Math Bridge: Battery Droop and Self-Heating

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Math BridgeSensorsStruggle-friendly runway

Why a depleted battery can hide self-heating

One thread from terminal voltage to current, squared heating, and the signal that the ADC loses.

Phoebe, the physics guidePhoebe guides
The one targetScale self-heating when the supply droops.
The chapter case3.6 V fresh, 3.0 V depleted, 0.10 °C fresh rise.
What it buys youSeparate a smaller error from a weaker signal.

A field team faces an unresolved physical question: Why a depleted battery can hide self-heating They must answer it before changing depleted battery voltage 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 depleted battery voltage. The middle card applies this page's relationship. The green card is heating-power ratio. 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.

Depleted battery voltage changes heating-power ratio An input card leads through the page relationship to the heating-power ratio result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The signal contains one voltage ratio; Joule heating contains that ratio twice because power uses I².

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for depleted battery voltage is 3.

  2. 2

    Name the relationship. Iratio=Vold/Vfresh; Pratio=Iratio²; ΔTold=ΔTfreshxPratio

  3. 3

    Substitute the chapter fixture. Set depleted battery voltage to 3. The page ledger gives heating-power ratio as 0.6944 times.

  4. 4

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

Predict, then change depleted battery voltage

Try Predict the direction of heating-power ratio. Move one control, calculate, then check your prediction.

3
Chapter baseline
Heating-power ratio

Observe The signal contains one voltage ratio; Joule heating contains that ratio twice because power uses I². Reset the control to 3 and compare heating-power ratio.

Explain Only depleted battery voltage 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 depleted battery voltage moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only depleted battery voltage. It replays the chapter fixture; it is not a field prediction.

1. Two voltage drops can stack

The cell's open-circuit voltage falls as chemistry depletes. Under load, internal resistance subtracts another IR term: Vterm=Voc(SoC)−IRint.

Phoebe: Rest can remove load sag, but it cannot refill depleted chemistry. The sensor sees the terminal voltage produced by both effects.

2. Follow voltage into current

For excitation through a fixed series resistance, Ohm's law makes current proportional to the rail.

1

Fixed resistanceI=V/R, so Iold/Ifresh=Vold/Vfresh.

2

Heating powerP=I²R, so Pold/Pfresh=(Vold/Vfresh)².

3

Temperature riseΔT=P/δ, so ΔT scales by the same squared ratio.

3. Name the trade-off

Less voltage reduces self-heating, but it also reduces the excitation signal linearly. A smaller temperature bias does not prove that the whole measurement improved.

heating ratio=(Vold/Vfresh)²; signal ratio=Vold/Vfresh

4. Try the depleted rail

Iratio=Vold/Vfresh; Pratio=Iratio²; ΔTold=ΔTfresh×Pratio

TryMove the depleted rail while the fresh 3.6 V reference and 0.10 °C rise stay fixed.

Excitation-current ratio
Heating-power ratio
Self-heating rise
Signal retained

ObserveAt 3.0 V, current and signal are 83.3% of fresh, while power is 69.4% and the rise is 0.0694 °C.

ExplainThe signal contains one voltage ratio; Joule heating contains that ratio twice because power uses I².

Technical boundaries.

The widget assumes a fixed-resistor excitation and unchanged sensor resistance/dissipation constant.

load pulses
Needs separate evidence
temperature
Needs separate evidence
chemistry
Needs separate evidence
converter dropout
Needs separate evidence
ADC reference
Needs separate evidence
noise measured
Needs separate evidence

Use field evidence or a deeper model before release.

5. Work the chapter's ratio

3.0/3.6=0.8333
0.8333²=0.6944
0.10 °C×0.6944=0.0694 °C

The self-heating bias is smaller near end of life, but the signal carrying the reading has also fallen by 16.7%.

6. Decide what to record

A field-hardening record should distinguish open-circuit voltage, loaded terminal voltage, actual excitation current, self-heating estimate, ADC counts, noise, and whether the source is regulated. Tracking only the apparent temperature bias can reward the wrong design.

7. Check yourself

Why is the heating ratio squared?
Answer: Current follows voltage, and heating power is I²R, so the voltage ratio appears twice.
What signal remains at 3.0 V versus 3.6 V?
Answer: 3.0/3.6=83.3%.
What design removes this direct coupling?
Answer: A regulated current source keeps excitation current independent of battery rail while it remains in compliance.
Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

3.6 V
Voltage or voltage-step value
3.0 V rails
Voltage or voltage-step value
0.10 °C fresh rise
Temperature or angle value
0.694 power ratio
Percentage, ratio, or gain
0.0694 °C depleted rise
Temperature or angle value
83.3% signal
Percentage, ratio, or gain
fixed-resistor condition
Chapter input or worked result
regulated-source exception come from the chapter
Chapter input or worked result

The page does not model a complete cell discharge curve or promise measurement accuracy from voltage alone.

Return to the chapter with the maths unpacked →