Math Bridge: Battery Sag and Brownout

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Math BridgeReference ArchitecturesStruggle-friendly runway

How can a battery with charge left still brown out?

Stored charge and pulse delivery are different physical checks.

Phoebe, the physics guidePhoebe guides
The one targetTest one radio pulse against the pack's voltage floor.
The chapter case3.00 V, 1.50 Ω, a 2.70 V floor, and a 250 mA pulse.
What it buys youA brownout check that average mAh cannot answer.

A field team faces an unresolved physical question: How can a battery with charge left still brown out? They must answer it before changing radio pulse current in amperes 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 radio pulse current in amperes. The middle card applies this page's relationship. The green card is loaded voltage. 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.

Radio pulse current in amperes changes loaded voltage An input card leads through the page relationship to the loaded voltage result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The same IxR drop derived above drives the live answer. Lowering average duty cycle does not raise voltage during this one pulse.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for radio pulse current in amperes is 0.25.

  2. 2

    Name the relationship. Vsag=IR; Vload=Voc-IR; Icrit=(Voc-Vmin)/R

  3. 3

    Substitute the chapter fixture. Set radio pulse current in amperes to 0.25. The page ledger gives loaded voltage as 2.625 V.

  4. 4

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

Predict, then change radio pulse current in amperes

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

0.25
Chapter baseline
Loaded voltage

Observe The same IxR drop derived above drives the live answer. Lowering average duty cycle does not raise voltage during this one pulse. Reset the control to 0.25 and compare loaded voltage.

Explain Only radio pulse current in amperes 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 radio pulse current in amperes moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only radio pulse current in amperes. It replays the chapter fixture; it is not a field prediction.

1. Picture a hidden resistor

A real cell behaves like an ideal voltage source with internal resistance in series. Pulse current crosses that resistance first. The resulting voltage drop leaves less voltage at the radio.

Phoebe: Capacity says how much charge remains; resistance says how quickly it can leave.

2. Name the algebra moves

1

Find the dropVsag=IRint.

2

Subtract itVload=Voc−Vsag.

3

State survivalVload≥Vmin.

4

RearrangeIRint≤Voc−Vmin.

5

Divide by resistanceIcrit=(Voc−Vmin)/Rint.

3. Work the cold, aged pack

Icrit=(3.00−2.70)/1.50=0.200 A

A 30 mA radio has 6.67× current headroom against that ideal threshold. A 250 mA radio drops 0.250×1.50=0.375 V, leaving 2.625 V. Since 2.625 is below 2.70, the pulse browns out even if charge remains.

4. Try one controlled change

Vsag=IR; Vload=Voc−IR; Icrit=(Voc−Vmin)/R

TryMove only the pulse current. Open-circuit voltage, cold resistance, and the radio floor stay fixed.

Voltage sag
Loaded voltage
Critical current
Threshold / pulse
Above 2.70 V?

ObserveAt 250 mA, sag is 0.375 V and load voltage is 2.625 V. The 200 mA threshold is only 0.80× the requested pulse.

ExplainThe same I×R drop derived above drives the live answer. Lowering average duty cycle does not raise voltage during this one pulse.

Technical boundaries.

This is a static series-resistance model.

Cell
Voc and Rint vary with chemistry, temperature, age, state of charge, and pulse length
Path
Contacts, wiring, regulator response, and local capacitance remain
Radio
Current shape and brownout floor vary by mode and firmware

Measure the full pulse waveform on the cold, aged pack with production hardware.

5. Do not replace this with average current

Average current predicts charge use over time. Brownout depends on the largest short pulse and the instantaneous supply path. A device must pass both checks.

6. Carry the evidence

Record chemistry, series cell count, Voc, Rint versus temperature and age, pulse waveform, wiring and contact drop, regulator floor, bulk capacitor, brownout reset, recovery, and remaining capacity.

7. Check yourself

Why is the critical current 200 mA?
Answer: The pack may lose only 0.30 V; 0.30/1.50=0.200 A.
Why does 250 mA leave 2.625 V?
Answer: Its drop is 0.250×1.50=0.375 V, then 3.00−0.375=2.625 V.
Can a large remaining mAh value prove pulse survival?
Answer: No. It does not state internal resistance or loaded voltage.
Honesty boundary.

The page exposes one brownout mechanism; it is not a cell qualification.

1.50 Ω
Catalog-typical cold, aged pack assumption
200 mA
Static critical current
2.625 V
Static voltage under 250 mA

Go deeper in the chapter, then pulse-test the actual pack, regulator, capacitor, radio, and firmware.

Return to Energy Selection with the maths unpacked →