# Designing a Transformerless power supply

I am working on a transformerless PSU for maiking a 3nos.x3 watt led, each having forward voltage around 3.6v and current around 650-700ma.

I have worked on this kind of circuit:

I guess a 1uf 400vac cap at C1 will offer current around 70-80ma, but here I need a current output at around 8-9 time above this. Means around 600-700ma. I am assuming a Vout at 12v dc (where input is 220vac), where 3 of 3 watt led's are connected in series.

Can anyone explain how can I get such an output from this circuit, means what value of C1 and R1 should be used here. Is there any theoretical formula to calculate the output of such a capacitive circuit, or I should say how do I know what value of C1 can give me an output (Iout) of a specified value. Though I have used This, for calculating the same but I can't derive the relation with the led circuit. Any help is appreciated.

• Why not use a electronic wall wart? They are much safer as they provide galvanic separation and on top have a stable output voltage and lower output impedance which makes your LEDs have continuous brightness. – jippie Apr 1 '13 at 11:21
• Well, "wall warts", as I am assuming is a ac power adapter or a battery eliminator. They use a transformer to stepdown the voltage but the limitation here is that a transformer is quite high on wattage or power consumption and then the large size is a big issue. Basically I am looking for an idea to make a room lighting circuit, that can be housed in a CFL circuit housing. – Cyberpks Apr 1 '13 at 11:34
• Have you looked at modern (switched mode) wall warts lately? You can barely fit your proposed capacitor in them. They are electronic and don't use a conventional transformer. Also you are entirely ignoring the safety issues that come with your solution. Your LED's are directly connected to mains power and can therefore be lethal when touched. – jippie Apr 1 '13 at 12:13
• delete that earth connection, else you'l be popping RCDs (ELCBs) – Jasen Jun 11 '16 at 6:14

If C1 was directly connected across L and N and takes a current of 80mA when the input voltage is 220V AC, then that is the limit of this circuit. Using a "transformer" is the only way you can increase this current.

By "transformer" I mean a conventional magnetic coupled transformer or a switching regulator that may or may not use a transformer.

If you are intent on pursuing this approach you'd need to have a capacitor value of at least 10uF - at 50Hz it's impedance is 318 ohms and with 220V across it there will be a current of about 0.7A.

600~700mA is too big for this circuit.It is best to use a AC/DC SMPS with transformer.

there are some non-isolate DC-DC IC could be use for your device. it is not very expensive.

Capacitor step-down circuit also can do some change, let it output high voltage, such as 50~60V, and use a DC-DC to step-down.

• An addition to this answer: Also for LEDs you can use drivers like HV9910b. This chip can operate with input voltages up to 400VAC. – Mert Gülsoy Sep 9 '15 at 10:40