The Core Formula: Actual Accuracy Limit Factor (ALF)
When configuring a protective relay, engineers rely on a set class calculator—often built into software like ETAP, SKM, or SEL AcSELerator—to verify that the Current Transformer (CT) will not saturate during a maximum fault. If you are doing this manually on the bench or verifying software outputs, the foundational math relies on the Actual Accuracy Limit Factor (ALF) for IEC classes (5P, 10P, PX) or the Voltage Saturation limit for ANSI classes (C100, C200).
The most universally applied formula for determining if an IEC-class CT will maintain its accuracy under real-world connected burdens is the ALF scaling equation:
ALFactual = ALFrated × (Srated_burden / Sactual_burden)
Where the actual connected burden in Volt-Amps is calculated as:
Sactual_burden = Isn2 × (Rw + Rr)
Symbol Definition Table
| Symbol | Definition | Standard Unit |
|---|---|---|
| ALFactual | Actual Accuracy Limit Factor (multiple of rated current before saturation) | Dimensionless |
| ALFrated | Rated ALF from the CT nameplate (e.g., the '20' in 5P20) | Dimensionless |
| Srated_burden | Maximum burden the CT is rated to drive while maintaining ALFrated | Volt-Amps (VA) |
| Sactual_burden | Total real-world burden of wires and relay inputs | Volt-Amps (VA) |
| Isn | Rated secondary current of the CT (typically 1A or 5A) | Amperes (A) |
| Rw | Total loop resistance of the secondary wiring (out and return) | Ohms (Ω) |
| Rr | Resistance of the protective relay input burden | Ohms (Ω) |
Real-World CT Baseline Data
Before running your set class calculator, you need baseline parameters. Here is a data-dense reference table of typical nameplate values and internal resistances you will encounter in the field:
| CT Class | Isn (Sec) | ALFrated / Class | Srated (VA) | Typical Rct (Internal) | Max Symmetrical Fault (Sec) |
|---|---|---|---|---|---|
| IEC 5P10 | 1 A | 10 | 10 VA | 0.50 Ω | 10 A |
| IEC 5P20 | 1 A | 20 | 15 VA | 1.20 Ω | 20 A |
| IEC 5P20 | 5 A | 20 | 30 VA | 0.15 Ω | 100 A |
| ANSI C100 | 5 A | N/A (Voltage) | 10 VA (B-0.5) | 0.15 Ω | 100 A (at 20x) |
| ANSI C200 | 5 A | N/A (Voltage) | 25 VA (B-1.0) | 0.30 Ω | 100 A (at 20x) |
Rearranged Forms & Unit Pitfalls
A good set class calculator allows you to solve for any missing variable. Here are the algebraically rearranged forms you need for field sizing:
- Solving for Maximum Allowable Burden:
S_actual_burden = S_rated_burden × (ALF_rated / ALF_required) - Solving for Maximum Wire Resistance (Rw):
R_w = (S_actual_burden / I_sn^2) - R_r - Solving for Required Nameplate ALF (Sizing a new CT):
ALF_rated = ALF_required × (S_actual_burden / S_rated_burden)
- The Loop Trap: Rw is the loop resistance. If your relay is 100 feet away, the wire run is 200 feet (out and back). Forgetting to multiply single-phase distance by 2 is the #1 cause of field saturation.
- VA vs. Ohms: Relay manuals often list burden in VA (e.g., 0.5 VA). You cannot add VA directly to Ohms. You must convert relay VA to Ohms first using
R_r = VA / I_sn^2. - Milli-ohms: CT internal resistance (Rct) is often published in milli-ohms on test sheets. Divide by 1000 before plugging it into the formula.
Worked Example 1: Verifying an IEC 5P20 Feeder CT
Scenario: You are setting a feeder relay. The primary fault current is 12,000A. The CT ratio is 600:5. The CT nameplate reads 5P20, 30 VA. The connected electromechanical relay has a burden of 0.5 VA. You are using 10 AWG copper wire (1.0 Ω per 1000 ft) for a one-way distance of 150 feet.
Step 1: Determine Secondary Fault Current
I_sec_fault = 12,000A / (600 / 5) = 100 A
Required ALF to avoid saturation = 100 A / 5 A = 20
Step 2: Calculate Actual Connected Burden (Sactual)
Wire loop length = 150 ft × 2 = 300 ft.
R_w = (300 / 1000) × 1.0 Ω = 0.30 Ω
Relay resistance (R_r) = 0.5 VA / (5A)2 = 0.5 / 25 = 0.02 Ω
Total Resistance = 0.30 + 0.02 = 0.32 Ω
S_actual = (5A)2 × 0.32 Ω = 25 × 0.32 = 8.0 VA
Step 3: Calculate ALFactual
ALF_actual = 20 × (30 VA / 8.0 VA) = 20 × 3.75 = 75
Verdict: Your actual ALF is 75. The fault only demands an ALF of 20. The CT will not saturate. In fact, it is massively oversized for this burden, which is typical when 5A secondary systems are used with short, thick wire runs.
Worked Example 2: Sizing Wire for an ANSI Class C100 CT
Scenario: You have an ANSI Class C100, 5A CT. Internal resistance (Rct) from the test sheet is 0.15 Ω. The solid-state relay burden is 0.01 Ω. The maximum secondary fault current is 150A. What is the maximum allowable wire resistance to prevent saturation?
ANSI C100 means the CT can deliver 100V to a standard burden without exceeding 10% ratio error. The formula is V_class ≥ I_sec_fault × R_total.
Step 1: Set up the voltage equation
100V = 150A × (R_ct + R_w + R_r)
100V = 150A × (0.15 Ω + R_w + 0.01 Ω)
Step 2: Solve for R_w
100 / 150 = 0.16 + R_w
0.667 Ω = 0.16 Ω + R_w
R_w = 0.667 - 0.16 = 0.507 Ω
Step 3: Convert to Wire Length
If using 12 AWG THHN (loop resistance ≈ 1.98 Ω / 1000 ft):
Max Length = (0.507 Ω / 1.98 Ω) × 1000 ft = 256 feet.
Verdict: You can run up to 256 feet of 12 AWG wire. If the switchgear is 260 feet away, you must step up to 10 AWG or the relay will see a clipped, saturated waveform during a close-in fault.
When This Formula Applies (and When It Fails)
Understanding the boundaries of your set class calculator is what separates a technician from a protection engineer. The formulas above assume steady-state, purely symmetrical AC fault currents.
The DC Offset Trap
In the real world, faults are rarely perfectly symmetrical. Depending on the X/R ratio of the power system and the exact point-on-wave where the fault occurs, the fault current will have a decaying DC offset. This DC component drives the CT core into saturation much faster than the AC component alone.
If your system has a high X/R ratio (common in generator bays or heavily inductive industrial plants, typically X/R > 15), the steady-state ALF formula is insufficient. You must apply a Transient Dimensioning Factor (Ktd). According to IEEE C57.13 guidelines and IEC 60044-1, the required ALF becomes:
ALFrequired = (I_fault_sec / I_sn) × Ktd
Where Ktd typically ranges from 1.5 to 3.0 depending on the system time constant. If you ignore Ktd on a high-X/R bus, your relay might under-reach or fail to trip because the CT saturated during the first half-cycle of the transient.
Realistic Answer Magnitudes
When reviewing software outputs or manual calculations, use these sanity checks:
- ALFactual between 10 and 40: Normal, well-designed distribution system.
- ALFactual > 100: The CT is drastically oversized, or you are using a 1A secondary system with extremely short wire runs. While safe from saturation, you are wasting money on copper and core steel.
- ALFactual < 5: Immediate danger. The CT will saturate during normal load swings or minor downstream faults. Check if you accidentally used primary current instead of secondary current in your math.
For deeper dives into transient saturation and relay setting coordination, the Electrical Engineering Portal's CT fundamentals guide provides excellent visual examples of saturated waveforms and their impact on differential and overcurrent elements.
