Transformer Impedance (%Z) Explained

Percent impedance (%Z) is the percentage of rated primary voltage that must be applied to a transformer, with its secondary short-circuited, to drive rated current through the windings. A transformer marked 5.75% needs only 5.75% of its rated voltage to push full-load current through a dead short. That is exactly why it will pass a very large current if a real short happens at full voltage.

The value comes from the factory short-circuit (impedance) test and is stamped on the nameplate, usually at the transformer's reference temperature. It is one of the few figures that has to be specified deliberately. Left to the supplier, it gets chosen for the transformer's convenience, not the system's.

What %Z controls

A single number does three jobs, and they pull in opposite directions:

Lower impedance makes a stiffer source. That's good for voltage, and bad for everything downstream that has to interrupt or withstand a fault. Choosing %Z means choosing where on that tradeoff the system sits.

Fault current: the worked example

A 1000 kVA transformer, 480 V secondary, nameplate impedance 5.75%:

• Full-load current: 1000 × 1000 ÷ (1.732 × 480) = 1,203 A
• Available fault current: 1,203 ÷ 0.0575 = about 20,900 A, or roughly 21 kA at the secondary terminals
• Same transformer at 4.0% impedance: about 30,100 A, a 44% increase from a change that looks small on paper

These are infinite-source figures. They assume the utility can supply unlimited current into the primary, which over-states the real value. That makes them conservative, and so safe for a first pass. The Available Fault Current Calculator runs this method directly. A full short-circuit study adds the utility source impedance, cable impedance and motor contribution.

The tolerance nobody reads

The nameplate %Z is not exact. IEEE C57.12.00 allows the tested impedance of a two-winding transformer to differ from the specified value by up to ±7.5%. For fault-duty purposes, the low end is the one that matters. A 5.75% specification can legitimately arrive as 5.32%:

• 5.75 × 0.925 = 5.32%
• Fault current rises to about 22,600 A, roughly 8% above the nominal figure

If the switchgear's interrupting rating was chosen with little margin over the nominal fault current, that 8% can matter. Use the minimum impedance within tolerance for equipment duty, or the tested value once the unit's test report is available.

Voltage regulation

Impedance has a resistive part (%R) and a reactive part (%X), related by %Z² = %R² + %X². Their ratio (X/R) matters on its own: a higher X/R means a larger DC offset in the first cycles of a fault, which affects the asymmetrical current that breakers must be rated for. For voltage drop at full load, a common approximation is:

With the 5.75% transformer above, an assumed %R of 1.0 gives %X = √(5.75² − 1.0²) = 5.66%. At 0.85 power factor (sin φ = 0.53), regulation is about 1.0 × 0.85 + 5.66 × 0.53 = 3.8%. The secondary voltage sags almost 4% between no load and full load, before any cable voltage drop is added. The %R value here is an assumption for the example: use the test report's load-loss figures for a real unit.

Motor starting

A motor starting across the line draws locked-rotor current, commonly around six times its full-load current, and the bus dips in proportion to how stiff the source is. The transformer's short-circuit kVA is its rating divided by %Z: 1000 ÷ 0.0575 = 17,391 kVA. A motor with 1,000 kVA of locked-rotor demand then dips the bus by about 1,000 ÷ (1,000 + 17,391) = 5.4%. At 4% impedance the source is 25,000 kVA and the dip falls to about 3.8%. The Motor Starting Voltage Dip Calculator uses the same screening method.

Choosing %Z

Distribution-class transformers typically fall between about 4% and 8%, and large power transformers are often higher. Within a range, the choice follows the system:

• Fault duty is the constraint: if the downstream switchgear is existing, or a lower interrupting rating saves real cost, specify a higher %Z and accept more voltage drop.
• Voltage is the constraint: large motors, long feeders or voltage-sensitive loads argue for a lower %Z, with switchgear rated for the higher fault current that follows.
• Arc flash: lower fault current doesn't always mean lower incident energy. Incident energy also depends on how long the protective device takes to clear, and a lower current can slow it down. Check it in the arc-flash analysis rather than assuming.
• Paralleled transformers: units meant to share load in parallel need closely matched impedance, as well as matching ratios and vector groups. A mismatch makes the lower-impedance unit carry more than its share.

Converting %Z to ohms

Studies sometimes need the impedance in ohms, referred to one side of the transformer:

For the example transformer, referred to the 480 V side: 0.0575 × 480² ÷ 1,000,000 = 0.0132 Ω. Referred to the primary, use the primary voltage instead. The percentage is the same from either side, which is why nameplates state it that way.

Frequently asked questions