Ten-Minute State Budget Calculation Audit

Ten-Minute State Budget Calculation Audit

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

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Ada ADA · CALCULATION AUDIT

Ten-Minute State Budget Calculation Audit

The chapter’s ten-minute cycle spends 599 s asleep at 10 uA, 0.7 s of active work at 20 mA, and 0.3 s of radio at 120 mA — totalling 55.99 mA-s, an average of about 93 uA. The radio supplies 64.3% of the charge while occupying just 0.05% of the cycle. This audit works the ten-minute state budget to show short high-current states can dominate charge, yet long sleep still earns real savings.

Companion to the chapter Power Consumption Analysis — every number here comes from that chapter.

Use the chapter’s measured ten-minute cycle. Convert every state to the same unit, multiply current by time, and divide by the full cycle length only at the end.

See the relationship before changing it

The figure reads from left to right. The blue card is radio time. The middle card applies the page rule. The green card is average current. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

Radio time changes average current An input card leads through the rule average = (5.99 + 14 + 120 x radio seconds) / 600 x 1000 to the average current result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. A short radio burst can own most of the charge even when it owns little time.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 0.3 s per 600 s.

  2. 2

    Name the relationship. average = (5.99 + 14 + 120 x radio seconds) / 600 x 1000

  3. 3

    Substitute with units. (5.99 + 14 + 120 x 0.3) / 600 x 1000 = 93.32 uA

  4. 4

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

Predict, then change radio time

Try Predict the direction of average = (5.99 + 14 + 120 x radio seconds) / 600 x 1000. Test another radio time, then compare average current.

0.3 s per 600 s
Chapter baseline
Average current

Observe A short radio burst can own most of the charge even when it owns little time. Reset radio time to 0.3 and compare average current.

Explain A short radio burst can own most of the charge even when it owns little time.

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 radio time moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only radio time. It is not a field prediction.

TryTen-Minute State Budget” from the shown inputs: The chapter’s ten-minute cycle spends 599 s asleep at 10 uA , 0.7 s of active work at 20 mA , and 0.3 s of radio at 120 mA — totalling 55.99 mA-s , an average of about 93 uA . Use Check derivation.
ObserveThe displayed ledger resolves 599 s, 10 uA, 0.7 s, 20 mA, 0.3 s at full precision. This audit works the ten-minute state budget to show short high-current states can dominate charge, yet long sleep still earns real savings. Check derivation shows this.
ExplainThe physics point is the time integral: short high-current states can dominate charge, but long sleep states still earn real savings when they span almost the whole cycle. Check derivation confirms it.

Formula

State charge = current x duration.

Average current = total charge / cycle time.

  • Sleep: 10 uA = 0.010 mA, so 0.010 mA x 599 s = 5.99 mA-s.
  • Active work: 20 mA x 0.7 s = 14.0 mA-s.
  • Radio: 120 mA x 0.3 s = 36.0 mA-s.
  • Total charge: 5.99 + 14.0 + 36.0 = 55.99 mA-s.
  • Average current: 55.99 mA-s / 600 s = 0.0933167 mA, so the report cycle averages about 93.3 uA.
  • Radio share: 36.0 / 55.99 = 0.643, so the radio contributes about 64.3% of the charge while occupying 0.3 / 600 = 0.0005, or 0.05%, of the cycle time.
  • Active-current improvement: 5 mA x 0.7 s = 3.5 mA-s saved, which lowers the average by 3.5 / 600 = 0.005833 mA, or 5.83 uA.
  • Sleep-baseline improvement: 3 uA = 0.003 mA, so 0.003 mA x 599 s = 1.797 mA-s saved, which lowers the average by 1.797 / 600 = 0.002995 mA, or about 3.00 uA.

The physics point is the time integral: short high-current states can dominate charge, but long sleep states still earn real savings when they span almost the whole cycle.

Technical boundaries
The ten-minute budget deliberately does not simulate boot spikes, retry traffic, regulator loss, voltage sag, cell aging, or temperature; it integrates the three stated current-duration rectangles over one fixed cycle.

Work the audit first, then check the displayed derivation.

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