1. Part Three – Relay Input Sources
Wei-Jen Lee, Ph.D., PE
Professor of Electrical Engineering Dept.
The Univ. of Texas at Arlington

2. Introduction
Protective relays require reasonably accurate reproduction of the normal, tolerable, and intolerable conditions in the power system fro correct sensing and operation.
This information input from the power system is usually through current and voltage transformers.
Some exceptions, such as temperature and vibration relays, which receive their information from other type of transducers.

3. Introduction

4. Introduction

5. Introduction

6. Equivalent Diagram for Instrument Transformers

7. Current Transformers Typical CT ratio 50:5 100:5 150:5 200:5 250:5
300:5
400:5
450:5
500:5
600:5
800:5
900:5
1000:5
1200:5
1500:5
1600:5
2000:5
2400:5
2500:5
3000:5
3200:5
4000:5
5000:5
6000:5

8. Current Transformers
Current transformer performance on a symmetrical AC component
If the CT does not saturate, it is reasonable to assume that Ie is negligible.
However, the CT excitation current is never equal to zero. Thus, it must be checked to assure that it does not cause intolerable errors.
This can be done by one of the three methods: 1) classic transformer formula, 2) CT performance curves, or 3) ANSI/IEEE accuracy classes for relaying.

9. Current Transformers Current transformer performance check
Classic analysis
The load of the CT consists of secondary resistance Rs, the impedance of he connecting leads Zld, and the equipment (relays and such) Zr. The voltage required by the burden (load) is

10. Current Transformers Current transformer performance check
CT characteristic curves
The calculation of the performance with the equivalent circuit of Fig. 5.6a is difficult.
ANSI/IEEE (C53.71) classifies CTs that have significant leakage flux within the transformer core as class T (class H before 1968)
The class C (class L before 1968) is CTs constructed to minimize the leakage flux in the core that can represented by the Fig. 5.6b.

11. Current Transformers Current transformer performance check
CT characteristic curves
The knee or effective point of saturation is defined by ANSI/IEEE standard as the intersection of the curve with 45o tangent line.
However, the IEC defines the knee as the intersection of straight lines extended from the nonsaturated and saturated parts of the exciting curve.
The IEC knee is higher voltage than the ANSO curve.

12. Current Transformers Current transformer performance check
CT characteristic curves - Typical overcurrent ratio curve for class T CT

13. Current Transformers Current transformer performance check
CT characteristic curves - Typical excitation curves for a multiratio class C CT

14. Current Transformers Current transformer performance check
ANSI/IEEE standard accuracy class
In many applications, the use of ANSI/IEEE accuracy class designation is adequate to assure satisfactory relay performance.
Manufacturer’s test curves must be used for Class T CTs.
For class C, the designations are followed by a number indicating the secondary terminal voltage (Vgh) that the transformer can deliver to a standard burden at 20 times the rated secondary current without exceeding the 10% ratio correction.

15. Current Transformers Current transformer performance check
ANSI/IEEE standard accuracy class

16. Current Transformers Current transformer performance check
ANSI/IEEE standard accuracy class
For relaying, the voltage classes are 100, 200, 400, and 800, corresponding to standard burdens of B-1, B-2, B-4, and B-8, respectively.
These burdens are at 0.5 power factor.
If the current is lower, the burden can be higher in proportion.
The lower voltage classes of 10, 20, and 50 with standard burdens of B-0.1, B-0.2, and B-0.5 at 0.9 power factor are primary for metering service and should be used very cautiously for protection.

17. Current Transformers Current transformer performance
Two similar CTs connected in the primary circuit, with the same ratio and their secondaries in series, will increase the accuracy capability.
Two similar CTs, with their secondaries in parallel, provides an overall lower ratio with higher-ratio individual CTs and their correspondingly higher accuracy rating.

18. Current Transformers Current transformer performance
ANSI classification does not provide actual value of error.
Also, the accuracy class only applies to the full winding and reduces proportionally when lower taps are available and used.

19. Current Transformers Current transformer performance
IEC specifies the accuracy of the current transformers as: XX VA Class YY P ZZ
where
XX: Continuous VA (2.5, 5, 10, 15, and 30)
YY: Accuracy class (5 and 10%)
P: For protection
ZZ: Accuracy limit factor (5, 10, 15, 20, and 30)
Rated secondary amperes: 1, 2, and 5 A

20. Current Transformers
Secondary burdens during faults

21. Current Transformers
Secondary burdens during faults

22. Current Transformers
Secondary burdens during faults

23. Current Transformers
CT selection and performance evaluation for phase faults
Assumption
Imax load = 90A, Imax fault= 2500A, and Imin. fault = 350A.
CT ratio selection
The conventional practice, over many years, has been that the secondary current should be just under 5A for the maximum load. Therefore, select CT ratio of 100/5 in this case.

24. Current Transformers
CT selection and performance evaluation for phase faults
Select relay tap for the phase-overcurrent relay
The tap should be higher than 4.5 A
Small tap 5 is selected
Minimum fault of 350/20=17.5 A, and 17.5/5=3.5 times the minimum relay pick up. This is desirable for any possible fault restriction.
If tap 6 is selected, then the margin above the load is greater (6/4.5=1.33), but a smaller margin (17.5/6=2.9) above the relay pick up.

25. Current Transformers
CT selection and performance evaluation for phase faults
Determine the total secondary burden
The total connected secondary load determination must include all of the impedance between the CTs and the equipment in the phase circuit.
Assume tap 5 is used and the burden is 2.64 VA at 5A and 580 VA at 20X. Also, the lead from the CT to relay is 0.4W.
The total secondary impedance at pick up:
Relay burden 2.64/52 = 0.106W
Lead resistance = 0.40W
Total impedance to CT terminals = 0.506W at 5A

26. Current Transformers
CT selection and performance evaluation for phase faults
Determine the total secondary burden
The total secondary impedance at 20X:
Relay burden 580/ = 0.058W
Lead resistance = 0.40W
Total impedance to CT terminals = 0.458W at 100A
It is frequently practical to add the burden impedance and the current algebraically (they should be combined phasorally)
Burdens are generally near unity power factor; hence Is tends to be near unity power factor. Ie (the excitation current) is around 90o lagging. Combine Is and Ie at right angle is a good approximation.

27. Current Transformers
CT selection and performance evaluation for phase faults
Determine the CT performance
When using a class T CT
Use the provided curves.
Use actual relay burden (0.506 or in this case)
When using a class C CT and performance by ANSI/IEEE standard
If a 600/5 multiratio CT with C100 rating is selected.
Vgh = (2500/20)*0.458 = V
The C /5 CT on the 100/5 tap can only develop,
Vgh = (100/600)*100 = V
This is not a good selection

28. Current Transformers
CT selection and performance evaluation for phase faults
Determine the CT performance
When using a class C CT and performance by ANSI/IEEE standard
An alternative is to use the 400/5 tap on the 600/5 C100 CT.
The secondary CT current at maximum load is (90/80=1.125A)
Tap 1.5 is selected.
Assume the relay burden at 100 A is 1.56W.
Total CT burden is equal to 1.96W (1.56W+0.40W)
Vgh = (2500/80)*1.96=61.25 V
The CT capability on the 400/5 tap is
Vgh = (400/600)*100 = 66.7 V

29. Current Transformers
CT selection and performance evaluation for phase faults
Determine the CT performance
When using a class C CT and performance with CT excitation curve

30. Current Transformers
CT selection and performance evaluation for phase faults
Determine the CT performance
When using a class C CT and performance with CT excitation curve
An alternative is to use the 400/5 tap on the 600/5 C100 CT.
The CT secondary resistance is 0.211W (Check figure)
Total impedance to excitation point ef.= 2.171W
The 1.5 A to develop voltage in the relay is
Vef = 1.5 * = 3.26 V
Ie = A
The pick up current is either120 or A, which is much smaller than 350 A.

31. Current Transformers
CT selection and performance evaluation for phase faults
Determine the CT performance
When using a class C CT and performance with CT excitation curve
For maximum fault current
Vef = (2500/80)*2.171 = 67.84
Ie = 0.16 A
Although this is near the knee of the saturation curve, the excitation current does not significantly decrease the fault current to the relay.

32. Current Transformers
Performance evaluation for ground relays

33. Current Transformers
Effect of unenergized CTs on performance

34. Current Transformers Effect of unenergized CTs on performance
Fault happens at phase A. Unfaulted CTs do not have current.
Assumption:
100:5 tap of a C100, 600:5 multiratio CT is used.
The secondary resistance of the CT, the leads, and the phase relay is 0.63 W.
The ground relay has 16 W on its 0.5A tap at 68o lag.
8 V (0.5A*16 W) will be established at the ground relay.
This voltage, less the small drop through the phase relay circuit, will appear across the secondary of the CTs at phase B & C

35. Current Transformers Effect of unenergized CTs on performance
It will have 0.38A flows through Ze of the Phase B & C CTs.
The current at phase A CT secondary terminal is 1.26 A or 1.24 A.
The exciting current of phase A CT is 0.41 A
It requires 33.4 A (( )*20) primary current to pick up the ground relay.
Only 10 A primary current required to pick up the ground relay if the exciting currents were neglected.

36. Current Transformers
Effect of unenergized CTs on performance

37. Current Transformers
Flux summation

38. Current Transformers
CT performance on the DC component

39. Current Transformers
Saturation on Symmetrical AC current input

40. Current Transformers
Saturation by the DC offset of the Primary AC current

41. Voltage Transformers
Voltage Transformer (VT) or Capacitor Coupled Voltage Transformer (CCVT)
Typical output voltage is 120 V line-to-line or 69.3 V per phase.

42. Voltage Transformers Typical VT ratio 1:1 2:1 2.5:1 4:1 5:1 20:1 40:1
60:1
100:1
200:1
300:1
400:1
600:1
800:1
1000:1
2000:1
3000:1
4500:1

43. Voltage Transformers
Internal configuration of CCVT

44. New Development Low signal level optical CT
Hall Effect current transducer