Abstract of National Electrical Code for Transformer’s Protection
- According to NEC 450.4, “each transformer 600 volts, nominal, or less shall be protected by an individual over current device installed in series with each ungrounded input conductor.
- Such over current device shall be rated or set at not more than 125% of the rated full-load input current of the auto transformer.
- Further, according to NEC Table 450.3(B), if the primary current of the transformer is less than 9 amps, an over current device rated or set at not more than 167% of the primary current shall be permitted. Where the primary current is less than 2 amps, an over current device rated or set at not more than 300% shall be permitted.
- Example: Decide Size of circuit breaker (over current protection device) is required on the primary side to protect a 75kva 440v-230v 3ø transformer.
- 75kva x 1,000 = 75,000va
- 75,000va / (440V x √3) = 98.41 amps.
- The current (amps) is more than 9 amps so use 125% rating.
- 98.41 amps x 1.25 = 123amps
- Use 125amp 3-pole circuit breaker (the next highest fuse/fixed-trip circuit breaker size per NEC 240.6).
- The over current device on the primary side must be sized based on the transformer KVA rating and not sized based on the secondary load to the transformer.
NEC, Code 450.3B:(Calculate over current Protection on the Secondary)
- According to NEC Table 450.3(B), where the secondary current of a transformer is 9 amps or more and 125% of this current does not correspond to a standard rating of a fuse or circuit breaker, the next higher standard rating shall be required. Where the secondary current is less than 9 amps, an over current device rated or set at not more than 167% of the secondary current shall be permitted.
- Example: Decide Size of circuit breaker (over current protection device) is required on the secondary side to protect a 75kva 440v-230v 3ø transformer.
- We have Calculate the secondary over current protection based on the size of the transformer, not the total connected load.
- 75kva x 1,000 = 75,000va
- 75,000va / (230V x √3) = 188.27 amps. (Note: 230V 3ø is calculated)
- The current (amps) is more than 9 amps so use 125% rating.
- 188.27 amps x 1.25 = 235.34 amps.
- Therefore: Use 300amp 3-pole circuit breaker (per NEC 240.6).
NEC, Section 450-3(a):(Transformers over 600 volts, Nominal)
- For primary and secondary protection with a transformer impedance of 6% or less, the primary fuse must not be larger than 300% of primary Full Load Amps (F.L.A.) and the secondary fuse must not be larger than 250% of secondary F.L.A.
NEC, Section 450-3(b):(Transformers over 600 volts, Nominal)
- For primary protection only, the primary fuse must not be larger than 125% of primary F.L.A.
- For primary and secondary protection the primary feeder fuse must not be larger than 250% of primary F.L.A. if the secondary fuse is sized at 125% of secondary F.L.A.
NEC, Section 450-3(b):(Potential (Voltage) Transformer)
- These shall be protected with primary fuses when installed indoors or enclosed
NEC, Section 230-95(Ground-Fault Protection of Equipment).
- This section show that 277/480 volt “wye” only connected services, 1000 amperes and larger, must have ground fault protection in addition to conventional over current protection.
- The ground fault relay (or sensor) must be set to pick up ground faults which are 1200 amperes or more and actuate the main switch or circuit breaker to disconnect all ungrounded conductors of the faulted circuit.
NEC, Section 110-9 – Interrupting Capacity.
- Any device used to protect a low voltage system should be capable of opening all fault currents up to the maximum current available at the terminal of the device.
- Many over current devices, today, are used in circuits that are above their interrupting rating.
- By using properly sized Current Limiting Fuses ahead of these devices, the current can usually be limited to a value lower than the interrupting capacity of the over current devices.
NEC, Section 110-10 – Circuit Impedance and Other Characteristics.
- The over current protective devices, along with the total impedance, the component short-circuit withstand ratings, and other characteristics of the circuit to be protected shall be so selected and coordinated so that the circuit protective devices used to clear a fault will do so without the occurrence of extensive damage to the electrical components of the circuit.
- In order to do this we must select the over current protective devices so that they will open fast enough to prevent damage to the electrical components on their load side.