CURRENT TRANSFORMERS (CTs) PERFORMANCE ANALYSIS

 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:

•  STEADY-STATE 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.    

Fig. 1 Circuit Model R-L for Fault Current Analysis

Assuming that the pre-fault 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,

The first term of i(t) in the Eq. 1 indicates that it has a sinusoidal part and represents the STEADY-STATE AC component of the solution; and, the second term 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.

So, we can re-write i(t) as follow:

i(t) = I STEADY-STATE AC + I TRANSIENT DC

Where:

 

• STEADY-STATE 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 Steady-State Symmetrical Fault Current.

 

•         TRANSIENT DC ANALYSIS

As we have already know the fault current consists of a steady-state 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 STEADY-STATE AC + I TRANSIENT DC

Then,

UNDER THIS SCENARIO THE MAXIMUM ASYMMETRICAL SHORT CIRCUIT CURRENT I(t) TAKES PLACE AS IS DEPICTED IN FIGURE 3.

Fig. 3 Maximum Asymmetrical Fault Current.

When in the equation 1 at t=0 the angle (ϴ - Ҩ) is equal to + π/2   the total current would be equal to:

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) STEADY-STATE 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 steady-state response and the transient DC (dynamic) response under short circuit conditions.

THE STEADY-STATE 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 C57.13-2008 section 6.4.1, it effectively describes the CT steady-state 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_{Sat =} 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_{Sat /} 1.1 = 0.909 . I_Sat                                                                   (Eq. 11)

Then, we can write the equation 9 as follows:

 (I_Sat/ I_{S RATED}) _{. (}Z_{B +} R_CT)/( Z_{B STD +} R_CT) ≤ 20/0.909             (Eq. 12)

Then, defining I_Sat 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 steady-state 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, line-to-ground 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 REAL-WORLD, 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 re-write 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 three-phase fault current that can be applied to this CT without exceeding 10 ratio error.

SOLUTION

In order to determine the maximum three-phase 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 worst-case that means  a completely offset waveform, any current fault other than the worst-case will result in less saturation.

Author: BSEE Hugo E Reyes

-          

-           REFERENCES

1.       ARIANA HARGRAVE, MICHAEL J. THOMPSON, AND BRAD HEILMAN. BEYOND THE KNEE-POINT: A PRACTICAL GUIDE TO CT SATURATION.

2.       IEEE STD C37.110-2007. IEEE GUIDE FOR THE APPLICATION OF CURRENT TRANSFORMERS USED FOR PROTECTIVE RELAYINNG PURPOSES.

3.       IEEE STD C57.13-2008. 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