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Selecting MCCB and MCB for MDB and SDB

Short answer

Procedure: calculate each circuit load current, select MCB ≥1.25×Ib with appropriate curve (B/C/D), size SDB incomer with simultaneity factor, MDB feeders one size higher for discrimination, MDB incomer matching transformer LT current, verify breaking capacity against fault level, ensure Ib ≤ In ≤ Iz, add RCCB for earth leakage.

With suitable example, explain the detail procedure for selection of MCCB and MCB for MDB and SDB in industrial electrification.

MCCBs (moulded case circuit breakers, about 16–1600 A, adjustable trips, high breaking capacity) are used as MDB incomers and feeders; MCBs (up to 63–125 A, fixed trips, 6–10 kA) are used in SDBs for final circuits. Both are selected by rated current, breaking capacity, number of poles and tripping characteristic.

Procedure

  1. Load current of each final circuit (Ib): I = P/(√3·V·η·pf) for 3-phase, P/(V·pf) for 1-phase.
  2. MCB for each outgoing circuit: In ≥ 1.25 × Ib, next standard rating (6, 10, 16, 20, 25, 32, 40, 50, 63 A). Curve: B for lighting, C for mixed/socket, D for motors and transformers.
  3. SDB incomer: total SDB current × simultaneity factor Ks, × 1.25 → MCB (up to 63 A) or MCCB.
  4. MDB outgoing feeders: MCCB ≥ SDB incomer rating (one size higher for discrimination).
  5. MDB incomer: sum of feeders × Ks of MDB × 1.25 → MCCB/ACB; it should match the transformer LT full-load current (I = kVA·1000/(√3·400)).
  6. Breaking capacity: Icu ≥ prospective fault current at that point (e.g. near a 250 kVA, 4.5 % impedance transformer, Isc ≈ 361/0.045 ≈ 8 kA → 10 kA MCB/MCCB at SDB, 25–36 kA MCCB at MDB).
  7. Poles: SP for single-phase, TP for motors, TPN/4P for 3-phase 4-wire incomers.
  8. Coordination: Ib ≤ In ≤ Iz (cable rating); upstream larger than downstream for selectivity.
  9. Earth leakage: RCCB (30 mA) for socket circuits, 100–300 mA at SDB incomer.

Example (400 V, 3-phase)

  • SDB-1: three motors, 7.5 kW each, η = 0.88, pf = 0.85.
  • SDB-2: 15 kW lighting and sockets, pf 0.9, balanced.
Motor Ib = 7500 / (√3 × 400 × 0.88 × 0.85) = 14.47 A
Motor MCB: 1.25 × 14.47 = 18.09 → 20 A TP, D curve

SDB-1: 3 × 14.47 × 0.9 = 39.08 A
       1.25 × 39.08 = 48.84 → 50 A TP MCB incomer
SDB-2: 15000 / (√3 × 400 × 0.9) = 24.06 A
       1.25 × 24.06 = 30.07 → 32 A TPN MCB incomer

MDB feeders: 63 A MCCB (to SDB-1), 40 A MCCB (to SDB-2)
MDB demand  = (39.08 + 24.06) × 0.9 = 56.82 A
MDB incomer = 1.25 × 56.82 = 71.02 → 80 A TPN MCCB, 25 kA
LocationDeviceRating
SDB-1 outgoingsTP MCB, D curve3 × 20 A
SDB-1 incomerTP MCB50 A
SDB-2 incomerTPN MCB + RCCB32 A
MDB feedersMCCB63 A, 40 A
MDB incomerTPN MCCB80 A, 25 kA

✓ Answer checked by Claude Opus 5.5 (high effort)

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