You could easily provide far more useful information than you have. If you are serious about getting good answers you can help yourself by giving as complete and accurate a description of the environment as possible subject to secrecy constraints. If secrecy is not an issue then tell us what you are doing.
In the comments you say "5 to 10 meters" but there is no indication that you needed that sort of separation in the question.
Why do you need contactless power transfer?
Sounds like a quad (or other) copter. If so, you could use downwash from the props to drive a "wind turbine". If it is electrically powered why not use the main supply?
If "gas" powered then an alternator (magnets moving past coils) could easily be driven from a motor.
Piezo tends to be hard to get substantial power from BUT a magnet vibrating in a coil can easily meet your needs. Mechanical resonance can be employed to achieve high amplitude (with the system moving into resonance as the driving frequency moves, as long as it does not move too fast.
You can calculate the approximate mass and distance needed from a vibrating energy harvester (regardless of how it is then turned to electrical power) by calculating the energy required to accelerate the mass from rest to peak velocity and back again 2F times per second.
At frequency \$f\$ one cycle takes \$1/f\$ second = \$t\$.
The mass accelerates from rest to \$V\$ peak in \$t/4 = 1/4f\$.
If total stroke is \$d\$ the it accelerates as above over \$d/2\$.
For linear velocity ramp (which you will not have) mean velocity is \$distance/time\$.
$$V_{mean} = \dfrac{d/2}{1/4f} = 2fd$$
Peak velocity will be double the mean (for linear ramp) = \$4fd\$.
Energy in an accelerated mass will be
$$\dfrac{1}{2} \times m \times V^2$$
$$ = 8 \times f^2 \times d^2 \times m $$
That's a half cycle so you get two per cycle and f per second so
$$Power = \text{Energy per 1/2 cycle} \times 2f$$
$$Power = 16 \times f^3 \times d^2 \times m$$
E&OE
You can be ABSOLUTELY CERTAIN that this is wrong! Either because of a bad assumption and/or a calculation error or my having missed something. But it gives you a starting point to think on.
The above formula if correct (and it's not) gives the power delivered to a vibrating mass by a linear motion ramp and which is then removed by braking (alternator). Even as I write this I'm raising mental objections to what I've done but it gives you a starting point
Let's plug in some figures "just for fun". Say 200 Hz, 10mm full stroke. What mass do we need to get say 200 mW?
$$Power_{wrong} = 16 \times f^3 \times d^2 \times m$$
$$m = \frac{Power}{16 \times f^3 \times d^2}$$
$$ = \frac{0.200}{16 \times 200^3 \times 0.01^2}$$
$$ = \text{about 16 micro gram (too small)}$$
So there's something wrong above - but is it one of the rough assumptions or a straight error?
Light
150 mW needed.
At 15% overall efficiency you need about 1 Watt of light energy to be converted by a PV cell.
Full sunlight has a power density of \$\cong 1000 W/m^2\$ so using sunlight you'd need
$$1/1000 \, m^2 = 10 cm^2 = 1000 mm^2$$
That's about \$30 \times 30 mm\$.
A modern LED can produce 25% + light output/ DC in.
So a 5 Watt LED focused on whatever area of PV cell you choose to use would produce 150 mW+. Whether that is feasible over the distance concerned depends on factors as yet undisclosed to us.
Inductive power transfer
How large a device is the target?
How large may a target coil be?
You can achieve 10's of Watts at 10 metres at around 10% - 20 efficiency with larger coils than you probably want to use.
MIT 2007
Link to image
Image source
Another account here
MIT technology review - A wireless powered lightbulb - part 1 and
part 2
Note the distance versus efficiency curve at left in the photo.
Cornell paper
Georgia Tech report
Agilent Wireless power transfer
