Calculate Current Used: Diode vs. LDO for lowering voltages?
The power wasted with an LDO will be \$P = V×I = (V_{in}-V_{out})×I\$ [W]

The power wasted with a diode will be\$P = V×I = V_f×I\$ [W], (Vf means forward voltage). Depending on the diode you're using it can be everything from 0.2 V to 1.5 V and even higher if you're using a Zener or an LED. I'll use 0.6 V because it's probably something you're having in mind.

So the power wasted with a diode will approximately be \$0.6×I\$ [W]



Let's plug in your numbers and see which one is the most efficient, hint: they are equally bad.

LDO: \$(V_{in}-V_{out})×I = 3.7-3.1 = 0.6×I\$ [W]

Diode: \$0.6×I\$... wait a second, this is the same as for the LDO! That's because you're not using a switching power supply (SPS). 



What about using a resistor instead? Maybe that will be better, hint: it's not, it's equally as bad as the above options.

If you know that your system will pull 1 mA, then what is the resistance required to make the voltage across the resistor 0.6 V?

\$U=R×I =&gt; R = \frac{U}{I} = \frac{0.6}{0.001} = 600Ω\$

The power wasted will be \$P = V×I = 0.6×I\$, oh no, the \$0.6×I\$ [W] again. Argh! 



Size wise a resistor may be smallest, next is probably a diode, third an LDO. If I were you I'd use an LDO because that is the only one that makes sure that it's 3.1 V at the output, the diode and resistor are just passive, there's no active regulation happening with them.



If you'd want to go for efficiency then a buck converter would make you happy. Sadly they are a little bit larger than resistors/diodes/LDO's because they require an inductor, also the noise from the switching might be too nasty for your load. 

Let's assume you'd go for some buck converter (one kind of a switching power supply) that's 90% efficient while supplying 1 mA, it's like a dream come true. Everything just matches and the noise isn't affecting your load at all. I can see a rainbow. 

Then the energy wasted would be \$P = P_{waste}×(1-P_{efficiency}) = 0.6×I×(1-0.9) = 0.06×I\$ [W]

Wasting \$0.06×I\$ [W] is much better than wasting \$0.6×I\$ [W], assuming your numbers and rainbows.