2.2 Instrument Transformer Burdens, Saturation Curves & Ratio Correction Factors

ANSI/IEEE Relaying Accuracy Classes (C vs. T Class)

Protective relaying CTs must maintain transformation fidelity under severe fault currents that reach 10 to 30 times normal load current. The standard ANSI/IEEE C57.13 classification system describes a CT's capability to deliver secondary voltage without excessive ratio error.

C-Class (Calculated) vs. T-Class (Test)

• C-Class (Calculated Accuracy): Designates CTs with low internal leakage flux, such as bushing or toroidal (donut) CTs with uniformly distributed secondary windings. For C-class CTs, the secondary ratio error can be accurately calculated directly from the manufacturer's secondary excitation curve.
• T-Class (Test Accuracy): Designates CTs with significant internal leakage flux, such as wound-primary CTs. Because leakage flux varies non-linearly with burden and current, performance cannot be purely calculated and must be empirically determined through physical laboratory testing.

Decoding the ANSI C-Rating (e.g., C100, C200, C400, C800)

An ANSI CT rating consists of a letter and a voltage rating:

1. The Letter 'C': Indicates that the ratio error will not exceed 10% at any secondary current from 1 to 20 times rated secondary current ($I_s = 20 \times 5\text{ A} = 100\text{ A}$) with standard burden connected.
2. The Voltage Number ($V_t$): The secondary terminal voltage that the CT can deliver to a standard burden at $100\text{ A}$ secondary current while remaining within the 10% ratio error limit.

$\text{Maximum Allowable External Burden } Z_b = \frac{V_{\text{rating}}}{100\text{ A}}$

Multi-Ratio CT Tap Derating Rule

Many substation CTs are multi-ratio (e.g., MR 1200:5 with taps at 300:5, 400:5, 600:5, 800:5, 900:5, 1000:5, and 1200:5). When a lower tap is selected, the available secondary turns ($N_{tap}$) are reduced. Because induced voltage is proportional to turns, the effective C-voltage rating derates proportionally with the tap ratio:

$V_{\text{tap}} = V_{\text{full}} \times \left( \frac{N_{\text{tap}}}{N_{\text{full}}} \right)$

For example, if a 1200:5 C800 CT is tapped at 600:5, its effective rating drops to:

$V_{\text{tap}} = 800\text{ V} \times \left( \frac{600}{1200} \right) = 400\text{ V} \implies \text{C400}$

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Secondary Excitation Curves & The Knee-Point Voltage

A CT's magnetic performance is characterized by its Secondary Excitation Curve, plotted on logarithmic scales showing secondary excitation RMS voltage ($V_s$) versus secondary excitation current ($I_e$).

Knee-Point Voltage ($V_k$)

• ANSI/IEEE Definition: The point on the log-log excitation curve where the tangent makes a $45^\circ$ angle with the coordinate axes.
• IEC Definition: The point where a 10% increase in secondary voltage causes a 50% increase in excitation current ($I_e$).

Below the knee-point, the core operates in its linear magnetic region where excitation current $I_e$ is negligible (< 0.05 A), ensuring faithful reproduction of primary current. Above the knee-point, the iron core saturates, drawing massive magnetizing current and clipping secondary current peaks.

Required Knee-Point Voltage Sizing Equation

To prevent saturation during symmetrical fault conditions, the CT knee-point voltage must exceed the total internal and external voltage drop under maximum primary fault current:

$V_k \ge I_{s\text{(max)}} \times Z_{\text{total}} = \left( \frac{I_{f\text{(primary)}}}{N_s} \right) \times (R_{ct} + R_{\text{leads}} + Z_{\text{burden}})$

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Burden Calculations in Protection & Metering Circuits

The burden of an instrument transformer is the total opposing impedance connected across its secondary terminals, expressed in ohms ($\Omega$) or volt-amperes (VA at rated secondary current, where $\text{VA} = I_s^2 Z_b = 25 \times Z_b$).

Components of Total Secondary Loop Burden ($Z_{\text{total}}$)

$Z_{\text{total}} = R_{ct} + R_{\text{lead\_loop}} + Z_{\text{device}}$

1. Internal Secondary Resistance ($R_{ct}$): The DC resistance of the CT secondary copper winding itself, typically $0.1\ \Omega$ to $0.8\ \Omega$. (Must be included when evaluating internal induced EMF $E_s$).
2. Lead Wire Resistance ($R_{\text{lead_loop}}$): The resistance of the control cable connecting the CT junction box to the control panel:
   • For a phase-to-phase or single-phase fault, current must travel out and back: $R_{\text{lead_loop}} = 2 \times L \times r_{\text{wire}}$.
   • For a balanced three-phase fault, return currents cancel in the neutral: $R_{\text{lead_loop}} = 1 \times L \times r_{\text{wire}}$.
   • For a single line-to-ground fault through a residual ground relay in the neutral leg, the neutral conductor carries $3I_0$, making effective neutral resistance $3 \times R_{\text{neutral}}$.
3. Connected Device Impedance ($Z_{\text{device}}$): Modern microprocessor relays present very small burdens ($< 0.05\ \Omega$), whereas electromechanical induction-disc relays presented significant inductive burdens ($0.5\ \Omega$ to $3.0\ \Omega$).

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CT Saturation Under Fault Conditions

Asymmetrical Faults and DC Offset

When a short circuit occurs at an instant other than the voltage peak, Faraday's Law requires an asymmetrical DC offset current to maintain flux continuity. This DC component decays exponentially with the primary system time constant:

$\tau = \frac{L}{R} = \frac{X}{\omega R} = \frac{X}{2\pi f R}$

The DC component acts as a unidirectional current that continuously injects flux of one polarity into the core without alternating reversal. To avoid saturation during the initial cycles of an asymmetrical fault, IEEE C37.110 recommends an over-sizing factor:

$V_k \ge I_{s\text{(fault)}} \times \left( 1 + \frac{X}{R} \right) \times (R_{ct} + R_{\text{leads}} + Z_{\text{burden}})$

In systems with high $X/R$ ratios (e.g., $X/R = 20$ near generation stations), the required CT saturation voltage is 21 times larger than for a pure AC fault of identical steady-state RMS magnitude!

Remanent Flux ($\Phi_r$)

When a circuit breaker clears a heavy fault, the sudden interruption leaves residual magnetic flux (remanence) trapped in the iron core, which can persist indefinitely at 60% to 80% of saturation flux density. If a subsequent fault occurs with polarity matching the remanent flux, the CT enters deep saturation within 0.25 to 0.5 cycles, severely blinding differential and overcurrent relays.

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Ratio Correction Factor (RCF) & Transformer Correction Factor (TCF)

In high-accuracy revenue metering, instrument transformer non-idealities are adjusted mathematically:

1. Ratio Correction Factor (RCF): The ratio of the true transformation ratio to the marked (nominal) ratio:

$\text{RCF} = \frac{\text{True Ratio}}{\text{Marked Ratio}} = \frac{I_p / I_s}{\text{Nominal Ratio}}$

$\text{True Primary Current } I_p = I_s \times \text{Marked Ratio} \times \text{RCF}$

1. Phase Angle Error ($\beta$): The angular displacement (in minutes or milliradians) by which the secondary current phasor leads the reversed primary current phasor due to core excitation.
2. Transformer Correction Factor (TCF): For active power revenue metering at load power factor angle $\theta$:

$\text{TCF} = \text{RCF} - \frac{\beta}{2600} \tan \theta \quad (\text{for lagging PF with } \beta \text{ in minutes})$

$\text{Billed Power } P_{\text{true}} = P_{\text{meter}} \times (\text{PT Ratio} \times \text{CT Ratio}) \times \text{TCF}$

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Worked Calculation Example

Problem Statement

A 1200:5 multi-ratio CT rated C400 is connected on its 800:5 tap to protect an outdoor 13.8 kV feeder. The secondary winding resistance at the 800:5 tap is $R_{ct} = 0.28\ \Omega$. The CT is connected to a microprocessor relay ($Z_{\text{relay}} = 0.04\ \Omega$) located in a control house 350 ft away via #10 AWG solid copper wire ($r = 1.24\ \Omega / 1000\text{ ft}$).

The maximum symmetrical 3-phase bus fault current is $16{,}000\text{ A}$ primary with an $X/R$ ratio of $8$.

Determine:

1. The derated ANSI C-voltage rating at the 800:5 tap.
2. The total secondary loop resistance ($R_{\text{total}}$) for a 3-phase fault (assuming no neutral return drop).
3. The secondary fault current ($I_s$) and the required secondary terminal voltage ($V_s$) under symmetrical conditions.
4. Whether the CT will saturate under symmetrical fault conditions.
5. The required voltage including DC offset ($1 + X/R$) and evaluate whether saturation will occur during an asymmetrical fault.

Step-by-Step Solution

Step 1: Calculate Derated C-Rating at 800:5 Tap $V_{\text{tap}} = 400\text{ V} \times \left( \frac{800}{1200} \right) = 266.67\text{ V} \implies \text{C266.67}$

Step 2: Calculate Lead and Total Loop Resistance $\text{One-way distance} = 350\text{ ft}$ $R_{\text{lead}} = 350\text{ ft} \times \left( \frac{1.24\ \Omega}{1000\text{ ft}} \right) = 0.434\ \Omega$ $R_{\text{total}} = R_{ct} + R_{\text{lead}} + Z_{\text{relay}} = 0.28 + 0.434 + 0.04 = 0.754\ \Omega$

Step 3: Calculate Secondary Fault Current and Symmetrical Voltage $\text{CT Ratio} = \frac{800}{5} = 160$ $I_s = \frac{I_{\text{fault}}}{\text{Ratio}} = \frac{16{,}000\text{ A}}{160} = 100.0\text{ A}$ $V_{s\text{(sym)}} = I_s \times R_{\text{total}} = 100.0\text{ A} \times 0.754\ \Omega = 75.4\text{ V}$

Step 4: Evaluate Symmetrical Saturation $\text{Available Voltage } V_{\text{tap}} = 266.67\text{ V} > 75.4\text{ V}$ Conclusion: The CT operates well below its knee point and will NOT saturate during a symmetrical fault.

Step 5: Evaluate Asymmetrical Fault Saturation with DC Offset $V_{\text{req(asym)}} = V_{s\text{(sym)}} \times \left( 1 + \frac{X}{R} \right) = 75.4\text{ V} \times (1 + 8) = 75.4 \times 9 = 678.6\text{ V}$ $\text{Since } V_{\text{req(asym)}} (678.6\text{ V}) > V_{\text{tap}} (266.67\text{ V})$ Conclusion: The CT will heavily saturate on the initial cycles of an asymmetrical fault due to the DC offset. To prevent saturation, either the CT ratio must be increased to 1200:5 ($V_{\text{tap}} = 400\text{ V}$) or a higher C-rating (e.g., C800) must be specified.

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Common Exam Traps & Pitfalls

1. Forgetting Tap Derating: Never use the full C-rating on a tapped multi-ratio CT. A C800 CT tapped at half ratio is only a C400 CT.
2. Ignoring Internal CT Secondary Resistance ($R_{ct}$): When calculating the induced EMF ($E_s$), the internal winding resistance $R_{ct}$ is in series with the external burden and must always be added to the lead and relay impedances.
3. Loop Resistance Multiplier Confusion: For single-phase or line-to-line faults, multiply lead wire length by 2 ($2 \times L$). For balanced three-phase faults, lead resistance is single-conductor length ($1 \times L$) because neutral current is zero.