Parallel Operation of Transformers – Conditions, Load Sharing and Calculation

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Running two or more transformers in parallel is the standard way to add capacity to an existing installation, to provide redundancy, or to improve efficiency by switching units in and out with the load. It works reliably — but only when a specific set of conditions is met. Ignoring any one of them produces circulating currents, uneven loading, or in the worst case, damage to both machines.

The Four Conditions

Condition Requirement Consequence if Violated
Vector group Must be identical, or compatible after reconnection Large circulating current even at no load; potential damage within seconds
Voltage ratio Must match, ideally within 0.5 % Circulating current proportional to the voltage difference, present at all times
Impedance voltage Should match within about 10 % Uneven load sharing; one unit overloads before the other reaches rated capacity
Phase sequence Must be identical Effectively a short circuit between phases on closing
Absolute vs practical: vector group and phase sequence are absolute — a mismatch makes parallel operation impossible. Voltage ratio and impedance are practical limits: small differences are tolerable, but they degrade performance in proportion to the mismatch.

Vector Group Compatibility

The vector group determines the phase relationship between HV and LV windings. Two transformers with different clock numbers produce LV voltages that are out of phase with each other, and connecting them in parallel places that phase difference across the two windings.

A 30° phase difference between two 400 V systems creates a voltage difference of roughly 207 V across the connection — driven only by the transformer impedances, which are deliberately low. The resulting current is limited only by the sum of the two impedances and can exceed the short-circuit rating.

Combination Result
Dyn11 + Dyn11 Permitted directly
Dyn11 + Yd11 Permitted — same clock, different connection
Dyn11 + Dyn5 Only after reversing two phases on one unit (shifts clock by 6)
Yyn0 + Yyn6 Only after reversing all three phases on one unit
Dyn11 + Dyn1 Not possible — 60° difference cannot be corrected by reconnection
Dyn11 + Yyn0 Not possible — 30° difference cannot be corrected

The full notation and clock convention are covered in Transformer Vector Groups Explained.

Load Sharing Between Transformers

When two transformers of different rating operate in parallel, they share load in inverse proportion to their impedances — not in proportion to their kVA ratings. The unit with lower impedance takes a larger share.

S1 / S2 = (kVA1 / Uk1) / (kVA2 / Uk2)
S = load carried by each unit  •  kVA = rated power  •  Uk = impedance voltage (%)

This is why matching impedances matters. Two transformers of the same rating but different impedance will not share load equally, and the total usable capacity is less than the sum of the two ratings.

Worked Example — Matched Impedances

Two units in parallel: 1000 kVA at 6 % and 630 kVA at 6 %, total connected load 1400 kVA.

Ratio of shares: (1000 / 6) : (630 / 6) = 166.7 : 105 = 1.587 : 1

Unit 1 carries: 1400 × 166.7 / 271.7 = 859 kVA (86 % of rating)

Unit 2 carries: 1400 × 105 / 271.7 = 541 kVA (86 % of rating)

Both units loaded equally in percentage terms — the ideal case.

Worked Example — Mismatched Impedances

Now the same two units with their standard impedance values — 1000 kVA at 6 % and 630 kVA at 4 %, as given in the table above.

Ratio of shares: (1000 / 6) : (630 / 4) = 166.7 : 157.5 = 1.058 : 1

Unit 1 carries: 1400 × 166.7 / 324.2 = 720 kVA (72 % of rating)

Unit 2 carries: 1400 × 157.5 / 324.2 = 680 kVA (108 % of rating)

The smaller unit is overloaded by 8 % while the larger runs at 72 %.

To keep the 630 kVA unit within its rating, total load would have to be limited to roughly 1300 kVA — meaning 330 kVA of installed capacity cannot be used. This is why transformers intended for parallel operation are often specified with matched impedance rather than the standard value for their rating.

Circulating Current from Voltage Ratio Mismatch

If the two transformers produce slightly different secondary voltages — because of different turns ratios, or because they are on different tap positions — a current circulates between them even with no external load.

Ic = ΔU / (Z1 + Z2)
Ic = circulating current  •  ΔU = voltage difference  •  Z = transformer impedance

Because transformer impedances are low by design, even a small voltage difference produces a significant current. Two 1000 kVA units at 6 % impedance with a 1 % voltage difference circulate roughly 8 % of rated current continuously — current that produces losses and heating without delivering any useful power.

Common cause: transformers left on different tap positions after commissioning. Always verify that paralleled units are on the same tap before closing the coupling.

Fault Level with Units in Parallel

Parallel operation increases the available short-circuit current at the common busbar, because both transformers contribute to a downstream fault. For two identical units the fault level approximately doubles.

Configuration Fault Current at 400 V
1 × 1000 kVA at 6 % 24.1 kA
2 × 1000 kVA at 6 % in parallel 48.2 kA
3 × 1000 kVA at 6 % in parallel 72.2 kA
Design implication: switchgear breaking capacity must be selected for the maximum number of transformers that can be paralleled simultaneously, not for a single unit. See

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Established in 1992, DATSAN combines over 30 years of engineering experience with modern manufacturing and testing capabilities. Our products are designed in accordance with IEC 60076, ANSI C57 and relevant international standards, with Ecodesign-compliant options available upon request.

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