Power Management in Embedded Systems Questions and Answers

Answer: C

The energy stored is half the capacitance times the square of the voltage, so nine joules at three volts and four at two, a difference of five. A harvesting design's burst budget follows directly from that difference and the permitted droop.

Answer: D

Correct answer: 2.7 V

\[ V = E - IR = 3.0 - (0.010 \times 30) = 3.0 - 0.3 = 2.7\ \mathrm{V}. \] A 300 millivolt sag from a modest 10 mA pulse is why a coin cell needs a bulk capacitor across it: the capacitor supplies the pulse and the cell recharges it slowly, so the terminal voltage never falls to where the brown-out detector resets the device.

Answer: A

A design asleep at microamperes and awake at milliamperes needs both figures and their durations, and a meter reporting one number a second gives neither reliably. A current probe with a logging instrument, or a coulomb counter, is what such a measurement needs.

Answer: B

Integrating over hours or days captures every wake-up, every retry and every rare event that a short capture would miss, and the total charge is exactly what the cell must supply. It says nothing about the peaks, which still need a fast measurement.

Answer: C

A hundred-ohm shunt gives a readable millivolt at ten microamperes and a whole volt at ten milliamperes, which the design cannot tolerate. Instruments for this work switch ranges automatically or use several shunts, which is why they are not ordinary meters.

Answer: D

Leakage doubles about every ten degrees, so a part measured at twenty and deployed at sixty may draw sixteen times the sleep current. A pin left floating or a pull-up left enabled is the other usual culprit and is worth auditing at the same time.

Answer: A

An enabled converter, a running timer, a pull-up against a driven pin or a floating input each costs tens of microamperes. Reaching the datasheet figure requires every block and every pin to be put into a defined low power state, which is a checklist rather than a single setting.

Answer: B

A 220 mAh cell at two microamperes gives 110 000 hours of load, some twelve and a half years, but a self-discharge of one or two per cent a year exhausts it in about the same time whatever the load. Predicted lives beyond ten years are governed by chemistry, not by the design.

Answer: C

The reason is correct as far as it goes and the assertion holds only where sleeping is nearly free, which is the race-to-idle case. On a part where leakage dominates, running slowly at a reduced voltage wins instead, so neither statement settles the matter alone.

Answer: D

Gating the clock stops the switching and therefore the dynamic power, exactly as the reason says, but the transistors are still powered and still leak. Removing the leakage needs the supply to be cut, which is power gating.