> Do they use these drivers in smaller led
> flashlights to increase voltage and produce higher
> lumens in a more compact form? For shorter periods
> of time of course.
> Good way to get more light more compact.
Yes. In most of the better ones with the higher output CREE and similar LEDs anyway. The junky LED lights like those at the Dollar Store are just driving the LEDs direct usually. I have tiny little AAA and AA sized lights that easily put out way more light than the big old 3-cell Maglights. Especially so if they run the 3.5v lithium cells. The driver boards generally are sized to easily fit into whatever battery form is to be used.
Also, I should have given a better example of where the Batteriser thing is intended to work.
Say you have a remote control or other device which requires between 1.5v and 1.1v to operate (I'm making these numbers up, I have no idea what a remote actually uses). At the point that the battery level can deliver between 1.5v and 1.1v the regulated circuit does little to nothing for you. It's effectively a wash between the regulated and unregulated. The point that it comes into play is when the voltage the battery can deliver drops below 1.1v. Without it, the device would appear to be effectively dead. With it, it will boost whatever power is available up to 1.5v allowing it to continue to work. That continues down to 0.6v.
So it's not really able to deliver a difference of the full "80%" left in the battery Between 1.5v and 1.1v, it's consuming slightly more power due to the circuit (but well designed circuits are fairly efficient so that's fairly negligible). So that's basically a wash between the two. It only works down to 0.6v. The difference would be whatever the time/power/percentage however you want to calculate it between 1.1v and 0.6v. i.e., For that device/example, it's leaving about "50%" that it can't use either below 0.6v and it doesn't really matter from 1.5v - 1.1v, so it's the ~0.5v range of difference or about 30%-ish in terms of the approximately longer time that it would continue to operate. (None of this is technically exactly correct but basically that's the magnitude of the difference.) Then you'd have to calculate the cost for that difference given the cost of batteries vs the device and whatever variance there may be for better or worse in whatever specific device you might use it.
Hopefully that all makes sense.