Protective Relays

Each relay's operating time is found from its plug setting multiplier (PSM) using the given curve, then multiplied by the time multiplier setting (TMS). R2 (upstream) must operate 0.5 s after R1.

Data: CT ratio 1000/5 (ratio 200); relay rated current 5 A; fault current = 5000 A; R1: PS = 100%, TMS = 0.3; R2: PS = 125%; discrimination margin = 0.5 s.

Relay current

Both relays see the same fault current (the fault is beyond R1 and fed through R2):

Relay current = 5000 / 200 = 25 A (secondary)

Relay R1

Pick-up current = 100% × 5 = 5 A
PSM = 25 / 5 = 5
Time at TMS = 1 (from table, PSM 5) = 4 s
Actual time t₁ = 0.3 × 4 = 1.2 s

Relay R2

Pick-up current = 125% × 5 = 6.25 A
PSM = 25 / 6.25 = 4
Time at TMS = 1 (from table, PSM 4) = 5 s
Required time t₂ = t₁ + 0.5 = 1.2 + 0.5 = 1.7 s
TMS of R2 = 1.7 / 5 = 0.34

Results

Answer: R1 operates in 1.2 s, R2 operates in 1.7 s, and the TMS of R2 is 0.34.

The feeder relay time is found from its PSM; the transformer relay plug setting must be above the transformer overload current, and its TMS gives 0.5 s more time than the feeder relay for the same fault.

Data: 20 MVA, 11 kV, 30% overload; transformer CT 1000/5; feeder CT 400/5; feeder relay PS 125%, TMS 0.3; fault current 5000 A; margin 0.5 s; relay rated current 5 A.

i) Operating time of feeder relay

CT ratio = 400/5 = 80
Relay current = 5000 / 80 = 62.5 A
Pick-up = 125% × 5 = 6.25 A
PSM = 62.5 / 6.25 = 10
Time at TMS 1 (table, PSM 10) = 3 s
Feeder relay time = 0.3 × 3 = 0.9 s

ii) Settings of transformer relay

Plug setting: the relay must not trip on the allowed 30% overload.

Full-load current = 20 × 10⁶ / (√3 × 11 × 10³) = 1049.73 A
Max load (130%)   = 1.3 × 1049.73 = 1364.65 A
CT ratio = 1000/5 = 200
Secondary current = 1364.65 / 200 = 6.82 A
Required PS > 6.82 / 5 = 136.5%

Choose the next standard plug setting: PS = 150% (pick-up 7.5 A secondary = 1500 A primary).

Time multiplier setting:

Relay current = 5000 / 200 = 25 A
PSM = 25 / 7.5 = 3.333
From table (linear interpolation between
PSM 3 → 6 s and PSM 5 → 4.1 s):
t = 6 − (3.333 − 3)/(5 − 3) × (6 − 4.1)
  = 6 − 0.1667 × 1.9
  = 5.683 s  (at TMS = 1)

Required time = 0.9 + 0.5 = 1.4 s
TMS = 1.4 / 5.683 = 0.2463

Choose TMS = 0.25 (next practical setting). Then actual time = 0.25 × 5.683 = 1.42 s, giving a margin of 0.52 s ≥ 0.5 s.

Results

Answer: feeder relay time = 0.9 s; transformer relay PS = 150% and TMS ≈ 0.25 (calculated 0.246), operating in about 1.42 s. (If the feeder CT ratio is not given, as in the 2072 paper, 400/5 is assumed.)

Percentage (biased) differential protection of an alternator compares the current entering and leaving each phase winding; it trips when the difference exceeds a fixed percentage of the through current. It protects the stator against phase-to-phase and phase-to-earth faults (Merz-Price scheme with bias).

Why "percentage" (biased) is needed

In a simple differential (circulating current) scheme, the two CTs of a phase never match perfectly. During heavy external faults, CT saturation and ratio errors make a spill current flow in the relay, which can cause a false trip. A restraining (bias) coil carrying the through current raises the operating current in proportion to the load, so the relay stays stable on external faults but is sensitive to internal faults.

Arrangement

 Neutral side                     Line side
   CT1  ┌──── stator winding ────┐  CT2
 ──(○)──┤                        ├──(○)──► to bus
   │ I₁                        I₂ │
   │                              │
   └──R/2──┬───────────┬──R/2────┘
           │ restraining coil R  │
           └────► O ◄─────────────┘
           operating coil (mid-point)
 (one such set per phase; star point of
  CTs connected to the relay neutral)

• Identical CTs are placed at the neutral end and the line end of each phase winding.
• CT secondaries are connected so that their currents circulate in the pilot wires.
• The restraining coil is connected in the pilot wires (carries (I₁ + I₂)/2).
• The operating coil is connected between the mid-point of the restraining coil and the CT star point (carries I₁ − I₂).

Operation

• Normal load or external fault: I₁ = I₂, so the operating coil current is nearly zero. Any small spill current due to CT mismatch is overcome by the large restraining torque.
• Internal fault: current in the faulty winding differs at the two ends (or one end gets reversed current when the generator is in parallel with others). A large differential current I₁ − I₂ flows in the operating coil, its torque exceeds the restraining torque and the relay trips the generator breaker and field breaker.

Operating condition:

|I₁ − I₂| > K × (|I₁| + |I₂|)/2  + I₀
K = bias slope (typically 5–20%), I₀ = minimum pick-up

Characteristic

 Operating  │        /  Trip zone
 current    │       /
 |I₁−I₂|    │      /  slope K
            │     /
            │____/   No-trip (restrain) zone
            └──────────────────────
              through current (I₁+I₂)/2

Features

• Protects about 85–90% of the winding against earth faults (when neutral is earthed through resistance, faults near the neutral give little current); a separate restricted earth fault or 100% stator earth fault relay covers the rest.
• Fast, selective; unaffected by load or external faults.
• Needs CTs of identical characteristics; pilot wires are short since both CTs are near the machine.

An impedance relay is a distance relay that measures the ratio of voltage to current (V/I = Z) at the relay point and operates when this impedance falls below a set value. Since line impedance is proportional to length, it measures the distance to the fault.

Universal torque equation

For a general electromagnetic relay with current, voltage and directional elements:

T = K₁ I² + K₂ V² + K₃ V I cos(θ − τ) + K₄

where K₁, K₂, K₃ are constants, θ is the angle between V and I, τ is the maximum torque angle, and K₄ is the mechanical restraining (spring) torque.

Torque equation of impedance relay

In the impedance relay the current element produces operating torque and the voltage element produces restraining torque; there is no directional element (K₃ = 0):

Put K₂ = −K₂ (restraint), K₃ = 0:
T = K₁ I² − K₂ V² − K₄

At the balance point (T = 0), neglecting spring torque K₄:

K₁ I² = K₂ V²
V² / I² = K₁ / K₂
Z = V / I = √(K₁ / K₂) = constant (setting Zs)

The relay operates when operating torque > restraining torque:

K₁ I² > K₂ V²   ⟹   Z = V/I < √(K₁/K₂) = Zs

Operating principle

• Construction: a balanced beam; the current coil (fed by CT) pulls one end down, the voltage coil (fed by PT) pulls the other end.
• Normal condition: voltage is high and current is normal, so Z_seen (load impedance) is large; voltage pull wins and contacts stay open.
• Fault within the zone: voltage at the relay falls and current rises, so Z_seen = impedance up to the fault point < Zs; current pull wins and the relay trips.
• Fault beyond the zone: Z_seen > Zs, so the relay does not operate.

        Voltage coil       Current coil
          (PT) ▼              ▼ (CT)
     ──────────────△───────────────
               pivot     ▲ trip contacts

Operating characteristic (R-X diagram)

Since operation depends only on |Z|, the characteristic is a circle centred at the origin with radius Zs.

          X
          │   .-~~~-.
          │ /  Trip  \
        ──┼─(  zone   )── R
          │ \  Z<Zs  /
          │   '-...-'
          │ outside: no trip

Properties for feeder protection

• Non-directional: it operates for faults in all four quadrants (behind the relay too), so it is used with a directional unit (mho or directional relay) in series.
• Affected by arc resistance moderately and by power swings, since the circle covers a large area.
• Used in three zones with time steps: Zone 1 ≈ 80–90% of the line (instantaneous), Zone 2 ≈ 120–150% (about 0.3–0.5 s), Zone 3 covers the next line (about 1 s).
• Its reach does not depend on fault current level, so it gives faster and more selective protection than overcurrent relays on long lines.

i) Carrier current protection

Carrier current protection is a unit protection for long transmission lines in which information about the currents at both ends is sent over the power line itself using a high-frequency carrier signal (50–500 kHz). It allows high-speed, simultaneous tripping at both ends for faults anywhere on the line, where pilot wires would be too long.

Equipment at each end

• Transmitter and receiver (carrier set) producing and detecting the HF signal.
• Coupling capacitor: connects the carrier set to the HV line and blocks power-frequency voltage.
• Line trap (wave trap): a parallel LC circuit tuned to the carrier frequency, in series with the line; it confines the carrier to the protected section and passes 50 Hz.
• Relays: directional or distance relays and a fault detector.

 Bus A                                        Bus B
  │  CB  Line trap ═══════ Line ═══════ Line trap  CB  │
  ├──□───[LC]──┬────────────────────────┬──[LC]───□──┤
               │                        │
          Coupling cap.            Coupling cap.
               │                        │
       Transmitter/Receiver      Transmitter/Receiver
               │                        │
            Relays ───── trip ───── Relays

Directional comparison (blocking) scheme

1. Directional relays at both ends see the fault direction.
2. For an internal fault, power flows into the line at both ends; both relays see "forward" and no blocking carrier is sent; both breakers trip.
3. For an external fault, the relay at the end nearest the fault sees power flowing out of the line; it starts its transmitter and sends a blocking signal to the remote end, which prevents tripping.

Phase comparison scheme: the phase of currents at the two ends is compared via the carrier; in-phase (internal fault) → trip; 180° out of phase (through fault) → block.

Advantages: fast (1–3 cycles), simultaneous tripping at both ends, permits high-speed auto reclosing, no separate communication channel needed; the carrier can also be used for telephony and telemetry.

ii) Bus-bar protection

Bus-bar protection detects faults on the busbars of a substation and trips all breakers connected to the faulty bus section. Busbar faults are rare but severe, so the scheme must be fast and very stable for external faults.

Methods

1. Back-up protection by remote relays: the bus is cleared by the time-delayed relays of the connected circuits; simple but slow.
2. Fault bus (frame leakage) protection: used for metal-clad switchgear; the frame is insulated from earth and earthed through a single CT and relay; an earth fault to the frame trips all breakers.
3. Differential protection (most common): based on Kirchhoff's current law: the sum of currents entering a healthy bus is zero.

   Feeder1     Feeder2     Feeder3    Source
     │CT          │CT        │CT       │CT
 ════╪════════════╪══════════╪═════════╪════ Bus
     │            │          │         │
     └─────┬──────┴────┬─────┴────┬────┘
           │  CT secondaries paralleled
           └────────[ R ]─────────┘
                differential relay

• CTs of the same ratio are placed in every circuit connected to the bus, and their secondaries are connected in parallel with the relay.
• Normal load or external fault: ΣI = 0, so no current flows in the relay.
• Bus fault: all circuits feed the fault; ΣI = total fault current, and the relay trips all breakers on that bus.
• High impedance differential scheme is widely used: a high-resistance relay with a stabilising resistor prevents false tripping when a CT saturates during a heavy external fault. Biased (percentage) differential relays are also used.
• Large stations divide the bus into zones with separate protection, plus a check zone to avoid false trips.

An electromagnetic relay works by the force produced by an electromagnet energised by current or voltage from CTs and PTs. There are two main types: attracted armature relays and induction relays.

1. Attracted armature relays

They work on electromagnetic attraction: force F ∝ I², so they operate on both AC and DC. They are fast (5–50 ms) and are used as instantaneous relays.

a) Hinged armature type

     ┌──────┐   armature (hinged)
     │ coil │ ┌───────────────────┐
     │ (I)  │ │                   ●─ contacts
     │ core │ └─┬─────────────────┘
     └──────┘   spring ↑

The coil current produces flux; when it exceeds pick-up, the armature is pulled towards the core against a spring and closes the contacts.

b) Plunger type (solenoid): a plunger is drawn into a solenoid when current exceeds the setting.

c) Balanced beam type: a beam pivoted at the centre is pulled by two coils (e.g. current coil on one side, voltage coil on the other); it trips when one force exceeds the other. Used as an impedance (distance) relay.

Applications: instantaneous overcurrent and undervoltage relays, auxiliary and tripping relays, differential relays, impedance relays.

2. Induction relays

They work like an induction motor/energy meter: two fluxes displaced in time and space induce eddy currents in a rotating disc or cup and produce torque T ∝ φ₁ φ₂ sin α. They work only on AC.

a) Shaded pole type: the pole face is partly covered by a copper ring (shading ring), giving two fluxes with a phase difference.

b) Watt-hour meter (wattmetric) type: two separate electromagnets (E-shaped upper and U-shaped lower) on either side of the disc; upper coil can be fed by voltage and lower by current, so it can act as a directional relay.

     ┌──── E-magnet (upper coil) ─────┐
     ════════════ disc ═══════════════ → contact
     └──── U-magnet (lower coil) ─────┘ + brake magnet

c) Induction cup type: a hollow aluminium cylinder (cup) rotates inside a four- or eight-pole stator; very low inertia and high torque, so it is fast and sensitive. Used for directional, mho and reactance relays.

Applications: IDMT overcurrent and earth fault relays (induction disc), directional overcurrent relays, distance relays (mho, reactance), reverse power relays.

Summary

An IDMT (Inverse Definite Minimum Time) characteristic is a relay time-current characteristic in which the operating time is approximately inversely proportional to fault current at low multiples of setting, but becomes almost constant (a definite minimum time) at high currents. It is usually obtained with an induction disc relay.

IDMT characteristic

• At currents just above pick-up the time is long; as current increases the time falls inversely.
• At high currents (PSM > about 10–20) the electromagnet saturates, torque stops increasing, and the time levels off to a definite minimum.
• This gives discrimination between relays in series and protects against both overloads and heavy faults.
• Standard (IEC 60255) normal inverse curve: t = TMS × 0.14 / (PSM^0.02 − 1) (e.g. about 3 s at PSM 10 for TMS = 1).

 Time
  │\
  │ \
  │  \   inverse region
  │   \
  │    `-.
  │       `--.______________  definite
  │                           minimum time
  └──────────────────────────── PSM
  1      5      10      20

Construction of induction disc (non-directional) relay

• Upper E-shaped electromagnet with a primary winding (fed by the CT) on the central limb; the primary has tappings brought to a plug setting bridge to set the pick-up current (50–200%).
• A secondary winding on the central limb is connected to windings on the two outer limbs of the upper magnet.
• Lower U-shaped electromagnet with no winding of its own.
• An aluminium disc pivoted between the two magnets, carrying a moving contact on its spindle.
• A control spiral spring gives restraining torque; a permanent brake magnet gives eddy current damping so the disc speed is proportional to torque.
• The disc's backward stop position is adjustable by the time multiplier setting (TMS 0.1 to 1), which changes the angle the contact must travel.

   CT ─► plug setting bridge
           │
     ┌─────┴──────┐ upper E-magnet
     │ prim+sec   │ (secondary feeds
     │  windings  │  outer limbs)
     └─┬──┬──┬────┘
   ════════════════ Al disc ═══ spindle ─► contact
     └──┬────┬──┘   lower U-magnet   │
                                      brake magnet

Working principle

1. The CT current in the primary winding produces flux φ₁ in the upper magnet.
2. This flux induces an emf in the secondary winding, whose current produces flux φ₂ through the outer limbs and the lower magnet. φ₂ lags φ₁ in time and is displaced in space.
3. The two fluxes induce eddy currents in the disc, giving a driving torque T ∝ φ₁ φ₂ sin α ∝ I² (below saturation).
4. When the current exceeds the plug setting, the driving torque exceeds the spring torque and the disc rotates; the brake magnet controls the speed.
5. After travelling the distance set by the TMS, the moving contact closes the trip circuit, energising the breaker trip coil.
6. Larger currents give larger torque and faster rotation, so the time is inverse; at heavy currents the core saturates and the time approaches a definite minimum.

Applications: overcurrent and earth fault protection of distribution feeders, transformers and motors, and back-up protection of lines.

In a time-graded scheme the relay farthest from the source (B) is set fastest, and relay A behind it is set slower by a discrimination margin for every fault in B's zone, so that A acts as back-up.

Assumptions: relay rated current 5 A; load current is below the CT primary rating, so PS = 100% is used on both relays (pick-up = CT primary current); discrimination margin = 0.5 s (breaker time + relay overshoot + safety); TDS steps of 0.01; the curve between table points is interpolated linearly.

Source ─[A]──── AB ────[B]──── BC ──── C
       300/5           200/5
      4000 A          3000 A        2000 A

Relay B (last relay, set fastest)

PS = 100% → pick-up = 200 A primary
Fault at B (3000 A): PSM = 15 → t = 2.2 s (TDS 1)
Fault at C (2000 A): PSM = 10 → t = 2.8 s (TDS 1)
Choose minimum TDS = 0.1
t_B (fault at B) = 0.1 × 2.2 = 0.22 s
t_B (fault at C) = 0.1 × 2.8 = 0.28 s

Relay A

PS = 100% → pick-up = 300 A primary
Grading point: fault at B, 3000 A
(maximum fault current in B's zone)
PSM_A = 3000/300 = 10 → t = 2.8 s (TDS 1)
Required t_A = 0.22 + 0.5 = 0.72 s
TDS_A = 0.72 / 2.8 = 0.257 → choose 0.26

Actual operating times of relay A (TDS = 0.26)

Interpolations used: PSM 13.33 → 2.8 − (3.33/5)(0.6) = 2.40 s; PSM 6.67 → 3.9 − (1.667/3)(0.75) = 3.483 s.

Settings

Answer: relay B: PS 100%, TDS 0.1; relay A: PS 100%, TDS 0.26. Relay A operates in 0.624 s for a fault at A in its own zone; as back-up it operates in 0.728 s for a fault at B and 0.906 s for a fault at C, always at least 0.5 s after relay B.

A Buchholz relay is a gas-actuated relay fitted in the pipe between the main tank and the conservator of oil-immersed transformers (above about 500 kVA). It detects internal faults by the gas and oil surge they produce.

How it protects

Any fault inside the tank (winding or core) produces heat; the heat decomposes the oil and insulation into gas (H₂, CO, C₂H₂, etc.). The gas rises towards the conservator and passes through the relay.

  Main tank ──pipe──► [ Buchholz relay ] ──► Conservator
                       ┌──────────────┐
                       │ float1 → alarm│ (upper)
                       │ flap/float2 → │
                       │   trip        │ (lower)
                       └──────────────┘

1. Incipient (minor) faults such as insulation breakdown between core laminations, core-bolt insulation failure, poor joints, inter-turn faults or loss of oil: gas forms slowly and collects at the top of the relay. The oil level in the relay falls, the upper float tilts and closes the alarm circuit.
2. Severe faults such as phase-to-earth or phase-to-phase winding faults: a large volume of gas is produced suddenly and forces a surge of oil towards the conservator. This surge deflects the lower float/flap, which closes the trip circuit and opens the transformer circuit breakers.
3. Oil leakage: if oil level falls, first the alarm and then the trip float operate.
4. The gas collected can be drawn off through a cock and analysed: its colour and composition indicate the type of fault (e.g. white: paper insulation; yellow: wood; dark grey: oil).

The relay therefore gives early warning of slowly developing faults that differential protection cannot see, and trips quickly on major internal faults.

A Buchholz relay is a gas-operated relay mounted in the pipe connecting the main tank of an oil-immersed transformer to the conservator. It protects against all internal faults that produce gas or an oil surge.

Construction

• A domed oil-filled chamber in the inclined connecting pipe.
• Upper float (hinged) with a mercury switch connected to an alarm circuit.
• Lower float or baffle flap in the path of oil flow, with a mercury switch connected to the trip circuit of the transformer breakers.
• A gas release cock at the top and an inspection window.

                     gas cock
                        │
  Tank  ════════╗  ┌────┴───────┐  ╔════════ Conservator
   oil  ───────►╠══│ (o)─upper  │══╣ (pipe rises
                ║  │   float→A  │  ║  2-3° slope)
                ║  │ [flap]─lower│  ║
                ║  │   float→T  │  ║
                   └────────────┘
          A = alarm switch, T = trip switch

Working

1. Normal condition: the relay is full of oil; both floats stay up and their contacts are open.
2. Incipient faults (core insulation failure, inter-turn short, hot spot, bad joint): heat slowly decomposes oil; gas bubbles rise to the relay chamber and collect at the top. Oil level falls, the upper float tilts and closes the alarm circuit.
3. Severe faults (earth fault, phase-to-phase winding fault): gas is formed rapidly; the pressure drives a surge of oil through the pipe towards the conservator. This surge tilts the lower float/flap and closes the trip circuit, disconnecting the transformer from both sides.
4. Loss of oil (leak): oil level falls first below the upper float (alarm) and then the lower float (trip).
5. The collected gas is sampled; its colour and analysis tell the nature of the fault.

Advantages

• Simplest form of transformer protection; no CTs or external supply needed.
• Detects incipient faults at an early stage, before they become serious, and gives an alarm.
• Detects inter-turn and core faults that differential protection may not see.
• Gas analysis helps to locate the fault type.
• Also protects against loss of oil.

Limitations

• Usable only for oil-immersed transformers with a conservator.
• Protects only against faults inside the tank; it does not cover bushings, cables or external connections.
• Slow: minimum operating time is about 0.1 s (average 0.2 s), so differential protection is still needed for fast clearance.
• Can maloperate on oil surges due to heavy through faults, earthquakes or mechanical shocks, or air entering after oil filling.
• Economical only for transformers above about 500 kVA.

For an earth fault at a point x (fraction of winding from the neutral), the voltage driving the fault current is x × phase voltage, and the current is limited mainly by the neutral earthing resistance. The relay operates only if this current exceeds the setting; faults close to the neutral give too little current, so that part of the winding is unprotected.

Data: 5000 kVA, 6600 V, star; R_n = 6.5 Ω; winding impedance 0.5 + j2 Ω per phase; relay operates at 30% of full-load current.

Step 1: Full-load and operating currents

I_FL = 5000 × 10³ / (√3 × 6600) = 437.39 A
I_op = 0.30 × 437.39 = 131.22 A
V_ph = 6600 / √3 = 3810.51 V

Step 2: Unprotected portion (usual textbook method)

Neglecting the impedance of the short faulted part of the winding (small compared with 6.5 Ω):

x × V_ph / R_n = I_op
x = I_op × R_n / V_ph
  = 131.22 × 6.5 / 3810.51
  = 0.2238

So 22.38% of the winding (from the neutral end) is unprotected.

Step 3: Including the winding impedance (check)

The faulted part has impedance x(0.5 + j2) Ω, so:

x × 3810.51 = 131.22 × |6.5 + x(0.5 + j2)|
Solving numerically: x = 0.2283

This gives 22.83% unprotected, only slightly more; the winding impedance has little effect because R_n dominates.

 Neutral ─R_n(6.5Ω)─ N ──x──●F── (1−x) ── line
                             │
                            earth

Answer: about 22.4% of the winding (22.8% if the winding impedance is included) near the neutral is unprotected; about 77.6% is protected.

An earth fault at fraction x of the winding from the neutral drives a current x·V_ph through the neutral resistance (and the faulted part of the winding). The relay sees this current and operates only if it exceeds the setting.

Data: 13.8 kV, 125 MVA, star; X_s = 1.4 pu; R negligible; R_n = 2 Ω; setting = 10% of full-load current.

Step 1: Basic quantities

V_ph = 13 800 / √3 = 7967.43 V
I_FL = 125 × 10⁶ / (√3 × 13 800) = 5229.62 A
I_op = 0.10 × 5229.62 = 522.96 A
Z_base = 13.8² / 125 = 1.5235 Ω
X_s = 1.4 × 1.5235 = 2.1329 Ω

Step 2: Unprotected fraction (neglecting winding reactance)

x × V_ph / R_n = I_op
x = 522.96 × 2 / 7967.43 = 0.1313
Protected = 1 − 0.1313 = 0.8687 → 86.87%

Step 3: Including the reactance of the faulted part

The faulted part has reactance x·X_s (as an approximation, since synchronous reactance does not divide exactly in proportion to turns):

x × 7967.43 = 522.96 × |2 + j x × 2.1329|
Solving numerically: x = 0.1326
Protected = 86.74%

The reactance changes the answer only slightly because x is small.

Answer: about 86.9% of the winding is protected against earth faults (86.7% if winding reactance is included); the 13.1% nearest the neutral is unprotected.

For an earth fault at fraction x of the winding from the neutral, the fault current is x·V_ph / R_n (winding impedance neglected). The relay operates when the corresponding CT secondary current reaches the setting.

Data: 20 MVA, 11 kV, star; R_n = 5 Ω; relay setting 1.5 A; CT 1000/5 (ratio 200).

Step 1: Primary operating current

I_op (primary) = 1.5 × 200 = 300 A
V_ph = 11 000 / √3 = 6350.85 V

Step 2: Percentage of winding protected

x × V_ph / R_n = I_op
x = 300 × 5 / 6350.85 = 0.2362   (unprotected)
Protected = 1 − 0.2362 = 0.7638 → 76.38%

Step 3: Earthing resistance to protect 90%

To protect 90%, only x = 0.10 may be unprotected:

0.10 × V_ph / R = 300
R = 0.10 × 6350.85 / 300 = 2.117 Ω

Any resistance larger than this leaves more than 10% unprotected, so 2.117 Ω is the limiting (maximum permissible) value of earthing resistance; a lower resistance protects more winding but gives larger earth fault currents.

Answer: 76.38% of the winding is protected; earthing resistance must be 2.117 Ω (not more) to protect 90% of the winding.

For an earth fault at fraction x of the winding from the neutral, the fault current is x·V_ph / R_n (winding impedance neglected). The relay operates when the CT secondary out-of-balance current reaches 1.8 A.

Data: 10 kV, star; R_n = 10 Ω; setting 1.8 A; CT 1000/5 (ratio 200).

Step 1: Primary operating current

I_op = 1.8 × 200 = 360 A
V_ph = 10 000 / √3 = 5773.50 V

(i) Percentage unprotected

x × V_ph / R_n = I_op
x = 360 × 10 / 5773.50 = 0.6235

So 62.35% of the winding (from the neutral) is unprotected and only 37.65% is protected. The large earthing resistance limits the fault current too much.

(ii) Earthing resistance to protect 80%

Unprotected part allowed = 20% (x = 0.2):

0.2 × V_ph / R = 360
R = 0.2 × 5773.50 / 360 = 3.208 Ω

Answer: (i) 62.35% of the winding is unprotected; (ii) the earthing resistance must be reduced to 3.21 Ω (or less) to protect 80% of the winding.

In a differential scheme, an earth fault at fraction x of the winding from the neutral gives a fault current x·V_ph / R_n. The relay sees this as out-of-balance current; the winding near the neutral where the current is below the setting is unprotected.

Data: 11 kV, 100 MVA; R_n = 6 Ω; CT 1000/5 (ratio 200); relay setting 1 A; winding impedance neglected.

I_op (primary) = 1 × 200 = 200 A
V_ph = 11 000 / √3 = 6350.85 V

x = I_op × R_n / V_ph
  = 200 × 6 / 6350.85
  = 0.1890   (unprotected)

Protected = 1 − 0.1890 = 0.8110

 N ──R_n=6Ω── neutral ──[ 18.9% ]──[   81.1%   ]── line
                        unprotected   protected

Answer: about 81.1% of the generator winding is protected; the 18.9% nearest the neutral is unprotected.

A voltage-monitored (voltage-controlled) overcurrent relay is used on generators because the sustained fault current of a generator (limited by synchronous reactance) can be less than full-load current, so a plain overcurrent relay would never operate. When the terminal voltage collapses during a fault, its plug setting is reduced so that it can pick up.

Assumptions

• Base 150 MVA (generator 120 MW / 0.8 = 150 MVA); prefault voltage 1.0 pu; three-phase fault just beyond R2; only this generator feeds the fault.
• Relay times are of the order of seconds, so the sustained current with X_s = 175% is used (AVR action neglected).
• R1 CT ratio 8000/1 A (generator rated current 7873 A); R2 CT 200/1 A as given; relay rated current 1 A.
• Table interpolated linearly; for PSM below 2 the IEC normal inverse curve t = 0.14/(PSM^0.02 − 1), which matches the table (gives 10.0 s at PSM 2 and 2.97 s at PSM 10), is used.

Fault current

X_total = X_s + X_T = 1.75 + 0.10 = 1.85 pu
I_f = 1 / 1.85 = 0.5405 pu
I_base(11 kV)  = 150×10⁶/(√3×11×10³)  = 7872.96 A
I_base(132 kV) = 150×10⁶/(√3×132×10³) = 656.08 A
I_f at 11 kV  = 0.5405 × 7872.96 = 4255.7 A
I_f at 132 kV = 0.5405 × 656.08  = 354.6 A
Generator terminal voltage = I_f × X_T = 0.5405 × 0.10
                           = 0.054 pu (< 0.7 pu)

Relay R2 (feeder)

Relay current = 354.6 / 200 = 1.773 A
Pick-up = 50% × 1 = 0.5 A
PSM = 1.773 / 0.5 = 3.546
t (TDS 1) = 6 − (3.546 − 3)/(5 − 3) × (6 − 4.1) = 5.481 s
t_R2 = 0.3 × 5.481 = 1.644 s

Relay R1 (generator, voltage monitored)

Relay current = 4255.7 / 8000 = 0.532 A
At PS 100%: PSM = 0.532/1 = 0.53 < 1
            → would never operate
Voltage 0.054 pu < 0.7 pu
→ PS reduced to 40% of 100% = 40%
Pick-up = 0.4 A
PSM = 0.532 / 0.4 = 1.33
t (TDS 1) = 0.14 / (1.33^0.02 − 1) = 24.48 s
t_R1 = 0.1 × 24.48 = 2.45 s

Result

Answer: R2 operates in about 1.64 s and R1 in about 2.45 s, so R1 correctly backs up R2 with about 0.8 s margin. Without voltage monitoring, R1 (PSM 0.53) would not operate at all. (Times depend on the assumed R1 CT ratio.)

Restricted earth fault (REF) protection is a unit protection that responds only to earth faults inside the generator stator winding (the zone between the neutral CT and the line CTs). It is not affected by external earth faults.

Connection

• Three line CTs (one per phase at the generator terminals) have their secondaries connected in parallel (residual connection); their sum is the residual current 3I₀ of the line side.
• One CT in the neutral-to-earth connection of the solidly grounded star point, of the same ratio.
• The residual circuit of the line CTs and the neutral CT are connected so that their currents circulate; an earth fault relay (high impedance relay with stabilising resistor R_st and non-linear resistor/metrosil) is connected across them.

         Generator stator         Line CTs
   R ────┐  ┌────────────────(CT)───────── R
   Y ────┤  │   ┌───────────(CT)───────── Y
   B ────┤  │   │  ┌────────(CT)───────── B
         │  │   │  │         │││
      star point             └┴┴─ parallel ──┐
         │                                   │
       (CT) neutral CT                       │
         │       │                           │
        ═╧═      └──────────┬────────────────┘
       earth                │
                       [R_st]──[ 64 ]── REF relay
                            │
                        common return

Operation

• Normal or external earth fault: the residual current of the line CTs equals the current in the neutral CT; the two secondary currents circulate and no current flows in the relay.
• Internal earth fault: fault current flows through the neutral CT but not through the line CTs (or flows in opposite sense if other sources feed it); the unbalance current passes through the relay and it trips the generator breaker and field.
• Since the neutral is solidly grounded, earth fault current is large and almost the whole winding (about 90–95%) is protected.

The universal torque equation is the general expression for the torque of an electromagnetic relay that has current, voltage and directional (voltage–current) elements together with a spring:

T = K₁ I² + K₂ V² + K₃ V I cos(θ − τ) + K₄

• K₁ I²: torque of current element
• K₂ V²: torque of voltage element
• K₃ V I cos(θ − τ): torque of directional element (θ = angle between V and I, τ = maximum torque angle)
• K₄: mechanical (spring) torque Constants are given sign (+ operating, − restraining) and some may be zero to get each relay type. In the derivations below the spring torque K₄ is neglected.

(i) Impedance relay

Operating torque by current, restraining by voltage: K₃ = 0, K₂ negative.

T = K₁ I² − K₂ V²
At balance (T = 0): K₁ I² = K₂ V²
V/I = Z = √(K₁/K₂) = constant
Operates when Z < √(K₁/K₂)

Since Z² = R² + X² = constant, the characteristic on the R-X diagram is a circle centred at the origin. It is non-directional.

        X
        │  .-~~~-.
        │ /       \
     ───┼(    O    )─── R
        │ \       /
        │  '-...-'   trip inside circle

(ii) Reactance relay

Operating torque by current, restraining by a directional element with τ = 90°: K₂ = 0, K₃ negative.

T = K₁ I² − K₃ V I cos(θ − 90°)
  = K₁ I² − K₃ V I sin θ
At balance: K₁ I² = K₃ V I sin θ
(V/I) sin θ = K₁/K₃
Z sin θ = X = K₁/K₃ = constant
Operates when X < K₁/K₃

The characteristic is a straight line parallel to the R axis at X = K₁/K₃. It measures only reactance, so it is not affected by arc resistance; it needs a directional (mho) starting unit.

        X
        │
  ──────┼────────────── X = K₁/K₃
        │   trip below line
     ───┼────────────── R

(iii) Mho (admittance) relay

Operating torque by the directional element, restraining by voltage: K₁ = 0, K₂ negative.

T = K₃ V I cos(θ − τ) − K₂ V²
At balance: K₃ V I cos(θ − τ) = K₂ V²
V/I = (K₃/K₂) cos(θ − τ)
Z = (K₃/K₂) cos(θ − τ)
Operates when Z < (K₃/K₂) cos(θ − τ)

This is the equation of a circle passing through the origin with diameter K₃/K₂ along the line at angle τ. The relay is inherently directional.

        X
        │    .-~-.
        │  /   τ  \     diameter K₃/K₂
        │ (  ↗     )    at angle τ
        │  \     /
        O───'-.-'────── R
     (circle passes through origin)

Summary

In transformer differential protection the CTs on the star side are connected in delta and the CTs on the delta side are connected in star, to correct the 30° phase shift and to block zero-sequence current. Then the currents in the pilot wires from both sides must be equal.

Data: 33/6.6 kV, star/delta; LV (delta side) CTs 300:5, connected in star.

Step 1: Pilot current from LV side

Take LV line current = 300 A (CT rated current):

LV CT secondary = 5 A
(star-connected, so pilot current = 5 A)

Step 2: Corresponding HV line current

I_HV = 300 × 6.6/33 = 60 A

Step 3: HV CT ratio (CTs in delta)

The line current leaving a delta of CTs is √3 times the CT secondary current. For a pilot current of 5 A:

CT secondary current = 5/√3 = 2.887 A
HV CT ratio = 60 : 2.887 = 103.92 : 5

 HV (star)            LV (delta)
 CTs in delta         CTs in star
 60 A → 2.887 A       300 A → 5 A
 line out = 5 A  ──── pilots ──── 5 A

Answer: HV CT ratio = 60 : 5/√3, i.e. about 103.9 : 5 (≈ 20.78 : 1), with HV CTs connected in delta.

For differential protection of a star/delta transformer, the CTs on the star (HV) side are connected in delta and those on the delta (LV) side in star. This corrects the 30° phase shift and keeps zero-sequence current out of the relay. The CT ratios are chosen so that equal currents flow in the pilot wires.

Data: 33/6.6 kV, star/delta; LV CTs 400/5 (star connected).

Step 1: Pilot current from LV side

Take LV line current = 400 A, so the LV CT secondary (and pilot) current = 5 A.

Step 2: HV line current for the same load

I_HV = 400 × 6.6/33 = 80 A

Step 3: HV CT ratio

HV CTs are in delta; the line current from the delta is √3 × CT secondary current. For a pilot current of 5 A:

CT secondary = 5/√3 = 2.887 A
HV CT ratio = 80 : 5/√3 = 80 : 2.887
            = 138.56 : 5   (≈ 27.71 : 1)

Connection

   33 kV star ─── HV CTs (delta) ─┐
                                  ├─ pilots + relay
   6.6 kV delta ─ LV CTs (star) ──┘

Answer: HV CT ratio = 80 : 5/√3 ≈ 138.6 : 5, HV CTs connected in delta and LV CTs in star.

In the Merz-Price (circulating current) scheme for a star/delta transformer, the HV (star) side CTs are connected in delta and the LV (delta) side CTs in star, so that the pilot currents from both sides are equal and in phase.

Data: 66 000/6600 V, star/delta; LV CTs 400/5.

LV line current = 400 A → LV CT secondary (pilot) = 5 A

HV line current = 400 × 6600/66 000 = 40 A

HV CTs in delta: pilot current = √3 × CT secondary
CT secondary = 5/√3 = 2.887 A

HV CT ratio = 40 : 5/√3 = 40 : 2.887
            = 69.28 : 5   (≈ 13.86 : 1)

Answer: HV CT ratio = 40 : 5/√3 ≈ 69.3 : 5, with HV CTs connected in delta and LV CTs in star.

For differential protection, CTs on the delta winding side are connected in star, and CTs on the star winding side are connected in delta. This compensates the 30° phase shift of the transformer and prevents zero-sequence currents (from earth faults on the earthed star side) from causing false trips.

Data: 11/132 kV, delta (LV) / star (HV); LV CTs 200/1.

CT ratio on HV side

LV line current = 200 A → LV CT secondary (pilot) = 1 A
                (LV CTs in star)

HV line current = 200 × 11/132 = 16.67 A

HV CTs in delta: pilot current = √3 × CT secondary
CT secondary = 1/√3 = 0.5774 A

HV CT ratio = 16.67 : 0.5774 = 28.87 : 1

Connection of CTs

• LV (11 kV, delta winding): CTs of 200/1 connected in star.
• HV (132 kV, star winding): CTs of 16.67 : 1/√3 (≈ 28.87/1) connected in delta.
• Pilot wires join corresponding phases of the two CT groups; the operating (and restraining) coils of the differential relays are connected between the pilots and the star point.

 11 kV delta ──[CTs 200/1, star]──┐
                                   ├─ pilots ─ 87 relays
 132 kV star ──[CTs 28.87/1, delta]┘

Answer: HV CT ratio = 16.67 : 1/√3 ≈ 28.87 : 1; LV CTs in star, HV CTs in delta.

HV is star connected, so the HV CTs are connected in delta; LV is delta connected, so the LV CTs are connected in star. The LV CT ratio is chosen so that pilot currents balance at the nominal ratio 132/33 kV. A tap change upsets this balance and a spill current flows.

Data: 10 MVA, 132/33 kV; HV CTs 75/1 (in delta); through fault = 5 × full load; tap set at 128 kV.

Step 1: LV CT ratio

For HV current I_H, the delta of HV CTs gives a pilot current √3 I_H / 75. The LV CTs (ratio N:1, star) give I_L / N, with I_L = 4 I_H (132/33).

I_L / N = √3 I_H / 75
N = 75 × (I_L/I_H) / √3 = 75 × 4 / √3 = 173.2
LV CT ratio = 173.2 : 1

Step 2: Fault currents with 128 kV tap

Full-load current on the LV (33 kV) side:

I_FL(LV) = 10 × 10⁶ / (√3 × 33 × 10³) = 174.95 A
Through-fault LV current = 5 × 174.95 = 874.77 A
HV current with tap at 128 kV = 874.77 × 33/128 = 225.53 A

Step 3: Pilot currents

From LV CTs: 874.77 / 173.2 = 5.051 A
From HV CTs: √3 × 225.53 / 75 = 5.208 A
Out-of-balance (relay) current = 5.208 − 5.051 = 0.158 A

Answer: LV CT ratio = 173.2 : 1 (star connected). For the through fault with the tap at 128 kV, the pilot currents are about 5.05 A (LV side) and 5.21 A (HV side), so an unbalanced current of 0.158 A circulates through the relay. A percentage-biased relay is needed so that this does not cause tripping. (If "5 times full load" is taken on the HV side, the corresponding values are 5.05 A, 4.90 A and 0.153 A.)

HV is delta connected, so its CTs are connected in star; LV is star connected, so its CTs are connected in delta. The relay current is the difference of the CT output line currents.

Data: 125 MVA, 220/132 kV; HV CTs 400/5 (ratio 80); LV CTs 1200/5 (ratio 240).

i) CT output line currents at full load

I_HV = 125 × 10⁶ / (√3 × 220 × 10³) = 328.04 A
I_LV = 125 × 10⁶ / (√3 × 132 × 10³) = 546.73 A

HV CTs (star): output = 328.04 / 80 = 4.100 A
LV CTs (delta): CT secondary = 546.73 / 240 = 2.278 A
                output line = √3 × 2.278 = 3.946 A

ii) Relay current at 15% overload

Relay current at full load = 4.100 − 3.946 = 0.1548 A
At 115% load = 1.15 × 0.1548 = 0.178 A

iii) Minimum relay setting for 25% overload

The relay must not operate up to 125% load:

Setting ≥ 1.25 × 0.1548 = 0.1935 A

Answer: (i) 4.10 A (HV) and 3.95 A (LV); (ii) 0.178 A; (iii) minimum relay setting ≈ 0.194 A (about 3.9% of 5 A).

HV is delta connected, so its CTs are connected in star; LV is star connected, so its CTs are connected in delta. The relay current is the difference between the CT output line currents caused by the CT ratio mismatch.

Data: 150 MVA, 220/110 kV; HV (delta) CTs 400/5 (ratio 80); LV (star) CTs 1600/5 (ratio 320).

(i) CT secondary currents at full load

I_HV = 150 × 10⁶ / (√3 × 220 × 10³) = 393.65 A
I_LV = 150 × 10⁶ / (√3 × 110 × 10³) = 787.30 A

HV CTs (star): secondary = 393.65 / 80 = 4.921 A
LV CTs (delta): secondary (phase) = 787.30 / 320 = 2.460 A
                output line current = √3 × 2.460 = 4.261 A

(ii) Relay current at full load

I_relay = 4.921 − 4.261 = 0.659 A

(iii) Relay current setting for 90% loading

At 90% load all currents scale by 0.9:

I_relay = 0.9 × 0.659 = 0.593 A

The relay setting must therefore be above 0.593 A so that it does not operate at this loading (in practice a percentage bias is used to cover this mismatch).

Answer: (i) HV CT secondary 4.92 A, LV CT secondary 2.46 A (line output 4.26 A); (ii) relay current 0.659 A; (iii) 0.593 A at 90% load, so the setting must exceed this value.

In the combined phase and earth fault protection scheme, a single set of CTs on each side of the generator winding feeds a differential (Merz-Price) scheme for phase faults and, through the residual (neutral) connection of the same pilots, an earth fault relay. This saves CTs and covers both types of fault.

Connection diagram

 Neutral-side CTs      Stator         Line-side CTs
  (CT)──────────────── R ───────────────(CT)
  (CT)──────────────── Y ───────────────(CT)
  (CT)──────────────── B ───────────────(CT)
   │││   star pt                         │││
   │││     │                             │││
   │││   R_n (earthing)                  │││
   │││     ⏚                             │││
   ││└───── pilot B ──────┬───────────────┘││
   │└────── pilot Y ──────┼──┬─────────────┘│
   └─────── pilot R ──────┼──┼──┬───────────┘
                         [P][P][P]  phase relays
                          │  │  │  (87)
                          └──┴──┴───[E]── neutral
                                  earth fault
                                  relay (64)

Arrangement

• Identical CTs at the neutral end and the line end of each phase; the two CT groups are each connected in star.
• Pilot wires join corresponding phases, forming a circulating-current loop.
• Two (or three) phase-fault relays (P) are connected between the pilots and a common star point.
• An earth-fault relay (E) is connected between this common point and the star point of the CTs (neutral pilot), so it carries the residual current.

Operation

• Normal and external faults: currents at both ends are equal, the secondary currents circulate and no current flows in any relay.
• Phase-to-phase fault inside the winding: the out-of-balance current returns through the phase relays of the faulted phases, which trip.
• Earth fault inside the winding: the unbalance current returns through the earth-fault relay (with a lower setting), giving sensitive earth fault protection.
• The relays trip the generator breaker, field breaker and prime mover.

Both are differential (unit) protection schemes that compare quantities at the two ends of a protected zone. In the circulating current scheme the CT secondary currents circulate through the pilots, and the relay is connected across the pilots at the equipotential point. In the balanced (opposed) voltage scheme the CT secondary voltages are opposed in series through the pilots, so no current flows normally, and relays are connected in series with the pilots.

Circulating current:          Balanced voltage:
 CT1 ──── pilot ──── CT2       CT1 ──R1── pilot ──R2── CT2
  │        │          │         │                      │
  └──── pilot ────────┘         └────── pilot ─────────┘
          [R] (shunt)             relays in series;
     relay across pilots          CTs oppose each other

Summary: in the circulating current scheme the CTs work as normal current sources in a closed loop, while in the balanced voltage scheme the CTs work like open-circuited voltage sources whose emfs cancel, so they must be air-gapped to avoid saturation and overheating.

A unit protection scheme protects a clearly defined zone (a generator, transformer, busbar or line section) by comparing electrical quantities at all the boundaries of the zone. It operates only for faults inside its zone and is inherently selective, so it can trip instantaneously without time grading. Differential protection (Merz-Price) is the main example.

Circuit diagram (circulating current differential)

           ┌──── Protected zone ────┐
  I₁ ──(CT1)───── equipment ─────(CT2)── I₂
         │                         │
      i₁ │      pilot wire         │ i₂
         ├─────────────────────────┤
         │           │             │
         │        [ Relay ]        │
         │     (i₁ − i₂ flows)     │
         ├─────────────────────────┤
              pilot wire

Operating principle

• Identical CTs are placed at both ends (or all terminals) of the zone, and their secondaries are connected through pilot wires so that their currents circulate.
• The relay is connected across the pilots at the equipotential point and carries the difference of the secondary currents (i₁ − i₂).
• Normal load or external (through) fault: current entering = current leaving (I₁ = I₂), so i₁ = i₂, no current flows in the relay and it does not operate, however large the through current.
• Internal fault: current entering ≠ current leaving (fault current is fed from one or both ends), so i₁ − i₂ flows in the relay. When it exceeds the setting the relay trips the breakers at all ends of the zone.
• To avoid false trips due to CT mismatch and saturation, percentage bias or high-impedance relays are used.

Types

• Circulating current (Merz-Price) differential.
• Balanced (opposed) voltage scheme (Translay, for feeders).
• Pilot wire and carrier (phase comparison) protection for lines.
• Restricted earth fault and frame leakage protection.

Applications

• Generators: stator phase and earth faults (biased differential).
• Transformers: differential protection with CTs in star/delta to correct phase shift.
• Busbars: high-impedance differential protection.
• Feeders and lines: pilot wire for short lines, carrier current/fibre-optic current differential for long lines.
• Large motors.

Merits and demerits

• Fast, sensitive and absolutely selective; no time grading needed.
• Does not give back-up for adjacent zones, so separate back-up (e.g. overcurrent) protection is required.
• Needs pilot or communication links and matched CTs.

Percentage bias is the setting of a differential relay in which the operating current needed to trip is not fixed but is a fixed percentage of the through (restraining) current. The relay operates only when

|I₁ − I₂| ≥ K × (I₁ + I₂)/2 + I_min

where K is the bias (slope), typically 10–40%.

Need for bias

In a simple differential relay, the CTs at the two ends are never perfectly matched. During heavy through faults the CTs may saturate unequally, and the transformer tap changer and magnetising current also cause mismatch. The resulting spill current may exceed a fixed setting and cause a false trip. With bias, the operating level rises as the through current rises, so the relay remains stable.

How it is achieved in practice

• The relay has a restraining (bias) coil and an operating coil.
• The restraining coil is connected in the pilot wire circuit, with its mid-point tapped; it carries the average through current (I₁ + I₂)/2 and produces restraining torque.
• The operating coil is connected between the mid-point of the restraining coil and the CT star point; it carries the differential current (I₁ − I₂) and produces operating torque.
• In an induction or balanced-beam relay the two coils act on the same moving element in opposite directions. The ratio of turns of the restraining and operating coils sets the percentage slope.
• In static/numerical relays the same rule is applied in software, often with a dual slope (low slope at low current, steeper slope at high current) and harmonic restraint (2nd harmonic for inrush, 5th for over-excitation) in transformers.

 CT1 ──────┬────R/2────●────R/2────┬────── CT2
           │      restraining      │
           │           │ mid-point │
           │       [Operating]     │
           │           │           │
           └───────────┴───────────┘

Characteristic

 Operating │          /  trip
 current   │        /
 I₁ − I₂   │      /  slope = K %
           │    /
           │__/      restrain (no trip)
           └──────────────────────
             through current (I₁+I₂)/2

When a percentage differential relay is preferred

• When the protected unit carries heavy through-fault currents that may saturate CTs (large generators, transformers).
• When CT ratios cannot be matched exactly (e.g. transformer ratios not matching standard CT ratios).
• On transformers with on-load tap changers, where the ratio changes with tap position.
• Where magnetising inrush or long pilot leads cause unbalance.
• In general, whenever high sensitivity to internal faults must be combined with high stability for external faults.

Time-current grading means setting each IDMT relay so that the relay nearest the fault trips first, and every relay towards the source waits one grading margin longer.

Assumptions: the given curve is for TMS = 1; the TMS of relay Z is 0.1; grading margin between adjacent relays = 0.5 s (CB time + relay overshoot + errors); CT secondary = 5 A; linear reading of the curve between points.

Formulas

Pick-up current (primary) = Current setting × CT primary
PSM = Fault current / Pick-up current
t = TMS × t(TMS=1, read from curve)

Relay Z (farthest from source)

Pick-up = 1.00 × 200 = 200 A
PSM     = 2000 / 200 = 10   → t(TMS=1) = 2.8 s
t_Z     = 0.1 × 2.8 = 0.28 s

Relay Y (back-up for Z)

For a fault at Z (2000 A) Y must wait 0.5 s longer than Z.

Pick-up = 1.25 × 200 = 250 A
PSM (fault at Z) = 2000 / 250 = 8 → 3.0 s
Required t_Y = 0.28 + 0.5 = 0.78 s
TMS_Y = 0.78 / 3.0 = 0.26
Fault at Y (4000 A): PSM = 4000/250 = 16 → 2.2 s
t_Y = 0.26 × 2.2 = 0.572 s

Relay X (back-up for Y)

Pick-up = 1.25 × 400 = 500 A
PSM (fault at Y) = 4000 / 500 = 8 → 3.0 s
Required t_X = 0.572 + 0.5 = 1.072 s
TMS_X = 1.072 / 3.0 = 0.357
Fault at X (5000 A): PSM = 5000/500 = 10 → 2.8 s
t_X = 0.357 × 2.8 = 1.0 s

Grading summary

Source ──X──────Y──────Z────── load
time:   1.0 s   0.57 s  0.28 s
        ← each relay 0.5 s slower than next →

Answer: TMS_Z = 0.1, TMS_Y = 0.26, TMS_X ≈ 0.357 (set 0.36); operating times for faults at their own points are 0.28 s, 0.572 s and ≈ 1.0 s. A fault at Z is cleared by Z in 0.28 s; Y would back up at 0.78 s and X still later, so discrimination is achieved.

Both relays are IDMT relays graded by time: R1 (33 kV, near fault) is the primary relay and R2 (132 kV side of the transformer) is its back-up, set to operate 0.5 s later.

Data: If = 2000 A at 33 kV, CT of R1 = 200/1, CT of R2 = 100/1, plug setting 100% for both (pick-up = 1 A secondary), TMS of R1 = 0.2, margin = 0.5 s.

Relay R1

Pick-up (primary) = 1.0 × 200 = 200 A
PSM = 2000 / 200 = 10
t (TMS=1) from table = 2.8 s
t_R1 = 0.2 × 2.8 = 0.56 s

Fault current seen by R2

The current on the 132 kV side is reduced by the transformer ratio (ignoring magnetising current):

I(132 kV) = 2000 × 33/132 = 500 A

Relay R2

Pick-up (primary) = 1.0 × 100 = 100 A
PSM = 500 / 100 = 5
t (TMS=1) from table = 3.9 s
Required t_R2 = t_R1 + margin = 0.56 + 0.5 = 1.06 s
TMS_R2 = 1.06 / 3.9 = 0.272

Answer: t_R1 = 0.56 s, t_R2 = 1.06 s, TMS of R2 = 0.272 (in practice the next available step, about 0.275, is chosen).

R1 on the 11 kV feeder is the primary relay and R2 on the 132 kV side is the back-up relay that must wait for the grading margin of 0.75 s.

Data: If = 3500 A at 11 kV; CT of R1 = 200/1, plug 125%; CT of R2 = 100/1, plug 150%; TMS of R1 = 0.25. Times between table points are found by linear interpolation.

Relay R1

Pick-up = 1.25 × 200 = 250 A
PSM = 3500 / 250 = 14
Between PSM 13.6 (2.3 s) and 15 (2.0 s):
t = 2.3 − (14 − 13.6)/(15 − 13.6) × (2.3 − 2.0)
  = 2.3 − 0.2857 × 0.3 = 2.214 s   (TMS = 1)
t_R1 = 0.25 × 2.214 = 0.554 s

Relay R2

I(132 kV) = 3500 × 11/132 = 291.67 A
Pick-up   = 1.5 × 100 = 150 A
PSM       = 291.67 / 150 = 1.94 ≈ 2
t (TMS=1) ≈ 10 s  (start of the given curve)
Required t_R2 = 0.554 + 0.75 = 1.304 s
TMS_R2 = 1.304 / 10 = 0.13

Note: PSM of R2 is just below 2, so the relay works at the very steep start of its curve; the curve is taken as 10 s here. In practice the 150% setting is a little high for this fault level and a lower plug setting on R2 would give more reliable back-up.

Answer: t_R1 ≈ 0.554 s, t_R2 ≈ 1.304 s (TMS of R2 ≈ 0.13).

R1 (33 kV) is the primary relay; R2 (132 kV) backs it up with a grading margin of 0.6 s.

R1: pick-up = 1.0 × 200 = 200 A
    PSM = 2700/200 = 13.5
    From graph (between 10→3 s, 14→2.4 s): t ≈ 2.475 s
    t_R1 = 0.15 × 2.475 = 0.371 s

R2: I(132 kV) = 2700 × 33/132 = 675 A
    pick-up = 1.0 × 100 = 100 A
    PSM = 675/100 = 6.75
    From graph (between 6→4 s, 10→3 s): t ≈ 3.81 s
    t_R2 = 0.371 + 0.6 = 0.971 s
    TMS_R2 = 0.971 / 3.81 = 0.255

Answer: t_R1 ≈ 0.37 s and t_R2 ≈ 0.97 s, with TMS of R2 ≈ 0.255. These values depend on how the graph is read, so they are approximate (±0.05 s).

Differential (Merz-Price) protection of a transformer compares the currents entering and leaving the transformer. For a normal load or an external fault these currents balance, so no current flows in the relay. For an internal fault they do not balance, and the difference current operates the relay.

      HV side                     LV side
 ──CT1──┬──[ POWER TRANSFORMER ]──┬──CT2──
   (Y)  │      (Δ / Y)            │  (Δ)
        i1                        i2
        │      ┌──────────┐      │
        └──────┤ OC  │ RC ├──────┘
     pilot     │  relay   │   pilot
               └──────────┘
   OC = operating coil (carries i1 − i2)
   RC = restraining coil (carries (i1+i2)/2)

Operation

1. CTs are placed on both sides; their secondaries are joined by pilot wires so that, under healthy conditions, the secondary currents circulate and the relay operating coil carries almost zero.
2. External fault / load: I1 and I2 rise together, the circulating currents stay equal, the relay does not trip (stable).
3. Internal fault (winding to earth, phase-to-phase, turn-to-turn): extra current flows from one or both sides into the fault, the secondary currents differ, and (i1 − i2) flows in the operating coil. When it exceeds the setting, the relay trips the HV and LV breakers.

Special points for transformers

• CT ratios: chosen so the secondary currents are equal, i.e. inverse of the voltage ratio.
• CT connections: CTs on the star side are connected in delta and CTs on the delta side in star, to correct the 30° phase shift and to block zero-sequence current for external earth faults.
• Percentage (biased) relay: the restraining coil makes the trip setting rise with through current, so CT mismatch, tap changing and CT saturation do not cause false tripping.
• Magnetising inrush: at switching-on, a large inrush current flows on one side only. It is rich in second harmonic, so harmonic restraint is used to block tripping.

Use: standard main protection for transformers above about 5 MVA.

An IDMT overcurrent and earth-fault scheme on a transformer uses CTs and relays on each side separately. By connecting the same CTs (on both sides) together through pilots and placing the relay in the difference path, the scheme becomes a differential (Merz-Price) scheme.

Conventional IDMT O/C and E/F scheme

HV CTs ─► 2 × O/C relays + 1 × E/F relay (residual)
LV CTs ─► 2 × O/C relays + 1 × E/F relay (residual)
Each relay sees only its own side current

These relays are graded with downstream relays, so they are slow (0.5–2 s) and cannot tell an internal fault from an external one.

Steps to convert into differential protection

1. Use CTs on both sides of the transformer and choose their ratios so that secondary currents are equal at full load (CT ratios in inverse proportion to the voltage ratio).
2. Correct the phase shift: connect CTs on the star winding in delta and CTs on the delta winding in star (for a Δ/Y transformer). This also removes zero-sequence current, so external earth faults do not cause tripping.
3. Join the CT secondaries by pilot wires so that currents circulate between the two CT groups.
4. Shift the relays from the CT secondaries into the spill (difference) path between the pilots – one relay per phase.
5. Replace plain O/C elements with biased (percentage) relays having operating and restraining coils, and add harmonic restraint against magnetising inrush.
6. Earth-fault element: a restricted earth-fault (REF) relay can be formed from the three line CTs and a neutral CT of the star winding.

CT(Δ-conn)─┐  pilot   ┌─CT(Y-conn)
           ├──[87]────┤
           │  spill   │
HV ═══[ TRANSFORMER Δ/Y ]═══ LV

Advantages over other schemes

Other advantages: zone is clearly defined by CT positions; it protects the windings and the bushings/connections between CTs; works for any loading level. O/C and E/F relays are then kept only as back-up.

The relays are graded so that C (farthest from source) trips first, B backs up C, and A backs up B, with a fixed time margin.

Assumptions: curve given for TMS = 1; TMS of relay C = 0.1 (fastest relay); grading margin = 0.5 s; CT secondary 5 A; linear interpolation of the table.

Pick-up = Current setting × CT primary
PSM = I_fault / Pick-up ;  t = TMS × t(curve)

Relay C

Pick-up = 0.5 × 200 = 100 A
PSM = 2000/100 = 20 → 2.2 s
t_C = 0.1 × 2.2 = 0.22 s

Relay B

Pick-up = 0.75 × 300 = 225 A
Fault at C: PSM = 2000/225 = 8.89
  t = 3.1 − (8.89−8.8)/(10−8.8) × 0.3 = 3.078 s
Required t_B = 0.22 + 0.5 = 0.72 s
TMS_B = 0.72/3.078 = 0.234
Fault at B: PSM = 3000/225 = 13.33 → 2.499 s
t_B = 0.234 × 2.499 = 0.585 s

Relay A

Pick-up = 1.0 × 400 = 400 A
Fault at B: PSM = 3000/400 = 7.5
  t = 3.8 − (7.5−6.5)/(8.8−6.5) × 0.7 = 3.496 s
Required t_A = 0.585 + 0.5 = 1.085 s
TMS_A = 1.085/3.496 = 0.310
Fault at A: PSM = 4000/400 = 10 → 2.8 s
t_A = 0.310 × 2.8 = 0.869 s

Answer: TMS_C = 0.1, TMS_B ≈ 0.234, TMS_A ≈ 0.31; times for faults at their own sections are 0.22 s, 0.585 s and 0.869 s. If a different TMS for C or margin is used, the same steps apply.

An induction type directional overcurrent relay operates only when the current exceeds its setting and flows in a chosen direction (e.g. towards a fault in the protected line). It has two elements on one case: a directional (power) element and a non-directional induction overcurrent element.

        CT                       PT
        │                        │
   ┌────┴─────────────┐   ┌──────┴──────┐
   │ Directional elem │   │ voltage coil│
   │ (wattmetric disc)│◄──┤             │
   │  current coil    │   └─────────────┘
   │  contacts 1-2 ───┼──┐
   └──────────────────┘  │ (in series with
   ┌──────────────────┐  │  lower coil of O/C)
   │ O/C element      │◄─┘
   │ upper coil (CT)  │
   │ lower coil       │
   │ disc + spring    │──► trip contacts
   └──────────────────┘

Construction

1. Directional element: like an induction wattmeter. Upper magnet has a voltage coil (from PT); lower magnet has a current coil (from CT). An aluminium disc/cup lies between them. Torque T = V·I·cos(θ − τ), where τ is the maximum torque angle.
2. Overcurrent element: a wattmeter-type induction disc relay. The upper magnet carries the main current coil (with plug setting tappings) and a secondary winding; the secondary winding feeds the lower magnet coil through the contacts of the directional element.
3. Control spring, brake (permanent) magnet for damping, and a disc spindle carrying the moving contact; time-setting by changing the backstop (TMS).

Principle of operation

• Normal power flow: directional torque is reverse, its contacts stay open, the lower coil of the O/C element is open, so the O/C disc has no torque – no trip even if current is high.
• Fault with power flow in the tripping direction: directional disc closes its contacts in a few cycles; the lower coil gets current, the O/C disc develops torque. If current exceeds pick-up, the disc rotates and closes the trip contacts after a time decided by current magnitude and TMS.

IDMT characteristic and how it is obtained

IDMT (Inverse Definite Minimum Time): operating time is inversely proportional to current at low multiples of setting, and tends to a definite minimum time at high multiples (PSM about 10–20).

 time│\
     │ \
     │  \__
     │     \____
     │          ‾‾‾‾‾‾‾‾  definite minimum
     └─────────────────── PSM
       2   5   10   20

How it is obtained in the induction relay:

• Driving torque ∝ I², so at low currents the disc speed rises with current and time falls (inverse part).
• The electromagnet core is designed to saturate at higher currents (about 10× setting). Beyond saturation the flux and torque hardly rise, so time stays nearly constant – the definite minimum part.
• The brake magnet gives damping torque ∝ speed, giving a steady disc speed.
• Standard IDMT (3-s or 1.3-s) curve: t = 0.14 × TMS / (PSM^0.02 − 1).

An overcurrent relay operates when the current through it exceeds a preset value (pick-up). By their time-current characteristic they are classified as below.

• Instantaneous + IDMT units are often combined in one relay to clear close-in faults fast.
• Earth-fault relays are overcurrent relays fed from residual CT current, with a lower setting (20–40%).

A distance relay measures the impedance (or a part of it, R or X) between the relay point and the fault, using the voltage and current at the relay: Z = V/I. Since line impedance is proportional to length, the relay operates when the measured impedance falls below the impedance of the protected section – i.e. when the fault lies within a set distance.

Principle

• Operating torque from current, restraining torque from voltage.
• Balanced-beam form: K₁I² > K₂V² → trips when V/I < √(K₁/K₂) = Z_set.
• During normal load, V is high and I moderate, so Z is large → no trip. During a fault within the zone, V falls and I rises → Z small → trip.

Types

Stepped zones

A──────────B──────────C
Zone 1: 80–90% of AB  instantaneous
Zone 2: 100% AB + 20–50% BC  t ≈ 0.3–0.5 s
Zone 3: AB + BC + 25% beyond  t ≈ 1 s

Applications

• Main protection of transmission and sub-transmission lines (66 kV and above), where grading of overcurrent relays is difficult.
• Reactance relays for short lines (arc resistance does not affect them).
• Impedance relays for medium lines; mho relays for long lines (less affected by power swings).
• Back-up protection of adjacent lines (zone 2, zone 3), and protection of generators against loss of field.

Advantages: fast and selective; reach is almost independent of source impedance and fault level.

An impedance relay is a distance relay that compares the voltage and current at the relay location and operates when the ratio V/I (the impedance seen up to the fault) is less than its setting Z_s.

Working principle (balanced beam type)

   PT ── voltage coil ┐       ┌ current coil ── CT
                      ▼       ▼
        restraining ◄─[ beam ]─► operating
                       pivot
                     trip contact

• Current coil gives operating torque ∝ I²; voltage coil gives restraining torque ∝ V².
• Torque equation (neglecting spring): T = K₁I² − K₂V². Relay trips when K₁I² > K₂V², i.e. Z = V/I < √(K₁/K₂) = Z_s.
• On the R–X diagram the characteristic is a circle of radius Z_s centred at the origin. Points inside the circle → trip. It is non-directional, so it is used with a directional unit.
• Under normal load, Z is large (outside circle); during a fault within reach, Z drops inside the circle and the relay trips.

Application area

• Protection of medium-length transmission lines, with three zones (zone 1 instantaneous for 80–90% of line).
• Back-up protection of adjacent line sections.
• Phase-fault protection where overcurrent grading is difficult (interconnected networks).
• Not ideal for short lines (arc resistance error) or very long lines (power-swing tripping).

Relay 1 (near fault F) is the primary relay; relay 2 backs it up with a 0.5 s margin.

Data: CT 300/1 for both, plug setting R1 = 125%, R2 = 150%, TMS of R1 = 0.4, If = 3000 A.

Relay 1

Pick-up = 1.25 × 300 = 375 A
PSM = 3000/375 = 8 → t(TMS=1) = 3.15 s
t1 = 0.4 × 3.15 = 1.26 s

Relay 2

Pick-up = 1.5 × 300 = 450 A
PSM = 3000/450 = 6.67 → t(TMS=1) = 3.5 s
Required t2 = 1.26 + 0.5 = 1.76 s
TMS2 = 1.76 / 3.5 = 0.503

Answer: operating time of relay 1 = 1.26 s; TMS of relay 2 ≈ 0.503 (≈ 0.5), giving t2 = 1.76 s.

Relay 1 is nearer the fault, so it trips first; relay 2 must wait 0.4 s more. Times between table points are found by linear interpolation.

Relay 1

Pick-up = 1.25 × 250 = 312.5 A
PSM = 3000/312.5 = 9.6
Between 9.23 (2.9 s) and 10 (2.8 s):
t = 2.9 − (9.6 − 9.23)/(10 − 9.23) × 0.1 = 2.852 s
t1 = 0.18 × 2.852 = 0.513 s

Relay 2

Pick-up = 1.5 × 250 = 375 A
PSM = 3000/375 = 8
Between 5 (3.9 s) and 9.23 (2.9 s):
t = 3.9 − (8 − 5)/(9.23 − 5) × 1.0 = 3.191 s
t2 = t1 + 0.4 = 0.913 s
TMS2 = 0.913 / 3.191 = 0.286

Answer: t1 ≈ 0.513 s, t2 ≈ 0.913 s (TMS of relay 2 ≈ 0.286).

Relay 1 (nearest the fault) operates first; relay 2 is the back-up and must operate 0.3 s after relay 1.

Data: CT 300/1 for both, plug setting R1 = 125%, R2 = 150%, TMS of R1 = 0.2, IF = 6000 A.

Relay 1

Pick-up current = 1.25 × 300 = 375 A
PSM = 6000/375 = 16
t(TMS=1) = 2.2 s (from table)
t1 = 0.2 × 2.2 = 0.44 s

Relay 2

Pick-up current = 1.5 × 300 = 450 A
PSM = 6000/450 = 13.33
t(TMS=1) = 2.8 s (from table)
t2 = t1 + margin = 0.44 + 0.3 = 0.74 s
TMS2 = 0.74 / 2.8 = 0.264

Answer: relay 1 operates in 0.44 s and relay 2 in 0.74 s (TMS of relay 2 ≈ 0.264).

Relay A is the back-up for relay B. It must not trip until relay B has operated and breaker B has cleared the fault, allowing for errors in both relays.

Data: CT 800/5 for both; relay B: PS 125%, TMS 0.2; relay A: PS 200%; If = 16000 A; CB time 0.2 s; relay error 0.15 s each.

Relay B

Pick-up = 1.25 × 800 = 1000 A
PSM = 16000/1000 = 16 → t(TMS=1) = 2 s
t_B = 0.2 × 2 = 0.4 s

Required time of relay A

Relay B may be 0.15 s slow and relay A may be 0.15 s fast, so:

t_A = t_B + CB time + error(B) + error(A)
    = 0.4 + 0.2 + 0.15 + 0.15 = 0.9 s

TMS of relay A

Pick-up = 2.0 × 800 = 1600 A
PSM = 16000/1600 = 10 → t(TMS=1) = 3 s
TMS_A = 0.9 / 3 = 0.3

Answer: minimum TMS of relay A = 0.3 (operating time 0.9 s). Any smaller TMS could let relay A trip before breaker B has cleared the fault.