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# Transformer Protection – Abstract from NEC

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## NEC, Code 450.4

### Calculate overcurrent Protection on the Primary

According to NEC 450.4, “each transformer 600 volts, nominal, or less shall be protected by an individual overcurrent device installed in series with each ungrounded input conductor.”

Such overcurrent 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 overcurrent 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 overcurrent device rated or set at not more than 300% shall be permitted.

### Example

Decide Size of circuit breaker (overcurrent 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.

123 amps x 1.25 = 112.76 amps

Use 125 amp 3-pole circuit breaker (the next highest fuse/fixed-trip circuit breaker size per NEC 240.6). The overcurrent 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 overcurrent 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 overcurrent device rated or set at not more than 167% of the secondary current shall be permitted.

### Example

Decide size of circuit breaker (overcurrent protection device) is required on the secondary side to protect a 75kva 440v-230v 3ø transformer. We have Calculate the secondary overcurrent 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 overcurrent 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 overcurrent 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 overcurrent devices.

## NEC, Section 110-10

### Circuit Impedance and Other Characteristics

The overcurrent 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 overcurrent protective devices so that they will open fast enough to prevent damage to the electrical components on their load side.

Originally published at Electrical Notes & Articles

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### Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

1. Abozer
Dec 05, 2017

Very good information

2. Salim
Feb 23, 2017

Thank you alot

3. Henry Sánchez
May 12, 2016

Dear Engineer: Thank you for sharing your knowledge to design professionals. It is very valuable material published, and clarifies any doubts that arise.

4. Jerry Solar
May 11, 2015

Thank you for your posting, just one comment: Regarding the overcurrent protection on the secondary side, the next available standard ampere rating for the calculated 235.34 amps will be 250 amperes (as per NEC article 240.6), not 300 amperes.
Thanks,
Jerry

5. papuchanda
Sep 27, 2013

thanx………………….for this excellent article.
why does the voltage level e.g..11kv,33kv, 132kv,220kv,400kv are maintained ?

6. shivakumar5065
Jun 16, 2013

Thanks for your posting and sharing the knowledge, it is very helpful to me and others….
My question is what will happen if we use fuse more than what you have mentioned (ex:125% or 250%), i am curious to know about the damage (if any) to the euipment

7. JigKhatri
Oct 08, 2012

your all articles are excellent dear. thank you for posting .

“The current (amps) is more than 9 amps so use 125% rating.

123 amps x 1.25 = 112.76 amps”

But it should be :
98.41 amps x 1.25= 123 amps
So select the next available size of 125 A breaker.

Keep posting and share knowledge. doing wonderful job here.

Thanks

Jignesh