Power Management

Assume DVFS can lower the supply voltage in proportion to frequency. When frequency is halved, roughly what happens to dynamic power and to the dynamic energy for a fixed amount of computation?

  1. APower falls to one-eighth and energy to one-quarter
  2. BBoth power and energy halve
  3. CPower falls to a quarter and energy is unchanged
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Answer: A. Power falls to one-eighth and energy to one-quarter

Power is proportional to V²f, so halving both V and f gives (1/2)² × 1/2 = 1/8. The computation takes twice as long, so energy is 1/8 × 2 = 1/4. Equivalently, since energy per switching event is proportional to CV², energy scales with the square of voltage. In practice the voltage range is limited and static power exists, so the reduction is not this ideal.

  • A. Correct. Power scales with V²f and energy with V².
  • B. Voltage is also lowered, so the reduction is larger.
  • C. Lowering voltage also reduces energy.
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More Power Management questions

  1. Which describes static (leakage) power in CMOS circuits?
  2. How does leakage power change as chip temperature rises, and what problem can result?
  3. Which describes power gating?
  4. Which correctly pairs a caveat of power-gated blocks with its countermeasure?
  5. Which correctly compares clock gating and power gating?