What is needed to connect 100 LEDs to AC?

I'm planning on doing a project for which I need about 500 LEDs connected to AC, to make it more simple I will start with 100.

Referring to the attached schematic by Dipankar (just in case see the link - **broken link removed**), I would like to know what capacitor and resistor should I use if I want to make analogical LED Tube but with 100 LEDs(see schematic in the link).Knowing how intricate it would be to solder all this, combination of several 30 LED segments as suggested by Dipankar is not an option.

AC ir european 220V
I will use 5mm LEDs with 3.2 to 3.8V forward voltage and maximum continous forward current of 30mA.

Thank you for any suggestions!

P.S.-Are there any formulas by which i can calculate the sizing of the capacitor and the resistor depending on the quantity of LEDs i use?

So,you are not going to use his 30 LED model & instead how would you connect your 100 LEDs. Can u show a blue-print??

There is on blue-print to show. I'm going to use the model given with 100 LEDs instead of 30, as you said. I need to know how to change the capacitor and the resistors size. I'm a complete novice thereby I assume that by increasing the quantity of LED's i might decrease the size of the resistor and increase/decrease the capacitance.

The capacitor and resistor work together to limit the current to the LEDs. The resistor *can* be omitted but I would not recommend it because it's the only thing that saves a total meltdown if the capacitor goes short circuit (the usual failure mode).

The calculation is best done in two stages, first find the total resistance needed to limit the current, for sake of calculation assume the RMS value of the AC is DC. The resistance calculation is the voltage you need to drop divided by the current through the LEDs. The voltage drop is the AC voltage minus all the forward voltage drops of the LEDs added together. You will find the Vf in the LED data sheet and the total will be that multiplied by the number of LEDs facing one direction in the chain (count parallel diodes as one Vf because only one at a time will be conducting). So for example, if Vf of each LED is 2V and you have 30 steps along the chain, the total Vf will be 60V and the amount you need to drop will be 220 - 60 = 160V. At 30 mA the total resistance needed will be 160V/0.03A = 5333 Ohms.

Some of that resistance can be made using a real resistor, some can be made using the reactance of the capacitor but their total should be 5333 Ohms. It's up to you to decide how much is from the resistor and how much from the capcitor. A larger resistor buys you a greater safety factor but also generates more heat, for example, if you ONLY used a resistor it would dissipate about 5W (W= V*I = 160 * .03 = 4.8W) so to allow for a safety margin you would probably use a 10W rated part and have to accept it got quite hot. If you only used a capacitor it would run cold but if it shorted out the whole 220V would appear across the LEDs and destroy them instantly.

Lets suppose you use a 1K resistor, the remaining 4,333 Ohms should be made up from the reactance of the capacitor at AC frequency which I assume is 50Hz as you are in Europe. The formula to work out the capacitor value is 1/(2*pi*f*Xc) so using 4333 Ohms as the Xc value should be 0.7346uF. The nearest standard value would be 0.82uF which gives an Xc of 3881 Ohms so you could use that and slightly increase the resistor to say 1.5K.

I would strongly advise you to add a fuse in the circuit!

Brian.

Thank you very much Brian, this was very detailed, simple and useful!