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Selecting main panel board and DB rating
Short answer
The rating of an MDB/SDB is based on its incomer MCCB/MCB. List loads, calculate circuit currents, apply simultaneity factors for demand, and select incomer rating ≥1.25× demand; e.g., a 56.82 A demand leads to an 80 A TPN MCCB. Cables are chosen to match or exceed the incomer rating.
Explain the procedure for selection of rating of main panel board and distribution board with suitable example.
The rating of a panel board (MDB) or distribution board (SDB) means the rating of its incomer (MCCB/MCB), busbar and incoming cable. It is fixed from the design current of the loads it feeds, after applying utilization and simultaneity (diversity) factors and a safety margin.
Procedure
- List the loads on each DB: kW, pf, efficiency, single or three phase. Balance single-phase loads on R, Y, B.
- Find the current of each outgoing circuit: I = P/(√3·V·η·cosφ) for 3-phase, I = P/(V·cosφ) for single phase.
- Select outgoing MCBs: In ≥ 1.25 × Ib (next standard size: 6, 10, 16, 20, 25, 32, 40, 50, 63 A); C curve for general loads, D curve for motors.
- SDB demand: add circuit currents and multiply by the simultaneity factor Ks (0.9 for 2–3 circuits, 0.8 for 4–5, 0.7 for 6–9, 0.6 for 10 or more).
- SDB incomer: In ≥ 1.25 × SDB demand; MCB up to 63/100 A, MCCB above.
- MDB demand: add the SDB demands, apply Ks of the MDB, add spare for future.
- MDB incomer: MCCB/ACB ≥ 1.25 × MDB demand; check that it does not exceed the transformer LT full-load current, and its breaking capacity ≥ prospective fault current.
- Busbar: rating ≥ incomer rating (copper about 1.2–1.6 A/mm²). Keep 20 % spare ways.
- Coordination: Ib ≤ In ≤ Iz (cable rating), and upstream device larger than downstream for discrimination.
Example
A factory, 400 V, 3-phase: SDB-1 feeds 3 motors of 7.5 kW (η = 0.88, pf = 0.85); SDB-2 feeds 15 kW balanced lighting and sockets at pf 0.9.
Motor current = 7500 / (√3 × 400 × 0.88 × 0.85)
= 14.47 A
Motor MCB = 1.25 × 14.47 = 18.09 A → 20 A (D curve)
SDB-1 demand = 3 × 14.47 × Ks 0.9 = 39.08 A
SDB-1 incomer = 1.25 × 39.08 = 48.84 A → 50 A TP MCB
SDB-2 current = 15000 / (√3 × 400 × 0.9) = 24.06 A
SDB-2 incomer = 1.25 × 24.06 = 30.07 A → 32 A TPN MCB
MDB demand = (39.08 + 24.06) × Ks 0.9 = 56.82 A
MDB incomer = 1.25 × 56.82 = 71.02 A → 80 A TPN MCCB
| Board | Demand (A) | Incomer | Outgoings |
|---|---|---|---|
| SDB-1 | 39.08 | 50 A TP MCB | 3 × 20 A TP MCB |
| SDB-2 | 24.06 | 32 A TPN MCB | SP MCBs 6–16 A + RCCB |
| MDB | 56.82 | 80 A TPN MCCB | 63 A, 40 A MCCB/MCB + spares |
Cables are then chosen so that their rating is above the protective device rating (e.g. MDB–SDB-1 with 50 A MCB → 10 mm² Cu), and checked for voltage drop.
✓ Answer checked by Claude Opus 5.5 (high effort)
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