The CTS (Current Transformer Saturation) formula calculates the maximum secondary voltage a current transformer (CT) can produce before its magnetic core saturates and distorts the output waveform. In steady-state AC conditions, the fundamental CTS formula is V_sat = I_sec_fault × (R_ct + R_b). If the calculated voltage exceeds the CT's ANSI/IEEE C-class rating (e.g., C200 means 200V), the CT will saturate, potentially blinding your protection relays during a fault.
Whether you are sizing wire for a new substation panel or troubleshooting a relay that failed to trip, mastering this formula is non-negotiable for power systems work. Below, we break down the variables, outline the assumptions, and walk through real-world calculations with strict unit tracking.
The Core CTS Formula and Symbol Definitions
The steady-state saturation voltage equation defines the boundary where the CT core can no longer linearly step down the primary current. The formula is expressed as:
V_sat = I_sec_fault × (R_ct + R_b)
Every symbol in this equation represents a specific physical parameter of the CT and its connected secondary circuit. Here is the exact spec-sheet definition for each variable:
| Symbol | Parameter | Unit | Description & Typical Values |
|---|---|---|---|
| V_sat | Saturation Voltage | Volts (V) | Maximum RMS voltage across the secondary terminals before saturation. For a C400 CT, this is 400V. |
| I_sec_fault | Secondary Fault Current | Amperes (A) | Primary fault current divided by the CT ratio. Typical values range from 5A to 120A. |
| R_ct | CT Secondary Winding Resistance | Ohms (Ω) | DC resistance of the CT's internal copper windings. Typically 0.1Ω to 1.5Ω at 20°C. |
| R_b | Total External Burden Resistance | Ohms (Ω) | Sum of lead wire resistance (round-trip) and relay burden. Typically 0.2Ω to 2.0Ω. |
Assumptions, Limits, and Unit Mistakes
When the Formula Applies (and Its Assumptions)
The standard CTS formula assumes steady-state symmetrical AC fault current. It calculates the RMS voltage required to drive the secondary current through the total circuit resistance. It assumes the burden is purely resistive (which is generally true for modern microprocessor relays and copper wire, where reactance X is negligible compared to R). If your circuit has high inductive burden (like old electromechanical plunger relays), you must use impedance (Z) instead of resistance (R), calculating the vector sum.
Which Unit Mistakes Break the Calculation
The most common ways engineers and technicians invalidate this formula on the bench or in the field:
- Mixing Primary and Secondary Current: Plugging the 12,000A primary fault current directly into the formula instead of scaling it down by the CT ratio (e.g., 12,000A / 600:5 = 100A secondary).
- Forgetting the Round-Trip Lead Length: Calculating wire resistance using the one-way physical distance from the CT to the relay. The current must return, so you must double the physical length for the resistance calculation.
- Ignoring Temperature Derating: Copper resistance increases by roughly 20% when a panel heats up from 20°C to 75°C. Using 20°C NEC Chapter 9 Table 8 values for a hot outdoor cabinet will under-predict V_sat.
What a Realistic Answer Magnitude Looks Like
For standard ANSI C-class protection CTs, V_sat will typically fall between 50V and 800V. A standard C200 CT is rated for 200V. If your calculation yields 250V, the CT will saturate, and the secondary current waveform will clip, delaying relay operation. If your calculation yields 4,500V, you have likely forgotten to divide the primary current by the CT ratio.
Rearranged Forms of the CTS Formula
In the field, you rarely just solve for V_sat. Usually, you know your CT class and need to find out how much wire you can run, or what the maximum fault current can be. Here are the algebraically rearranged forms:
- Solving for Maximum Allowable Burden (R_b):
R_b = (V_sat / I_sec_fault) - R_ct
Use this to determine the maximum wire length and relay burden you can add to an existing CT. - Solving for Maximum Secondary Fault Current (I_sec_fault):
I_sec_fault = V_sat / (R_ct + R_b)
Use this to find the fault current threshold where your CT will begin to saturate. - Solving for Maximum CT Winding Resistance (R_ct):
R_ct = (V_sat / I_sec_fault) - R_b
Use this during procurement to specify the maximum allowable internal resistance when ordering new CTs.
Worked Examples with Unit Tracking
Let's apply the formula to two common scenarios. We will track units at every step to ensure dimensional consistency. For wire resistance, we reference baseline values from the NETA ATS testing standards and NEC Chapter 9, Table 8.
Problem 1: Calculating Saturation Voltage for an Existing Circuit
Scenario: You have a 600:5 CT with a C200 rating. The maximum primary fault current is 12,000A. The CT internal resistance (R_ct) is 0.5Ω. The connected microprocessor relay has a burden (R_r) of 0.1Ω. The one-way wire run is 200 feet using 12 AWG THHN copper. Will this CT saturate?
- Calculate Secondary Fault Current (I_sec_fault):
I_sec_fault = I_primary × (Secondary Ratio / Primary Ratio)
I_sec_fault = 12,000 A × (5 / 600) = 100 A - Calculate Lead Wire Resistance (R_leads):
12 AWG copper at 20°C is 1.588 Ω per 1,000 ft.
Round-trip distance = 200 ft × 2 = 400 ft.
R_leads = (400 ft / 1,000 ft) × 1.588 Ω = 0.6352 Ω - Calculate Total External Burden (R_b):
R_b = R_r + R_leads = 0.1 Ω + 0.6352 Ω = 0.7352 Ω - Calculate Total Circuit Resistance:
R_total = R_ct + R_b = 0.5 Ω + 0.7352 Ω = 1.2352 Ω - Apply the CTS Formula for V_sat:
V_sat = 100 A × 1.2352 Ω = 123.52 V
Verdict: The required voltage is 123.52V. Since the CT is rated C200 (200V), it will not saturate under steady-state conditions. However, note that if this panel operates at 75°C, the 12 AWG resistance jumps to ~1.98 Ω/kft, pushing V_sat closer to 145V. It remains safe, but the margin shrinks.
Problem 2: Sizing Maximum Wire Length for a C400 CT
Scenario: You are installing a new 1200:5 C400 CT. The maximum secondary fault current is calculated at 80A. The CT nameplate lists R_ct as 0.8Ω. The relay burden is 0.2Ω. You are pulling 10 AWG copper wire (0.9989 Ω/kft at 20°C). What is the maximum one-way physical distance you can run the wire without saturating the CT?
- Determine Maximum Total Resistance Allowed:
R_total_max = V_sat / I_sec_fault = 400 V / 80 A = 5.0 Ω - Isolate Maximum Allowable External Burden (R_b):
R_b_max = R_total_max - R_ct = 5.0 Ω - 0.8 Ω = 4.2 Ω - Isolate Maximum Allowable Lead Resistance (R_leads):
R_leads_max = R_b_max - R_r = 4.2 Ω - 0.2 Ω = 4.0 Ω - Calculate Round-Trip Wire Length:
Length_round_trip = (R_leads_max / Resistance_per_kft) × 1,000
Length_round_trip = (4.0 Ω / 0.9989 Ω) × 1,000 = 4,004.4 ft - Calculate One-Way Physical Distance:
Distance_one_way = 4,004.4 ft / 2 = 2,002.2 ft
Verdict: You can run the 10 AWG wire up to 2,002 feet one-way. In practice, you would derate this by 20% for temperature and voltage drop margins, limiting the physical run to roughly 1,600 feet. For deeper insights into transient saturation limits over long runs, refer to the IEEE C37.110 Guide for Application of Current Transformers.
Frequently Asked Questions
How does DC offset affect the CTS formula?
The standard CTS formula only accounts for symmetrical AC RMS current. In real-world faults, especially near generators or transformers, the fault current contains a decaying DC offset. This DC component drives the CT core into saturation much faster than AC alone. To account for this, protection engineers apply a transient dimensioning factor (often denoted as K or X/R ratio multiplier) to the formula: V_sat_transient = I_sec_fault × (R_ct + R_b) × (1 + X/R). If your system has a high X/R ratio (e.g., 20), the required V_sat can be 21 times higher than the steady-state calculation, forcing you to use much larger CTs or air-gapped cores.
What is the difference between the CTS formula and the knee-point voltage?
The CTS formula calculates the voltage required by the circuit. The knee-point voltage (V_k) is a physical property of the CT itself, defined by IEC 61869-2 (Class PX). V_k is the point on the excitation curve where a 10% increase in voltage causes a 50% increase in magnetizing current. When designing to IEC standards, you use the CTS formula to find the required circuit voltage, and then ensure the CT's physical V_k is greater than or equal to that calculated value (usually with a safety factor of 2). ANSI C-class ratings (like C200) are roughly analogous but defined by a 10% ratio error limit at 20 times nominal current.
Why does the CTS formula sometimes fail to predict actual saturation?
If your math says the CT shouldn't saturate, but your relay shows clipped waveforms, the culprit is usually residual magnetism (remanence) in the CT core. If a previous fault or a DC hi-pot test left the core partially magnetized, the available flux swing before saturation is drastically reduced. The standard CTS formula assumes the core starts at zero flux. To fix this, you must either demagnetize the CT, use a core with an air gap (which lowers remanence but increases magnetizing current), or oversize the CT rating by a factor of 2 to 3 to absorb the remanence penalty.
