Power Management in Embedded Systems Questions and Answers

Answer: A

The same current flows in and out, so the power ratio is the voltage ratio: dropping 5 V to 3.3 V is at best sixty-six per cent efficient. That is why a linear part is used for small drops and small currents and a switcher for anything else.

Answer: B

A part sleeping at two microamperes behind a regulator drawing fifty is a design consuming fifty-two, and every microampere saved in the microcontroller is then wasted. Low quiescent regulators, or running the part directly from the cell, are the answers.

Answer: C

A cell with tens of ohms of series resistance loses hundreds of millivolts at a ten milliampere pulse, and the brown-out detector may trip long before the cell is exhausted. A bulk capacitor across the cell supplies the pulse and is the standard remedy.

Answer: D

The capacitor is recharged slowly between pulses through the cell's resistance and delivers the burst at low impedance, so the terminal voltage holds up. Sizing it from the pulse charge and the permissible droop is a short calculation worth doing.

Answer: A

A lithium coin cell loses a per cent or two a year, which is a few microamperes' worth spread over the year; a design drawing less than that is limited by the chemistry rather than by its own load. Predicted lives beyond about ten years are therefore not credible.

Answer: B

A small indoor cell yields microwatts in ordinary office light, so the design must accumulate energy in a capacitor and act in short bursts. Matching the burst's energy to what can be gathered between bursts is the whole of such a design.

Answer: C

Correct answer: 25 \( \mu \)W

\[ \frac{50\ \mu\mathrm{J}}{2\ \mathrm{s}} = 25\ \mu\mathrm{W}. \] Energy divided by the interval gives the average power, and the figure is worth setting against the sleep current: at 3 volts, 25 microwatts is about 8 microamps, so a design sleeping at 1 microamp is dominated by its wake-ups rather than by its sleep.

Answer: D

Five milliamperes at three volts is fifteen milliwatts, and for four milliseconds that is sixty microjoules. Comparing two implementations by energy per execution rather than by current is what reveals which is actually cheaper.

Answer: A

Correct answer: about the same, the power having halved and the time doubled

Dynamic energy per operation is \( CV^{2} \) per transition and does not depend on frequency, so halving the clock halves the power and doubles the time, leaving the energy essentially unchanged. Slowing down therefore saves nothing by itself - and can cost, since leakage is paid for twice as long. What saves energy is lowering the *voltage*, which the reduced frequency may permit.

Answer: B

Energy per operation goes as the square of the voltage, and 1.8 over 3.0 squared is 0.36; the frequency change leaves the energy alone. This is the whole argument for voltage scaling and why it is preferred to frequency scaling on its own.