3
\$\begingroup\$

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:

enter image description here

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.

\$\endgroup\$
4
  • \$\begingroup\$ 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. \$\endgroup\$ – jippie Apr 1 '13 at 11:21
  • \$\begingroup\$ 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. \$\endgroup\$ – Cyberpks Apr 1 '13 at 11:34
  • 3
    \$\begingroup\$ 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. \$\endgroup\$ – jippie Apr 1 '13 at 12:13
  • \$\begingroup\$ delete that earth connection, else you'l be popping RCDs (ELCBs) \$\endgroup\$ – Jasen Jun 11 '16 at 6:14
1
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ An addition to this answer: Also for LEDs you can use drivers like HV9910b. This chip can operate with input voltages up to 400VAC. \$\endgroup\$ – Mert Gülsoy Sep 9 '15 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.