Electronic – arduino – Powering and controlling multiple 12V LED strips with Arduino LilyPad USB

arduinoled striplilypadmosfetpower

I'm trying to use an Arduino Lilypad USB to control several separate 12V white LED strips. I am however struggling to work out how to power them.

The LilyPad only operates at 3.3V and I thought a MOSFET would work, but I'm not sure how to do this when I want to control multiple strips. This will all be as part of a costume headdress, so the power supply needs to be quite compact and 'wearable'.

Any assistance would be hugely appreciated.

Thanks,
R

EDIT: to answer the comment on there not being enough info: Yes, I want to turn the strips off in patterns using the LilyPad, so each strip needs to be on a different pin. There will be 4 strips, each of 18 white LEDS. At the moment I have one strip working attached to a 2A MOSFET, 9V battery and the LilyPad. My question is then how I can best connect the other strips and their respective MOSFETs to the LilyPad and the same 9V battery, or is there in fact a better way of doing this.
Regarding size, everything except the strips will need to fit on some kind of fabric pouch on the back of the head.
My plan is just to hack it together and see, but any actual knowledge or advice would be very helpful.

EDIT 2 (current situation): The circuit is working but the lights aren't as bright as they could be. I'm using FQP2N60C MOSFETs, but have no resistors in the circuit. I've also simplified the design, so it just uses LilyPad 2 pins; 3 LED strips are connected in series on one and 4 LED strips are connected in series on the other. I'm also using one 9V and 2*AA batteries to power the strips. Any advice on which factor could be making the lights not so bright would be really helpful. Thanks!

Best Answer

Simply use a 4 pin opto coupler PC817. It has 4 pins. Connect pin1 to the pin of your lilypad, pin2 is gnd. Apply 12 v on pin 4. Connect a pull down 10k resistor on pin3 and check the output from this pin, you will get 11.7-11.8v which I think is sufficient for you.