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.
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Step 1: STEP-1

Picture of 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.

Step 2: STEP-2

Picture of STEP-2
In this step you solder the LED's as shown in the circuit diagram.

EXTRA care should be taken to maintain the POLARITY of the LED's.

For this reason see the placement of the LED's as shown in the diagram

with the long and short legs of the LED's.

Long leg = + (plus) . Short Leg = -.(minus)

Step 3: STEP-3

Picture of STEP-3
In step-3 you can see from where the base of the light is taken

and how the Capacitor and Resistor are fixed to it.

The base of he light is made of the lower portion of a fused CFL tube light.

The two leads goes to the bottom of the base.

Step 4: STEP-4

Picture of STEP-4
In Step-4 you can see the reverse side of the Board, and how the LED's are soldered.

Each legs of the 4-LED's are soldered in the middle and so on......................
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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!

Dipankar (author)  samanthakate1 year ago

What you should be interested in is the current through your
LED's, and that can be calculated this way:

For each cycle (+ or -), you have 15 of them. So that means
together they 'need' 15 * 3.3v or about 50v.

Now the 220 vac puts out 220 * 1.4 or 308v

So, the "leftover" volts the cap need to handle is
308-50, or 258volts peak.

Now the REAL maths: the resistance of the .22uF cap:

1 / (2 * pi * freq * C) which works out to 14.5K-ohms for
50c/s. (that's why we need AC)

Now current, i = V/R, so for we have 258/14500 = 18mA, which
is also what is allowed to go through the LED's in your circuit. Now remember
that this is only for half the time, so it's REALLY safe for the LED.

But fortunately, we have another set of LED's lighting in the
OTHER half of the time, plus our eyes have a thing called persistence of vision
which remembers the brightest light for about 1/20 of a second, and so your
light 'appears' to be almost as bright as a LED that is constantly lit at 17mA,
which is about 75% of maximum brightness.

foobear1 year ago
Why is it 220v? Is that a european thing?
Dipankar (author)  foobear1 year ago
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
(removed by author or community request)
It is a common 5mm LED which works on 3.5 volts.
trekman5 years ago
Where do you get the Sun-mica Lamination Board? What store sells it? Thanks, Great Instructable!
Dipankar (author)  trekman5 years ago
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
Dipankar (author)  trekman5 years ago
(I think you might be talking about a plastic laminate that is glued on top. Dose that sound right?)
You are RIGHT
Would .22uf @ 250 Volt work? NO NO NO NO NOT SAFE
For 220 volts 400
For 110 volt 250.
fohio Dipankar3 years ago

It is hard to find nonpolarized capacitor .22uF @ 400 for 220, can I use .22uF @630 volts for 220 ?

Dipankar (author)  fohio3 years ago
Yes you can use a .22uf @ 360volts
fohio Dipankar3 years ago
excuse me, sir. You might read my post incorrectly ...I asked 630 volts , can I use it? Many thanks in advanced
Dipankar (author)  fohio3 years ago
Try it, it should work.
If the voltage is more no problem but it should be .22uf.
fohio Dipankar3 years ago
Thanks for your quick replies...I would try and post the result soon, again, thanks
fohio Dipankar3 years ago
Thanks for your quick replies...I would try it and post the result soon, again thanks.
fohio fohio3 years ago
please...answer ....anyone in advance
fizaro4 years ago
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?
Dipankar (author)  fizaro3 years ago
I am unable to follow your question. Please write in detail.
Dipankar (author)  fizaro3 years ago
I am unable to follow your question. Please write in detail.
sir your circuit works fine. but not able to download as pdf.
digikawa5 years ago
Tell me if and where I'm wrong !

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 ?
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.
Dipankar (author)  santolim4 years ago
You do not need to calculate anything.
Just use a non-polar Capacitor .33uF / 250 or 300 volts for 110 volt AC.
And .22uf / 400 volts Capacitor for 220 volt AC.
Dipankar (author)  digikawa5 years ago

This is from my friend QS who has explained it very well in his Instructable

V is shorthand for the Voltage for your country,

C is the value of the Capacitor,

R is the resistor value and n is the number of LEDs per side.

the * is used here for the multiply sign and / is to divide.

So 22200 * V means 22200 multiplied by 220


With a .47uF capacitor, you will be running the LED’s quite hot - around 80-degrees, so make sure they are in open air to cool them a bit. LED’s will slowly lose their brightness (in months) if they are run too hot. A .33uF should be chosen if you want the LED’s to last a few years. Remember to get at least 400V, non-polarised ones.


In your voltage and a .33uF (or .47uF), you can have 31 + 31 before the LED’s start looking 'dim'. You can have up to 100+100 before they stop working.


The formulas that you should use are:


C = 1 / (22200 * V) = 0.2uF   for 220vAC (non-polarized, rated for 350V)

R = 1.5K, 1/2W

n = max number of LED’s in each direction

  = .14 * V = 31 LED’s before it starts to be noticeably dim.


If you make C too large, you will be letting too much current through the LED’s and that will cause it to heat up too much, which will drastically shorten the life of the LED’s.

marlopax4 years ago
Hello Dipankar, I am in a similar type issue here: 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 MarlöPax
Dipankar (author)  marlopax4 years ago
Send your E-Mail ID for me to send the CIRCUIT.
fizaro Dipankar4 years ago
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?
Dipankar (author)  fizaro4 years ago
Send your E-Mail ID so that I can send the CIRCUIT.
fizaro Dipankar4 years ago
THIS IS MY EMAIL : thx be4 Plz mention the name of elements, coz I am novice and plz see the photographs which I posted on the above url. Thank You MarlöPax
fizaro4 years ago
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?
digikawa5 years ago
"This design uses 10 x 20mA LEDs in series, giving us a total of 200mA draw through each of the three circuits." FALSE !

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) ? "write a private message if you want explanations", 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.

PLEASE DON'T BUILD THIS unless you FULLY UNDERSTAND what you are doing (in which case you wouldn't build this...).
Dipankar (author)  digikawa5 years ago
This page is taken from Bowden's Hobby Circuits.

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. A larger capacitor will increase the current and a smaller one will reduce it. The capacitor must be a non-polarized type with a voltage rating of 200 volts or more.

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.

Dipankar (author)  digikawa5 years ago
You still don't get it Do you?
I have been using these LED;s as night lamp for the last 18 months and on problem.
I will explain later in details.
GSKhurana5 years ago
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.

Is there any way of recycling the ballast of CFL into this of such similar project.

rajibc465 years ago
How I could'nt understand how current will flow through the circuit to glow the LEDs if the LEDs are not connected in series.
Pl. send me an email at
Rajib Chatterje

TLLCUK5 years ago
This design uses 10 x 20mA LEDs in series, giving us a total of 200mA draw through each of the three circuits.

Using Ohms Law: R = V (volts) / I (current)
In this case R = 220V / 200mA = 1100 Ohms or 1.1K, close to the 1K chosen.

Substitute in the need to run 150mA LEDs, (the 8mm size has no factor).

R = 220 / ( 10 x 150mA )
R = 220 / 1.5A
R = 146.66666

Nearest preffered value is 150 ohm.
TLLCUK5 years ago
A bead of silicone mastic could be used to seal between the bottle and the lamp holder to prevent/reduce corrosion, if used outside.
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