Check the Active-Time and Charge Arithmetic

Check the Active-Time and Charge Arithmetic

Ada re-derives this chapter’s own numbers step by step, at full precision

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energy-power
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Ada ADA · CALCULATION AUDIT

Check the Active-Time and Charge Arithmetic

On an FPU-less core a software floating-point multiply costs about 50 cycles, while the same work in fixed-point costs about 3 cycles. The chapter pushes both through a 16 MHz core drawing 1.5 mA while active — a 10,000-multiply burst and a 1,000-tap filter running at 100 samples/s — on the claim that fewer cycles simply mean a shorter awake window. This audit checks that step: do those cycle counts really turn into the active-time and charge savings the chapter claims?

Companion to the chapter Fixed-Point Arithmetic — every number here comes from that chapter.

Ada: The energy claim is only credible if the cycle counts become time and charge with the units carried through. These checks use only the chapter's existing 16 MHz clock, 1.5 mA active current, 10000-operation burst, 1000-tap filter, 100 samples/s rate, 50-cycle software-float cost, and 3-cycle fixed-point cost.

See the relationship before changing it

The figure reads from left to right. The blue card is cycles per multiply. The middle card applies this page's rule. The green card is burst active time. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only cycles per multiply, so the numeric fixture does not switch without explanation.

Cycles per multiply changes burst active time An input card leads through the rule time = 10,000 multiplies x cycles / 16,000,000 Hz x 1,000 to the burst active time result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. Fewer processor cycles shorten the awake window for the same ten-thousand-operation burst.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 50 cycles.

  2. 2

    Name the relationship. time = 10,000 multiplies x cycles / 16,000,000 Hz x 1,000

  3. 3

    Substitute with units. 10,000 x 50 / 16,000,000 x 1,000 = 31.25 ms

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change cycles per multiply

Try Predict the direction of time = 10,000 multiplies x cycles / 16,000,000 Hz x 1,000. Test another cycles per multiply, then compare burst active time.

50 cycles
Chapter baseline
Burst active time

Observe Fewer processor cycles shorten the awake window for the same ten-thousand-operation burst. Reset cycles per multiply to 50 and compare burst active time.

Explain Fewer processor cycles shorten the awake window for the same ten-thousand-operation burst.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only cycles per multiply moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only cycles per multiply. It is not a field prediction.

TryRecompute Burst active time: float 10000 x 50 / 16000000 = 0.03125 s = 31.25 ms ; fixed 10000 x 3 / 16000000 = 0.001875 s = 1.875 ms .
ObserveTrack Burst charge: float 1.5 mA x 0.03125 s = 0.046875 mA-s ; fixed 1.5 mA x 0.001875 s = 0.0028125 mA-s .
ExplainExplain how cutting a 1,000-tap filter from 3.125 ms to 0.1875 ms per sample reduces daily compute charge from 11.25 to 0.675 mAh.

Ready: use the stated baseline inputs, then compare each displayed result.

  • Burst active time: float 10000 x 50 / 16000000 = 0.03125 s = 31.25 ms; fixed 10000 x 3 / 16000000 = 0.001875 s = 1.875 ms.
  • Burst charge: float 1.5 mA x 0.03125 s = 0.046875 mA-s; fixed 1.5 mA x 0.001875 s = 0.0028125 mA-s.
  • Once-per-second saving: 0.046875 - 0.0028125 = 0.0440625 mA-s/s, so the average-current saving is 0.0440625 mA and the daily charge saving is 0.0440625 x 24 = 1.0575 mAh/day.
  • Filter duty cycle: float 1000 x 50 = 50000 cycles and 50000 / 16000000 = 3.125 ms/sample; fixed 1000 x 3 = 3000 cycles and 3000 / 16000000 = 0.1875 ms/sample.
  • Filter daily compute charge: at 100 samples/s, float duty is 100 x 3.125 ms = 0.3125 s/s, so 1.5 x 0.3125 x 24 = 11.25 mAh/day; fixed duty is 100 x 0.1875 ms = 0.01875 s/s, so 1.5 x 0.01875 x 24 = 0.675 mAh/day.

Every number above is taken from the chapter’s own material and re-derived step by step.

Technical boundaries. This timing model deliberately does not simulate instruction mixes, memory stalls, overflow, or changing clock and current. It converts the fixed 50-cycle and 3-cycle multiply costs at 16 MHz and 1.5 mA into active time and charge.