Relay Coil and Contact Calculation Audit

Relay Coil and Contact Calculation Audit

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

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Ada ADA · CALCULATION AUDIT

Relay Coil and Contact Calculation Audit

The chapter switches off a 100 mH relay coil carrying 80 mA and the induced kick reaches 8000 V — enough to punch straight through a transistor rated for 60 V. The same page separates the contacts of a 50 mH vacuum-motor winding at 8 A and derates that inductive load up to a 20 A relay. This audit re-derives each coil, arc, and derating figure and asks whether the identical GPIO command can be electrically safe or unsafe depending only on the driver, flyback path, and contact rating.

Companion to the chapter Relays and Solenoids — every number here comes from that chapter.

— driver current, stored energy, arc stress, and derating, ~4 minutes

A relay or solenoid is a magnetic energy device before it is a software output. The mathematics checks current, energy, and contact stress; the physics explains why the driver, clamp, and relay rating must be selected from the switch-off case, not the GPIO command.

See the relationship before changing it

The figure reads from left to right. The blue card is coil current before switch-off. The middle card applies this page's rule. The green card is unclamped inductive kick. 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 coil current before switch-off, so the numeric fixture does not switch without explanation.

Coil current before switch-off changes unclamped inductive kick An input card leads through the rule kick voltage = 0.100 H x current / 0.000001 s to the unclamped inductive kick result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. More coil current stores more magnetic energy and raises the ideal unclamped kick for the same switch time.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 0.08 A.

  2. 2

    Name the relationship. kick voltage = 0.100 H x current / 0.000001 s

  3. 3

    Substitute with units. 0.100 H x 0.080 A / 0.000001 s = 8,000 V

  4. 4

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

Predict, then change coil current before switch-off

Try Predict the direction of kick voltage = 0.100 H x current / 0.000001 s. Test another coil current before switch-off, then compare unclamped inductive kick.

0.08 A
Chapter baseline
Unclamped inductive kick

Observe More coil current stores more magnetic energy and raises the ideal unclamped kick for the same switch time. Reset coil current before switch-off to 0.08 and compare unclamped inductive kick.

Explain More coil current stores more magnetic energy and raises the ideal unclamped kick for the same switch time.

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 coil current before switch-off moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only coil current before switch-off. It is not a field prediction.

The chapter's values

Use only values already introduced in this chapter: a 5 V relay coil at 80 mA, the 100 mH flyback example, the 50 mH vacuum-motor winding switched from 8 A in 5 ms, the 120 V supply case, and the inductive-load derating example.

power = voltage x current; stored_energy = 0.5 x L x I^2; kickback_voltage = L x (current / switch_time); required_rating = load_current / derating_factor

The worked checks

Check Arithmetic Review meaning
GPIO overload 5 V x 0.080 A = 0.40 W An 80 mA coil is a power load; the GPIO should command a driver, not source the coil current.
Stored coil energy 0.5 x 0.100 H x 0.080^2 = 0.00032 J = 0.32 mJ The flyback path must absorb the magnetic energy every time the coil is released.
Fast switch-off spike 0.100 H x (0.080 A / 0.000001 s) = 8000 V The chapter's example exceeds a 60 V transistor limit by about 8000 / 60 = 133 times before clamping.
Motor contact spike 0.050 H x (8 A / 0.005 s) = 80 V The 50 mH winding can add an 80 V inductive kick as contacts separate.
Contact voltage and arc power 120 V + 80 V = 200 V; 200 V x 8 A = 1600 W The arc concentrates kilowatt-scale instantaneous power at a tiny contact area.
Inductive derating 8 A / 0.4 = 20 A The 8 A inductive load needs about a 20 A relay under the chapter's 0.4 derating example.

Audit conclusion: the same command can be electrically safe or unsafe depending on the driver, flyback path, load type, and derated contact rating. Record the switch-off path and the load class before approving the relay.

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

TryRun Check audit for the 100 mH relay coil carrying 80 mA with a 1 microsecond current collapse.
ObserveThe unclamped kick reaches an ideal 8000 V, far beyond the contact-load row's derated current rating.
ExplainInductance resists the 1 microsecond current change, so a clamp must absorb stored energy before voltage or contact arcing becomes destructive.
Technical boundaries. The calculation assumes linear inductance and an imposed one-microsecond current collapse. It does not model diode or TVS clamp voltage, coil resistance, parasitic capacitance, arc duration, contact material, motor commutation, or transient thermal stress.
Audit result

The ideal unclamped result exceeds a 60 V transistor rating by orders of magnitude, establishing the need for a characterized flyback path.