Difference between Mosfet and Voltage Regulator

A level shifter is usually a part that converts digital signals from one logic standard to another. It might also be called a translator. For example, the MC14504B converts TTL logic signals to CMOS levels, and a MC10H607 converts PECL signals to TTL. A level shifter isn't meant to provide power, it can only source as much current as its target logic levels require.

The terms voltage regulator and dc-dc converter are somewhat overlapping. Classic linear regulators are almost always called regulators. Linear regulators can only be used to produce a lower voltage from a higher one. Switching supply circuits might be called regulators or dc-dc converters. (Purists might claim that the regulator is just one part of a dc-dc converter circuit. That is the regulator is what provides the feedback control, whereas the dc-dc converter is a complete circuit including external magnetics, switching transistors or diodes, etc.) Switching supply circuits include different types that are able to produce either lower or higher voltages from an input voltage.

To produce 2.5 V from 5 V, you can use either a linear regulator or a "buck" switching converter.

It'll work with 1Mohm and you could use a 10k or 100k with out problems BUT consider this scenario: -

The battery gets low and switches off the load. The dumping of the load will inevitably cause the battery voltage to rise again because it no longer is under a load condition. This rise then re-enables the load FET and you are into a possible problematic situation. You really should be "latching" the battery low condition with a flip-flop BUT now we're into trying to work out how this situation remedies itself - what mechanism is acceptable for resetting the power?

The low-battery comparator does have hysterisis but it's only 10mV - it maybe enough but i wouldn't count on it. I'm sure the battery could rise (under load to no-load conditions more than 5% and this would be 25mV).

Another option is to have a resistor from the LB output to the LBI - this will increase hysterisis but it's a suck it and see approach. Having said that it'll probably work. You'll need more like a 10k resistor on the LBO and possible no smaller that 100k as feedback to LBI. Your normal LBI resistors will need to be about the 10k area to prevent too much hysterisis too.

EDIT following disclosure of circuit Q1 won't work how you imagine it to - if the source was connected to 0V and the drain to the negative end of the load I believe this would work. The 1Mohm pull-up would keep the load connected via the FET to 0V. As you now have it this won't work. If you are intent on a high-side switch (maybe because the load HAS to be always connected to ground) then there is a way but it involves two fets somewhat along these lines: -

BUT.....

Replace the BC547 with an n-channel FET and get rid of the 10kohm. Ignore the 5.5V - read it as 5V - it's getting late over here and i didn't have the perseverence to fully correct the diagram in paint!!! The MCU line is your LBO. The high-side FET is P-channel - choose one with very low Rds(on) and one that has a "logic-level" gate drive voltage. The n-channel is less stringent - it can't be a bioplar given the high impedance pull-up on the LBO pin BUT, given that you may use a 10kohm (previouse pre-edit comments) it could be a bipolar (just about). Stick with an n-channel logic gate driven device to be sure!