Phoebe guidesA technician must decide whether accumulated sleep energy is safe before changing service life on the real device. The result is unresolved until the rule and units are checked. Predict the direction first.
See the relationship before changing it
The figure reads from left to right. The blue card is service life. The middle card applies this page's rule. The green card is accumulated sleep 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 model keeps those stated values fixed and changes only service life, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 5 years.
- 2
Name the relationship. sleep energy = 3 uA x 3.6 V x 24 h/day x 365 days/year x years
- 3
Substitute with units. 0.000003 x 3.6 x 24 x 365 x 5 = 0.473 Wh
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change service life
Try Predict the direction of sleep energy = 3 uA x 3.6 V x 24 h/day x 365 days/year x years. Test another service life, then compare accumulated sleep energy.
Observe A longer service promise accumulates more sleep energy even without extra readings. Reset service life to 5 and compare accumulated sleep energy.
Explain A longer service promise accumulates more sleep energy even without extra readings.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Begin with the physical story
Milliamp-hours measure charge, not energy. Voltage tells how much energy each unit of charge can deliver. A five-year claim then has two long clocks: the battery loses capacity by self-discharge, and the device keeps drawing sleep current between brief readings.
2. Put names and units on the maths
Keep the units beside every number. They are an error detector: only like units can be added or subtracted.
| Symbol | Meaning | Unit |
|---|---|---|
| C | cell charge rating | Ah |
| V | cell voltage | V |
| k | self-discharge fraction per year | fraction |
| E_usable | energy after leakage and derating | Wh |
3. Derive it with every move named
Convert charge units2.4 Ah is already in amperes × hours; if given mAh, divide by 1000.
Multiply by voltageE_cell = 2.4 × 3.6 = 8.64 Wh.
Apply repeated self-dischargeFive years keeping 99% gives 8.64 × 0.99^5.
Apply the design deratingMultiply by 0.90 for the chapter's 10% margin.
Price the long sleepE_sleep = 3 μA × 3.6 V × five years of hours.
Price the brief readingsReadings × 40 mA × 3.6 V × 0.2 s, then divide joules by 3600 for Wh.
Check voltage sag separatelyV_terminal = V_oc − IR_int = 3.6 − 0.040(10) = 3.2 V.
4. Reproduce the chapter's numbers
The cell stores 8.64 Wh nominally. After 1%/year self-discharge for five years and 10% derating, usable energy is 7.39 Wh. Sleep uses 0.473 Wh; 9,125 active readings use 0.0730 Wh; total 0.546 Wh is about 7.4% of usable energy.
The 40 mA burst through about 10 Ω internal resistance causes 0.4 V sag, from 3.6 V to 3.2 V. Plenty of watt-hours does not guarantee the radio avoids brown-out.
5. Try the formula
TryMove the service-life slider from one to seven years and compare usable energy with accumulated sleep and active cost.
ObserveObserve that self-discharge reduces what remains while sleep cost grows steadily; the short readings remain the smaller term in this stated duty cycle.
ExplainEvery readout comes from E_usable = 8.64(0.99)^t(0.90) and E = IVt, the same battery-ledger equations derived above.
This small widget varies one named input and holds the chapter constants fixed.
- The honesty boundary below names what it does not model
- Needs separate evidence
Use field evidence or a deeper model before release.
6. What the result buys you
The simple ledger shows that this stated duty cycle has substantial nominal margin, but it cannot certify a five-year product. Cold temperature, pulse capability, aging internal resistance, radio retries, regulator loss, sensing warm-up, manufacturing spread, and end-of-life voltage must be measured. Architecture uses the estimate to decide what to test, not to replace the test.
7. Check yourself
Try each question before revealing the answer.
1. How does 2.4 Ah at 3.6 V become 8.64 Wh?
Answer: Energy = charge × voltage = 2.4 × 3.6.
2. Why use 0.99^5 rather than subtracting one arbitrary lump?
Answer: The cell keeps 99% of the remaining capacity each year, so the loss compounds.
3. Why check voltage sag if the energy ledger has margin?
Answer: A brief current pulse can cross the radio brown-out threshold even when total stored energy remains.
These are the chapter inputs, worked results, and named teaching assumptions.
- 500 sensors
- Device, payload, or sample count
- five readings/day
- Frequency, sample rate, or event rate
- five years
- Time, interval, or service-life value
- then states the illustrative ER14505-class 2.4 Ah
- Named teaching assumption
- 3.6 V
- Voltage or voltage-step value
- 1%/year
- Time, interval, or service-life value
- 3 μA
- Current or responsivity value
- 40 mA for 0.2 s
- Time, interval, or service-life value
- 10 Ω
- Resistance or impedance value
- 10% derating assumptions
- Named teaching assumption
They are not a cell guarantee. Temperature, aging, retry rate, regulator efficiency, pulse recovery, safety, and actual discharge curves require datasheet and bench evidence.