Math Bridge: Pump surge and relay flyback

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Math BridgePrototypingStruggle-friendly runway

Why can a pump reset a board before firmware gets a vote?

Follow current through a real shared path, then see why a regulator loses control and an unclamped relay coil fights switch-off.

Voltage Vera, the prototyping guideVoltage Vera guides
The one targetTurn a motor reset into a rail-voltage ledger.
The chapter caseA 5 V relay, a 1.50 A pump surge, and a 2.00 Ω shared path.
What it buys youA measurement order that separates power faults from firmware faults.

A field team faces an unresolved physical question: Why can a pump reset a board before firmware gets a vote? They must answer it before changing relay current 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 relay current. The middle card applies this page's relationship. The green card is pump sag. 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.

Relay current changes pump sag An input card leads through the page relationship to the pump sag result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The board resets when the electrical state leaves its allowed region; firmware is downstream of that event.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for relay current is 1.5.

  2. 2

    Name the relationship. Icoil=5/71=70.4 mA ΔVpump=1.50x2.00=3.00 V Vrail=5.00-3.00=2.00 V Regulation threshold=3.30+0.30=3.60 V VL=0.3x0.0704/1 us=21,127 V

  3. 3

    Substitute the chapter fixture. Set relay current to 1.5. The page ledger gives pump sag as 3.00 V.

  4. 4

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

Predict, then change relay current

Try Predict the direction of pump sag. Move one control, calculate, then check your prediction.

1.5
Chapter baseline
Pump sag

Observe The board resets when the electrical state leaves its allowed region; firmware is downstream of that event. Reset the control to 1.5 and compare pump sag.

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

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

1. Start with current and resistance

Current is charge flowing each second. Resistance makes that flow spend voltage. A source wire, connector, breadboard rail, and ground return all contribute resistance even when no resistor symbol appears in the diagram.

Voltage Vera: Measure at the board while the pump starts; an idle reading cannot reveal inrush sag.

2. Name every algebra move

1

Find the coil currentI=V/R.

2

Find the shared-path dropΔV=IinrushRsource.

3

Subtract from the sourceVrail=Vsupply−ΔV.

4

Test regulationCompare Vrail with Vout+Vdropout.

5

Estimate unclamped kickVL≈LI/Δt.

3. Reproduce the chapter case

Icoil=5/71=70.4 mA
ΔVpump=1.50×2.00=3.00 V
Vrail=5.00−3.00=2.00 V
Regulation threshold=3.30+0.30=3.60 V
VL=0.3×0.0704/1 µs=21,127 V

The 2.00 V rail is 1.60 V below the regulator-input threshold and 0.43 V below the named 2.43 V brownout level. The huge flyback number describes an intentionally unclamped teaching estimate, not the voltage across a fitted diode.

4. Try the pump inrush

TryMove the pump inrush while the supply and shared path stay fixed.

Relay current
Pump sag
Shared rail
Regulator threshold
Regulation headroom
Brownout margin
Unclamped flyback
Radio-current difference
Extra radio sag
Combined rail

ObserveEvery extra 0.1 A removes another 0.20 V because the 2.00 Ω path has not changed.

ExplainThe board resets when the electrical state leaves its allowed region; firmware is downstream of that event.

Technical boundaries.

This is a lumped steady-path and unclamped-coil ledger, not a transient circuit simulator.

Inrush
Real pump current changes with load, driver, wiring, and time
Regulator
Dropout and brownout thresholds depend on the exact parts and configuration
Flyback
A diode, TVS, transistor avalanche, capacitance, and parasitics limit the real voltage

Capture the rail and switch node with suitable probes, then verify the fitted protection network.

5. Test the fault in the right order

Measure the 5 V input, 3.3 V rail, pump current, and reset line on one time base. Then separate the pump supply or shorten the return path and repeat the same start event.

6. Record the evidence state

Store the supply, cable, connector, pump load, relay part, probe points, time scale, minimum rail, protection parts, firmware hash, and the single change that removed or preserved the reset.

7. Check yourself

Why does 1.50 A through 2.00 Ω remove 3.00 V?
Answer: Ohm's law gives ΔV=IR=1.50×2.00=3.00 V.
Why is 2.00 V not enough for a 3.3 V regulator?
Answer: The input must exceed 3.3 V by its dropout allowance; this example needs about 3.60 V.
Does the 21,127 V estimate predict the clamped switch voltage?
Answer: No. It shows what the ideal unclamped LI/Δt arithmetic demands and why a real clamp is mandatory.
Honesty boundary.

The arithmetic reproduces the chapter's catalog-typical pump, relay, path, regulator, and radio example.

2.00 V
One shared-path estimate, not a measured board rail
21,127 V
An unclamped idealization, not a safe probe expectation
1.96 V
A stacked teaching result, not a complete power-distribution model

Correct, not complete: this ledger does not diagnose a particular reset without a captured waveform.

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