## Introduction

For supplying a load in excess of the rating of an existing transformer, two or more transformers may be connected in parallel with the existing transformer. The transformers are connected in parallel when load on one of the transformers is more than its capacity.

**The reliability is increased with parallel operation than to have single larger unit.**

The cost associated with maintaining the spares is less when two transformers are connected in parallel. It is usually economical to install another transformer in parallel instead of replacing the existing transformer by a single larger unit.

The cost of a spare unit in the case of two parallel transformers (of equal rating) is also lower than that of a single large transformer. In addition, it is preferable to have a parallel transformer for the reason of reliability.

**With this at least half the load can be supplied with one transformer out of service.**

## Condition for Parallel Operation of Transformer

For parallel connection of transformers, primary windings of the Transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.

- Same voltage and Turns Ratio (both primary and secondary voltage rating is same)
- Same Percentage Impedance and X/R ratio
- Identical Position of Tap changer
- Same KVA ratings
- Same Phase angle shift (vector group are same)
- Same Frequency rating
- Same Polarity
- Same Phase sequence

Some of these conditions are convenient and some are mandatory.

The * convenient conditions *are: Same voltage Ratio and Turns Ratio, Same Percentage Impedance, Same KVA Rating, Same Position of Tap changer.

The * mandatory conditions* conditions are: Same Phase Angle Shift, Same Polarity, Same Phase Sequence and Same Frequency. When the convenient conditions are not met paralleled operation is possible but not optimal.

### 1. Same voltage Ratio and Turns Ratio (on each tap)

If the transformers connected in parallel have slightly different voltage ratios, then due to the inequality of induced emfs in the secondary windings, a circulating current will flow in the loop formed by the secondary windings under the no-load condition, which may be much greater than the normal no-load current.

The current will be quite high as the leakage impedance is low. When the secondary windings are loaded, this circulating current will tend to produce unequal loading on the two transformers, and it may not be possible to take the full load from this group of two parallel transformers (one of the transformers may get overloaded).

Now when the secondary of these transformers are connected to same bus, there will be a circulating current between secondary’s and therefore between primaries also. As the internal impedance of transformer is small, a small voltage difference may cause sufficiently high circulating current causing unnecessary extra I^{2}R loss.

The ratings of both primaries and secondary’s should be identical. In other words, the transformers should have the same turn ratio i.e. transformation ratio.

### 2. Same percentage impedance and X/R ratio

If two transformers connected in parallel with * similar per-unit impedances* they will mostly share the load in the ration of their KVA ratings. Here Load is mostly equal because it is possible to have two transformers with equal per-unit impedances but different X/R ratios. In this case the line current will be less than the sum of the transformer currents and the combined capacity will be reduced accordingly.

A difference in the ratio of the reactance value to resistance value of the per unit impedance results in a different phase angle of the currents carried by the two paralleled transformers; one transformer will be working with a higher power factor and the other with a lower power factor than that of the combined output. Hence, the real power will not be proportionally shared by the transformers.

**The current shared by two transformers running in parallel should be proportional to their MVA ratings.**

**The current carried by these transformers are inversely proportional to their internal impedance.**

From the above two statements it can be said that impedance of transformers running in parallel are inversely proportional to their MVA ratings. In other words percentage impedance or per unit values of impedance should be identical for all the transformers run in parallel.

When connecting single-phase transformers in three-phase banks, proper impedance matching becomes even more critical. In addition to following the three rules for parallel operation, it is also a good practice to try to match the *X/R *ratios of the three series impedances to keep the three-phase output voltages balanced.

When single-phase transformers with the same KVA ratings are connected in a Y-∆ Bank, impedance mismatches can cause a significant load unbalance among the transformers

**Lets examine following different type of case among Impedance, Ratio and KVA.**

If single-phase transformers are connected in a Y-Y bank with an isolated neutral, then the magnetizing impedance should also be equal on an ohmic basis.

Otherwise, the transformer having the largest magnetizing impedance will have a highest percentage of exciting voltage, increasing the core losses of that transformer and possibly driving its core into saturation.

#### Case 1: Equal Impedance, Ratios and Same kVA

** **The standard method of connecting transformers in parallel is to have the same turn ratios, percent impedances, and kVA ratings. Connecting transformers in parallel with the same parameters results in equal load sharing and no circulating currents in the transformer windings.

**Example** Connecting two 2000 kVA, 5.75% impedance transformers in parallel, each with the same turn ratios to a 4000 kVA load.

- Loading on the transformers-1 =KVA1=[( KVA1 / %Z) / ((KVA1 / %Z1)+ (KVA2 / %Z2))]X KVAl
- kVA1 = 348 / (348 + 348) x 4000 kVA = 2000 kVA.
- Loading on the transformers-2 =KVA1=[( KVA2 / %Z) / ((KVA1 / %Z1)+ (KVA2 / %Z2))]X KVAl
- kVA2 = 348 / (348 + 348) x 4000 kVA = 2000 kVA
**Hence KVA1=KVA2=2000KVA**

#### Case 2: Equal Impedances, Ratios and Different kVA

This Parameter is not in common practice for new installations, sometimes two transformers with different kVAs and the same percent impedances are connected to one common bus. In this situation, the current division causes each transformer to carry its rated load. There will be no circulating currents because the voltages (turn ratios) are the same.

**Example** Connecting 3000 kVA and 1000 kVA transformers in parallel, each with 5.75% impedance, each with the same turn ratios, connected to a common 4000 kVA load.

- Loading on Transformer-1=kVA1 = 522 / (522 + 174) x 4000 = 3000 kVA
- Loading on Transformer-1=kVA2 = 174 / (522 + 174) x 4000 = 1000 kVA

From above calculation it is seen that different kVA ratings on transformers connected to one common load, that current division causes each transformer to only be loaded to its kVA rating. The key here is that the percent impedance are the same.

#### Case 3: Unequal Impedance but Same Ratios and kVA

Mostly used this Parameter to enhance plant power capacity by connecting existing transformers in parallel that have the same kVA rating, but with different percent impedances.

This is common when budget constraints limit the purchase of a new transformer with the same parameters.

We need to understand is that the current divides in inverse proportions to the impedances, and larger current flows through the smaller impedance. Thus, the lower percent impedance transformer can be overloaded when subjected to heavy loading while the other higher percent impedance transformer will be lightly loaded.

**Example **Two 2000 kVA transformers in parallel, one with 5.75% impedance and the other with 4% impedance, each with the same turn ratios, connected to a common 3500 kVA load.

**Loading on Transformer-1**=kVA1 = 348 / (348 + 500) x 3500 =**1436 kVA****Loading on Transformer-2=**kVA2 = 500 / (348 + 500) x 3500 =**2064 kVA**

It can be seen that because transformer percent impedances do not match, they cannot be loaded to their combined kVA rating. Load division between the transformers is not equal. At below combined rated kVA loading, the 4% impedance transformer is overloaded by 3.2%, while the 5.75% impedance transformer is loaded by 72%.

#### Case 4: Unequal Impedance and KVA Same Ratios

This particular of transformers used rarely in industrial and commercial facilities connected to one common bus with different kVA and unequal percent impedances. However, there may be that one situation where two single-ended substations may be tied together via bussing or cables to provide better voltage support when starting large Load.

If the percent impedance and kVA ratings are different, care should be taken when loading these transformers.

**Example** Two transformers in parallel with one 3000 kVA (kVA1) with 5.75% impedance, and the other a 1000 kVA (kVA2) with 4% impedance, each with the same turn ratios, connected to a common 3500 kVA load.

**Loading on Transformer-1=**kVA1 = 522 / (522 + 250) x 3500 = 2366 kVA**Loading on Transformer-2=**kVA2 = 250 / (522 + 250) x 3500 = 1134 kVA

Because the percent impedance is less in the 1000 kVA transformer, it is overloaded with a less than combined rated load.

#### Case 5: Equal Impedance and KVA Unequal Ratios

Small differences in voltage cause a large amount of current to circulate. It is important to point out that paralleled transformers should always be on the same tap connection. Circulating current is completely independent of the load and load division. If transformers are fully loaded there will be a considerable amount of overheating due to circulating currents.

The Point which should be Remember that circulating currents do not flow on the line, they cannot be measured if monitoring equipment is upstream or downstream of the common connection points.

**Example** Two 2000 kVA transformers connected in parallel, each with 5.75% impedance, same X/R ratio (8), transformer 1 with tap adjusted 2.5% from nominal and transformer 2 tapped at nominal. What is the percent circulating current (%IC)

- %Z1 = 5.75, So %R’ = %Z1 / √[(X/R)2 + 1)] = 5.75 / √((8)2 + 1)=0.713
- %R1 = %R2 = 0.713
- %X1 = %R x (X/R)=%X1= %X2= 0.713 x 8 = 5.7
- Let %e = difference in voltage ratio expressed in percentage of normal and k = kVA1/ kVA2
**Circulating current %IC = %eX100 / √ (%R1+k%R2)2 + (%Z1+k%Z2)2.**- %IC = 2.5X100 / √ (0.713 + (2000/2000)X0.713)2 + (5.7 + (2000/2000)X5.7)2
- %IC = 250 / 11.7 = 21.7

The circulating current is * 21.7% of the full load current*.

#### Case 6: Unequal Impedance, KVA and Different Ratios

** **This type of parameter would be unlikely in practice. If both the ratios and the impedance are different, the circulating current (because of the unequal ratio) should be combined with each transformer’s share of the load current to obtain the actual total current in each unit.

For unity power factor, 10% circulating current (due to unequal turn ratios) results in only half percent to the total current. At lower power factors, the circulating current will change dramatically.

**Example **Two transformers connected in parallel, 2000 kVA1 with 5.75% impedance, X/R ratio of 8, 1000 kVA2 with 4% impedance, X/R ratio of 5, 2000 kVA1 with tap adjusted 2.5% from nominal and 1000 kVA2 tapped at nominal.

- %Z1 = 5.75, So %R’ = %Z1 / √[(X/R)2 + 1)] = 5.75 / √((8)2 + 1)=0.713
- %X1= %R x (X/R)=0.713 x 8 = 5.7
- %Z2= 4, So %R2 = %Z2 /√ [(X/R)2 + 1)]= 4 / √((5)2 + 1) =0.784
- %X2 = %R x (X/R)=0.784 x 5 = 3.92
- Let %e = difference in voltage ratio expressed in percentage of normal and k = kVA1/ kVA2
**Circulating current %IC = %eX100 / √ (%R1+k%R2)2 + (%Z1+k%Z2)2.**- %IC = 2.5X100 / √ (0.713 + (2000/2000)X0.713)2 + (5.7 + (2000/2000)X5.7)2
- %IC = 250 / 13.73 = 18.21.

The circulating current is * 18.21% of the full load current*.

### 3. Same polarity

Polarity of transformer means the instantaneous direction of induced emf in secondary. If the instantaneous directions of induced secondary emf in two transformers are opposite to each other when same input power is fed to the both of the transformers, the transformers are said to be in opposite polarity.

Polarity of all transformers run in parallel should be same otherwise huge circulating current flows in the transformer but no load will be fed from these transformers.

If the instantaneous directions of induced secondary emf in two transformers are same when same input power is fed to the both of the transformers, the transformers are said to be in same polarity.

### 4. Same phase sequence

** **The phase sequence of line voltages of both the transformers must be identical for parallel operation of three-phase transformers. If the phase sequence is an incorrect, in every cycle each pair of phases will get short-circuited.

**This condition must be strictly followed for parallel operation of transformers.**

### 5. Same phase angle shift (zero relative phase displacement between the secondary line voltages)

** **The transformer windings can be connected in a variety of ways which produce different magnitudes and phase displacements of the secondary voltage. All the transformer connections can be classified into distinct vector groups.

* Group 1:* Zero phase displacement (Yy0, Dd0, Dz0)

*180° phase displacement (Yy6, Dd6, Dz6)*

**Group 2:***-30° phase displacement (Yd1, Dy1, Yz1)*

**Group 3:***+30° phase displacement (Yd11, Dy11, Yz11)*

**Group 4:**In order to have zero relative phase displacement of secondary side line voltages, the transformers belonging to the same group can be paralleled. For example, two transformers with Yd1 and Dy1 connections can be paralleled.

The transformers of groups 1 and 2 can only be paralleled with transformers of their own group. However, the transformers of groups 3 and 4 can be paralleled by reversing the phase sequence of one of them. For example, a transformer with Yd1 1 connection (group 4) can be paralleled with that having Dy1 connection (group 3) by reversing the phase sequence of both primary and secondary terminals of the Dy1 transformer.

We can only parallel * Dy1* and

*by crossing two incoming phases and the same two outgoing phases on one of the transformers, so if we have a DY11 transformer we can cross B&C phases on the primary and secondary to change the +30 degree phase shift into a -30 degree shift which will parallel with the Dy1, assuming all the other points above are satisfied.*

**Dy11**### 6. Same KVA ratings

If two or more transformer is connected in parallel, then load sharing % between them is according to their rating. If all are of same rating, they will share equal loads

Transformers of unequal kVA ratings will share a load practically (but not exactly) in proportion to their ratings, providing that the voltage ratios are identical and the percentage impedances (at their own kVA rating) are identical, or very nearly so in these cases a total of than 90% of the sum of the two ratings is normally available.

Transformers having different kva ratings may operate in parallel, with load division such that each transformer carries its proportionate share of the total load To achieve accurate load division, it is necessary that the transformers be wound with the same turns ratio, and that the percent impedance of all transformers be equal, when each percentage is expressed on the kva base of its respective transformer. It is also necessary that the ratio of resistance to reactant in all transformers be equal.

For satisfactory operation the circulating current for any combinations of ratios and impedance probably should not exceed ten percent of the full-load rated current of the smaller unit.

### 7. Identical tap changer and its operation

When the voltage tap need change all three tap changing switches should be operated identical for all transformers. The OL settings of the SF6 also should be identical. If the substation is operating on full load condition, tripping of one transformer can cause cascade tripping of all three transformers.

In transformers Output Voltage can be controlled either by Off Circuit Tap Changer (Manual tap changing) or By On – Load Tap Changer-OLTC (Automatic Changing).

*In the transformer with OLTC, it is a closed loop system, with following components:*

**1. AVR (Automatic Voltage Regulator)** – an electronic programmable device). With this AVR we can set the Output Voltage of the transformers. The Output Voltage of the transformer is fed into the AVR through the LT Panel. The AVR Compares the SET voltage and the Output Voltage and gives the error signals, if any, to the OLTC through the RTCC Panel for tap changing. This AVR is mounted in the RTCC.

**2. RTCC (Remote Tap Changing Cubicle)** – This is a panel consisting of the AVR, Display for Tap Position, Voltage, and LEDs for Raise and Lower of Taps relays, Selector Switches for Auto Manual Selection… In AUTO MODE the voltage is controlled by the AVR. In manual Mode the operator can Increase / decrease the voltage by changing the Taps manually through the Push Button in the RTCC.

**3. OLTC is mounted on the transformer** – It consists of a motor, controlled by the RTCC, which changes the Taps in the transformers.

Both the Transformers should have same voltage ratio at all the taps and when you run transformers in parallel, it should operate as same tap position. If we have OLTC with RTCC panel, one RTCC should work as master and other should work as follower to maintain same tap positions of transformer.

However, a circulating current can be flown between the two tanks if the impedances of the two transformers are different or if the taps of the on-load tap changer (OLTC) are mismatched temporarily due to the mechanical delay. The circulating current may cause the malfunction of protection relays.

###### References

- Say, M.G. The performance and design of alternating current machines.
- Application Guide, Loading of Transformer, Nashville, TN, USA.
- Toro, V.D. Principles of electrical engineering.
- Stevenson, W.D. Elements of power system analysis.
- MIT Press, Magnetic circuits and transformers, John Wiley and Sons.

AMIT PATEL

I am a diploma electrical engineer from R.C. Technical Institute, Saraspur, Ahmedabad.

I have >15 years power plant operation and maintenance experience in different locations in Gujarat and UAE.

if u can give me your mail ids.

JOSE LANDA

Thanks,

kasif jamil

i m working at 220 kv grid substation. there are two transformer of 100 mva rating running in parallel.their percentage impedance differs by one percent hence the load sharing of one of the transformer having lower impedance share ten percent more load current.But here i see the transformer shows different power factors although they are sharing a common load.As we know the transformer are not having their own power factor it depends on type of the load.So kindly clear my doubt that why both of the transformer are showing different power factor.

Bwalya Chisata

What’s the difference between restricted earth fault(RFE) & differential protection

Bwalya Chisata

What is the purpose of having a good earthing system in the substation

Andrew Bell

Can a single multi-function digital relay used to protect three 3-phase transformers connected as outlined below?

The three 3-phase transformers’ primary windings are connected to a common source bus-bar, controlled by one common breaker, and each transformer’s secondary is connected to separate 3-phase loads with individual breakers, .

Naif

thanks alot

andy

i have two microwave oven xfmrs. i chopped off the secondaries & rewound each with new

windings using AWG 6 stranded wire with teflon coat. i want to use these to make a home-

made welder. using only one wont do the trick. but if i somehow hook up both together to

get double the current i am sure this will work. how do i connect these secondaries & how

do i connect their primaries to 120 vac? also i will make sure that the physical aspects of

the windings remain identical upon each core. thank you!

Amit Chaurasia

What is the effect of transformer Vector group on the Operation of the motors connected to the transformer as load.

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Debashish Paul

its realy nice information………..

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rajneesh.hpsebl

You have mentioned that difference in phase voltage should be 0.4%. How it has been derived and whether the percentage is of primary voltage or secondary. Please explain. Thanking you

Wale Adereti

Thank you so much for making this article available.

I am presently working on a project (1500kVA+500kVA) typically as Case 2 Above, two transformers of different kVA ratings would deliver each based on kirchoff current law:

i total = i1 + i2, all other conditions fulfilled.

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Edvard

Excellent write up Jignesh! Thank you very much!

saad1966

Thanks so much this is very useful for me