Industrial Transformers: Connections, Fault Current, and Protection

Every plant depends on transformers that nobody looks at until one fails. A senior electrician reads a transformer differently from a helper. The numbers on the nameplate decide how much fault current the panel downstream must survive, what the secondary conductors can be, and whether a replacement unit is really a like-for-like swap. This covers the three-phase connections, the math, the code limits and the checks that go with that responsibility.

What You Will Be Able to Do

• Tell line current from winding current in delta and wye connections.
• Set a tap in the right direction to correct secondary voltage.
• Estimate available fault current from kVA and percent impedance.
• Size transformer overcurrent protection under NEC 450.3(B).
• Apply the secondary conductor and bonding rules to a separately derived system.
• Check a replacement transformer before it is installed.

Warning: Transformer work is done with both sides isolated, locked out and proven dead, by qualified people. A transformer can backfeed, so voltage on the secondary puts full voltage on the primary. Readings on energized equipment follow your energized work program.

Section numbers follow recent National Electrical Code (NEC) editions. Your state or city may enforce a different edition, with changes. The adopted edition is the rule, and engineering decisions on a plant system belong to the engineer of record.

Delta and Wye Windings

Two facts drive all three-phase transformer math.

• In a wye connection, line current equals winding current, and line voltage is 1.732 times winding voltage.
• In a delta connection, line voltage equals winding voltage, and line current is 1.732 times winding current.

That is why a winding can fail and leave you with readings that look wrong. You are measuring the line, and the winding sees something else.

Common Connections

A delta primary has no neutral, so you do not pull one to it. A delta-wye also shifts the secondary 30 degrees from the primary. Two transformers can run in parallel only if their voltage ratio, phase shift and polarity match and their impedances are close, so a delta-wye cannot parallel a delta-delta.

Worked Example: Line and Winding Currents

A 150 kVA delta-wye transformer steps 480 volts down to 208Y/120. A kVA, or kilovolt-ampere, is the unit for transformer size.

1. Primary line current: 150,000 / (480 x 1.732) = 180 amps.
2. Primary winding current, since the primary is delta: 180 / 1.732 = 104 amps.
3. Secondary line current: 150,000 / (208 x 1.732) = 416 amps.
4. Secondary winding current, since the secondary is wye: also 416 amps.
5. Secondary winding voltage: 208 / 1.732 = 120 volts.

If a clamp meter on the feeder shows 180 amps, each delta winding is carrying only 104.

Setting Taps

Taps change the turns ratio in small steps, usually 2.5 percent. They correct a supply that sits steadily high or low. They do not fix a sag or a load swing.

The direction trips people up, because you change the primary tap to match the actual supply.

Worked Example: Correcting Low Voltage

A plant's 480-volt supply actually measures 456 volts. The transformer is on its 480-volt tap, and the secondary reads low.

1. Secondary now: 456 x (208 / 480) = 198 volts.
2. Move to the tap marked 456 volts, which is 5 percent below nominal.
3. Secondary after the change: 456 x (208 / 456) = 208 volts.

A lower primary tap raises the secondary voltage. Match the tap to what the supply actually delivers under load, and measure at more than one time of day. Taps are changed with the transformer de-energized, locked out and proven dead. Change all three phases the same way.

Percent Impedance and Fault Current

The %Z on the nameplate tells you how much the transformer limits fault current. A low %Z lets more current through.

The simple calculation assumes the source behind the transformer can supply unlimited current, so it gives a worst-case figure. It is called the infinite bus method.

Maximum fault current = full-load current / (%Z / 100)

Worked Example: Fault Current at the Secondary

A 1,000 kVA, 480-volt transformer has a %Z of 5.75.

1. Full-load current: 1,000,000 / (480 x 1.732) = 1,203 amps.
2. Divide by 0.0575: about 20,900 amps.

Running motors add a little more for the first few cycles, and the cable between the transformer and the panel reduces it. Real studies include both. This quick figure tells you whether the equipment is in the right range.

Why a Replacement Changes Everything

Suppose that same transformer fails, and the replacement has a %Z of 4.0.

1. 1,203 / 0.04 = about 30,100 amps.
2. A panel with 22,000-amp breakers was fine before, and it is now underrated.
3. The arc flash labels are no longer valid, because fault current and clearing times have changed.

NEC 110.9 requires interrupting devices to be rated for the current available at their line terminals. A replacement with the same kVA and voltage is not automatically the same transformer.

Protecting the Transformer

NEC 450.3(B) covers transformers up to 1,000 volts. It sets the largest device you may use, as a percent of rated current. There are two ways to protect it.

Where 125 percent falls between standard sizes, a table note lets you use the next higher size. That note does not apply to the 250 percent limit, so you round down. Smaller currents have higher percentages, so read the table for a control transformer.

Worked Example: A 150 kVA Unit

Using the same 480 to 208Y/120 transformer:

Primary only: 180.4 x 1.25 = 225.5 amps. The next standard size is 250 amps.

Primary and secondary: primary maximum 180.4 x 2.50 = 451 amps, so 450 amps. Secondary maximum 416.4 x 1.25 = 520.5 amps, so the next standard size, 600 amps, is the ceiling.

These protect the transformer. They do not protect the conductors. Conductors have their own rule, and both must pass.

Secondary Conductors

Transformer secondary conductors are a tap rule of their own, in NEC 240.21(C). On a wye secondary, the primary device cannot protect them, since the current ratio does not carry through a four-wire secondary. That protection is allowed only for a single-phase two-wire secondary or a delta-delta three-wire secondary.

Each rule has more conditions on routing and termination, so read them in full. The next-size-up allowance of 240.4(B) does not apply here. The conductors must be at least as large as the device they end at.

Worked Example: The 25-Foot Check

A 250-amp primary device feeds the 480 to 208 volt transformer, and the secondary runs 20 feet to a 450-amp main.

1. One-third of 250 is 83.3 amps.
2. Voltage ratio: 480 / 208 = 2.31.
3. 83.3 x 2.31 = 192 amps minimum from this rule.
4. The conductors must also be good for the 450-amp device they end at.

The larger figure wins, so the conductors need an ampacity of at least 450 amps.

Grounding a Separately Derived System

A transformer secondary has no direct connection to the supply conductors, so it is a separately derived system. You have to create its neutral-to-ground bond. NEC 250.30 requires a system bonding jumper in one place: at the transformer, or at the first disconnect, or between them. Not both.

With a bond in two places, neutral current splits and flows on the conduit, the equipment grounding conductor (EGC) and the building steel. Symptoms include a clamp reading on a grounding conductor, stray voltage and interference on sensitive equipment. A grounding electrode conductor also has to connect at the same point as the bond.

To check a suspect system with the power off, lift the bond at one location and test whether neutral and ground are still connected. Do this only under your plant's procedure.

Heat, Load and Harmonics

Heat is what ends a transformer's life. The main causes are overload, blocked ventilation, high ambient temperature, and distorted current from electronic loads.

Nonlinear loads such as drives, computers and LED drivers add heating that a clamp reading of load does not show. A K-rated transformer is built to survive this heat. It does not remove the harmonics. A unit that runs hot at moderate load is telling you to measure true RMS current and harmonic content, not to add a fan.

Failure Clues

Energizing a transformer draws a brief inrush many times its rated current, which is why the code allows primary devices above full-load current. If a fuse blows on a unit that has run for years, suspect the transformer first. A qualified tester can use a turns ratio test, winding resistance test and insulation test to prove it.

Checking a Replacement

Before a new or spare unit goes in, compare it against the old nameplate:

1. kVA, primary and secondary voltage.
2. Connection type and phase shift.
3. %Z, and the fault current it creates.
4. Tap positions.
5. Temperature rise and K rating.
6. Physical clearance and ventilation.

If the %Z or kVA is different, tell whoever owns the short-circuit and arc flash study before you energize. That is a planning job, not a nameplate swap.

Common Mistakes

Check Your Understanding

1. A 300 kVA, 480-volt delta primary draws full load. What is the current in each delta winding?

Line current is 300,000 / (480 x 1.732) = 361 amps. Winding current is 361 / 1.732 = 208 amps.

2. A 500 kVA, 208-volt secondary has a %Z of 4.5. What is the worst-case fault current?

500,000 / (208 x 1.732) = 1,388 amps. 1,388 / 0.045 = about 30,800 amps.

3. Your secondary reads steadily high. Which way do you move the primary tap?

To a higher primary tap, which lowers the secondary. First confirm that the supply is steadily high, and that this is not a short event.

4. Under primary and secondary protection, a 75 kVA, 480-volt unit has a 90-amp primary. What is the largest primary device?

90 x 2.50 = 225 amps. The 250 percent limit has no next-size-up note, and 225 is a standard size.

5. A clamp around the bonding jumper at a panel fed by a transformer reads 18 amps, and another clamp at the transformer's bond reads 12 amps. What is the likely problem?

The neutral is bonded to ground at both points, so neutral current is splitting between the neutral and the grounding paths. Find which bond is the extra one, and remove it under your procedure.

Key Takeaways

• In a delta, line current is 1.732 times winding current. In a wye, line voltage is 1.732 times winding voltage.
• A lower primary tap raises the secondary voltage.
• Fault current is full-load current divided by %Z as a decimal, and a lower %Z means more fault current.
• NEC 450.3 protects the transformer. Article 240 protects the conductors. Both must pass.
• Bond a separately derived system in one place only.
• A replacement with a different %Z can invalidate panels and arc flash labels.