Math Bridge: Self-Discharge, Sag, and Cutoff

A field team faces an unresolved physical question: How can a battery hold charge yet fail during one radio pulse? They must answer it before changing internal resistance on the real device. Predict the direction first.

See the relationship before changing it

The chapter figure below shows: Self-discharge slowly reduces stored charge, while a radio pulse can temporarily pull terminal voltage below the device cutoff even when substantial charge remains.

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.

Two-panel battery comparison. The self-discharge panel shows an illustrative one-percent-per-year example retaining 98.01 percent after two years. The voltage-sag panel shows terminal voltage dipping below device cutoff during a radio burst and recovering afterward. The takeaway distinguishes slow stored-charge loss from temporary load-induced voltage sag and notes that the two-percent example does not explain a full 35-point derating gap. — Self-discharge slowly reduces stored charge, while a radio pulse can temporarily pull terminal voltage below the device cutoff even when substantial charge remains.

Derive the baseline in four named moves

1
Name the input. The chapter baseline for internal resistance is 25.
2
Name the relationship. Retention after two years=0.99²=98.01% Shelf loss=1.99%≈2.00 points Share of 35-point gap=1.99/35=5.69% Fresh pulse: 3.0-0.015x25=2.625 V End-of-life linear check: 2.6-0.015x200=-0.400 V
3
Substitute the chapter fixture. Set internal resistance to 25. The page ledger gives fresh terminal voltage as 2.625 V.
4
Read the result. Keep V beside the value. Use it only inside the technical boundary on this page.

Predict, then change internal resistance

Try Predict the direction of fresh terminal voltage. Move one control, calculate, then check your prediction.

Observe Predict a decrease in fresh terminal voltage as the input increases over this range. Reset to the chapter baseline and compare.

Explain Only internal resistance moves here. The other chapter fixtures remain fixed.

Source: content/energy-power/energy-aware-sources.qmd — Phoebe's Field Notes: What Is Actually Inside That 65% Usable Fraction. Use the chapter equation and fixed ledger above; only the named input varies.

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 internal resistance moves. Field effects named in the page's technical boundary stay fixed.

1. Start with the physical story

Self-discharge removes stored charge slowly even when the node is off. Internal resistance causes a voltage drop only while current flows. A cell can therefore show plenty of remaining charge and still fall below the device cutoff during a radio burst.

Battery Bruno: Keep the units beside every number. They show whether a result is charge, energy, voltage, time, or a ratio.

2. Name every algebra move

Compound shelf retentionQ(t)=Q0(1−r)^t.

Find shelf lossSubtract retained percent from 100%.

Compare with the derating gapDivide shelf-loss points by the 35-point gap.

Find pulse dropΔV=I×Rint.

Find terminal voltageVterm=Voc−ΔV.

Test the cutoffMargin=Vterm−Vcutoff.

3. Reproduce the chapter case

Retention after two years=0.99²=98.01%
Shelf loss=1.99%≈2.00 points
Share of 35-point gap=1.99/35=5.69%
Fresh pulse: 3.0−0.015×25=2.625 V
End-of-life linear check: 2.6−0.015×200=−0.400 V

A negative linear-model voltage is not a physical cell output. It is a warning that the requested pulse is outside this simple model and the cell will current-limit or collapse before reaching it.

4. Try one real input

TryMove the control and predict which outputs should change before reading them.

ObserveRaising internal resistance lowers pulse terminal voltage and cutoff margin. Retained charge does not move because shelf loss is a different mechanism.

ExplainCapacity and pulse capability can fail independently. A single usable-capacity percentage is useful for planning, but it must not hide the pulse-voltage check.

Technical boundaries.

This is a small formula ledger, not a complete source qualification.

Shelf rate: One percent per year is an illustrative primary-cell value.
Thevenin model: A single resistance omits nonlinear polarisation and current limiting.
Cutoff: The actual regulator and load waveform decide usable pulse margin.

Correct, not complete: the ledger separates two failure mechanisms; it does not qualify a cell without end-of-life pulse evidence.

5. Use the result in the design

Record self-discharge separately from cutoff, temperature, aging, and pulse resistance. Test the real current waveform at the lowest expected open-circuit voltage.

6. Record the evidence state

Keep chemistry, age, temperature, storage time, open-circuit voltage, internal resistance method, pulse current and duration, regulator cutoff, and recovery voltage.

7. Check yourself

Why does retained charge stay fixed when resistance moves?

Answer: Self-discharge and pulse resistance are separate inputs in this model.

What does the negative end-of-life result mean?

Answer: The linear model has left its valid range; the requested pulse cannot be sustained.

Does 2% shelf loss explain a 35% derating?

Answer: No. It explains only about 5.7% of that gap; other losses need separate evidence.

Honesty boundary.

The arithmetic uses the chapter's named or clearly labelled catalog-typical inputs.

Shelf rate: One percent per year is an illustrative primary-cell value.
Thevenin model: A single resistance omits nonlinear polarisation and current limiting.
Cutoff: The actual regulator and load waveform decide usable pulse margin.

Correct, not complete: the ledger separates two failure mechanisms; it does not qualify a cell without end-of-life pulse evidence.