This document discusses overcurrent protection and different types of overcurrent relays. It describes the causes and effects of overcurrent, and introduces overcurrent protection using fuses, circuit breakers and overcurrent relays. It explains the operating principles of different types of overcurrent relays including attracted armature, definite time, and inverse definite minimum time (IDMT) relays. Examples are provided to illustrate how to select settings for IDMT relays in a power system to achieve coordinated overcurrent protection.

1. CHAPTER 3
OVERCURRENT PROTECTION
OVERCURRENT PROTECTION

2. OVERCURRENT
CAUSES:
1 Failure of insulation flashover between
1. Failure of insulation – flashover between
phases caused by equipment failure,
lightning strikes, metal parts falling on to
live equipment
2. Mistake – connect portable earth to a live
busbar
busbar
EFFECTS:
1 Injuries to personnel
1. Injuries to personnel
2. Damage to equipment – melting of copper
parts, fires..

3. INTRODUCTION OF OVERCURRENT PROTECTION
INTRODUCTION OF OVERCURRENT PROTECTION
 Definition:
 A Protection Relay is a relay that responds to
 A Protection Relay is a relay that responds to
abnormal conditions in an electrical power system,
and controls a circuit breaker so as to isolate the
faulty section of the system.
 Overcurrent Protection is achieved by the use of fuses,
by direct-acting trip mechanisms on circuit breakers or
by relays.

4. FORMS OF OVERCURRENT PROTECTION
FORMS OF OVERCURRENT PROTECTION
 Overcurrent relay – trip CB or contactor
 Overcurrent relay trip CB or contactor
 Fuses – Good short circuit protection, cheap
but must be replace once it blow
but must be replace once it blow
 MCCBs/MCBs – Internal thermal element for
l d t ti & i t t g ti
overload protection & instantaneous magnetic
element for short circuit protection

5. RELAY FUNCTIONS
1. Can measure an electrical quantity, i.e:
l &
voltage & current
2. Send the signal to activate a sudden pre-
determined change or changes in one or
more electrical circuit, i.e: to trip a breaker
3. Receive a controlling signal & then relays
the signal to activate another device, i.e: to
g ,
reduce the speed of motor.

6. TYPES OF OVERCURRENT RELAY
BASIS PRINCIPLES: OPERATES WHEN IFAULT > IRELAY SETTING
FAULT RELAY SETTING
• Iron armature with coil carrying current from CT
• When Ifault > Irelay setting, the relay will pull of armature,
Instantaneous fault relay setting, y p ,
overcomes the spring force & closes contacts to trip
CB
• Operates typically 20ms – 40ms
Instantaneous
Relay
• Combination of instantaneous relay & timer
• Operate when I > I for preset time
Definite time • Operate when Ifault > Irelay setting for preset time
• Need settings for current and time delay
relay
• Consists of rotating aluminium disc driven by
electromagnet, which is energized by the CT current
• when I < Irelay setting , disc remains stationary
• when I > Irelay setting , disc moves, completes its travel,
Inverse Definite
Minimum Time relay setting
relay contact closes, CB trip
• I increase, disc rotates faster, operating time is
quicker
(IDMT) Relay

7. ATTRACTED ARMATURE (INSTANTANEOUS
RELAYS)
RELAYS)
ADVANTAGES
 ADVANTAGES:
i. Can be used on a.c and d.c systems
V f b f h h l h f l
ii. Very fast because of the short length of travel
 If time delay is required, then a timer is required. Once
the time is set the breaker will trip at the set time
the time is set, the breaker will trip at the set time
regardless of the current.
 This type of time delay is known as Definite Time Lag
(DTL)

8. CONSTRUCTION OF AN ATTRACTED ARMATURE
CONSTRUCTION OF AN ATTRACTED ARMATURE

9. INDUCTION DISC RELAYS
 PRINCIPLE OF OPERATION
i. The current flowing in the primary coil will produce a
primary magnetic flux
primary magnetic flux
ii. The primary flux will induce an emf in the secondary coil.
The emf in the secondary coil will cause a current to flow
through the winding of the lower magnet.
iii. The secondary current lags behind the secondary emf.
This current creates a magnetic field in the lower magnet.
g g
iv. Both lower magnet and primary magnetic flux will act on
the induction disc and cause it to rotate.
The torque created by the magnetic fields is counteracted
v. The torque created by the magnetic fields is counteracted
by the tension of a spiral spring. When the turning torque
overcomes the force of the spring, the relay will operate –
this determines the minimum operating current of the
this determines the minimum operating current of the
relay

10. CONSTRUCTION OF AN INDUCTION DISC RELAY
CONSTRUCTION OF AN INDUCTION DISC RELAY

11. CONT…INDUCTION DISC RELAYS
o Current Setting Adjustment
- Taps on the coil are used to adjust the operating current of the relay.
The taps are selected by the insertion of a single pin plug in the
The taps are selected by the insertion of a single pin plug in the
appropriate position of a ‘plug bridge’.
- The current setting of a relay is referred to as plug setting (PS). PS is
marked as a percentage i e: 50% 75% 100% 125% 150% 175% and
marked as a percentage, i.e: 50%, 75%, 100%, 125%, 150%, 175% and
200%.
o Time Setting
- As the disc rotates, a point on the disc will take a fixed time to move
from one position to another.
- In Figure above, the moving contact at Position A will take a longer time
to meet with the fixed contact than if it was at Position B.
- The position of the moving contact can be adjusting by turning the Time
Multiplier Setting (TMS) knob. The TMS varies from (0.1 - 1.0)

12. CONT…INDUCTION DISC RELAYS
o Instantaneous Trip
For very high currents the IDMT relay has an instantaneous
- For very high currents, the IDMT relay has an instantaneous
trip. The instantaneous trip is of the attracted armature
type.
- For a fault near the generating source, the current will be
very high. In this situation, the relay must trip
instantaneously.
instantaneously.

13. IDMT RELAY
IDMT RELAY
 The time for relay to give a trip signal depends
 The time for relay to give a trip signal depends
on:
 1 Magnitude of fault current
 1. Magnitude of fault current
 2. Current Setting (Plug Setting, PS)
 3 Time Multiplier Setting (TMS)
 3. Time Multiplier Setting (TMS)

15. EQUATION INVOLVED IN IDMT RELAY
SETTING
SETTING
%
100
% max

 load
I
PS %
100
%
 g
relayratin
I
ratio
CT
PS
Plug setting Multiplier :
g
relayratin
fault
I
ratio
CT
PS
I
PSM



ug se g u p e
14
0
TMS can be obtained through time/current characteristic curves or equations:
1
14
.
0
02
.
0


PSM
Tchar
T
char
operate
T
T
TMS 

16. QUESTION
QUESTION
 Figure below shows a radial system attached with IDMT relay at point A, B and C.
T bl b l i d t il th CT d l ti d t h i t R l t
Table below gives details on the CT and relay rating used at each point. Relay at
point B is set with Plug Setting (PS) of 75% and it will operate within 0.84 sec if a
three phase fault with fault current of 8100A occurs close to point C. On the other
hand, if a three phase fault with fault current of 10kA occurs near to point B, the
, p p ,
relay will operate within 0.58 sec. Determine the appropriate setting (PS and TMS)
for each relay by using 0.6 sec of time delay between relays. Use IDMT relay
characteristics for reference.
A B C
CT ratio 300/5 700/5 500/5
Relay Rating 5A 5A 5A

17. IDMT RELAY CHARACTERISTICS

18. FIGURE & TABLE
FIGURE & TABLE
A B C
CT ratio 300/5 700/5 500/5
R l R i 5A 5A 5A
Relay Rating 5A 5A 5A

19. ANSWER
ANSWER
The calculation must consider every relay such
The calculation must consider every relay such
as Relay A, B and C on the single diagram
circuit.
circuit.

20. RELAY C
RELAY C
Step 1: Find the load current, IL.
2.5
437 39
L d
S M
I A
  437.39
3 3(3.3 )
Load
Line
I A
V kV
Step 2: Find the Plug Setting, PS.
437.39
0.87475
500
Load
I
PS   
 Select 100%
0.87475
500
Re 5
5
Ratio
PS
CT layRating
 
 Select 100%

21. FAULT AT C (PRIMARY)
Step 3: Find the Plug Setting Multiplier (PSM)
8.1 8.1
16.2
500
Re 500
1 5
5
Fault
Ratio
I kA kA
PSM
PS CT lay Rating
   
   
Step 4: Find the operating times toperate
5
Step 4: Find the operating times, toperate
0.84 0.6 0.24 sec
operate relay B delay each relay
t t t
    
operate relay B delay each relay

22. FAULT AT C (PRIMARY)
FAULT AT C (PRIMARY)
Step 5: Find the TMS from the IDMT graph
Step 5: Find the TMS from the IDMT graph
, 0.1
T M S F rom curve T M S
 
 TMS also can obtain by formula if not state refer to the IDMT characteristics
Step 6: Conclude the final setting at relay
Step 6: Conclude the final setting at relay
Re [ 100%, 0.1]
Setting at lay C PS TMS
 

23. RELAY B
 Step 1: Find the load current, IL.
No calculation needed as PS for relay B has been given. (IL is detemined to
compute the value of PS)
 Step 2: Find the Plug Setting, PS.
Given PS = 75%

24. FAULT AT C (BACK UP)
Step 3: Find the Plug Setting Multiplier (PSM)
8100
Gi I A
8100
Fault
Given I A

4
.
15
700
8100


PSM
Step 4: Find the operating times toperate
5
5
700
75
.
0 

Step 4: Find the operating times, toperate
0 84sec
t t 0.84sec
operate relay B
t t
 

25. FAULT AT C (BACK UP)
FAULT AT C (BACK UP)
Step 5: Find the TMS from the IDMT graph
Step 5: Find the TMS from the IDMT graph
, 0.3
TMS Fromcurve TMS
 
 TMS also can obtain by formula if not state refer to the IDMT characteristics
Step 6: Conclude the final setting at relay
Step 6: Conclude the final setting at relay
Re [ 100%, 0.3]
Setting at lay B PS TMS
 

26. RELAY A
 Step 1: Find the load current, IL.
5
262.43
3 3(11 )
Load
Line
S M
I A
V kV
  
 Step 2: Find the Plug Setting, PS.
262.43
0 87
Load
I
PS   
Select 100%
0.87
300
Re 5
5
Ratio
PS
CT lay Rating
 
 Select 100%

27. FAULT AT B
 Given Fault Current
10
Fault
Given I kA

 Given operating time
0.58sec
operate
t 

28. FAULT AT B (BACK UP)
( )
 Step 3: Find the fault current
Step 3: Find the fault current
10
Load
Given I kA

 Transformer here, the current will CHANGE
,
3.3k
k k
3.3
10 3
11
Faultnew
k
I k kA
k
  

29. FAULT AT B (BACK UP)
Step 3: Find the Plug Setting Multiplier (PSM)
3
Gi I kA
3
Fault
Given I kA

3 3
10
300
R 300
Fault
I kA kA
PSM
PS CT l R i
   
Step 4: Find the operating times toperate
300
Re 300
1 5
5
Ratio
PS CT lay Rating
   
Step 4: Find the operating times, toperate
0.58 0.6 1.18sec
l B d l
t t t
    
0.58 0.6 1.18sec
operate relay B delay
t t t
 

30. FAULT AT B (BACK UP)
FAULT AT B (BACK UP)
Step 5: Find the TMS from the IDMT graph
Step 5: Find the TMS from the IDMT graph
, 0.4
TMS Fromcurve TMS
 
 TMS also can obtain by formula if not state refer to the IDMT characteristics
Step 6: Conclude the final setting at relay
Step 6: Conclude the final setting at relay
]
4
.
0
%,
100
[ 

 TMS
PS
A
relay
for
Setting ]
,
[
y
f
g

31. EXAMPLE 2
A 33kV power system shown below is installed with IDMT relays at each substation
to provide overcurrent protection scheme. As an engineer, you are required to
determine the Tap Setting (TS) and Time Dial Setting (TDS) for each relay so that
the protection system will function well according to the data given in Table below In
the protection system will function well according to the data given in Table below. In
the design, you have to ensure relay at substation C operates within 0.23 second.
You also should consider time discrimination between each location is 0.5 second.
33 kV C
B
A
Substation
Maximum Fault
CT Ratio
Relay Rating Maximum Load
Substation
Current (A)
CT Ratio
(A) Current (A)
A 5074 300/5 5 252
B 2975 300/5 5 148
C 1925 200/5 5 96

32. SOLUTION
%
48
100
96

PS
Relay C (Fault at C)
%
50
%
48
100
5
5
200






PS
PS
25
.
19
5
5
200
5
.
0
1925




PSM
14
0
23
.
0
5
 s
tope
1
0
23
.
0
298
.
2
1
25
.
19
14
.
0
02
.
0





TMS
tchar
]
1
.
0
%,
50
[
Re
1
.
0
298
.
2





TMS
PS
A
lay
for
Setting
TMS

33. RELAY B (FAULT AT C)
%
49
100
300
148



PS
83
12
1925
%
50
300




PSM
PS
73
.
0
5
.
0
23
.
0
83
.
12
5
5
300
5
.
0





s
t
PSM
67
.
2
1
83
.
12
14
.
0
73
.
0
5
.
0
23
.
0
02
.
0




t
s
t
char
ope
Setting for Relay B =
27
.
0
67
.
2
73
.
0


TMS
[PS = 50%, TMS = 0.27]

34. RELAY B (FAULT AT B)
PSM 83
.
19
300
2975


27
2
14
.
0
5
5
300
5
.
0 

t
TMS
t
ope
char 27
.
2
1
83
.
19 02
.
0




s
t
t
TMS
ope
char
61
.
0
27
.
2
27
.
0 




35. RELAY A (FAULT AT B)
%
84
100
300
252



PS
2975
%
100
300

 PS
92
.
9
5
5
300
1
2975




PSM
98
.
2
1
92
.
9
14
.
0
02
.
0



tchar
37
.
0
11
.
1
11
.
1
5
.
0
61
.
0





TMS
s
tope
Setting for Relay A =
[PS = 100%, TMS = 0.37]
98
.
2

36. OVERCURRENT RELAY CONNECTIONS
- The output of the current transformers are connected to the
coils of the respective overcurrent relays as shown figure
below
below.
- If there is a fault in any phase, the relay in that phase will
operate and trip the breaker.

37. THREE OVERCURRENT RELAYS & ONE
EARTH FAULT RELAY
EARTH-FAULT RELAY
 The protection scheme using three overcurrent
l & h f l i h i fi b l
relays & one earth-fault is shown in figure below.
 I thi t ti h th h t ill
 In this protection scheme, the phase currents will
still flow though the overcurrent relays. The fault
current due to an overload or short-circuit between
phases or between phases and neutral will be
detected by the overcurrent relays.
 The earth-fault relay will still monitor the sum of the
currents of the three phases and neutral
currents of the three phases and neutral.

38. DIAGRAM OF THREE OVERCURRENT RELAYS & ONE
EARTH-FAULT RELAY

39. TWO OVERCURRENT RELAYS & ONE EARTH-
FAULT RELAY
 Figure below shows that a protection scheme using
g p g
two overcurrent relays and one earth fault relay.
 In this scheme, an overload or short circuit between
t h ill b d t t d b th t
any two phases will be detected by the overcurrent
relays.
 The earth-fault relay is monitoring the sum of the
 The earth fault relay is monitoring the sum of the
currents in the three phases. If there is a short-circuit
between one of the phase and earth, the earth-fault
relay will detect the current imbalance and trip the
circuit breaker.

40. DIAGRAM OF TWO OVERCURRENT RELAYS
& ONE EARTH-FAULT RELAY

41. DIGITAL OVERCURRENT & GROUND
OVERCURRENT RELAY

42. OVER CURRENT RELAY FOR GENERATOR
OVER CURRENT RELAY FOR GENERATOR