Electronic – Switching pre regulator and linear regulators

Linear regulators work by effectively putting a controlled variable resistor between the source and load. All of the current for the load flows through this resistive element. And the voltage across it is equal to the difference between source voltage and load voltage. So the power dissipated is

\$P_{lin} = I_{load}\times{}(V_{src}-V_{load})\$.

Switching regulators work by changing the duty cycle of the current flow over a switching cycle, then averaging out the output using a filter. During part of the cycle a high current flows with a low voltage drop. During the other part of the cycle almost no current flows with a high voltage drop. Neither of these conditions dissipates much power as heat. Ideally the power lost becomes

\$P_{sw}= \mathrm{DC}(I_{on})(0\ \mathrm{V}) + (1-\mathrm{DC})(0\ \mathrm{A})(V_{off})\$,

which is, of course, 0 W. Typically much of the inefficiency in the real world is due to power lost during the very short switching inteval between the "on" and "off" parts of the cycle.

I believe I may need a buck-boost however, because the components will need at least 1.8 V and 3.3 V, and my 3.6 V battery could drain below that

First hit on Linear Technology might be useful: -

Try going to LT to get an idea and use this parameter search engine.

for the 1.8 V rail it's 1.154 mA

Use a linear regulator - it's just not worth the effort I believe to use a switcher from 3V3 to 1V8. However, if current was over 50mA I might start considering a switcher.

Generally, you can assume a buck regulator will give you about 90% power efficiency near full load and a buck-boost will be about 85%. Ball-park figures.