How much current will a buck regulator draw if I'm not drawing current from it?

I wish I could use the enable pin, but I'm probably going to be using off the shelf switching regulators designed for remote control vehicles, because I don't have any time left to get this board ready and it would take too long to design my own.

My board is basically an arduino clone specifically designed for controlling lots of LEDs, playing sound effects, and controlling servos. It's got an SD card on board for holding the sound effects and rather than programming it in C it'll run scripts also stored on the SD card.

I screwed up my power calculations on the linear regulator I was using several months ago and now at the last minute I need to find a solution. I'd wanted the board to be able to be powered off anything from 6V to 12V. But it turns out the board can only dissipate 2W, and a linear regulator powered by 12V would generate 21W of heat.

The obvious solution is to use a switching regulator, but that would mean I couldn't put the boards inside small handheld props, because I wouldn't be able to fit enough batteries to get to the 7.5-8V required. It would also take a lot of time to design, and the potential for error is high. In addition, for low quantities of boards, the parts would be expensive, adding around $1000 to my costs.

So, barring adding a switching regulator to my board, the next best solution is to use these ready made switching regulators. They're only around $4 each, and I can get ones which can either regulate down to a lower voltage or boost my input voltage, so I would actually have a wider range of input voltages than I originally had hoped for.

The downside is, there's no enable pin on them. Which means even if I switch the main board off using a mosfet, they'll still drain the battery slowly.

I'm going to need to find a solution to that, but I'm really out of time. Thankfully, most of my customers will be using these boards in these things:

And typically, they run on 12V so they can power an audio amplifier. They also tend to buy crappy amps that don't have power switches. So they will need a power switch on the battery anyway. But it will be on the back of the pack and hard to reach, and they want to be able to turn the sound and lights off from the handheld particle thrower thing hanging on the side of it. That's where my onboard mosfet would come in. It would turn off my board, making the pack appear to be off, but the amp and the voltage regulator would stay on, which is ideal anyway because the amp probably would cut out part of the startup sound when they flip the pack on if the amp was not already powered up.

But I really didn't want to have to have a big power switch on the battery. I mean in that case it's needed, but my board has a 3W audio amp onboard for smaller props, and I wanted the user to be able to use any cheap switch they like to switch power on and off on the prop. It kinda sucks that I thought I had this perfect little solution with a voltage regulator that had an enable pin, only to be hit with a heathy dose of reality at the last minute. Kinda screws up my board layout too. Now I've got everything tightly packed on one side of the board and this big bare section where my voltage regulator used to be. :confused:

I'm ALMOST tempted to try to stick that buck regulator I posted the datasheet for on there just so I feel like I have a complete product instead of one missing a chunk, but it's so expensive and requires two large caps and a large inductor, and it would take me two weeks to design it and figure out all the math to make sure I have it all correct. And I'd have to move a bunch of crap around on my board and make it larger so it would fit. Ugh.

Anyway, thanks for the help, it's appreciated.