A field team faces an unresolved physical question: Why can a battery test pass and the field unit still run short? They must answer it before changing average current 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 average current. The middle card applies this page's relationship. The green card is nameplate 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.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline for average current is 80.
- 2
Name the relationship. Enameplate=2.400x3.6=8.64 Wh ΔV=0.120x3=0.360 V; Vterm=3.24 V Qusable=0.8x2400=1920 mAh; Eusable=6.91 Wh Life=1920/0.080=24,000 h=2.74 years Pt=20-6=14 dBm; RF-stage ratio=3.98x
- 3
Substitute the chapter fixture. Set average current to 80. The page ledger gives nameplate energy as 8.64 Wh.
- 4
Read the result. Keep Wh beside the value. Use it only inside the technical boundary on this page.
Predict, then change average current
Try Predict the direction of nameplate energy. Move one control, calculate, then check your prediction.
Observe A long-run budget and a pulse-voltage check can fail independently; both must use the installed radio path. Reset the control to 80 and compare nameplate energy.
Explain Only average current moves here. The other chapter fixtures remain fixed.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Separate charge from energy
Milliamp-hours count charge. Watt-hours include the voltage that pushes that charge. A current pulse then changes terminal voltage through the cell's internal resistance.
2. Name every algebra move
Attach voltageE(Wh)=Q(Ah)×V.
Subtract pulse sagVterm=Voc−IRint.
Apply the derateQusable=fderateQnominal.
Divide by currentLife=Qusable/Iaverage.
Account for antenna gainPt=EIRP−G.
3. Reproduce the test-rig case
ΔV=0.120×3=0.360 V; Vterm=3.24 V
Qusable=0.8×2400=1920 mAh; Eusable=6.91 Wh
Life=1920/0.080=24,000 h=2.74 years
Pt=20−6=14 dBm; RF-stage ratio=3.98×
The service-life result belongs to the current trace and installed antenna used by the test, not merely to the cell label.
4. Try the sleep-average current
TryMove the measured average current while the cell and pulse case stay fixed.
ObserveLife changes inversely with average current, while the separate 120 mA pulse still sets sag.
ExplainA long-run budget and a pulse-voltage check can fail independently; both must use the installed radio path.
This is a bounded battery test ledger, not a cell-discharge model.
- Derate
- 80% is a stated design allowance, not a guaranteed capacity curve
- Resistance
- 3 Ω changes with temperature, age, state of charge, and pulse duration
- RF power
- Conducted-power ratio is not the same as total radio-current ratio
Use the intended cell, antenna, enclosure, temperature, firmware, and current trace in the release test.
5. Recreate field stress
Run cold, aged, weak-link, retry, and update cases. Verify minimum terminal voltage and average current from the same firmware and antenna configuration.
6. Preserve the test state
Store cell lot, temperature, resistance method, firmware, current trace, RF configuration, derate, cutoff voltage, and the requirement ID.
7. Check yourself
Why is 2400 mAh not yet an energy figure?
Why can 6.91 Wh still fail a transmit pulse?
Does 6 dBi guarantee 3.98× lower battery current?
The arithmetic reproduces the chapter's catalog-typical test-rig values.
- 2.74 years
- Charge-only result at 80 µA after an explicit 80% derate
- 3.24 V
- One pulse estimate using a fixed 3 Ω resistance
- 3.98×
- RF-stage output ratio, not a service-life multiplier
Correct, not complete: this ledger does not qualify field battery life.
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