <h2><font><font face="comic Sans MS" size="6"><br /> </font></font></h2> <style type="text/css"><![CDATA[p.MsoNormal, li.MsoNormal, div.MsoNormal { margin: 0.0in; font-size: 12.0pt; font-family: Times New Roman; } h3 { margin-right: 0.0in; margin-left: 0.0in; font-size: 13.5pt; font-family: Times New Roman; font-weight: bold; } p { margin-right: 0.0in; margin-left: 0.0in; font-size: 12.0pt; font-family: Times New Roman; } div.Section1 { page: Section1; } ]]></style> <p class="MsoNormal" style="text-align: justify;"><br /> <span style="font-size: 14.0pt;">The circuit below illustrates powering a LED (or two) from the 120 volt AC line using a capacitor to drop the voltage and a small resistor to limit the inrush current. Since the capacitor must pass current in both directions, a small diode is connected in parallel with the LED to provide a path for the negative half cycle and also to limit the reverse voltage across the LED. A second LED with the polarity reversed may be substituted for the diode, or a tri-color LED could be used which would appear orange with alternating current. The circuit is fairly efficient and draws only about a half watt from the line. The resistor value (1K / half watt) was chosen to limit the worst cases inrush current to about 150 mA which will drop to less than 30 mA in a millisecond as the capacitor charges. This appears to be a safe value, I have switched the circuit on and off many times without damage to the LED. The 0.47 uF capacitor has a reactance of 5600 ohms at 60 cycles so the LED current is about 20 mA half wave, or 10 mA average. <b><span style="color: red;">A larger capacitor will increase the current and a smaller one will reduce it. </span></b>The capacitor must be a non-polarized type with a voltage rating of 200 volts or more. </span></p> <p style="text-align: justify;"><span style="font-size: 14.0pt;">The lower circuit is an example of obtaining a low regulated voltage from the AC line. The zener diode serves as a regulator and also provides a path for the negative half cycle current when it conducts in the forward direction. In this example the output voltage is about 5 volts and will provide over 30 milliamps with about 300 millivolts of ripple. Use caution when operating any circuits connected directly to the AC line. </span></p> <p />
You still don't get it Do you?<br /> I have been using these LED;s as night lamp for the last 18 months and on problem.<br /> I will explain later in details.<br />
Excellent instructable. Though the circuit can confuse a beginner, but subsequent Questions and answers clarify those quite amply. Dipankar thanks a ton for all the effort.<br /> <br /> Is there any way of recycling the ballast of CFL into this of such similar project.<br /> <br />
Dipankar,<br /> How I could'nt understand how current will flow through the circuit to glow the LEDs if the LEDs are not connected in series. <br /> Pl. send me an email at rajibc46@gmail.com<br /> Rajib Chatterje<br /> <br />
This design uses 10 x 20mA LEDs in series, giving us a total of 200mA draw through each of the three circuits.<br /> <br /> Using Ohms Law: R = V (volts) / I (current)<br /> In this case R = 220V / 200mA = 1100 Ohms or 1.1K, close to the 1K chosen.<br /> <br /> Substitute in the need to run 150mA LEDs, (the 8mm size has no factor).<br /> <br /> R = 220 / ( 10 x 150mA )<br /> R = 220 / 1.5A<br /> R = 146.66666<br /> <br /> Nearest preffered value is 150 ohm.<br />
A bead of silicone mastic could be used to seal between the bottle and the lamp holder to prevent/reduce corrosion, if used outside.<br />
At this stage, a small stepdown transformer could be used and placed inside.<br /> It would make the circuit much safer as the output voltage will be at DC and will be more energy effecient and run cooler.<br /> Some adjustments to the Resistor values will be required and the capacitor will no longer be needed.<br />
Those run pretty damn bright, but would there be a better way to defuse the light maybe?
You can also lightly sand the the LEDs with fine wet and dry sand paper or anything with light abrasive properties.<br /> Some plastics can also be wiped over with IPA or Acetone, which will cause the surface of the to melt slightly causing it to diffuse.&nbsp; You should always try it on a sample piece first.<br />
YES, Just put a tracing paper between the LED's and the PVC cover, it will defuse the light. OR you can cut an milky translucent PVC bottle for the cover.
you could lightly spraypaint the inside of the bottle.
Hopefully not with the milk still in it though! But yes, the tracing paper, or maybe even colored tissue paper would work well. I want to put one of these together now, but I don't have strips of led's atm, probably have all the right resistors and capacitors on hand (actually I may need to go buy some higher voltage resistors).
As shown in the above drgs.LEDs are not connected in series.<br /> Will it if the are not conected in series ie&nbsp; - ve terminal of 1st LED will be connected with the +ve terminal of the 2nd LED.
very good I&nbsp;must trying, because I&nbsp;have lot of this broken PLC&nbsp;lamp.<br /> Arifjt (http://technosains.com<br />
How we can use 150ma ,8mm led in place of 20ma 5mm.what changes should we make in circuit? It may be a great idea using 30led x 25lumen=750lumen lamp.<br/>
Hi manish, Frankly I do not know, I have not used 150ma, 8mm LED's. Sorry.

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Bio: Now I am a retired person, who enjoys life and making small things to pass the time keep myself busy.
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