A field team has a real problem to settle: Why is 2000 mAh not yet a lifetime? They must decide what happens before they change average current on the device. Predict the direction first.
See the relationship first
The figure reads from left to right. The blue card is average current. The middle card uses this page's rule. The green card is nameplate energy. Follow the arrows: set the input, use the rule, then read the result and its unit.
The audit later on checks more than one number. Here, the added model uses the baseline named below and holds every other chapter value fixed. That sentence bridges the fixtures, so the numbers do not change without a reason.
Derive the baseline in four moves
- 1
Name the input. The chapter baseline for average current is 50.
- 2
Name the rule. Enameplate=2.000x3.6=7.20 Wh ΔV=0.030x3=0.0900 V; Vterm=3.51 V Qusable=0.80x2000=1600 mAh; Eusable=5.76 Wh t=1600/0.050=32,000 h=3.65 years
- 3
Put in the chapter value. Set average current to 50. The page rule gives nameplate energy as 7.20 Wh.
- 4
Read the result. Keep Wh next to the value. Use it only within the limits on this page.
Predict, then change average current
Try Predict what happens to nameplate energy. Move one control, calculate, then check your idea.
Observe The control changes the capacity/current division. The pulse uses a separate 30 mA input, so the two battery questions remain separate. Reset to 50 and compare nameplate energy.
Explain Only average current moves here. The other chapter values stay fixed.
Check yourself
What should you do before you trust the result?
What does this small model leave out?
1. Charge and energy are related, not identical
Milliamp-hours count charge. Watt-hours include voltage. A real cell's terminal voltage falls during a current pulse because current also crosses the cell's internal resistance.
2. Name every algebra move
Convert mAh to AhQAh=QmAh/1000.
Approximate energyEWh=QAhV.
Use Ohm's law inside the cellΔV=IRint; Vterm=Voc−ΔV.
Apply design deratingQusable=fQnominal.
Divide capacity by average currentth=Qusable/Iavg.
3. Reproduce the leaf-node case
ΔV=0.030×3=0.0900 V; Vterm=3.51 V
Qusable=0.80×2000=1600 mAh; Eusable=5.76 Wh
t=1600/0.050=32,000 h=3.65 years
These numbers form one bounded comparison, not a discharge curve.
4. Try average current
TryMove the duty-cycle average while cell and pulse assumptions stay fixed.
ObserveAt 50 µA the bounded runtime is 32,000 h. Doubling average current halves this simple runtime while pulse sag stays fixed.
ExplainThe control changes the capacity/current division. The pulse uses a separate 30 mA input, so the two battery questions remain separate.
The ledger uses fixed voltage, resistance, derating, and average current.
- Cell
- Temperature, age, self-discharge, rate effects, and capacity curves are omitted
- Load
- Startup, sensing, processing, receive windows, retries, and regulators need a state trace
- Cutoff
- Usable life ends at the system threshold, not necessarily zero stored charge
Measure pulse voltage and integrate the complete deployed load schedule.
5. Keep sag and life separate
A cell may have charge left and still reset the node during a pulse. Check pulse voltage against cutoff as well as the average-current lifetime.
6. Build the node record
Record cell chemistry and lot, voltage, capacity test conditions, internal resistance, cutoff, all state currents and durations, temperature, retries, margin, owner, and retest trigger.
7. Check yourself
Why is the nameplate energy 7.20 Wh?
Why does the pulse voltage fall by 0.0900 V?
Does 3.65 years predict a deployed node?
The 3.6 V, 2000 mAh, resistance, derating, and current values are the chapter's explicit typical example.
- 7.20 Wh
- Nameplate constant-voltage approximation
- 5.76 Wh
- Simple 80% design ledger
- 3.65 years
- Bounded result, not a field guarantee
Go deeper in energy duty cycling and replace typical inputs with measurements.
Packet Pete guides