Coil and Stall-Current Physics Calculation Audit

Ada re-derives this chapter’s own numbers step by step, at full precision

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ADA · CALCULATION AUDIT

Coil and Stall-Current Physics Calculation Audit

A 12 V motor whose windings measure 2 ohms draws 6 A at stall but settles to 1.5 A once 9 V of back-EMF builds — a 4× inrush — while a 20 mH coil at 0.5 A quietly stores 2.5 mJ. Those figures look reassuringly small on the page. This audit re-derives the stored-energy, stall-current, and back-EMF numbers and asks whether a supply sized for the tidy 1.5 A running current can survive the 6 A startup surge, and whether that stored energy has anywhere safe to go.

Companion to the chapter The Output Side of IoT — every number here comes from that chapter.

— stored magnetic energy, stall current, and back-EMF, ~4 minutes

Protection is not an accessory — it is arithmetic. Three equations decide whether the switch and supply survive: stored coil energy ½LI², Ohm's-law stall current V/R, and the back-EMF that reins it in. Every number below is one this chapter already stated.

See the relationship before changing it

The figure reads from left to right. The blue card is motor winding resistance. The middle card applies this page's rule. The green card is stall current. 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 motor winding resistance, so the numeric fixture does not switch without explanation.

Walk the arrows. More winding resistance lowers ideal stall current while back-EMF remains absent at startup.

Derive the baseline in four named moves

1
Name the input. The chapter baseline is 2 ohm.
2
Name the relationship. stall current = 12 V / winding resistance
3
Substitute with units. 12 V / 2 ohm = 6.00 A
4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change motor winding resistance

Try Predict the direction of stall current = 12 V / winding resistance. Test another motor winding resistance, then compare stall current.

Observe More winding resistance lowers ideal stall current while back-EMF remains absent at startup. Reset motor winding resistance to 2 and compare stall current.

Explain More winding resistance lowers ideal stall current while back-EMF remains absent at startup.

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 motor winding resistance moves here. Field effects named in the technical boundary stay fixed.

1. A coil stores energy in its magnetic field.

With 20 mH (0.020 H) carrying 0.5 A:

E = ½ × L × I² = 0.5 × 0.020 × 0.5² = 0.0025 J = 2.5 mJ

2. Before back-EMF builds, the motor is just its winding resistance.

A 12 V motor with 2 Ω windings, at the instant of startup or a stall, obeys plain Ohm's law:

I_stall = V / R = 12 / 2 = 6 A

3. Once spinning, back-EMF opposes the supply.

At 9 V of back-EMF only the difference drives current, so the surge is several times the running draw:

Physical quantity	Arithmetic shown	Result
Stored coil energy (½LI²)	0.5 × 0.020 × 0.5²	0.0025 J (2.5 mJ)
Stall / startup current (V / R)	12 / 2	6 A surge
Running current (back-EMF 9 V)	(12 − 9) / 2	1.5 A
Inrush ratio (surge ÷ running)	6 / 1.5	4×
Solenoid valve power	12 × 0.5	6 W

What this means for your design: 2.5 mJ sounds negligible until it has nowhere to go — open the switch with no flyback diode, snubber, or TVS and the coil forces that energy into a fast voltage spike that can punch through the driver transistor. And a supply sized for the tidy 1.5 A running current will collapse under the 6 A startup surge, four times higher, browning out the controller. Size the driver and supply to the stall current, and give every inductive load a safe path for its stored energy.

Every number above is taken from the chapter’s own examples and re-derived step by step.

TryPress Check audit after evaluating a 12 V motor with 2 ohm winding resistance and a 20 mH, 0.5 A coil.

ObserveThe motor row moves from 6 A stall to 1.5 A running, while coil storage reads 2.5 mJ.

ExplainBack-EMF reduces settled motor current, but it is absent at startup; supply and clamp sizing must therefore follow transient rather than nominal current.

Technical boundaries. The audit uses fixed resistance, back-EMF, inductance, and current. It does not model winding temperature, motor acceleration, inductance during commutation, driver limiting, supply impedance, clamp voltage, mechanical load, or thermal duration.

Audit result

12 V / 2 ohm = 6 A at stall; with 9 V back-EMF only 3 V remains across the winding, giving 1.5 A.