Current and Potential Transformers

Knee point voltage

The knee point voltage of a CT is the point on its excitation (magnetisation) curve where a 10 % increase in secondary e.m.f. causes a 50 % increase in exciting current (definition of IEC/IS). Beyond this point the core saturates, the exciting current rises rapidly and the CT can no longer reproduce the primary current faithfully.

 Vs (sec. emf)
  │            ___________ saturation
  │        _--
  │   Vk ●   knee point
  │     /
  │    /  linear region
  │   /
  └──┴────────────────── Ie (exciting current)

A protective CT needs a high knee point voltage so that it does not saturate during heavy fault current; it is chosen as Vk ≥ K × If(sec) × (R_CT + R_lead + R_relay).

Measuring CT vs protective CT

The knee point voltage (Vk) is the secondary e.m.f. on the CT excitation curve at which a 10 % increase in voltage produces a 50 % increase in exciting current. It marks the beginning of core saturation.

Excitation characteristic

 Vs (V)
  │                 ________ saturation
  │             _--     (flat)
  │        Vk ●         10% ↑V → 50% ↑Ie
  │         /│
  │        / │ ankle-to-knee
  │       /  │ (linear) region
  │      /   │
  │ ankle    │
  └─/────────┴──────────────── Ie (A)

• Ankle region: at very low flux, permeability is low; Ie relatively high.
• Linear region: Ie is small and proportional to Vs; CT is accurate (ratio error small).
• Knee point: beyond Vk the core saturates; a small increase in Vs needs a large Ie.
• Saturation region: most of the primary ampere-turns are used to magnetise the core; the secondary current waveform is distorted and lower than expected.

 i_s ideal   ╱╲    ╱╲
 i_s saturated ╱▔╲__╱▔╲__ (clipped)

Importance in CT selection for protection

1. A protective CT must remain unsaturated up to the maximum fault current, otherwise the relay sees less current and may operate slowly or not at all.
2. Required knee point: Vk ≥ K · I_f(sec) · (R_ct + 2R_lead + R_relay) where I_f(sec) is the maximum fault current referred to the secondary and K (typically 1.5–2, higher with DC offset) allows for transient saturation.
3. In differential and restricted earth-fault schemes, CTs on the two sides must have matched, high Vk (class PS/PX) so that one CT saturating during a through-fault does not create a false spill current; a stabilising resistor is chosen using Vk.
4. Measuring CTs are intentionally given a low knee point so that they saturate and protect meters during faults.
5. Higher Vk needs a larger core and more turns, so cost rises; the designer chooses the minimum Vk that meets the fault level and burden.

Example: For I_f(sec) = 20 × 1 A = 20 A, R_ct + leads + relay = 5 Ω and K = 2: Vk ≥ 2 × 20 × 5 = 200 V.

Typical knee point voltage

The knee point voltage (Vk) of a CT is the secondary e.m.f. at which a 10 % rise in voltage causes a 50 % rise in exciting current. It is the boundary between the linear part of the excitation curve and saturation.

 Vs │           ______ saturation
    │       Vk ●
    │        /
    │       /  linear (accurate) region
    └──────/──────────────── Ie

• Below Vk the CT transforms current accurately; above it the core saturates and secondary current is distorted and reduced.
• A protective CT must have Vk higher than the voltage needed to drive the maximum fault current through its burden: Vk ≥ K·I_f(sec)·(R_ct + R_lead + R_relay). Typical values are a few hundred volts for protection CTs and much lower for metering CTs.

Errors of CT and their causes

An ideal CT has I_s = I_p / K_n exactly in phase opposition. In practice part of the primary ampere-turns is used to supply the exciting current I₀, causing errors.

1. Ratio (current) error: difference between actual and nominal ratio, Current error % = (K_n·I_s − I_p) / I_p × 100 Cause: the loss component of I₀ (in phase with I_s) reduces I_s; also burden power factor.
2. Phase angle error (phase displacement, β): angle between the primary current and the reversed secondary current. Cause: the magnetising component of I₀ (in phase with flux) makes I_s lag/lead the ideal phasor.
3. Composite error: r.m.s. value of the difference between ideal and actual secondary current over a cycle, including harmonics caused by saturation (important for protective CTs).

Causes and factors increasing errors

• High exciting current due to poor core material or high flux density.
• High burden (impedance of relays + leads) – requires more e.m.f., more flux, more I₀.
• Burden power factor (inductive burden increases ratio error; resistive increases phase error).
• Saturation due to high fault current or DC offset; remanent flux in core.
• Low primary current (operating near ankle region, low permeability).
• Frequency different from rated.

Errors are reduced by using high-permeability cores (CRGO, nickel-iron), low flux density, low burden and turn compensation.

A practical CT does not give secondary current exactly equal to I_p/K_n because part of the primary ampere-turns is used to set up flux in the core (exciting current I₀ = magnetising component I_m + loss component I_w). This causes the following errors:

1. Ratio (current) error

The actual ratio R = I_p / I_s differs from the nominal ratio K_n.

 Current error (%) = (K_n·I_s − I_p) / I_p × 100
 Approximately   R ≈ K_n + (I₀ sin(δ+α)) / I_s

where δ is the burden phase angle and α the angle between flux and I₀. Mainly caused by the loss (in-phase) component of I₀.

2. Phase angle error (phase displacement)

The angle β between the primary current phasor and the reversed secondary current phasor.

 β ≈ (180/π) · I₀ cos(δ+α) / (n·I_s)   degrees

Mainly caused by the magnetising component of I₀. It matters for wattmeters, energy meters and directional/distance relays.

3. Composite error

The r.m.s. difference between (K_n × instantaneous i_s) and instantaneous i_p over one cycle, as a percentage of I_p r.m.s. It includes ratio error, phase error and harmonic distortion due to saturation. It is the main error specified for protective CTs (5P, 10P).

Factors affecting errors

• Exciting current (core material, flux density, core size).
• Burden magnitude and power factor.
• Saturation due to fault current or DC offset; remanent flux.
• Supply frequency.

Errors are reduced by low-loss high-permeability cores, low flux density, small burden and turns compensation (one or two secondary turns less than nominal).

Purpose of a CT

A current transformer steps down the high line current to a standard low value (1 A or 5 A) for meters and relays, and isolates them from the high-voltage circuit, so that standard low-rated instruments can be used safely.

Phasor diagram of CT

 Taking flux Φ as reference:
                     ▲ I_p (primary)
         I₀ ↗       /
           /       /  β (phase error)
   ────────●──────/──────────► Φ
  I_m along Φ,   /
  I_w leads Φ by 90°
           ─n·I_s (reversed sec. current)
 E_s lags Φ by 90°; I_s lags E_s by δ.
 I_p = n·I_s(reversed) + I₀ (phasor sum)

Where: Φ = core flux, E_s = secondary induced e.m.f., I_s = secondary current (lags E_s by burden angle δ), I₀ = exciting current = I_m (magnetising, in phase with Φ) + I_w (loss, leading Φ by 90°), n = turns ratio N_s/N_p.

The primary current I_p is the phasor sum of the reversed secondary current (n·I_s) and the exciting current I₀. Because of I₀, I_p is neither exactly n·I_s in magnitude nor exactly in phase with it.

Ratio error

The actual ratio R = I_p/I_s differs from the nominal ratio K_n:

 R ≈ n + I₀ sin(δ + α) / I_s
 Ratio error (%) = (K_n − R) / R × 100

It arises mainly from the loss component of I₀ and depends on burden.

Phase angle error

β is the angle between I_p and reversed I_s:

 β ≈ (180/π) · I₀ cos(δ + α) / (n·I_s)  degrees

It arises mainly from the magnetising component I_m. It is important for wattmeters, energy meters and directional relays.

Instrument transformers (current transformers and potential/voltage transformers) reduce the large currents and high voltages of the power system to standard low values (1 A or 5 A; 110 V or 110/√3 V) for meters and relays.

Need of instrument transformers

1. Isolation: meters, relays and operators are isolated from the high-voltage system, improving safety.
2. Standardisation: relays and meters of standard rating (5 A / 1 A, 110 V) can be used on any system voltage or current.
3. Economy: low-rated instruments with thin wiring are much cheaper than high-voltage, high-current instruments.
4. Remote location: relays and meters can be placed in a control room away from the switchyard.
5. Protection functions: CTs and VTs supply the current and voltage signals needed by overcurrent, differential, distance and directional relays.
6. Combination: several CT cores can serve separate metering and protection circuits.

Errors of CT

Because part of the primary ampere-turns supplies the exciting current I₀, the secondary current is not exactly I_p/K_n.

1. Ratio (current) error: (K_n·I_s − I_p)/I_p × 100 %. Caused mainly by the loss component of I₀; increases with burden.
2. Phase angle error (phase displacement): angle between I_p and reversed I_s, caused mainly by the magnetising component of I₀.
3. Composite error: r.m.s. difference between ideal and actual secondary current over one cycle including harmonics; important for protective CTs under saturation.

Errors increase with high burden, saturation (high fault current, DC offset), remanent flux, low primary current and poor core material.

In an ideal PT, V_s = V_p/K_n and exactly in phase opposition. In a practical PT the voltage drops in winding impedances and the no-load current cause ratio error and phase angle error.

Phasor relation

 V_p = (reversed) E_p + I_p(R_p + jX_p)
 E_s = V_s + I_s(R_s + jX_s)
 I_p = I₀ + I_s/n   (I₀ = I_m + I_w)

So the reversed V_s differs from V_p in magnitude (ratio error) and angle (phase error).

Causes of ratio error

1. No-load (exciting) current I₀ flowing through the primary resistance and leakage reactance produces a voltage drop.
2. Load current flowing through primary and secondary resistances (R_p, R_s) and leakage reactances (X_p, X_s) causes further drop.
3. Burden: higher burden (more VA) means more current and larger drops; ratio error increases.
4. Burden power factor: at lagging pf the reactive drops add more directly to the magnitude.
5. Supply voltage or frequency different from rated changes I₀.

 Ratio error (%) = (K_n − R)/R × 100,  R = V_p/V_s

Causes of phase angle error

1. Magnetising current I_m and loss current I_w through winding impedance shift the phase of E_p relative to V_p.
2. Resistive drops (I·R) of the windings, especially at unity-pf burden.
3. Leakage reactance drops at lagging pf burden.
4. Burden magnitude and power factor.

Methods to reduce errors

• Low flux density and good core material to reduce I₀.
• Low winding resistance and leakage reactance (thick conductors, interleaved windings).
• Turns compensation (slightly fewer primary turns).
• Keeping the burden within rating.

A capacitor voltage transformer (CVT) is a voltage transformer that uses a capacitive voltage divider to reduce a very high system voltage (66 kV and above) to an intermediate level (about 10–20 kV), followed by a small electromagnetic transformer that gives the standard secondary voltage (110/√3 V).

Construction

 Line (HV)
    │
   ═╪═ C1  (stack of capacitor units)
    │
    ├────── tap (≈ 10–20 kV)
    │        │
   ═╪═ C2   [L]  tuning reactor
    │        │
    │      ┌─┴─┐
    │      │IVT│ intermediate VT
    │      └─┬─┘
   ─┴─ E     ├── secondary to relays/meters
    │       FSC (ferro-resonance
   PLC     suppression circuit)
  (carrier tap)

1. Capacitor divider (C1, C2): many paper/polypropylene film, oil-impregnated capacitor elements stacked in series inside a porcelain or polymer insulator housing. C1 (high-voltage section) is much smaller than C2. Tap voltage V₂ = V₁ × C1/(C1 + C2).
2. Tuning (compensating) reactor L: connected in series with the IVT primary; it is tuned to resonate with (C1 + C2) at system frequency, i.e. ωL = 1/[ω(C1 + C2)], so that the output voltage is independent of burden and in phase with the line voltage.
3. Intermediate voltage transformer (IVT): a small oil-filled electromagnetic VT stepping down the tap voltage to 110/√3 V; provided with taps for ratio adjustment.
4. Ferro-resonance suppression circuit: a damping resistor/filter across the secondary to prevent sustained ferro-resonant oscillations.
5. Carrier (PLC) coupling: the low-voltage end of C2 is brought out through a drain coil so that the same capacitor stack can be used to couple power-line-carrier signals.
6. Base tank: houses reactor, IVT and terminal box; filled with oil.

CVTs are cheaper and smaller than electromagnetic VTs at EHV, but have poorer transient response.

A capacitive (capacitor) voltage transformer (CVT) steps down extra-high voltage using a capacitor divider followed by a small intermediate electromagnetic VT and tuning reactor.

Need of CVT in power system

1. Economy at high voltage: an electromagnetic VT for 132 kV and above needs very heavy insulation, so it becomes large and costly. A capacitor stack handles the high voltage cheaply; only the small IVT works at 10–20 kV. The higher the voltage, the larger the saving.
2. Voltage signal for protection and metering: distance relays, directional relays, synchronising, voltmeters and energy meters require a voltage proportional to line voltage (110/√3 V).
3. Power line carrier communication (PLCC): the same capacitor stack acts as the coupling capacitor for carrier signals (30–500 kHz) used for teleprotection, telephone and SCADA. One equipment serves two purposes.
4. Smaller size and weight: easier transport and installation in EHV switchyards.
5. Reduced risk of ferro-resonance with breaker grading capacitors compared with some inductive VTs (with suppression circuit).
6. Surge protection: the capacitance helps to slope steep wave fronts.

Working principle (brief)

 V_tap = V_line × C1/(C1 + C2)
 Tuning: ωL = 1 / [ω(C1 + C2)]

The tuning reactor compensates the capacitive source impedance so that the output is nearly independent of burden.

Limitations

• Poorer transient response: after a fault, the energy stored in capacitors and reactor causes transients that may affect high-speed distance relays.
• Accuracy is affected by frequency and temperature.

Therefore CVTs are standard for line voltage measurement at 66 kV and above.

Why voltage transformers are required in protection

A voltage (potential) transformer reduces the system voltage to a standard value (110 V line or 110/√3 V phase) for relays and meters.

1. Many relays need voltage as an input: distance (impedance) relays, directional overcurrent and directional earth-fault relays, under/over-voltage relays, under/over-frequency relays, reverse power and loss-of-excitation relays, and synchronism-check relays.
2. Relays cannot be connected directly to 11 kV–400 kV lines; the VT gives isolation and safety.
3. Standard low-voltage relays can be used for any system voltage.
4. Open-delta VT connection provides the residual voltage (3V₀) for earth-fault detection and directional earth-fault relays.

Construction of CCVT (coupling capacitor voltage transformer)

 HV line
   │
  ═╪═ C1 ┐
   │     │ capacitor stack in
  ═╪═ C2 ┘ porcelain housing
   ├──── tap V₂ (≈10–20 kV)
   │       │
   │      [L] tuning reactor
   │       │
   │     ┌─┴──┐
   │     │IVT │→ 110/√3 V to relays
   │     └────┘   + ferro-resonance damper
  [Drain coil]──► PLC carrier equipment
   │
  ═══ earth

1. Capacitor divider: series-connected oil-impregnated capacitor elements (C1 high-voltage part, C2 low-voltage part) in a porcelain/polymer column. Tap voltage = V × C1/(C1 + C2).
2. Tuning reactor (L): in series with the IVT primary; resonates with C1 + C2 at 50 Hz (ωL = 1/ω(C1+C2)) so the ratio and phase do not change with burden.
3. Intermediate VT: small oil-immersed transformer stepping the tap voltage to the standard secondary.
4. Ferro-resonance suppression circuit across the secondary.
5. Drain coil and carrier terminal: low impedance at 50 Hz, high at carrier frequency; allows the capacitor stack to couple PLC signals to the line (hence "coupling capacitor").
6. Oil-filled base tank with terminal box and protective spark gap.

CCVTs are used at 66 kV and above because they are cheaper than electromagnetic VTs and also serve carrier communication.

The accuracy class of a CT states the maximum permissible errors at specified currents when the CT supplies its rated burden. Values below are as per IS 2705 / IEC 60044-1 (now IEC 61869-2).

Measuring CTs

Classes 0.1, 0.2, 0.5, 1 (limits at 100–120 % of rated current), and classes 3 and 5 (at 50–120 %).

Composite error is not specified for measuring CTs; instead an instrument security factor (ISF, e.g. ≤ 5 or ≤ 10) ensures they saturate and protect meters at fault current.

Protective CTs

Designated as 5P and 10P, followed by the accuracy limit factor (ALF), e.g. 5P20 = class 5P accurate up to 20 × rated current.

• Standard ALFs: 5, 10, 15, 20, 30.
• Class PS (IS) / PX (IEC): special protection CTs for differential and restricted earth-fault schemes, specified by turns ratio error (≤ ±0.25 %), minimum knee point voltage, maximum exciting current at Vk/2 and secondary winding resistance, instead of error limits.

Typical use: 0.2/0.5 for revenue metering, 1 for indicating instruments, 5P for overcurrent and distance relays, PS for differential protection.

Applications of CTs and PTs in power system

Current transformers (CTs)

• Metering: ammeters, wattmeters, energy meters (kWh, kVArh) and tariff metering on HV and LV feeders.
• Protection: overcurrent, earth fault, differential (transformer, generator, busbar), distance and directional relays.
• Supplying current to power-factor meters, maximum demand indicators and SCADA transducers.
• Core-balance CTs for sensitive earth-fault detection on cables and motors.

Potential (voltage) transformers (PTs/VTs)

• Metering: voltmeters, wattmeters, energy meters, frequency and power-factor meters.
• Protection: distance relays, directional relays, under/over-voltage, under/over-frequency, reverse power relays.
• Synchronising of generators and synchro-check.
• Open-delta (broken-delta) connection for residual voltage in earth-fault detection.
• CVTs at EHV also couple power-line carrier signals.

Both provide isolation from high voltage and standardise signals to 1 A/5 A and 110 V.

Burden on CT

The burden of a CT is the load connected to its secondary: the total impedance of relay coils, meters and connecting leads. It is expressed in volt-amperes (VA) at rated secondary current and specified power factor (usually 0.8 lagging), or in ohms.

 Burden (VA) = I_s² × Z_burden
 e.g. 5 A CT, Z = 0.6 Ω → 5² × 0.6 = 15 VA

• Standard ratings: 2.5, 5, 10, 15, 30 VA.
• If the connected burden exceeds the rated burden, the CT needs more e.m.f., flux and exciting current, so ratio and phase errors increase and the CT may saturate during faults.
• For 1 A CTs the lead burden is 25 times smaller than for 5 A CTs for the same lead resistance, so 1 A CTs are preferred for long leads.

A current transformer reduces a large line current to a standard value (1 A or 5 A), while a potential transformer reduces a high voltage to a standard value (110 V). Their main distinguishing facts are:

Both provide isolation, standardise relay and meter inputs, and have ratio and phase angle errors.

A current transformer (CT) is an instrument transformer whose primary winding carries the line current in series with the circuit and whose secondary delivers a proportionally reduced current (1 A or 5 A) to meters and relays, with a known phase relation.

Working principle

It works on the principle of electromagnetic induction, like a transformer operating with its secondary nearly short-circuited.

     I_p (line current)
  ═══════════╦═══════════  1 or few turns
          ┌──╨──┐
          │core │  flux Φ
          └──╥──┘
        N_s turns
        ┌────╨────┐
        │  A / R  │ ammeter/relay
        └─────────┘ low impedance burden

• The primary current is fixed by the load of the line, not by the CT.
• It sets up primary ampere-turns N_p·I_p. The secondary current I_s produces opposing ampere-turns; only a small difference remains to magnetise the core:

 N_p·I_p ≈ N_s·I_s   →   I_s ≈ I_p × (N_p/N_s)

• Example: a 500/5 A CT gives 5 A at 500 A primary; turns ratio 1:100.
• The flux density is low in normal operation; if the secondary is opened, the full primary ampere-turns magnetise the core, causing saturation and dangerously high secondary voltage. Hence the secondary must never be open-circuited.

Types

Wound type, bar type, window/ring type, bushing type, core-balance CT.

Application areas

1. Metering: ammeters, wattmeters, energy meters, power-factor meters on HV/LV feeders and tariff metering.
2. Protection: overcurrent and earth-fault relays, differential protection of transformers, generators and busbars, distance and directional relays.
3. Core-balance CTs: earth-leakage and sensitive earth-fault protection of cables and motors.
4. Control and monitoring: SCADA transducers, maximum demand indicators, power quality meters.
5. Isolation: keeps meters and personnel away from high voltage.

Why the CT secondary must not be left open

• A CT primary is in series with the line, so its current I_p is fixed by the load, not by the CT secondary.
• Normally, secondary ampere-turns N_s·I_s nearly cancel primary ampere-turns; only a small exciting current magnetises the core and the flux density is low.
• If the secondary is opened, there is no opposing m.m.f. The entire primary current becomes magnetising current, so the flux rises to saturation.
• Consequences:
  1. Very high secondary voltage: the large secondary turns and rapid flux change (flat-topped flux, steep reversals near current zero) induce peak voltages of several kilovolts. This is dangerous to persons and can break down the secondary insulation.
  2. Core overheating: high flux causes large eddy-current and hysteresis loss, which can burn the insulation.
  3. Residual magnetism: the core may keep remanent flux, increasing future ratio and phase errors.
• Therefore the CT secondary must always be connected to a burden or short-circuited (using a shorting link or test block) before removing a meter or relay.

Why the PT secondary must not be short-circuited

(The correct practice is that the PT secondary must not be short-circuited.)

• A PT primary is connected across the line and works like a normal power transformer with a fixed primary voltage; the secondary feeds high-impedance loads and runs almost on open circuit.
• If the secondary is short-circuited, the only limit to current is the small leakage impedance of the windings, so a very large short-circuit current flows.
• This overheats and burns the windings within a short time and can damage the PT and cause a fault on the HV side.
• Hence the PT secondary must be left open (no-load condition is safe) and is protected by fuses or MCBs; the PT primary is also protected by HV fuses.

Given: N_p = 1 turn, N_s = 180 turns, ammeter resistance R_A = 0.15 Ω, full-scale primary current I_p = 900 A, system voltage 400 V.

(i) Maximum secondary current

For a CT, ampere-turns balance (neglecting exciting current):

 N_p · I_p = N_s · I_s
 I_s = I_p × N_p/N_s
     = 900 × 1/180
     = 5 A

So the ammeter must be a 0–5 A meter giving full-scale at 5 A.

With the ammeter connected, the CT terminal voltage is only

 V_s = I_s × R_A = 5 × 0.15 = 0.75 V
 Primary drop = 0.75/180 = 0.00417 V ≈ 4.17 mV

(ii) Secondary voltage when the ammeter is removed

When the ammeter is removed, the secondary is open-circuited. There is no secondary m.m.f. to oppose the primary, so the CT behaves like a step-up voltage transformer of ratio 1 : 180 connected in the line. In the limit (core not saturated, no other limit), the voltage across the single-turn primary can rise up to the phase voltage of the system.

Assumption: 400 V is the line voltage of a 3-phase system, so phase voltage = 400/√3 = 230.94 V.

 V_s(open) = V_p × N_s/N_p
           = (400/√3) × 180
           = 230.94 × 180
           = 41 569 V ≈ 41.6 kV

(If 400 V were applied directly across the primary, V_s = 400 × 180 = 72 kV.)

Answer: (i) I_s(max) = 5 A (terminal voltage 0.75 V with ammeter). (ii) On open circuit the secondary voltage can rise to about 41.6 kV (theoretical upper limit; 72 kV if 400 V across primary).

In practice core saturation limits the r.m.s. value, but very high peak voltages still appear; this shows why a CT secondary must never be opened under load and must be short-circuited before removing the ammeter.