INTRODUCTION
In this paper we are going to analyze the performance of CTs
used for relaying purposes applying ANSI/IEEE Standards. Analyzing CTs
performance is nothing but to evaluate CTs saturation and its effect upon CTs
ratios and CTs errors.
In order to achieve this goal we will analyze the CTs
performance based on:
 STEADYSTATE FAULT CURRENT, and
 TRANSIENT FAULT CURRENT
FAULT CURRENT BASIS
Having the next RL circuit in Figure 1, we can study how the
fault current in a power system behaves as a function of time.
Assuming that the prefault current through the circuit is zero, the current fault at t=0 would be equal to:
Or,
Where:
i(t): is
the instantaneous current after the switch is closed at t=0
Vm: is the maximum peak value of the voltage waveform
w : is the 2πf where f is the system frequency in Hertz (Hz)
t : is time in seconds
R : is the resistance in Ohms
L : is the inductance in Henries
Z : is the impedance in Ohms
ϴ : is the angle of the applied voltage in
radians
Ҩ
: is the characteristic angle of the system in radians {arctan (X/R)}
τ : is the time constant that is determined by L/R
ratio of the circuit
Also,
So, we can rewrite i(t) as follow:
i(t) =
I STEADYSTATE AC + I TRANSIENT DC
Where:
 STEADYSTATE CURRENT ANALYSIS
When in the equation
1 at t=0 the
angle (ϴ  Ҩ) is equal
to zero (or π) the
transient DC current would be equal to zero, so:
UNDER THIS SCENARIO THE TOTAL CURRENT I(t)
IS
IDEALLY SYMMETRICAL AS IS DEPICTED IN FIGURE 2.
Fig. 2 SteadyState Symmetrical
Fault Current.
 TRANSIENT DC ANALYSIS
As we have already know the fault current consists of a
steadystate AC component and a transient decaying DC offset, the worst case
analysis happens when the angle (ϴ  Ҩ) is equal to +/
π/2
If in the equation 1
at t=0 the angle (ϴ
 Ҩ) is
equal to  π/2
the total current would be equal to:
i(t) =
I STEADYSTATE AC + I TRANSIENT DC
Then,
Fig. 3 Maximum Asymmetrical
Fault Current.
UNDER THIS SCENARIO THE MAXIMUM ASYMMETRICAL SHORT
CIRCUIT CURRENT I(T) TAKES PLACE AS IS DEPICTED IN FIGURE 4.
Fig. 4 Maximum Asymmetrical
Fault Current.
IN SUMMARY, THE TRANSIENT DC (DC OFFSET) COMPONENT MAY VARIES
FROM ZERO UNTIL THE ABSOLUTE VALUE OF Vm/Z . DC OFFSET ALSO DEPENDS ON THE SUPPLY VOLTAGE AT t=0 AND THE
CHARACTERISTIC ANGLE OF THE SYSTEM (Ҩ).
CURRENT TRANSFORMER
(CT) STEADYSTATE PERFORMANCE
In relaying protection we need to know the CT performance
and to evaluate the CT saturation and its impact on CT ratio and CT errors. CT
performance analysis involves the steadystate response and the transient DC
(dynamic) response under short circuit conditions.
THE STEADYSTATE
PERFORMANCE IS RELATED TO THE RELAYING CT BEHAVIOR TO SYMMETRICAL SHORT CIRCUIT
CURRENTS.
When we are selecting a current transformer for relaying
purposes is very important to evaluate if the CT saturates or not under the
maximum fault current that may occurs in the place where it will be installed. When
a CT saturates, it provides an unreliable signal to the connected relay, as a
result CT saturation can lead in protective relay misoperation.
At this time we are going to analyze the performance of CTs used
for relaying purposes applying IEEE STD C57.132008 section 6.4.1, it
effectively describes the CT steadystate performance. The standard is based on
the ANSI CT accuracy letter designation codes “C” and “T” class and is given in
terms of the secondary terminal voltage, which is nothing but to the CT
secondary voltage that it will deliver to a standard secondary burden, such as
a protective relay, at 20 times rated secondary current, without exceeding a
10% ratio error.
The secondary terminal voltage rating are based on 5 A
nominal secondary current and a standard burden as follows:
For example, a CT class C400 means that the ratio error will not exceed 10% at any current from 1 to 20 times rated secondary
current with a standard 4 Ohms burden (20 x 4 Ohms x 5 A = 400 V).
However, if the short circuit current through a CT is higher
than 20 times its rated current, or the load or connected burden is higher than
the standard burden, the CT will probably exceed 10% ratio error as a result of
the CT saturation.
According to the ANSI STD we can write the next equation:
V_{STD} = 20 . I_{S RATED}
. Z_{B STD} (Eq.
3)
Where:
V_{STD} : is the secondary
terminal voltage rating
I_{S RATED} : is the rated secondary current
Z_{B STD}: Is the standard burden in Ohms
A simplified CT equivalent circuit is shown in the next
figure:
Fig. 5 Simplified CT
Equivalent Circuit.
Where,
V_{s }: is the secondary terminal voltage V_{s }in Volts_{}
V_{E}_{ : }_{is the excitation voltage in Volts}
I_{s }: is the secondary load current in Amps
I_{E}_{ }_{: }is the excitation current in Amps_{}
Z_{B} : is the burden in Ohms
R_{CT }: is the secondary winding resistance of
the CT in Ohms
Xm : represents
the varying reactance of the CT’s core in Ohms
Using the CT
equivalent circuit in fig. 5 and ANSI/IEEE voltage rating definition we will
be able to get that the ratio error will not exceed 10% due to the secondary
terminal voltage V_{s} is less than the
secondary terminal voltage rating, then:
V_{s < }V_{STD }(Eq. 4)_{}
Also, considering the voltage drops due to secondary winding resistance (R_{CT):}
V_{E }≤ V_{E STD } (Eq. 5)
Now, we can write the definition of V_{E STD }as:
V_{E STD = }20
. I_{S RATED} . (Z_{B STD + }R_{CT) }(Eq. 6)
And the
actual voltage V_{E }is:
V_{E }=
I_{s . (}Z_{B + }R_{CT) }(Eq.
7)
Then, using eq. 5, 6
and 7:
I_{s . (}Z_{B + }R_{CT) }≤ 20 . I_{S RATED}
. (Z_{B STD + }R_{CT) }(Eq.
8)
Then,
rearranging the equation 8:
(I_{s/} I_{S RATED) . (}Z_{B + }R_{CT)/(} Z_{B STD + }R_{CT) }≤ 20 (Eq. 9)
By
the other hand:
I_{S}_{at = }I_{p}_{/n = }I_{s + }I_{E }(Eq.
10)
And
also, for the worst scenario at the CT saturation point, the ratio error would
be 10% (I_{E }/ I_{s }≤ 0.1), then
we can find that:
I_{s }=_{ }I_{S}_{at / }1.1 = 0.909 . I_{S}_{at }(Eq.
11)
Then,
we can write the equation 9 as
follows:
(I_{S}_{at}_{/} I_{S RATED})_{ . (}Z_{B + }R_{CT)/(} Z_{B STD + }R_{CT) }≤ 20/0.909 (Eq. 12)
Then,
defining I_{S}_{at }as the maximum short circuit current (I_{F }_{SEC})_{ }that the CT can_{ }handle
before the saturation point:
(I_{F }_{SEC} _{/} I_{S RATED})_{.(}Z_{B + }R_{CT)/(} Z_{B STD + }R_{CT) }≤ 22 (Eq.
13)
We
have assumed that Z_{B} was purely inductive but unfortunately
modern digital relays are purely resistive, as a result the equation 13 can
produce important errors.
In
the equation 10 we assumed that ZB
was purely inductive but unfortunately modern digital relays are purely resistive,
as a result the equation 13 can produce important errors. For convenience, a
more conservative equation is used instead:
(I_{F }_{SEC} _{/} I_{S RATED})_{.(}Z_{B + }R_{CT)/(} Z_{B STD + }R_{CT) }≤ 20 (Eq.
14)
Finally,
it is very important to accommodate in an equation the primary fault current
and the CT ratio:
(I_{FAULT} _{/} I_{PRIM})_{.(}Z_{B + }R_{CT)/(} Z_{B STD + }R_{CT) }≤ 20 (Eq. 15)
Where:
I_{FAULT} : is the maximum fault current in primary in
Amperes
I_{PRIM
}: is the primary current rating of the
CT (e.g. for a 1500/5, I_{PRIM }is 1500 A)
Z_{B } : is the actual burden of the CT’s
secondary circuit, which includes both the impedance of the connected relay and
the impedance of the leads from the CT to the relay.
R_{CT } : is the internal resistance of the CT
secondary winding in Ohms
Z_{B STD}: is the
standard burden of the CT (e.g. for a C400 CT is equal to 8 Ohms)
EXAMPLE
A C400,
2,000/5 CT (I_{PRIM} = 2,000)
has a burden of 8 Ω (Z_{B}) and
the standard burden of a C400 CT is 4 Ω
(Z_{B STD}), we assumed
that the CT has a secondary winding resistance equal to zero (R_{CT} = 0).
SOLUTION:
We
need to calculate the maximum symmetrical short circuit current that the CT can
handle without exceeding a 10% ratio error, it would be:
I_{FAULT}
≤
20 . I_{PRIM . (}Z_{B STD }+ R_{CT})/(Z_{B + }R_{CT})
I_{FAULT}
≤
20 . 2000 . (4)/(8) = 20 kA
THEN,
THE MAXIMUM SYMMETRICAL SHORT CIRCUIT CURRENT ALLOWABLE IS 20 kA.
In the other hand, if for the same CT
the connected burden is 2 Ω (Z_{B}), the maximum symmetrical short circuit
current that the CT can handle without exceeding a 10% ratio error, it would
be:
I_{FAULT} ≤ 20 . 2,000 . (4)/(2) = 80 kA
However, the maximum allowable
current is really (when Z_{B STD }=_{ }Z_{B)}:
I_{FAULT} = 20 . 2,000 . (4)/(4) = 40 kA
For the same CT if we have a symmetrical
short circuit current of 25 kA the
maximum allowable burden (Z_{B}) without exceeding a 10% ratio error is:
Z_{B }=_{ }20 . (I_{PRIM/} I_{FAULT})_{ . (}Z_{B STD }+ R_{CT}) + R_{CT}
Z_{B
}=_{ }20 . (2,000_{/} 25,000)_{ . }(4+ 0) + 0 = 6.4 Ω
The equation (Eq. 15) just
can be considered for CT steadystate performance although it should not be
used in practice because is very infrequent that at the time any short circuit current
occurs its waveform is symmetrical.
By the other hand, linetoground faults are more likely to
be symmetrical faults, but in any three phase fault, all currents cannot be
zero at the same time in each phase so DC offset would be inevitable in at
least one phase. Finally, selecting CTs considering only symmetrical faults is not
recommended because it does not take into account the heavy CT saturation due
to Transient DC offset.
UNFORTUNATELY, SYMMETRICAL SHORT CIRCUIT CURRENTS AT THE TIME OF FAULT
INCEPTION IS UNCOMMON IN THE REALWORLD, SO THEY SHOULD NEVER BE USED
PRACTICALLY TO ANALYZE CTs PERFORMANCE.
CURRENT TRANSFORMER
(CT) TRANSIENT DC PERFORMANCE
We have already
learned that fault currents has an exponentially decaying part that is called
TRANSIENT DC COMPONENT (or DC OFFSET), this DC component eventually disappears
a few cycles after of the fault inception determined by the X/R ratio of the
power system, this DC offset can produce significant CT saturation that is
usually called ASYMMETRICAL SATURATION.
The transient DC component of an asymmetrical fault current
greatly increases the magnetic flux in the CT. At the time the DC offset is at
a maximum, the CT flux can increases to 1
+ X/R times the total flux, by that we can rewrite the equation 15 in
order to analyze CT saturation from fault current with DC Offset:
(I_{FAULT} _{/} I_{PRIM})_{. }{_{(}Z_{B + }R_{CT)/(}
Z_{B STD + }R_{CT) }} . (1 + X/R) ≤ 20 (Eq.
16)
EXAMPLE
A C800, 2000/5 CT with a R_{CT} equal to 0.5 Ω is
connected to a 1 Ω burden (relay and lead resistance), the X/R ratio is 12. We need
to calculate the maximum primary threephase fault current that can be applied
to this CT without exceeding 10 ratio error.
SOLUTION
In order to determine the maximum threephase fault current
we need to apply the equation 16:
I_{FAULT}
≤
20 . I_{PRIM . }{(Z_{B STD }+ R_{CT})/(Z_{B }+_{
}R_{CT})} / (1 + X/R)
Where:
I_{FAULT} : is the maximum fault current in primary in
Amperes to be determined
I_{PRIM
}= 2000
A (is the primary current rating of the CT)
Z_{B } = 1 Ω is (the actual burden of the CT’s secondary circuit)
R_{CT } = 0.5 Ω (is the internal resistance of the CT secondary winding)
Z_{B STD} = 8 Ω (is the standard burden of the CT)
X/R
= 12 ( is the X/R of the power system)
So,
I_{FAULT} ≤
20 . 2000 . {(8+0.5)/(1 + 0.5)} / (1 + 12) = 17.44 kA
THEN,
THE MAXIMUM ALLOWABLE FAULT CURRENT HAS TO BE LOWER THAN 17.44 kA TO AVOID CT
ASYMMETRICAL SATURATION
SUMMARY

Depending on the type of current faults there
are two different types of CT saturation:
 CT SYMMETRICAL SATURATION, and
 CT
ASYMMETRICAL SATURATION.
Symmetrical
saturation is caused by symmetrical short circuit currents while asymmetrical saturation
is caused by asymmetrical short circuit currents.
 IEEE defines the voltage rating as the CT
secondary voltage that the CT delivers when it is connected to a standard
burden at 20 times rated secondary burden without exceeding a 10% ratio error.

A criterion for avoid relaying CT saturation has
been shown using the next equation:
This
equation is quite conservative because represents the worstcase that means a completely offset waveform, any current fault
other than the worstcase will result in less saturation.
Author: BSEE Hugo E Reyes


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2. IEEE STD C37.1102007. IEEE GUIDE FOR THE
APPLICATION OF CURRENT TRANSFORMERS USED FOR PROTECTIVE RELAYINNG PURPOSES.
3. IEEE STD C57.132008. IEEE STANDARD
REQUIREMENTS FOR INSTRUMENTS TRANSFORMERS
4. MAULIO RODRÍGUEZ, ANALISIS DE SISTEMAS DE POTENCIA, 2DA. EDICIÓN 1992, EDITORIAL EDILUZ, UNIVERSIDAD DEL ZULIA.
2. STANLEY E. ZOCHOLL. ANALYZING AND APPLYING CURENT TRANSFORMERS. SCHWEITZER ENGINEERING LABOATORIES. 2004