Low-Power Transition Cost Calculation Audit

Low-Power Transition Cost Calculation Audit

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

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Ada ADA · CALCULATION AUDIT

Low-Power Transition Cost Calculation Audit

A BLE node that wakes once per second sleeps for 999.6 ms at 4.8 µA, then spends 0.3 ms active at 14.6 mA and 0.14 ms moving the radio into transmit at 7 mA. That last radio-transition term is easy to fold into the active bucket — and doing so quietly hides about 9.7% of the cycle's real average current. This audit keeps the transition as its own ledger line to show why it belongs there.

Companion to the chapter Low-Power Design Strategies — every number here comes from that chapter.

See the relationship before changing it

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

Radio transition time changes cycle average An input card leads through the rule average = 4.79808 + 4.38 + 7 x transition ms to the cycle average result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. A short transition still matters because its current is large beside the sleep term.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 0.14 ms.

  2. 2

    Name the relationship. average = 4.79808 + 4.38 + 7 x transition ms

  3. 3

    Substitute with units. 4.79808 + 4.38 + 7 x 0.14 = 10.16 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 transition time

Try Predict the direction of average = 4.79808 + 4.38 + 7 x transition ms. Test another radio transition time, then compare cycle average.

0.14 ms
Chapter baseline
Cycle average

Observe A short transition still matters because its current is large beside the sleep term. Reset radio transition time to 0.14 and compare cycle average.

Explain A short transition still matters because its current is large beside the sleep term.

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

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

TryTransition Cost” from the shown inputs: A BLE node that wakes once per second sleeps for 999.6 ms at 4.8 µA , then spends 0.3 ms active at 14.6 mA and 0.14 ms moving the radio into transmit at 7 mA . Use Check derivation.
ObserveThe displayed ledger resolves 999.6 ms, 4.8 µA, 0.3 ms, 14.6 mA, 0.14 ms at full precision. This audit keeps the transition as its own ledger line to show why it belongs there. Check derivation shows this.
Explainthe transition term is small in isolation (0.98 µA-s) but it sits right next to a sleep term of similar size (4.79808 µA-s), so leaving it out understates the cycle's average current by nearly a tenth. Label every state that has its own current and duration -- sleep, active, and transition -- and sum them separately before averaging. Check derivation confirms it.

Ada: Use the same one-second BLE cycle from the chapter above. Convert every duration to seconds, convert every current to microamps, then multiply current by time to get charge. Keep the radio-transition term on its own line instead of folding it into “active” – that is the whole point of this audit.

  • Sleep: 999.6 ms = 0.9996 s, so 4.8 uA x 0.9996 s = 4.79808 uA-s.
  • Active transmit and CPU: 14.6 mA = 14,600 uA and 0.3 ms = 0.0003 s, so 14,600 uA x 0.0003 s = 4.38 uA-s.
  • Radio transition: 7 mA = 7,000 uA and 0.14 ms = 0.00014 s, so 7,000 uA x 0.00014 s = 0.98 uA-s.
  • Total with transition: 4.79808 + 4.38 + 0.98 = 10.15808 uA-s; over the 1 s cycle, that averages to about 10.16 uA.
  • Total without transition: 4.79808 + 4.38 = 9.17808 uA-s; over the same 1 s cycle, that averages to about 9.18 uA.
  • Difference: 10.15808 - 9.17808 = 0.98 uA, and 0.98 / 10.15808 = 0.0965, so omitting the transition hides about 9.7% of the measured average current.

The engineering lesson is not that every BLE product has this exact budget. It is that short transition states can be large enough to change a release decision, so a current trace should label them instead of folding them into a vague active bucket.

What the audit buys you: the transition term is small in isolation (0.98 µA-s) but it sits right next to a sleep term of similar size (4.79808 µA-s), so leaving it out understates the cycle's average current by nearly a tenth. Label every state that has its own current and duration -- sleep, active, and transition -- and sum them separately before averaging.

Technical boundaries
The transition ledger deliberately does not simulate variable connection time, protocol retries, oscillator startup, regulator loss, voltage sag, or temperature; it integrates the chapter's fixed one-second BLE state sequence.

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.