Electrical – How to drive a high-side logic-level power MOSFET

Simplest way is to put something like a zener diode structure from the drain to the gate. That way when the zener diode voltage is exceeded by a bit the MOSFET turns on and clamps the voltage. Something like this:

^{simulate this circuit – Schematic created using CircuitLab}

If I read the datasheet correctly, the part you mention clamps at about -22 to -16V with a supply voltage 6~20V, so there's probably something a bit different going on- the zener is changing a logic level relative to ground (before a level shifter).

Holding the P-channel MOSFET closed (gate at 5V) will use hardly any power. It is basically the same as holding an N-channel MOSFET gate at 0V.

(See http://www.vishay.com/docs/91076/91076.pdf for a p-channel version of the IRF530 - same gate-source leakage current of +-100nA max)

To prevent powering the LED array via CLK or DAT you will need to either drive them low or set them to high-impedence when the MOSFET is off. Since CLK and DAT are both outputs, simply driving low is fine.

Note: This does not apply to your circuit here, but for inputs you can use a tristate buffer like the 74LVC1T45 or similar to prevent powering via signal pins... See FTDI (FT232RL) powers itself through RX/TX lines for the solution I used.

Edit:

Alternative solution with existing parts

Actually, you could just move the N-channel MOSFET to the low side and drive CLK and DAT to 5V when you want to turn the LEDs off and go to sleep. That would address the problem in your first idea.