Transformer sizing and protection

Transformers appear on licensing exams in three ways: choosing the kVA rating for a load, calculating the rated currents on each side, and applying the Code's rules for the overcurrent protection and conductors on the primary and secondary. All three depend on the formulas from lesson 2, and the protection rules have a structure that is easy to apply once you see it. The rules for transformers are in Section 26 of the CE Code, and for dry-type transformers of 750 V and less the key rules are Rule 26-254 for overcurrent protection and Rule 26-256 for conductor sizing.

The running example is the Gaetz Avenue strip mall. The mall is fed at 347/600 V three-phase, which suits its rooftop units and lighting, but the dental clinic needs 120/208 V for its receptacles, equipment and most of its lighting. Parkland Valley installed a dry-type step-down transformer in the clinic's electrical room, fed from the mall's 600 V distribution, with a 120/208 V panel on the secondary. As always, the Code's wording is summarised, not quoted, and you must check the details in your own copy.

Choosing the kVA rating

A transformer is rated by its apparent power in kVA, because it must carry the current regardless of the power factor of the load. To choose a rating, find the load in kVA, add any allowance for growth the design calls for, and choose the next standard size up. Common three-phase dry-type ratings are 15, 30, 45, 75, 112.5, 150, 225 and 300 kVA. The clinic's calculated load from lesson 5 was about 61.5 kW; with the equipment power factors it is about 62 kVA, so the next standard size is 75 kVA, which also leaves some room for future equipment.

Rated currents

The rated current on each side follows directly from the kVA rating and the voltage. For a three-phase transformer, divide by the line-to-line voltage times the square root of three; for a single-phase transformer, divide by the voltage. Write both currents down before you touch the protection rules, because every rule is expressed as a share of them.

The ratio of the two currents is the inverse of the voltage ratio: 600 / 208 is 2.88, and 208.2 / 72.2 is also 2.88. That is a quick check that both numbers are right.

How the protection rules are structured

Rule 26-254 gives two main ways to protect a dry-type transformer of 750 V or less. In the first, you protect the primary only, with an overcurrent device rated at not more than 125% of the rated primary current, and size the secondary conductors for at least 125% of the rated secondary current. The secondary then needs no separate overcurrent device for the transformer's sake. In the second, you protect the secondary with a device rated at not more than 125% of the rated secondary current, which allows the primary device to be much larger, up to 300% of the rated primary current. The larger primary device is useful because it rides through the high inrush current a transformer draws when it is energised.

Where 125% does not correspond to a standard overcurrent device rating, the rule permits the next higher standard rating in the cases it specifies. The rule also has conditions on the arrangement, including a limit on the combined length of primary and secondary conductors in some cases where the primary protection is above 125%. Those details are exam material; read them in your Code and add them to your lookup sheet.

Method 1 at the clinic

Method 2 at the clinic

Why does the second method allow such a large primary device? Because the secondary device now protects the transformer against overload, and the primary device's job is mainly to protect against faults in the transformer and the primary conductors. A device rated at 125% of primary current can trip on the inrush current when the transformer is switched on, which can be several times the rated current for a few cycles. Raising the primary device lets the transformer energise reliably.

Sizing the conductors

Rule 26-256 requires the primary and secondary conductors of a dry-type transformer to have an ampacity of at least 125% of the rated primary and secondary currents. The rule also allows conductors to be sized for the calculated demand load instead, in which case they must be protected according to the conductor protection rules in Section 14. For exam questions, assume the 125% method unless the question gives a demand load and asks you to use it.

Single-phase transformers

The same rules apply to single-phase transformers, with single-phase currents. At Hartley Farms, Parkland Valley installed a 25 kVA single-phase transformer to step 600 V down to 120/240 V for the shop's receptacles and lighting, fed from two lines of the farm's three-phase supply. The rated currents are 25 000 / 600 = 41.7 A on the primary and 25 000 / 240 = 104.2 A on the secondary.

Reading the question

Transformer protection questions are easy to get wrong for one reason: they are easy to read wrong. The question may tell you there is a secondary main breaker, which points to the second method, or it may say the transformer feeds a panel through a short tap with no secondary device, which points to the first. It may ask for the maximum primary device, the minimum conductor ampacity, or the rating you would actually choose, and those are three different answers. Denise's rule is to underline three things before calculating: whether there is secondary protection, which side the question asks about, and whether it wants a maximum, a minimum or a choice.

Marcus adds one from his industrial work: check whether the voltages are line-to-line. A question about a 600 V to 120/240 V single-phase transformer fed from two lines of a 347/600 V system uses 600 V for the primary current, not 347 V. A question about a transformer with a 347 V primary connected line to neutral uses 347 V. Using the wrong one changes the primary current by a factor of 1.73 and every answer after it.

Impedance and fault current

A transformer's nameplate impedance, given as a share of rated voltage, limits the fault current it can deliver on its secondary. A short circuit right at the secondary terminals draws the rated current divided by the per-unit impedance, assuming an infinite supply on the primary. That figure is used to check that the secondary panel and breakers have a high enough interrupting rating, which is a Code requirement for all overcurrent devices.

So the clinic's 120/208 V panel needs breakers with an interrupting rating comfortably above about 6 kA, which standard 10 kA breakers provide. Larger transformers, or ones with low impedance, can deliver much more, and the panel ratings must follow.

Other transformer considerations

Transformers feeding many electronic loads may need a K-factor rating to handle the extra heating from harmonic currents, the issue raised for neutrals in lesson 2. Transformers generate heat and need the ventilation and clearances their installation instructions and Section 26 require. Grounding and bonding of the secondary as a separately derived system is covered in MELC-104. For this course, the calculations are the focus: kVA, rated currents, protection and conductors.

Autotransformers, including the small buck-boost transformers used to correct a supply that is a little high or low, are a special case. Their windings are connected electrically between primary and secondary, so they are not separately derived systems, and Section 26 has its own requirements for them, including limits on where they may be used. A buck-boost transformer rated a few kVA can adjust the voltage of a much larger load, because only the buck or boost portion passes through the transformer's own windings. That surprising sizing is a favourite short-answer exam question, and the arithmetic is simple: the transformer's kVA is the load current times the buck or boost voltage, not the load voltage. Boosting a 208 V supply by 32 V for a 50 A load needs a transformer of only about 50 × 32 = 1.6 kVA per phase.

import math

def rated_currents(kva, v_pri, v_sec, phases=3):
    k = math.sqrt(3) if phases == 3 else 1
    return kva * 1000 / (k * v_pri), kva * 1000 / (k * v_sec)

def protection(kva, v_pri, v_sec, phases=3):
    ip, isec = rated_currents(kva, v_pri, v_sec, phases)
    return {
        'rated primary A': round(ip, 1),
        'rated secondary A': round(isec, 1),
        'primary only: max primary device A (125%)': round(1.25 * ip, 1),
        'with secondary device: max primary A (300%)': round(3.0 * ip, 1),
        'max secondary device A (125%)': round(1.25 * isec, 1),
        'min conductor ampacity pri / sec': (round(1.25 * ip, 1), round(1.25 * isec, 1)),
    }

for k, v in protection(75, 600, 208).items():
    print(f'{k:45s} {v}')

rated primary A                               72.2
rated secondary A                             208.2
primary only: max primary device A (125%)     90.2
with secondary device: max primary A (300%)   216.5
max secondary device A (125%)                 260.2
min conductor ampacity pri / sec              (90.2, 260.2)

Change the transformer to 112.5 kVA or to single-phase 25 kVA at 600 to 240 V and compare with the worked examples.