how to put 2/4/6 LED to 220V circuit without using the transformer ?

as i want to inset 4 LEDs inside a tube where there is no sufficient place to keep transformer.I want a circuit which can replace the use of transformer,that may be using resistor ,capacitor or so on.

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ajaysaini5 years ago
This is the circuit for 5 LED using 16v/1W Zener.

Change D5 Zener Diode and number of LED's as per requirement.

Follow the Rule of (3.5*No. of LED's=Zener Diode Voltage)
pandyaketan6 years ago
See my projects. They are very simple, but well researched. Hope they help...

mnphysicist8 years ago
4 diodes, a resistor, and a cap is all you need.... However, designing non-isolated, capacitive power supplies is not for the inexperienced, nor is it prudent to do so without testing to verify the design is safe. You have 2 concerns, fire hazards, and electrical shock... and the problem is, its easy to whip up a circuit, its tricky to make sure that its not going to explode, catch on fire, or cause a shock over the long haul. For test purposes, typically one exposes the design to a series of fast transients, such as might occur due to a motor, or other related gear, or worst case, a nearby lightning strike. The 2 key things to keep in mind, are electrical/fire rated insulation and components rated for use on the mains. A 250VAC capacitor unless specifically designed for such power supplies could either explode, or catch on fire. A typical homebrew enclosure unless flamability and shrapnel aspects are considered is likely to fail... and that is not cool except in the lab :)... even the enclosures we math model do not always pass.... and we have 1" thick clear polycarbonate blast shields specifically for transient testing, just because of that. So going forward, what is the ID of the tube you wish to use? There may be solutions out there, but we really need to know how much space you have to work with. For safety purposes, what is the wall thickness and material of the tube?
pyper8 years ago
Ive been thinking about this for a while and come up with a few unworkable solutions, but I think I'm finally onto a winner. A capacitive ballast feeding a bridge rectifier with a small decoupling / tank cap. Then use the semi-smooth DC at about 30v-ish to run a single chip no-external-components type voltage regulator, which are quite tolerant about input voltage, then use that with a conventional resistor-LED setup.. You'll have to do the math yourself, it's late and I don't know exactly what you want.
Peter-F8 years ago
I have a 120V version of what you want- bought retail - USA, though 3 LED's in a circuit of Rs, 1Ns & Cs.... MFG Brand : FEIT Electric model# : C7/LED/WG WHITE UL#E170906 228-6-32 Busted one... opened it & found 3 surface-mount (blue) LED's, 2 resistors, 4 diodes (4001 series as rectifier), and a capacitor. Lost that circuit, and the intact version hides the Capacitor. This should be rather easy to find, but maybe not for 220V values... and I have hears using a capacitor as a voltage reducer can be tricky... so be careful (and don't seek advice from me... no experience in this field!) best of luck!
merrills8 years ago
If you're in the UK, could you perhaps modify a mains LED bulb, like the GU10 for your needs?
Goedjn8 years ago
Do you have an available ground, or just a high and a low?
Depending on what you're trying to accomplish, couldn't you
just ground one lead of the LED, and tape the other end along
the 220V line (outside the insulation) and pick up your
power inductively? (I mean, technically, that *IS* a
transformer, but still...)
You need ~15mA through the LED and ~3V across it in forward direction. Things to think about: Resistors are for ~300V peak maximum -> take some in series to be safe. LED is not safe to block 300V reverse -> add a 1N400x in series A voltage divider would need a large capacitor: 10k + 34µF A large resistor dumps a lot of power -> needs to be a ~5W type my suggestion is: 1x1N400x silicon diode reverse parallel to the led for reverse blocking 2x1k resistors (0.5W) in series 1x220nF in series
arzthaus8 years ago
you will need a capacitor. Dont use a resistor, the amount of energy that it dissipates will be ridiculous. For more info, look at thisthis
vlotti8 years ago
Try the modern approach, look for a leftover cellphone charger. A small one preferably, like those slender Nokia things. they are switched mode, with that very small print and tiny transformer inside that delivers a low voltage at 100 milliamps. Then connect the LEDS in a configuration, such that only a small resistor of 50 or 100 Ohms is sufficient as current limiter... Ivo
Bruwer8 years ago
This is quite a dangerous thing to do. Any break in the circuit will leave the full 220V across that break.

However, it can be done easily. You can use a 12K ohm 5 watt resistor but note that the resistor will get quite hot.

............12K....... |................................................. |

Or you can use a 0.33 uF capacitor instead of the resistor. BUT it must be a 600V version to allow for the peak voltage AND for any transient voltages that WILL occur on the mains voltage
jeff-o8 years ago
You could use a resistor-capacitor voltage divider. It's the same as an RC low pass filter. LINK

By my calculations, for an output of about 14V and an input of 220V, you will need a 260k resistor and a 1uF capacitor.
jeff-o jeff-o8 years ago
Scratch that! I forgot to properly calculate the radian frequency. I also forgot that you're on 50Hz instead of 60. The new values are R=10M, C=490uF

Hmmm, it will be difficult to get exactly 14V with values that huge... are you sure you can't just add a dozen more LEDs? ;)
Plasmana8 years ago
A resistor will work, can you tell me what are the LED's forward voltage and I can do the calculations and find your desired resistor for you.
You're going to need a HUGE resistor. LEDs operate (depending on the color of the LED) from 1.7v to 3.5v or so. Resistor value will need to be in the range of 10kohm or higher. Not only that, but the resistor will have to be able to dissipate large amounts of power, normal 1/4, 1/2 or even 1 Watt resistors won't do it. It'll be more like 5 watt, and the resistor will get quite hot. Maybe even hot enough to burn you or start a fire. In short, use a transformer. Step it down to 12v or 6v or so. Since you're taking the space for a transformer, you can add a rectifier (either simple diode or bridge rectifier), some smoothing caps and a regulator.