Differential Protection
A differential relay responds to vector difference between two or more similar electrical quantities.
Requirements:
(a) It must have at least two actuating quantities, say I, and I₂
(b) The actuating quantities should be similar in nature i.e. current/current or voltage/voltage.
(c) The relay responds to the vector difference between the two quantities, i.e. to (𝐼𝐼1 − 𝐼𝐼2 ), which includes
magnitude and/or phase angle difference. When this vector difference exceeds a predetermined amount,
the relay operates
The differential protection is frequently called unit protection. The vector difference is achieved by suitable
connection of CT and PT secondary. Most differential relay applications are of the current differential type.
Applications:
•
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•
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protection of generators and generator-transformer unit
protection of transformer
protection transmission line by pilot wire scheme
protection of x-mission line by phase comparison carrier current
protection of large motors
bus zone protection.
circulating current differential protection (Mertz Price Protection)
A simple example of circulating current differential relay is shown in the Fig. 1.
Fig. 1: A simple differential relay application
The block represents the system element (transformer, motor, generator, bus etc.) to be protected. Two
suitable CTs are connected series as shown in the figure with the help of the pilot wires. The relay operating
coil is connected between the mid points (equipotential point) of the pilot wires. The secondary current of
the CTs will circulate through the combined impedance of the pilot wires and the CTs. When the operating
coil is not connected between the equi-potentail points, even though the current through each CT is same,
the burden on the two CTs is unequal. This causes the heavily loaded CTs to saturate during through faults,
thereby causing dissimilarities in the characteristic of two CTs which results in mal-operation of the relay.
The relay could be any of the ac type that has been discussed but preferably attracted armature type.
When there is no internal fault, the current entering the protected element is equal in magnitude and phase
to current leaving the protected element. The CTs are of such a ratio that during normal operating condition
or for external faults (through faults) the secondary current of CTs are equal. These currents say 𝐼𝐼1 and
𝐼𝐼2 circulate in the pilot wires. The polarities of CTs are such that 𝐼𝐼1 and 𝐼𝐼2 } are in the same direction during
normal operation or through fault condition. The differential current (𝐼𝐼1 − 𝐼𝐼2 ) which flow through relay coil
is zero. So, the relay does not operate
Fig 2: Differential current for internal fault
If the fault occurs somewhere in the protected zone, the direction of fault current in the circuit is shown in
Fig.2. The differential current, (𝐼𝐼1 − 𝐼𝐼2 ) through the relay coil is not equal to zero i.e., (𝐼𝐼1 − 𝐼𝐼2 ) ≠ 0. If the
operating torque due to this differential current exceeds the restraining torque the relay will operate.
Drawback of the simple circulating current differential relay
The above form of protection was assumed on the fact that the two CTs used are identical. But in practice it
is not true. CT of the type normally used do not transform their currents very accurately under transient
condition in particular This is due to the fact that the short circuit current is offset i.e., it contains de
components. Suppose the two CTs under normal conditions differ in their magnetic properties slightly in
terms of different amounts of residual magnetism or in terms of unequal burden on two CTs, one of them
will saturate earlier during short circuit current flow (having high degree of offset) and thus two CTs will
transform their primary current differently even for
a through fault condition. This effect is more
pronounced especially when the scheme is used for
the protection of transformers.
To accommodate these features, Mertz-Price
protection is modified by biasing the relay. This
commonly known as biased differential or
percentage differential protection as shown in Fig. 3.
The relay consists of an operating coil and a restraining coil. The operating coil is connected to the midpoint
of the restraining coil. Under through fault condition due to dissimilarities in CTs the differential current in
the operating coil is (𝐼𝐼1 − 𝐼𝐼2 ) and the equivalent current in the restraining coil is (𝐼𝐼1 + 𝐼𝐼2 )/2.
The torque developed by the operating coil is proportional to the ampere-turn i.e., T ∝ (𝐼𝐼1 − 𝐼𝐼2 ). No, where
No is the number of turns in the operating coil. The restraining torque (due to restraining coil) is
𝑇𝑇𝑟𝑟 ∝
At balance.
(𝐼𝐼1 +𝐼𝐼2 )𝑁𝑁𝑟𝑟
2
, where N, is the number turns of restraining coil
(𝐼𝐼1 − 𝐼𝐼2 )𝑁𝑁0 =
𝑜𝑜𝑜𝑜,
(𝐼𝐼1 − 𝐼𝐼2 )
(𝐼𝐼1 + 𝐼𝐼2 )
(𝐼𝐼1 +𝐼𝐼2 )𝑁𝑁𝑟𝑟
2
=
𝑁𝑁𝑟𝑟
𝑁𝑁0
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
= 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
This means that as the differential current is increased, so do the restraining current. The operating
characteristic is shown in Fig. 4 below.
It is clear from the characteristic that except for the effect of the control spring at low currents, the ratio of
the differential operating current to the average restraining current is a fixed percentage. This is why it is
called percentage differential relay. The relay described is called current balance relay.
Drawbacks of simple circulating current differential relay
1. Two CTs are not exactly identical.
2. Two CTs do not transform their currents very accurately particularly during transient condition
because of presence of DC offset in the fault current.
3. Two CTs may differ in their magnetic properties slightly in terms of different amount of residual
magnetism or in terms: unequal burden on the CTs.
4. This problem is very acute when this scheme is used in transformer protection.
Distance Protection
Distance protection is a general term which is given to a group of non-unit protective systems in which
relaying devices measure the impedance or reactance of the line fault of which are proportional to the
distance between the measuring point and the fault.
Actuating quantity: Voltage and Current
(The torque produced is such that when V/I ratio reaches below a set value, the relay operates)
The ratio of V/I is measured at the relay location i.e., at the location of CT and PT.
If the fault is nearing to the relay location, the voltage at the relay point is lesser and opposite happens if
the fault is farther from the relay location.
Hence each value of V/I measured from the relay location corresponds to distance between relaying point
and the fault. Hence such protection is called distance protection.
Applications
1. Long high voltage transmission lines
2. Becoming increasingly popular, in a modified form, on lines operating of 66 kV, 33 kV or even 11 kV
lines
Modern distance relay can provide very high-speed protection and it is simple of apply. It can be used as a
primary and back up protection.
A distance relay is one whose operation is based on the measurement of impedance, reactance or
admittance between relay location and the fault point.
Distance relay can be classified as
1. Impedance relay
2. Reactance relay
3. Mho relay
Impedance relay
The ratio of voltage and current across a branch gives the impedance of the branch
𝑉𝑉
= 𝑍𝑍 = 𝑅𝑅 + 𝑗𝑗𝑗𝑗
𝐼𝐼
The current gives the operating torque and the voltage gives the restraining torque.
The general torque equation of an impedance relay is
𝑇𝑇 = 𝐾𝐾1 𝐼𝐼 2 − 𝐾𝐾2 𝑉𝑉 2 − 𝐾𝐾3
Where, T= net torque
𝐾𝐾1 𝐼𝐼 2 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
𝐾𝐾2 𝑉𝑉 2 = 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
𝐾𝐾3 = 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
At threshold, T= 0
𝐾𝐾2 𝑉𝑉 2 = 𝐾𝐾1 𝐼𝐼 2 − 𝐾𝐾3
𝑜𝑜𝑜𝑜,
𝑉𝑉 2 𝐾𝐾1
𝐾𝐾3
=
−
;
2
𝐼𝐼
𝐾𝐾2 𝐾𝐾2 𝐼𝐼 2
𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑖𝑖𝑖𝑖 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑖𝑖. 𝑒𝑒. , 𝐾𝐾3 = 0
𝑉𝑉
𝐾𝐾1
= 𝑍𝑍 = � = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐼𝐼
𝐾𝐾2
Therefore, the relay will operate if the fault impedance 𝑍𝑍𝑓𝑓 (𝑖𝑖. 𝑒𝑒. ,
𝐾𝐾
�𝐾𝐾1 , if 𝑍𝑍𝑓𝑓 is greater then Z, then the relay will not operate.
2
𝑉𝑉𝑓𝑓
𝐼𝐼𝑓𝑓
) is less then a certain value of Z given by
The characteristic in terms of V and I is shown below:
Fig 1: operating characteristic of an impedance relay
By adjustment, the slope of the operating characteristic can be changed so the relay will respond to all
values of impedance less then any desired upper limit.
A much more useful way of showing the operating characteristic of impedance relay is by means of
impedance diagram or R-X diagram as shown below. Since the relay operates for certain value, less then
the set value of Z, the operating characteristic is a circle of radius Z
Fig 2: R-X diagram of an impedance relay
If 𝑍𝑍𝑓𝑓 < 𝑍𝑍 the relay will operate; and if 𝑍𝑍𝑓𝑓 > 𝑍𝑍 the relay will not operate, this is a rule regardless of phase
angle between V and I.
At very low current where the operating characteristic of Fig 1 departs from a straight line because of the
control spring, the effect on fig2 is to make the radius of the circle smaller.
Disadvantages of impedance relay
1. It is non directional; it will see fault from in front of and behind the relaying point and therefore
requires a directional element to give correct discrimination.
2. It is affected by are resistance in the fault path
3. It is highly sensitive to power swings, because of the large area covered by the impedance circle
To make an impedance relay directional, a directional unit is incorporated with the impedance relay.
Three Zone Distance Protection
Correct co-ordination between distance relays on a power system is obtained by controlling the reach
settings and tripping times of the various zones of measurements. A conventional distance protection will
comprise an instantaneous directional zone-1 protection and one or more-time delayed zones. Typical three
zone distance protection scheme is shown in the figure which consists of two-line sections AB and CD.
The protection scheme is divided in three zones. Say for relay at A, the three zones are Zia, Z2a and 23a. The
Zia represents the reach setting of relay at A for the instantaneous zone-1 protection and corresponds to
approximately 80% impedance (length) of the line AB. No intentional time lag is provided for this zone. The
ordinate shown corresponding to Zia gives the operating time when the fault takes place in this zone. It is to
be noted here that the first zone is extended only up to 80% and not 100% length of the line AB as the relay
impedance measurement will not be very accurate towards the end of the line especially when the current
is offset.
Second zone, Z2a for relay at A covers remaining 20% length of the line AB and 20% of the adjoining line i.e.,
line CD. In case of a fault in this section relay at A will operate when the time elapsed corresponds to the
ordinate Z2a. The main idea of the second zone is to provide protection for the remaining 20% section of the
line AB. In case of an arcing fault in the section AB which adds to the impedance of the line as seen by the
relay at A, the adjustment is such that the relay at A will see that impedance in second zone and will operate.
This is why the second zone is extended into the adjoining line. The operating time of the second zone is
normally about 0.2 to 0.5 seconds.
The 3rd zone unit at relay A provides back up protection for faults in the line CD. That means, if there is a
fault in the line CD and if for some reasons the relay at C fails to operate then relay at A will provide backup
protection. The delay time for the 3rd zone is usually 0.4 to 1.0 second.
In case the feeder is being fed from both the ends and say the fault takes place in the second zone of the
line AB (20% of the line AB), the relay at B will operate instantaneously (because it lies in the first zone of
BA) whereas the fault lies in the second zone of the relay at A. This is undesirable from stability point of view
and it is desirable to avoid this delay. This is made possible when the relay at B gives an inter trip signal to
the relay at A in order to trip the breaker quickly rather than waiting for zone 2 tripping.
Reactance Relay
In this relay the operating torque is proportional to the square of the current (𝜏𝜏 ∝ 𝐼𝐼 2 ) and the restraining
torque is proportional to the product of voltage and current and sine of the angle between them.
The reactance type distance relay has reactance measuring unit. The reactance measuring unit has an over
current element developing positive torque and a directional element (VIsin𝜃𝜃) which either gives a positive
or negative torque. Hence, the reactance relay is an overcurrent relay with directional restraint.
The directional element is so designed that its maximum torque angle is 90° i.e., 𝜏𝜏 = 90° in the torque
equation
Where,
𝑇𝑇 = 𝐾𝐾1 𝐼𝐼 2 − 𝐾𝐾2 𝑉𝑉𝑉𝑉 cos(𝜃𝜃 − 𝜏𝜏)
T = net torque
I= current to rely coil
V= voltage to relay coil
Θ = phase angle between V4I
𝜏𝜏= angle of the maximum torque
According to the design of the reactance relay
cos( 𝜃𝜃 − 𝜏𝜏) = cos(𝜃𝜃 − 90°) = sin 𝜃𝜃
∴ 𝑛𝑛𝑛𝑛𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 , 𝑇𝑇 = 𝐾𝐾1 𝐼𝐼 2 − 𝐾𝐾2 𝑉𝑉𝑉𝑉 sin 𝜃𝜃
on the verge of relay operation, T=0
∴ 𝐾𝐾1 𝐼𝐼 2 = 𝐾𝐾2 𝑉𝑉𝑉𝑉 sin 𝜃𝜃
𝑜𝑜𝑜𝑜,
𝑍𝑍 sin 𝜃𝜃 =
For the operation of the relay,
𝐾𝐾1
,
𝐾𝐾2
𝑍𝑍𝑓𝑓 <
𝑜𝑜𝑜𝑜,
𝑉𝑉
𝐾𝐾1
sin 𝜃𝜃 =
𝐾𝐾2
𝐼𝐼
𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝐾𝐾1
𝐾𝐾2
This means for the operation of the relay the reactance seen by relay should be smaller than the reactance
for which the relay has been set. The characteristic of the relay is shown below,
The important point about this characteristic is that the resistive Component of the impedance has no effect
on the operation of the relay. It responds only to the reactance component of the impedance. we can also
say that the setting of the reactance relay does not vary in presence of arc resistance, because it is designed
to measure only the reactive component of the line. This relay will operate for all impedances whose heads
lie below operating characteristic whether below or above the R-axis. This relay, as can be seen from the
characteristic is a non-directional relay.
Ground fault protection (Earth fault protection)
When the fault current flows through earth return path, the fault is called earth fault (E/F). The other faults
which do not involved ground are called phase faults. Since earth faults are relatively frequent, E/F
protection is necessary in most cases. When separate E/F protection is not economical, the phase relays
sense the E/F currents. However, such protection lacks sensitivity. More sensitive protection against E/Fs
can be obtained by using a relay which responds only to the residual current of the system, since the residual
component exists only when the fault current flows to the earth. The E/F current relay is, therefore,
completely unaffected by the load currents whether balanced or not.
Low current setting is permissible to increase the relay sensitivity for the E/Fs. But this low current setting
may be limited in magnitude by neutral earthing impedance or by earth contact resistance.
Residual connection of CTs to E/F relay
Fig. (a): E/F relay connected in residual circuit
In the absence of E/F, the vector sum of three-line currents is zero. Therefore, the vector sum of three
secondary currents is also zero.
𝐼𝐼𝑎𝑎𝑎𝑎 + 𝐼𝐼𝑏𝑏𝑏𝑏 + 𝐼𝐼𝑐𝑐𝑐𝑐 = 0 = 𝐼𝐼𝑟𝑟𝑟𝑟 (𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
The E/F relay is connected in such a way that 𝐼𝐼𝑟𝑟𝑟𝑟 flows through it as shown in the fig. (a). So in the absence
of earth fault the relay will not operate. However, in the presence of earth fault
𝐼𝐼𝑎𝑎𝑎𝑎 + 𝐼𝐼𝑏𝑏𝑏𝑏 + 𝐼𝐼𝑐𝑐𝑐𝑐 ≠ 0
and if the residual current is above the pickup value, the E/F relay will operate. In this scheme E/F at any
location near or away from the CTs location can cause the residual current to flow. Hence the protected zone
is not defined. Such protection is called unrestricted E/F protection. For selectivity directional E/F protection
is necessary.
E/F relay connected in neutral to earth circuit fig (b)
𝐼𝐼𝑛𝑛 = 𝐼𝐼𝑎𝑎 + 𝐼𝐼𝑏𝑏 + 𝐼𝐼𝑐𝑐 , the magnitude of E/F current is
dependent on the type of earthing (resistance,
reactance or solid) and the location of fault.
In this type of protection, the zone of the protection
cannot be accurately defined. The protected area is
not restricted to the transformer/generator
winding alone. The relay senses the E/Fs beyond the transformer/generator winding. Hence such protection
is called unrestricted E/F protection.
Combined E/F and phase fault protection:
It is convenient to incorporate phase fault relays and E/F relay
in a combined phase and earth fault protection.
Three phase fault O/C relays are provided on three phases and
the E/F relay is residually connected as shown in fig.(c)
Instead of using three O/C relay for three phases, very
often phase fault O/C relays are provided on only two
phases since these will detect any interphase fault; the
connection to the E/F relay are unaffected by this
consideration. The arrangement is illustrated in fig.(d).
OVERCURRENT RELAYS
The operating time of all O/C relays tends to become asymptotic to a definite minimum value with increase
in current. This is inherent in electromechanical relays due to saturation of the magnetic circuit. So, by
varying the point of saturation different characteristic are obtain; these are:
1) Definite time → 𝐼𝐼𝐼𝐼 = 𝑡𝑡
2) Inverse Definite Minimum Time (IDMT) → 𝑡𝑡 =
3) Very inverse → 𝑡𝑡 =
13.5
𝐼𝐼−1
.14
𝐼𝐼 .02 −1
80
4) Extremely inverse → 𝐼𝐼 2 𝑡𝑡 = 𝐾𝐾 ; 𝑡𝑡 = 2
5) Instantaneous O/C relays
𝐼𝐼 −1
These characteristics are obtained by induction disc or induction cup relay
As we have seen earlier that the torque of these relays is proportional to 𝜑𝜑1 𝜑𝜑2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. where 𝜑𝜑1 and 𝜑𝜑2 are
the two fluxes cutting the disc or cup and 0 is the angle between them if both fluxes are produced by the
same quantity, current or voltage operated relays, then below saturation the torque is proportional to 𝐼𝐼 2 ,
the coil current. i.e., 𝑇𝑇 = 𝐾𝐾𝐾𝐾 2
If the core is made to saturate at very early stage with the result that by increasing 1, k decreases so that the
time of operation remains the same over the working range. This type of characteristic is shown by the curve
(1) and is known as definite time.
if the core is made to saturate at a later stage, the characteristic assumes the shape of carve (2), known as
IDMT. The time current characteristic is inverse over some range and the after saturation assume the definite
time form. At low values of operating current, the shape of the curve is determined by the effect of the
restraining force of the control spring, while at high values, the effect of saturation predominates. Different
time setting multiplier (TSM) are obtained by varying the
travel of the disc required to close the contacts. The higher the TSM, the greater will be the spring restraining
force. As the disc moves in the tripping direction, winding up the spring, more and more conducting metal
of the disc comes into play in the active air-gap of the electromagnet to increase the electric torque, thus
compensating the increasing the spring torque.
If the saturation occurs at a still later stage the characteristic become very inverse, shown in curve (3). The
curve (4) shows extremely inverse characteristic. The equation describing the curve (4) is approximately of
the form 𝐼𝐼 2 𝑡𝑡 = 𝐾𝐾, where I is the operating current and t is the operating time.
Instantaneous O/C relay is one in which no intentional time delay is provided for the operation. The time of
operation of such relays is approximately 0.01 sec. This characteristic is achieved by hinged attracted
armature relay. This instantaneous relay is more effective when the impedance 𝑍𝑍𝑠𝑠 between the source and
the relay is small compared with the impedance 𝑍𝑍1 of the section to be protected. With so fast in operation
it is likely that the relay may operate on transients beyond the normal range of setting.
A relay is said to over reach when it operates at a value, lower than its setting. The relay is set for symmetrical
currents but responds to both symmetrical and offset current waves that persist for a few cycles. The over
reach depends on the design of the relay as well as on the parameters of the power system on which it is
used. The X/R ratio of the system from source to the fault controls the degree of offset and the rate of
decrement of the current wave and the ratio 𝑍𝑍𝑠𝑠 /𝑍𝑍1 determines the degree of over reach which will occur
Non-directional Time and current grade schemes
Among the various possible methods used in correct relay coordination, the following methods are well
known:
1) Time grading
2). Current grading
3) Combination of current and time grading
The common aim of all three methods is to give correct discrimination. That means, each one must select
and isolate only the faulty section, leaving the rest of the power system undisturbed. All are used for the
protection of radial feeders and each has its application depending on the power system operating condition.
Discrimination by time (time grade system)
In this method an approximate time interval is given by each of the relays controlling the circuit breakers in
a power system to ensure that the breaker nearest to the fault opens first A simple radial distribution system
is shown in the following fig. I to illustrate the principle.
Fig. 1: Radial system with time discrimination
Circuit breaker protection is provided at H. J, K and L. i.e., at the in-feed end of each section of the power
system. Each protection unit comprises a definite time delay O/C relay in which the operation of the current
sensitive element simply initiates the time delay element. Provided the setting of the current element is
below the fault current value this element plays no part in the achievement of the discrimination. The
operating time of the relay is independent of the level of over current.
It is the time delay element, therefore, which provides the means of discrimination. The relay at H is set at
the shortest time delay possible to allow a fuse to blow for a fault on the secondary side of the transformer
G. Typically, a time delay of 0.25 sec. is adequate. If the fault occurs at F, the relay at H will operate in 0.25
sec. and the subsequent operation of the CB at H will clear the fault before the relays at J, K and L have time
to operate.
The main disadvantage of this method of discrimination is that the longest fault clearance time occurs for
faults in the section closest to the power source, where the fault level (MVA) is highest
Current graded system (Discrimination by current)
Discrimination by current relies on the fact that fault current varies with the position of the fault because of
the difference in impedance values between the source and the fault. Hence, the relays controlling the
various CBs are set to operate at suitably tapered values of current such that only the relay nearest to the
fault trips its breaker. Figure-B 2 illustrates the method.
For a fault at F1, the system short-circuit current is given by:
I=6350/(Zs+ZL1)
Where, ZS = source impedance =112/250=0.482 Ohm
ZL1=cable impedance between C and B = 0.24 Ohm
Hence, I=6350/0.725=8800 Ampere
So, a relay controlling the circuit breaker at C and set to operate at a fault current of 8800A would in theory
protect the whole of the cable section between C and B. However, there are two important practical points
that affect this method of co-ordination:
It is not practical to distinguish between a fault at F1 and a fault at F2, since the distance between these
points may be only a few meters, corresponding to a change in fault current of approximately 0.1%.
In practice, there would be variations in the source fault level, typically from 250MVA to 130MVA. At this
lower fault level the fault current would not exceed 6800A, even for a cable fault close to C. A relay set at
8800A would not protect any part of the cable section concerned.
Discrimination by current is therefore not a practical proposition for correct grading between the circuit
breakers at C and B. However, the problem changes appreciably when there is significant impedance
between the two circuit breakers concerned. Consider the grading required between the circuit breakers at
C and A in Figure B. Assuming a fault at F4, the short-circuit current is given by:
I=6350/(ZS+ZL1+ZL2+ZT)
Where, ZS = source impedance=112/250=0.485 Ohm
ZL1 = cable impedance between C and B = 0.24 Ohm
ZL2 = cable impedance between B and 4MVA transformer = 0.04 Ohm
ZT = transformer impedance=0.07 (112/4) =2.12 Ohm
Hence, I=6350/2.885=2200 Ampere
For this reason, a relay controlling the circuit breaker at B and set to operate at a current of 2200A plus a
safety margin would not operate for a fault at F4 and would thus discriminate with the relay at A. Assuming
a safety margin of 20% to allow for relay errors and a further 10% for variations in the system impedance
values, it is reasonable to choose a relay setting of 1.3 x 2200A, that is, 2860A, for the relay at B. Now,
assuming a fault at F3, at the end of the 11kV cable feeding the 4MVA transformer, the short-circuit current
is given by:
I=6350/(ZS+ZL1+ZL2)
Here, we assume Fault level at 250MVA.
I=6350/(0.485+0.24+0.04)=8300 Ampere
Now, we assume source fault level at 130MVA
I=6350/(0.93+0.214+0.04)=5250 Ampere
For either value of source level, the relay at B would operate correctly for faults anywhere on the 11kV cable
feeding the transformer.
Current-time grading using IDMT relays (Discrimination by both time and Current)
Each of the two methods described so far has a
fundamental disadvantage. In the case of
discrimination by time alone, the disadvantage is
due to the fact that the more severe faults are
cleared in the longest operating time. On the other
hand, discrimination by current can be applied only
where there is appreciable impedance between the
two circuit breakers concerned.
It is because of the limitations imposed by the
independent use of either time or current coordination that the inverse time overcurrent relay
characteristic has evolved. With this characteristic,
the time of operation is inversely proportional to the
fault current level and the actual characteristic is a
function of both ‘time’ and ‘current’ settings. Figure
9.3 shows the characteristics of two relays given
different current/time settings. For a large variation
in fault current between the two ends of the feeder,
faster operating times can be achieved by the relays
nearest to the source, where the fault level is the
highest. The disadvantages of grading by time or
current alone are overcome.
relay current setting
Relay current setting refers to the process of adjusting the current threshold at which a relay will activate
and trip, typically expressed as a percentage of the rated current.
It is very important to choose the current setting adjustment at a value in excess of the nominal rating of the
plant to be protected in order to maintain system stability under overloading and transient condition. For
example, a load margin of 30% is chosen for plant having a nominal load current of 250 A. The overload
rating of the plant will be 250 (1+.3) 325 A. Assuming 200 A CT primary current, the
plug setting should be set as 175% to give relay current setting of 200*1.75 =350 A which is greater than 325
A, the overload rating.
Time Setting Multiplier (TSM)
The time setting multiplier (TSM) in a protective relay adjusts the time it takes for the relay to operate,
influencing how quickly it responds to a fault.
For proper coordination between the relays, connected in series, on a radial feeder, the TSM of the relay
farthest from the source should be set at a minimum value so that it operates at minimum possible time.
The TSM of the succeeding relay towards the source should be increased for selective operation. The time
grading should be done at a maximum fault current because it will automatically give higher selectivity at
minimum fault current as its characteristic curve is more inverse in the lower current region,
The University of Jordan
EE482 Power System Analysis 2
School of Engineering
Electrical Engineering Department
Tutorial #3: IDMT OC Relay
Question # 1:
The current plug (tap) settings (CTS) of a GEC 5-A overcurrent relay can be varied from 1 A to 12 A and the
TMS can be varied from 0.5 to 10 as shown in Fig. 1. If the input current to the overcurrent relay is 10 A,
determine the relay operating time for the following current tap setting (CTS) and time dial setting (TDS):
(a) CTS = 1.0 and TDS = ½;
(b) CTS = 2.0 and TDS = 1.5;
(c) CTS = 2.0 and TDS = 7;
(d) CTS = 3.0 and TDS = 7; and
(e) CTS = 12.0 and TDS = 1.
Use the overcurrent relay characteristics
 A

t p  TDS   p
 B 
 Ir 1

tp is the pickup or operating time
Ir is the ratio of |If|/|Ip|
A = 28.2, B = 0.1217 and p = 2.0
Fig. 1
Solution:
(a) CTS=1.0, then Ir=|If|/|Ip|=10. Therefore,
 28.2

t p  0.5   2
 0.1217   0.2033 sec
 10  1

(b) CTS=2.0, then Ir=|If|/|Ip|=5. Therefore,
 28.2

t p  1.5   2
 0.1217   1.945 sec
5

1


(c) CTS=2.0, then Ir=|If|/|Ip|=5. Therefore,
 28.2

tp  7 2
 0.1217   9.0769 sec

 5 1
(d) CTS=3.0, then Ir=|If|/|Ip|=3.33. Therefore,
 28.2

tp  7
 0.1217   20.418 sec
2
 3.33  1

(e) CTS=12.0, then Ir=|If|/|Ip|<1. Therefore, the relay does not operate.
1
Question # 2:
The calculated short-circuit current through a feeder is 1200 A. An overcurrent relay of rating 5 A is connected
for the protection of the feeder through a 600/5 A CT as shown in Fig. 2.
600/5 A
Syst em
5A
IDM T
If = 1200 A
PS = 50%, TMS= 0.8
Fig. 2
Calculate the operating time of the relay when it has a plug setting (PS) of 50% and time multiple setting
(TMS) of 0.8. The characteristic of the relay is as follows:
PSM
1.3
2
4
6
10
20
Time (sec)
30
10
6.5
3.5
3
2.2
Solution:
I pickup  PS  I rated  0.5  5  2.5 A
I f  relay 
PSM 
If
CTR
I f  relay
I pickup

1200
 10 A
600 / 5

10
4
2.5
 operating time at TMS = 1 is 6.5 s
Actual operating time t p  6.5  TMS  t p  6.5  0.8  5.2 s
Question #3:
Figure 3 shows a radial distribution system having identical IDMT overcurrent at A, B and C. For a time delay
step (t) of 0.5 s, calculate the time multiplier settings (TMS) at A and B.
A
Syst em
B
C
200/5 A
200/5 A
100 /5 A
5A
IDM T
5A
IDM T
5A
IDM T
CS = 5 A
CS = 2.5 A
CS =2 .5 A
TM S =?
TMS =?
TMS =0.1
If = 1000 A
Fig. 3
The characteristic of the IDMT relay is as follows:
PSM
2
3
5
10
20
Time (sec)
10
6
4.5
3
2
2
Solution:
For relay C,
TMSC = 0.1, I C  pickup  2.5 A
I C  relay 
I Cf
CTRC

I C  relay
1000
50
 50 A PSM C 

 20
100 / 5
I C  pickup 2.5
 operating time at TMS = 1 is 2 s
Actual operating time of relay C is t p  2  TMS  t p  2  0.1  0.2 s
For relay B,
I B  pickup  2.5 A
If
I C  relay 

CTRC
1000
 25 A
200 / 5
PSM C 
I C  relay
I C  pickup

25
 10
2.5
 operating time at TMS = 1 is 3 s
Actual operating time of relay B is tp = 0.2 + 0.5 = 0.7 s = 3  TMS
 TMS B 
0.7
 0.233
3
For relay A,
I A pickup  5 A
I A relay 
If
CTR A
I A relay
25
1000
 25 A PSM A 

5
200 / 5
I A pickup
5

 operating time at TMS = 1 is 4.5 s
Actual operating time of relay A is tp = 0.2 + 0.5+0.5 = 1.2 s = 4.5  TMS
TMS A 
1.2
 0.266
4.5
Question #4:
A 20 MVA Transformer which is used to operate at 30% overload feeds an 11 kV busbar through a circuit
breaker (CB) as shown in Fig. 4. The transformer CB is equipped with a 1000/5 CT and the feeder CB with
400/5 CT and both CTs feed IDMT relays having the following characteristics
PSM
2
3
5
10
15
20
Time (sec)
10
6
4.1
3
2.5
2.2
The relay on the feeder CB has PS = 125% and TMS = 0.3. If a fault current of 5000 A flows from the
transformer to the feeder, determine
a. operating time of feeder relay.
b. Suggest suitable PS and TMS of the transformer relay to ensure adequate discrimination of 0.5 s
between the transformer relay and feeder relay.
11 kV
20 MV A
Syst em
1000/5
A
400/5 A
5A
IDM T
5A
IDM T
PS = ?
TMS = ?
PS = 125%
TMS = 0.3
If = 5000 A
Fig. 4
3
Solution:
For Feeder relay
TMSFeeder = 0.3, PS = 125%  I Feeder  pickup  PS  I rated  1.25  5  6.25 A
I Feeder  relay 
If
CTR Feeder

I Feeder  relay
5000
62.5
 62.5 A PSM Feeder 

 10
400 / 5
I Feeder  pickup 6.25
 operating time at TMS = 1 is 3 s
Actual operating time of the Feeder relay is t p  3  TMS  t p  3  0.3  0.9 s
For Transformer relay, I Transformer  pickup  PS  I rated
Transformer overload current, I T  1.3 
I Transformer  relay 
I Transformer overload
CTRTransformer

S rated
3VLL
 1.3 
20  10 6
3  11  10 3
 1365 A
1365
 6.825 A
1000 / 5
Since the transformer relay must not operate to overload current, PSTransformer 
I Transformer -relay
I relay-rated
6.825
 1.365 or 136.5% , the PS are restricted to standard values in steps of 25%, so the
5
nearest value but higher than 136.5% is 150%  PS Transformer  150%
PS Transformer 
I Transformer  pickup  PS Transformer  I rated  1.5  5  7.5 A
PSM Transformer 
I f Transformer  relay
I Transformer  pickup

5000 /(1000 / 5) 25

 3.3
7.5
7.5
Operating time corresponding to PSM Transformer  3.3 and TMS=1 from the PSM-time curve is
tp = 5.6 s, Actual operating time of transformer relay is tp = 0.9 + 0.5 = 1.4 s = 3  TMS
 TMS Transformer 
1.4
 0.25
5.6
4