Fuse melt current calculation method - Database & Sql Blog Articles

Probe current voltage pin 420*4450 head diameter 5.0 over current current and voltage pin

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The fuse current includes two key aspects: the rated current of the fuse tube and the rated current of the melt. These should not be confused, as they serve different purposes in circuit protection.

1. For a single motor start-up, the melt current (IRN) must satisfy IRN ≥ Iq/K (A), where: - IRN is the rated melt current (A) - Iq is the motor starting current (A) - K is a factor Alternatively, IRN = (1.5–2.5) IH (A), where: - IH is the rated motor current (A) For 380V AC systems: IRN = 7PH (A) For 220V AC systems: IRN = 12PH (A) Where PH is the motor rated power (kW).

2. When multiple main motors are involved, the total fuse rating is calculated as: - IRN = (1.5–2.5) IHmax + I(n-1) (A) Where: - IHmax is the maximum motor or group rated current during simultaneous start-up - I(n-1) is the sum of the rated currents of other motors Another formula: IRN = I(nl) + Iqmax / 2.5 (A) Or IRN = K1 [Iqmax – I(nl)] (A), with K1 typically around 0.4, though some sources suggest 0.9–1.

3. For electric lamp circuits, the fuse rating is generally: - IRN = (1.3–1.5) × watt-hour meter rated current (A) This is usually installed on the meter outlet.

4. For branch lighting circuits, the fuse rating is based on the working current of all lights on that branch (A).

5. For control transformers, the fuse rating is: - IRN ≥ PH + 0.1Pq/U2 (A) Where: - PH is the control transformer’s rated capacity (VA) - Pq is the starting capacity of the largest appliance or sum of several appliances (VA) - U2 is the secondary voltage of the control transformer (V)

6. For transformers without control circuits, the fuse rating is: - IRN ≥ 0.4 [Iq + IH(n-1)] (A) Where: - Iq is the starting current of the largest appliance - IH(n-1) is the sum of the rated currents of other appliances

7. For load balancing control circuits, the fuse rating is: - IRN ≥ IH (A) Where IH is the sum of the rated currents of all electrical loads (A).

8. For high-voltage fuses: - If the transformer capacity is below 100kVA, IRN = (2–2.5) × high-side rated current (A) - If the capacity is above 100kVA, IRN = (1.5–2) × high-side rated current (A)

9. For low-voltage side fuses, the rating is set based on the low-voltage side rated current. For example, if the transformer has a rated current of 910A, and the distribution aluminum busbar is 80×8 mm², the allowable current is 1320A. With a temperature correction factor of 0.88, the actual allowable current becomes 1320 × 0.88 = 1160A, which is sufficient for the transformer's rated current.

If the calculated fuse current does not match standard values, choose the nearest one. For example, for an S7-50/10-0.4 transformer, the high-voltage side current is 2.89A. Using formula (13): (2–2.5) × 2.89 = 5.78–8.23A. Standard fuse ratings include 2, 3, 5, 7.5, 10, 15, 20, etc. In this case, a 10A fuse would be appropriate.

Therefore, a 10A high-voltage fuse can be safely used in this scenario.