The answer first: five conditions, one impedance rule
You can parallel two distribution transformers safely only when all five of these hold at once. Skip one and the two units stop sharing load — they start circulating current between themselves, overheat, and can trip protection even at zero external load:
- Same voltage ratio — within ±0.5% (IEC 60076-1 / GB/T 17468).
- Compatible vector group — identical phase displacement, e.g. Dyn11 with Dyn11, never Dyn11 with Yyn0 (a 30° shift acts as a short circuit).
- Impedances within 10% of each other — the tighter the better; this is what decides load sharing.
- Same phase sequence (and phase rotation) on the LV and MV sides.
- Capacity ratio no greater than 3:1 — a 1,000 kVA unit should not run beside a 200 kVA unit.
The one number that governs everything else is impedance voltage (Z%). Load is shared in inverse proportion to Z%: the transformer with the lower impedance carries the larger share, and if the two Z% values differ by more than 10%, the smaller-impedance unit is overloaded long before the pair reaches nameplate capacity. This guide gives the formulas, a worked example, and the circulating-current check you must run before connecting anything in parallel. If you are still choosing the units themselves, start from our Transformer Selection Guide; if the question is the voltage class, see How to Choose the Right Voltage Class.
Condition 1 — Voltage ratio: keep it within ±0.5%
Two transformers with different no-load ratios will fight each other. With no load connected, the higher-ratio unit drives a circulating current through the lower-ratio unit, limited only by the two impedances in series. The circulating current is:
Ic = ΔU% ÷ (Z1% + Z2%) × In
where ΔU% is the no-load voltage difference and In is rated current. A 1% ratio mismatch between two units with Z = 5% each produces a circulating current of 1 ÷ (5 + 5) = 10% of rated current — before you connect a single load. That current flows continuously, heats both windings, and eats into the capacity you bought. This is why IEC 60076-1 and GB/T 17468 cap the ratio difference at ±0.5% for parallel operation, and why the tap changer must be set to the same tap position on both units.
Condition 2 — Vector group: identical, not just “similar”
This is the condition that causes the most catastrophic mistakes. The vector group encodes the phase displacement between primary and secondary voltage — Dyn11 gives +30° (LV leads HV by 30°), Yyn0 gives 0°, Dyn1 gives −30°. Parallel units must share the same displacement:
| Vector group pair | Phase displacement | Can they be paralleled? |
|---|---|---|
| Dyn11 + Dyn11 | 0° | Yes |
| Dyn1 + Dyn1 | 0° | Yes |
| Dyn11 + Dyn1 | 60° | No — heavy circulating current |
| Dyn11 + Yyn0 | 30° | No — effectively a short circuit |
| Yyn0 + Yyn0 | 0° | Yes (but weak on unbalanced load) |
A 30° displacement is not a “small mismatch” — it behaves like a dead short across the secondaries. The current is limited only by the (small) impedances, typically 5–10 times rated current, so the protection trips immediately (if you are lucky) or the windings fail first. Always read the vector group off the nameplate, not off the purchase order.
Condition 3 — Impedance: the 10% rule and how load is shared
Load sharing between paralleled transformers is set by their impedances, not by their kVA ratings. For two equal-rated units A and B:
SA ÷ SB = ZB% ÷ ZA%
For unequal ratings the share is proportional to rating divided by impedance:
SA ÷ SB = (SAn ÷ ZA%) ÷ (SBn ÷ ZB%)
Put an 800 kVA unit (Z = 4.5%) beside a 400 kVA unit (Z = 4.5%) and a 1,200 kVA load splits as 800 + 400 — in proportion to rating, because the impedances match. Now change the 400 kVA unit to Z = 6% and the same load splits as (800 ÷ 4.5) ÷ (400 ÷ 6) ≈ 2.67, so the 800 kVA unit carries roughly 873 kVA (109% of rating — overloaded) while the 400 kVA unit idles at about 327 kVA. The tighter you keep Z%, the closer load sharing tracks nameplate; the common design rule (IEC and most utilities) is to keep the two impedances within 10% of each other and the capacity ratio within 3:1. For the full economics of which efficiency grade to buy, see our S11 vs S13 vs S20 loss-grade comparison.
Conditions 4 and 5 — Phase sequence and capacity ratio
Phase sequence: the LV of every unit must be connected so that phases R-S-T (A-B-C) match across all units. Getting one unit’s phase rotation reversed creates a circulating current just like a vector-group error. Confirm rotation with a phase-sequence meter before closing the paralleling breaker, and again after any cable re-termination.
Capacity ratio ≤ 3:1: even with matched impedance, a very large unit beside a very small one is unstable — the small unit’s impedance dominates the fault path, protection coordination becomes unreliable, and a transient can push the small unit into overload. Keep the largest unit no more than roughly three times the smallest.
A worked example: 800 kVA + 400 kVA in parallel
| Parameter | Unit A | Unit B |
|---|---|---|
| Rating | 800 kVA | 400 kVA |
| Voltage | 10 ÷ 0.4 kV | 10 ÷ 0.4 kV |
| Vector group | Dyn11 | Dyn11 |
| Impedance Z% | 4.5% | 4.5% |
| Tap position | 3 (rated) | 3 (rated) |
With matched impedance and vector group, a 1,200 kVA load splits 800 + 400 kVA in proportion to rating. The circulating-current check: at ΔU% = 0.5% worst case, Ic = 0.5 ÷ (4.5 + 4.5) = 5.6% of rated — acceptable. Had Unit B been a Yyn0 unit, the 30° shift would produce a circulating voltage of 2 × sin(15°) ≈ 0.52 per unit, divided by the 9% combined impedance ≈ 5.8 times rated current flowing between the two units with nothing connected — a dead short that trips protection instantly. Had Unit B been Z = 6%, load would split 2.67:1 and Unit A would run at roughly 109% of rating while Unit B idled at 82%. For the full chain downstream of the transformer — breakers, cables and protection settings — see How to Size a Distribution System.
When to parallel — and when to buy one bigger unit
Parallel operation buys you three things: N+1 redundancy (one unit down, the other carries the critical load), flexible growth (add a unit as load rises instead of replacing), and lower no-load losses at light load (switch one unit out at night). It costs you a more complex protection scheme, the circulating-current risk above, and slightly higher combined footprint and no-load losses when both run. As a rule of thumb:
- If you need redundancy or phased growth, parallel two or three matched units.
- If you just need more kVA and the load is steady, one larger unit is usually cheaper and simpler to protect — a single S13-M unit up to 5,000 kVA is available from our oil-immersed range.
Final checklist before you close the paralleling breaker
- Voltage ratios matched within ±0.5%, same tap position on every unit.
- Vector groups identical (Dyn11 + Dyn11, not Dyn11 + Yyn0).
- Impedances within 10% of each other; capacity ratio ≤ 3:1.
- Phase sequence and rotation confirmed with a phase-sequence meter.
- Protection settings coordinated for the combined fault level of both units.
Send us the nameplate details of the units you plan to parallel — rating, voltage ratio, vector group and Z% — and we will run the load-sharing and circulating-current check for you, then quote matched units with test reports so the pair behaves as designed from the first energization. Two nameplates in, one parallel system out.