Electronic – Headphone sensitivity/efficiency

Here's a relatively simple solution, but not necessarily a cheap one (depending on your budget).

An oscilloscope is not likely to give you the accuracy you asked for because they typically use an 8-bit ADC, giving 0.2% in measurement uncertainty just from the sampling digitization.

Instead, consider simply using two benchtop multimeters, like Agilent 34401A's. I haven't looked at other models, but the 34401A can measure current at the accuracy you need (for example, 0.05% of reading + 0.005% of full-scale on the 100 mA range).

They can be triggered externally at 300 readings / second (for 5-1/2 digit resolution), so that gets you a sample window much shorter than your load switching cycle. If you hook the meters up to measure the input and output current, then trigger them simultaneously you'll be able to compare the results to determine the efficiency (assuming the input and output voltages are holding constant).

If the input and output voltages are also changing, you may need 4 multimeters to get all the information you need.

If you can synchronize your measurement to the moments when the load switches, you only need half as many multimeters, because you can first measure how the input current & voltage change in response to the load switch and then move the meters around and measure how the output responds to the switching event.

Using a lower voltage for the LED is mostly fine. You will waste less power in the resistor, but you trade this for less tolerance in matching the resistor and LED. The edge case is obviously where the resistor has to be 0 Ohms, and the forward voltage of the LED is 3.3V -- at this point, the U-I curve becomes fairly steep, so manufacturing tolerances will heavily affect light output.

Whether a linear regulator or a buck converter is more efficient for the processor and accelerometer depends mostly on the currents involved -- for small currents linear tends to be better.

You can usually mix 5V and 3.3V components with resistor dividers on the data lines (these are high-impedance, so the divider works), but you need a proper regulator for 3.3V power, and if you design that to be efficient, you are likely to be better off using a single voltage everywhere.

If all you need is full brightness, and cost isn't that much of an issue, the Allegro A4490 could work for you:

• using the voltage across the series resistor as the feedback voltage (\$\frac{800\mathrm{mV}}{20\mathrm{mA}} = 40\mathrm{\Omega}\$) gets you efficient current regulation
• turning the channels on with the high-impedance enable inputs reduces the current draw of the MCU, so you can use a linear regulator here (ideally, the MCU does not drive anything).

The downside here is BOM -- you need three resistors, inductors and flyback diodes as external components for that IC, and since your current demands are fairly small, it is most likely that there are better ICs out there.