Here, we will look at getting our LEDs to work without a transformer and build a simple light that is integrated into an expansion bar.

WARNING: For countries with 110v mains, we'd be working with voltages of 150 volts! For Europe and other countries, we are talking about 300 volts or more! At these levels, electricity is lethal! Do not continue unless you are comfortable with working with high voltages and are aware of the precautions to take!

AC supplies are quoted in rms (root-mean-square) values. The PEAK voltage is sqrt(2) * Vrms, which is about 1.4 * Vrms

**Signing Up**

## Step 1: Some background

150 / 20mA = 7500-ohms (we should subtract the voltage of the LED from 150v first, but the difference is minor)

7500-ohms? Not too bad... But then let's consider the power rating of this resistor, using the Power rule: P = (V

^{2) / R, we get:}

150 * 150 / 7500 = 3 watts, and that's a pretty hefty resistor. Brits with 240v mains will need a 17000-ohm resistor rated for almost 7-watts. And these will be running HOT!

Fortunately, by substituting a capacitor for the resistor, we can get the same reduction on voltage without the (or as much) heat. Capacitors delay the phase angle of AC which we can use to oppose itself, much like receding waves on the shore cancelling out some of the force of incoming ones.

hello sir, how to connect only one 1watt or 5 watt led on direct 220 volt ac. plz help me

Special Thanks for clearing so many queries with just a few instuctables. Very well explained. When I went hunting for LEDs in the market I came across 5mm and 10mm LEDs. Do they use the same current as the regular ones? Cause I cant really figure out how to do the calculations without the values

The LED current should be well advertised by the vendor, maybe a query or two would clear things up.

At the risk of oversimplifying, when it comes to discrete (2-lead) LEDs, 20mA would be the norm. Certainly go with that for the 5mm and even 8mm ones. Some 10mm feature a larger 'dome' to focus the light into a smaller area so they can advertise a higher lux output at the same current.

A perhaps semi-useful guideline will be to look at the die size - for white LEDs, that's the yellowish blob under the 'dome'. Larger dies could mean better current capacity. Also the thickness of the leads may likewise give some clues.

Again, thank you for your mail and I am glad you find my 'ibles helpful.

Hey thanks very much for the overview. I'm wondering about the size of the package.. I want to add an LED to power tools that don't have them. I was hoping for a cheap efficient way that also fit in a small size. The 1k ohm resistor isn't a problem, and I was considering using a diode to half the voltage. An SMD capacitor of .47uf/200v would keep the package small; potentially fitting in the tool body. Could a thyristor be used instead of the diode to further drop the voltage? Any input on these tweaks greatly appreciated!

Not too sure about the availability of a non-polarised 0.47uF/200v cap in smd format. They are necessarily quite large to prevent arc-over.

There's no need to use tricks with diodes and such -- which only work with incan lighting, not LEDs, btw -- just DECREASE the value of the capacitor to make the LEDs dimmer. Brighter? Add more LEDs in series without changing the C.

thanking you in advance.

minimumnumber of LEDs for this circuit is 2, although you can substitute a 1N4001 diode for one of them. Then use R=1K and C=0.2uF (or 0.22uF) as stated in Step 2.LED strips are organized in modules, each has 3 led in series with a resistor, then all the modules are paralel. in terms os voltage they the led strip is 12vDC

can we build a simple Capacitor circuito to feed the all strip?

Paulo

As noted, the design here is not very efficient and is suitable for low power, multiple-LED applications. The "LED strip" will further cut into this efficiency as they lose another 20% in the resistor. Better to invest in a 12-volt wall-wart for $5 on eBay and use that.

However, it is completely feasible to cut 15 pieces off the strip and (re-) connect them in series. Then they will run off (a rectified) 120vac mains.

thanks!

marC:)

I tested with 0.22uF 630V non-polar Caps, one 1K ½W resistor, two white led wired in parallel, it work, so my query is, if I would like to add another 2 x 7 LED in parallel, do I still need another set of C & R ? Is there any limitation number of parallel row added to the circuit ? Thx in advance : )

I just built & replaced one of my table lamp with 2 rows of 7 White LED & it works : )

My question is, if I wish to add another 2 rows of 7 White LED, do I still need another RC set or I just parallel connect to the existing circuit ? Is there any limitation ?

Btw, if I want to filter out the flicker effect, add one Bridge Rectifier to it correct ? (400V rated @ 1.5A) ?

"Decide the end you wish to work on - whether you want the light to shine UP or DOWN.

Then plug it in where you plan to use it and mark the top and bottom."Some electricians (at least in the U.S.) always install receptacles with the ground up, others with the ground down, and many like to mix it up. It would be a shame to go through whole project and then find out your UP light was shining DOWN!

Once again, nice job, and definitely on my to-do list.

For higher power lights, this is not a good circuit, take a look at some of my earlier answers and check out both parts 1 and 4 of this series

A well-matched transformer system can be 80% efficient and off-line (Switched) circuits even higher; using a capacitor (or any other method) to limit current flow is much less efficient, although it slowly increases with the number of LEDs.

At the point where the total Vf of the LEDs equal the equivalent DC voltage, about 52 for 117vac, we can direct connect the LEDs and the efficiency then becomes 100%.

The image below is a light just announced by Phillips which uses a cluster of 96 LEDs to operate directly off rectified 230v mains.

Please reread the values for 220vac carefully! The values you supplied will seriously overdrive the LEDs and permanently damage them. The capacitor value is calculated as: ( 0.02 / ( 2 * pi * f * ((V * 1.4)-Vf) ), so it's not a linear relationship and certainly not directly proportional to the number of LEDs you want to use!

For 220v supplies, you can have up to 15 LEDs in each branch

without changing the basic values, so you can easily have your 20 by putting 10 in series, in 2 opposing chains. See my response tolurkingdevilfor an explanation of this.at 150-volts(for 110v mains), so an extra LED represents less than 2.5% of the total, which is not easily detected. And that is also why I placed a limit of 8 LEDs total in a single branch - by then the drop will be 26-volts, or 17%, which will become noticeable.erm..

can tell me why we need to add the 1k resistor ?

thanks

it, the capacitor, protects the rest of the components.I once tried this same circuit without the resistor. The capacitor was fine, but a string of LEDs got burnt out.

And it turns out, the capacitor is identical to what you show.

Oh well, 3 pairs are pretty good. Its a good first try for me.

Turbonut48

What do you think may have caused this? Cheap LEDs?

Turbonut48

Another reason may be the use of a polarized or out of spec capacitor. Make sure it is a 0.47mF (or .47uF or 470nF, which are all equivalent) non-polarized unit.

Many thanks.

Xc= 1 / (2 * pi * f * C)

Therefore the Capacitance for any desired Reactance, by solving for C,is

C = 1 / (2 * pi * f * Xc) in Farads

For f=60Hz and Xc=6500-ohms, I get 408E-9 or 0.408uF on my calculator, while f=50Hz and Xc=16000-ohms, yield a C of 198.8E-9 or about 0.2uF

Hope this helps!

if ur c is .4uF for 110v, then ur reactance would be Xc = 6.63k ohms right??? so

using the formula for the impedance above Z=1k + j6.63k

= 6.7k< -81.42 ohms right???

so if u calculate the total current I = (110 v/.707)/(6.7k<-81.42 ohms)

will give us

=(155.58<0 volts)/(6.7k<-81.42 ohms)

I = 23.22 < 81.42 mA...am i right????

pls correct me if im wrong hhehehehe i just want to confirm i really like to play with these things to but im just confused heehehe and what would be the voltage drop across the resistor????hahhhahahhahapls reply tnx....

^{2}+ Xc^{2}) since Xc is kinda imaginary number