Math Bridge: Cellular Attach Pulse Sag

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Math BridgeCellular IoTStruggle-friendly runway

How can a 96.5%-full battery still fail an attach?

Keep charge capacity and pulse delivery separate, then turn current and internal resistance into terminal voltage and brownout margin.

Radio Remi, the guideRadio Remi guides
The one targetSeparate remaining mAh from attach-pulse voltage.
The chapter case300 mA attach, 3.6 V cell, 1.5 Ω buffered or 15 Ω bare.
What it buys youCatch brownout risk that an energy ledger cannot see.

A field team faces an unresolved physical question: How can a 96.5%-full battery still fail an attach? They must answer it before changing battery internal resistance 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 battery internal resistance. 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.

Battery internal resistance 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 retained-charge outputs do not move with resistance because self-discharge and pulse sag are different mechanisms. An HLC buffer helps the pulse; antenna improvement can shorten search time but does not lower cell resistance.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for battery internal resistance is 1.5.

  2. 2

    Name the relationship. Vload=Voc-IattachRint; Cretained=C0(1-k)^t

  3. 3

    Substitute the chapter fixture. Set battery internal resistance to 1.5. The page ledger gives loaded voltage as 3.15 V.

  4. 4

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

Predict, then change battery internal resistance

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

1.5
Chapter baseline
Loaded voltage

Observe The retained-charge outputs do not move with resistance because self-discharge and pulse sag are different mechanisms. An HLC buffer helps the pulse; antenna improvement can shorten search time but does not lower cell resistance. Reset the control to 1.5 and compare loaded voltage.

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

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

1. Separate two battery questions

Capacity asks how much charge remains. Pulse delivery asks whether the cell can hold voltage while a large current flows. A pack can answer the first question well and the second badly.

Radio Remi: Search and attach can be the highest-current state even when transmit dominates the energy story.

2. Name the algebra moves

1

Retain chargeCretained=C0(1−k)^years.

2

Apply Ohm's lawVsag=Iattach Rinternal.

3

Subtract the sagVload=Voc−Vsag.

4

Compare with cutoffmargin=Vload−Vcutoff.

3. Read a negative margin

margin = 3.6 V − (0.300 A)(Rint) − 2.7 V

The largest resistance that preserves a 2.7 V cutoff is (3.6−2.7)/0.300=3.00 Ω. Above it, the first-order loaded voltage crosses the stated threshold even if most mAh remain.

4. Try one controlled change

Vload=Voc−IattachRint; Cretained=C0(1−k)^t

TryIncrease only internal resistance from a buffered 1.5 Ω toward a bare-cell 15 Ω case. Current, open-circuit voltage, cutoff, capacity, age, and self-discharge stay fixed.

Attach sag
Loaded voltage
Cutoff margin
Maximum resistance
Retained charge
Retained fraction
Self-discharge loss

ObserveAt 1.50 Ω, sag is 0.45 V and Vload=3.15 V, leaving 0.45 V above cutoff. Five-year self-discharge still leaves 2,896 mAh, or 96.5%. At 15 Ω the first-order sag is 4.50 V: the pulse request exceeds what the 3.6 V source can supply.

ExplainThe retained-charge outputs do not move with resistance because self-discharge and pulse sag are different mechanisms. An HLC buffer helps the pulse; antenna improvement can shorten search time but does not lower cell resistance.

Technical boundaries.

V=Voc−IR is a first-order source model.

Resistance
Changes with chemistry, temperature, age, state of charge, and pulse duration
Attach current
300 mA is catalog-typical, not a module guarantee
Negative voltage
Signals an impossible demand and brownout, not a physical negative terminal voltage

Validate the actual pack and modem with cold, aged pulse testing and captured voltage traces.

5. Reproduce both worked cases

At 1.50 Ω, Vsag=0.300×1.50=0.450 V and Vload=3.6−0.45=3.15 V. At 15.0 Ω, the requested sag is 4.50 V, so attach cannot be sustained. Separately, 3000(0.993)⁵=2,896 mAh: only 104 mAh lost to self-discharge.

6. Carry the evidence forward

Record attach current and duration, search duration, sleep current, state charge, cell and buffer part numbers, internal resistance across temperature and age, minimum modem voltage, antenna/link evidence, and attach retry traces.

7. Check yourself

Why does 96.5% retained charge not prove attach success?
Answer: Capacity measures stored charge; attach success also needs the source to hold voltage during a high-current pulse.
What does the 3.00 Ω threshold mean?
Answer: In this simple 3.6 V, 300 mA, 2.7 V model, it is the largest resistance that leaves zero cutoff margin.
What can a better antenna change?
Answer: It may reduce search or retry time and therefore charge, but it does not repair battery internal resistance.
Honesty boundary.

The page exposes why energy and power gates are both required. It does not qualify a battery-modem pair.

96.5%
Five-year 0.7%/year self-discharge illustration
0.45 V
Buffered-case first-order sag
4.50 V request
Model signal that the bare-cell pulse is infeasible

Release needs measured pulse delivery, full-state energy, RF conditions, temperature, and ageing evidence.

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