Math Bridge: Power-Island Energy

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Why is one milliamp-hour not one energy amount?

Charge must cross a named voltage before it becomes energy.

Phoebe, the physics guidePhoebe guides
The one targetTurn each island's measured charge into a reviewable energy term.
The chapter case1 mAh at 1.20 V versus 1 mAh at 3.30 V.
What it buys youA load ledger that does not mistake mAh for energy.

A field team faces an unresolved physical question: Why is one milliamp-hour not one energy amount? They must answer it before changing main power island charge in milliamp-hours on the real device. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is main power island charge in milliamp-hours. The middle card applies this page's relationship. The green card is main-island energy. 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 added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Main power island charge in milliamp-hours changes main-island energy An input card leads through the page relationship to the main-island energy result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The E=QV terms track load energy. The sampling formula is separate: moving data without the CPU does not remove the bandwidth requirement.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for main power island charge in milliamp-hours is 1.

  2. 2

    Name the relationship. Eready=QreadyVready; Emain=QmainVmain; Eload=Eready+Emain; fs,min=2fmax

  3. 3

    Substitute the chapter fixture. Set main power island charge in milliamp-hours to 1. The page ledger gives main-island energy as 3.30 mWh.

  4. 4

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

Predict, then change main power island charge in milliamp-hours

Try Predict the direction of main-island energy. Move one control, calculate, then check your prediction.

1
Chapter baseline
Main-island energy

Observe The E=QV terms track load energy. The sampling formula is separate: moving data without the CPU does not remove the bandwidth requirement. Reset the control to 1 and compare main-island energy.

Explain Only main power island charge in milliamp-hours moves here. The other chapter fixtures remain fixed.

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 main power island charge in milliamp-hours moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only main power island charge in milliamp-hours. It replays the chapter fixture; it is not a field prediction.

1. Start with charge, power, and energy

Current is charge flow. Power is current times voltage. Energy is power added over time. When a rail voltage is nearly constant, those facts join as energy = charge × voltage.

Phoebe: The same bucket of charge does more work when it falls through a larger voltage.

2. Name the algebra moves

1

Start with powerP = IV.

2

Add power over timeE = ∫Pdt.

3

Hold voltage near constantE = V∫Idt.

4

Name accumulated chargeQ = ∫Idt, so E = QV.

5

Sum railsEload = ΣQiVi.

3. Work the chapter islands

1 mAh × 1.20 V = 1.20 mWh; 1 mAh × 3.30 V = 3.30 mWh

The equal-charge energy ratio is 3.30/1.20 = 2.75. Those are load-side values. Battery energy must divide each term by its converter efficiency and add quiescent and transition energy. The 2.00 kHz vibration path also needs at least 4.00 kHz sampling.

4. Try one controlled change

Eready=QreadyVready; Emain=QmainVmain; Eload=Eready+Emain; fs,min=2fmax

TryChange only charge used on the 3.30 V main rail. The retained 1 mAh, both rail voltages, and vibration band stay fixed.

Ready-island energy
Main-island energy
Total load energy
Equal-charge rail ratio
Minimum sample rate

ObserveAt equal 1 mAh charges, the two rails deliver 1.20 and 3.30 mWh, 4.50 mWh total, while the path still needs 4.0 kHz.

ExplainThe E=QV terms track load energy. The sampling formula is separate: moving data without the CPU does not remove the bandwidth requirement.

Technical boundaries.

The widget shows load-side rail energy, not a complete battery model.

Converters
Efficiency varies with load and state
Overhead
Quiescent, wake, leakage, and transition energy remain
Sampling
Anti-alias filtering, DMA limits, memory traffic, and clock accuracy remain

Measure the real state sequence at battery and island boundaries before approving runtime.

5. Keep the boundary in the unit

mAh × V gives mWh. It does not give battery mWh unless that voltage is the battery terminal and all conversion losses are already included.

6. Carry the evidence

Record island charge, rail voltage, converter input and output energy, efficiency at each state, quiescent current, wake count, transition time, sample rate, sensor bandwidth, DMA traffic, and battery sag.

7. Check yourself

Why is 1 mAh at 3.30 V larger than 1 mAh at 1.20 V?
Answer: Energy is charge times voltage, so the ratio is 3.30/1.20 = 2.75.
Why is the sum still load-side energy?
Answer: It is measured after the converters; their losses and overhead have not been added.
Why must a 2.00 kHz band use at least 4.00 kHz?
Answer: Nyquist requires a sample rate at least twice the highest signal frequency.
Honesty boundary.

The page repairs a unit boundary; it does not predict battery life.

1.20 mWh
Retention-rail load energy
3.30 mWh
Main-rail load energy
4.00 kHz
Ideal Nyquist minimum

Go deeper in the chapter, then measure converter losses, state overhead, and timing on the chosen SoC.

Return to SoC Architecture with the maths unpacked →