Differential Protection is
based on the fact that any fault within electrical equipment will result into the
difference in current entering it and the current leaving it. Thus by comparing
the two currents either in magnitude or in phase or both we can determine a
fault and issue a trip decision if the difference exceeds a predetermined set
value.

In this post we will discuss
about the Differential Protection of a Transformer. Consider
an ideal transformer with the CT connections, as shown in figure below.

Suppose that current rating
of primary winding = 100A

Current rating of secondary
winding = 1000A.

Then if we use 100/5 and
1000/5 CT on the primary and secondary winding respectively, then under normal operating
conditions the both CT currents will be 5 A in magnitudes.

By connections the primary and
secondary CTs with due care to the dots (In actual CT polarity marking is by
Terminal P1 and P2. Current is supposed to flow from P1 to P2), a circulating
current can be set up as shown by dotted line.

No current will flow through
the branch having overcurrent current relay in normal condition as

Differential current

**I**= 5-5 = 0 A_{d}
Now if an internal fault
occurs within the Transformer like interturn short etc., then the normal mmf balance
is upset i.e. under this condition, the CT secondary currents of primary and secondary
side CTs will not match. The resulting differential current will flow through
overcurrent relay. If the pickup setting of overcurrent relay is close to zero,
it will immediately pick up and initiate the trip decision.

In practice, the transformer
is not ideal. A differential current always flows through the overcurrent
relay. Therefore overcurrent relay pick up is adjusted above the no load
current value.

In Differential Protection
of Transformer CT matching is an important thing to care care else differential
protection won’t be efficient and reliable.

Let the transformer turns
ratio given N1/N2 and the corresponding CT ratio be given by 1/n1 and as 1/n2.

Therefore,

Current in CT - 1 primary = I

_{1}_{}

Current in CT - 1 secondary
=I

_{1}/n_{1}_{}

Current in CT - 2 primary =
N

_{1}I_{1}/N_{2}_{}

Current in CT - 2 secondary
= N

_{1}I_{1}/N_{2}n_{2}_{}

In normal operating
condition of Transformer, the differential current through the Relay should be
zero. Therefore,

Current in CT - 1 secondary
= Current in CT - 2 secondary

_{}
I

_{1}/n_{1}= N_{1}I_{1}/N_{2}n_{2}_{}

N

_{2}n_{2}= N_{1}n_{1}_{}

Thus CT should be selected
in such a way that their Turn Ratio satisfies

**N**

_{2}n_{2}= N_{1}n_{1}

_{}
In case, Transformer Tap is
used then nominal Tap position should also be taken into account.

When dealing with three
phase transformers, the transformer connections like Y-Y or delta – delta connection
play a great role in determining CT secondary interconnections to establish
circulating current scheme. This is because of the phase shifts typically of
the order of +- 30° that result in the line currents when we move from primary
to secondary side of the power transformer.

*If transformer winding are connected in Y configuration then use DELTA configuration for corresponding CT secondary interconnections and vice-versa. "*

**Thank you!**

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