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Comment by falcolas | Hacker News Reader

Thankfully, a circuit for dropping the 5v input to 3.3v is really easy to build. It only requires a pair of capacitors and a 3.3v regulator, only about $2 of materials total.

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You can also use a voltage divider if you are willing to suffer a bit of leakage current. That's just 2 resistors.

The linear regulators we are talking about are basically voltage dividers with feedback. They aren't significantly more efficient than your design, though they are ton less noisy.

Because of that feedback, you get to use your load as the lower half of the divider, so the total current drawn from the source is the load current plus a few microamps for the regulator's internals.

In the passive voltage divider circuit, if the load varies its current draw, you have to have a resistor in parallel with the load so that the total current through both the load and the resistor doesn't vary significantly (where "significant" is derived from your tolerance for voltage variation).

While at peak current, the efficiency of a voltage divider is the same as a linear regulator, at lower currents a voltage regulator has significant benefits. This is because the quiescent current used by an unloaded voltage regulator can be very low (500nA for the TPS783!), while for a voltage divider to be sufficiently "stiff" its quiescent current should be about 10X the max load current. As a lot of microcontroller applications alternate between a low sleep current (several uA) and bursts of high run current (several mA), a voltage divider is significantly (>10000X!) more power hungry then a linear regulator in practice, even if the worst case efficiency for both is closer.

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