7.2 The Ib ≤ In ≤ Iz Coordination Rule

The Ib ≤ In ≤ Iz Coordination Rule

Quick Answer: For ordinary overload coordination taught with AS/NZS 3000 / AS/NZS 3008: Ib ≤ In ≤ Iz. Ib = design current of the circuit, In = rated current of the protective device, Iz = current-carrying capacity of the conductors after derating. The left inequality stops chronic nuisance tripping on normal load; the right inequality stops the cable cooking under currents the device still allows.

The Three Currents You Must Name

Every cable-and-protection question on the capstone eventually collapses to three numbers. Write them on the page before you touch a table.

Iz is not the raw catalogue CCC for a cable lying on a cool bench. Chapter 4 taught derating; this chapter uses the result. If grouping and roof insulation cut a tabulated 32 A down to 22 A, then Iz = 22 A for coordination — not 32 A.

Why Ib ≤ In

The protective device must be able to carry the intended design current continuously without operating. If In < Ib, the breaker or fuse is smaller than the load you claim the circuit will supply. Outcomes:

• Nuisance tripping or fuse melting in normal service.
• Pressure to “upsize the breaker” without revisiting cable size — which creates the next inequality’s trap.
• A design that was never honest about load.

So In is chosen at least equal to Ib (and in practice often the next standard rating above Ib, subject to other rules). Lighting and socket circuits, fixed appliances and motor circuits all start from a clear Ib before In is selected.

Note: motor starting and inrush are handled with curve type and sometimes dedicated motor protection (Section 7.3), not by violating Ib ≤ In for continuous rating. A 10 A design current still wants In ≥ 10 A; a Type C or D characteristic may be chosen so start current does not instantaneous-trip.

Why In ≤ Iz

This is the inequality candidates forget under time pressure.

The protective device’s In is the continuous current it is designed to carry indefinitely without operating. Conductors on the load side (and generally the protected run) must be able to carry at least that continuous current without exceeding their temperature limits. If Iz < In:

• Currents between Iz and In can flow for a long time without the device treating them as overload.
• The cable is the weak thermal link: insulation temperature rises above the design limit while the breaker “sees” a current it considers acceptable.
• You have not provided effective overload protection of the conductors relative to their installed capacity.

Hence In ≤ Iz: choose a cable (or installation method) so that after derating, capacity is at least the protective device rating used for that coordination check.

Together: Ib ≤ In ≤ Iz.

The Classic Trap: “Breaker Matches the Load”

Exam and site story:

1. Load calculation gives Ib = 18 A.
2. Candidate selects a 20 A MCB (In = 20 A) — so far Ib ≤ In looks fine.
3. Candidate grabs 2.5 mm² because “that’s what we always use,” without checking the installed method.
4. After grouping and thermal insulation, Iz = 17 A.
5. Now In (20 A) > Iz (17 A) — coordination fails.

The trap was sizing the breaker to the load only. Correct practice:

1. Establish Ib.
2. Select In ≥ Ib (standard rating).
3. Select cable / method so Iz ≥ In (read AS/NZS 3008, apply derating).
4. Still check voltage drop, fault withstand and disconnection (Chapters 4–5 themes).

If Iz cannot reach In, increase conductor size, improve installation method/grouping, or redesign — do not pretend derating does not apply.

Worked Teaching Example A — Final Subcircuit

Given (illustrative numbers — verify live AS/NZS 3008 tables on assessment day):

• Single-phase final subcircuit, Ib = 16 A continuous design current.
• Proposed MCB In = 20 A.
• Copper thermoplastic twin-and-earth, clipped direct, no extra derating → suppose tabulated CCC for 2.5 mm² ≈ 27 A (illustrative) → Iz ≈ 27 A.

Check: 16 ≤ 20 ≤ 27 → passes continuous coordination. Still verify voltage drop and short-circuit withstand before locking the design.

Change only the installation method: same 2.5 mm² in a hot, grouped roof-space route. After factors, Iz = 18 A. Now 16 ≤ 20 ≤ 18 fails on In ≤ Iz. Upsize to a size whose derated Iz ≥ 20 A, or revise enclosure/grouping. The load did not change — Iz did.

Worked Teaching Example B — Refusing to Upsize In Blindly

Ib = 28 A. Someone proposes In = 32 A on a cable whose derated Iz = 30 A. Ib ≤ In holds (28 ≤ 32), but In ≤ Iz fails (32 ≰ 30). Illegal “fix”: leave the 32 A breaker and hope. Legal paths: larger cable / better method so Iz ≥ 32 A, or a redesign that uses a lower In only if Ib and application still allow (usually you do not drop In below a proper Ib). Never “fix” coordination by ignoring Iz.

Relationship to Maximum Demand and Submains

For consumer mains and submains, Ib is often the diversified maximum demand (Chapter 6) or the design current of that distribution run. The same rule applies: main switch / protective device In and mains conductor Iz must satisfy Ib ≤ In ≤ Iz (alongside distributor rules and voltage-drop limits). Large MCCBs do not exempt you from the inequality; they only change how In is labelled on the trip unit.

What the Rule Does Not Replace

Ib ≤ In ≤ Iz is continuous overload coordination. It does not by itself prove:

• Short-circuit breaking capacity (Icu/Ics vs prospective fault current).
• Instantaneous curve suitability for motor starting.
• Discrimination between upstream and downstream devices.
• Earth-fault loop impedance / disconnection times.
• Voltage drop ≤ 5% end-to-end.

Pass the inequality, then keep walking the selection checklist.

Capstone Open-Book Habits

1. Write Ib, In, Iz with units (A) and state the derating assumptions for Iz.
2. Show the two comparisons explicitly — markers reward visible logic.
3. Quote that Iz comes from AS/NZS 3008 after factors, not from memory of “2.5 mm² = 20 A always.”
4. If a question gives only load and breaker rating, ask yourself what is missing: Iz.

Bridge Forward

With continuous coordination understood, Section 7.3 explains why two breakers of equal In behave differently on start current and fault current (curves), and whether the device can interrupt the fault at all (breaking capacity).