Could you power a computer from your voice via a microphone?
Typical speakers are about 1% to 5% efficient. Suppose a microphone is about the same. Someone shouting at a distance of a meter is supposedly about 85dBa; 0 dBa (0 dB SIL) is defined as 10⁻¹² W/m², so this is about 3 mW/m². The shout is maybe spread over an area of a square meter, so it’s probably about 3mW in total, and you could probably capture a significant fraction of that if you had the microphone actually in your mouth. 3mW is plenty of power to run certain microcontrollers.
There’s another issue of impedance and voltage matching. Piezo and electret microphones are essentially capacitive, making them very high impedance at audio frequencies, while dynamic microphones are essentially inductive, which seems more circuit-friendly. But since we’re dealing with an audio-frequency AC waveform, you ought to be able to step it up or down with a steel-core transformer.
Otherwise, though, I think you could probably just feed a low-capacitance capacitor with the rectified dynamic mic, run a high-efficiency boost or buck converter from it to step the voltage up or down, and feed a low-voltage, high-capacitance capacitor with that to run the microcontroller.
For example, the LTC3388 is designed for this kind of application, accepting voltages of 2.7 to 20 V, with 720 nA quiescent current up to 820 nA at 20 V; different configurations manage output voltages down to 1.2 V; the efficiency is about 90% with low input voltages, dropping to about 70% for 20 V input, unless the load is under 100 μA. 100 μA at 1.8 V (typical for low-power microcontrollers) would be 180 μW, which would allow a one-second 3-mW burst of sound to power the microcontroller for 16⅔ s. Various low-power microcontrollers manage a bit under a nanojoule per instruction (like 300 picojoules to 900 picojoules per instruction), so 180 μW is about 360 000 instructions per second, comparable to a Commodore 64.
I have the intuition that for this kind of thing, it’s important to be able to draw down the charge stored on the input side fast enough to follow the sound waveform on its way down, to ensure that the available energy there is successfully harvested. The LTC3388 can source up to 50 mA, which would be 90 mW at 1.8 V, which would be 4.5 mA at 20 V or 10 mA at 9 V. If the sound wave had most of its energy in a 200 Hz component, it falls from 9 V to 0 in 0.74 ms; without getting into actually calculating derivatives precisely, that would require that the input capacitor be no larger than about 0.8 μF.
(We shouldn’t really try to analyze it linearly into sinusoids, because both the rectifier charging the input capacitor and the buck converter draining it are nonlinear, but that should give us a ballpark.)
I don’t know how much input capacitance we need, or if we need any, but it seems like the LTC3388 lets its inductor current ramp up to 150 mA before shutting the gate, so we probably need enough energy stored to get a 22 μH inductor up to 150 mA. LI²/2 is then about a quarter of a microjoule. A capacitor that charges up to a quarter of a microjoule at, say, 7 volts would be .25 J / (7 V)² = 5000 pF. So probably anything in the middle of the 5000 to 800 000 pF range would work.
My intuition is that if the input capacitor is ever above the rectified signal, no current will flow, so no energy is harvested; while if it’s far below the rectified signal, then most of the voltage has to be dropped by the rectifier itself rather than the capacitor, and that dropped voltage also represents inefficiency. So I think you want to ensure that the input capacitor stays just below the constantly changing supply voltage, at least until the output capacitor is fully charged.
I don’t think the LTC3388 will do that, at all, and I don’t know what will.
One thing that might help would be an inductor in series with the rectifier. This would keep a significant voltage from ever being dropped across the rectifier, instead charging the inductor until the capacitor is ready to accept its energy. But maybe that will make it hard to ramp up harvesting when there’s a rapid rise in voltage?