CT Saturation & Burden Calculator

How the CT saturation result is calculated

The calculator follows the nonlinear excitation model used in the IEEE PSRC CT saturation calculator theory. It treats the CT secondary as an ideal current source feeding winding resistance, lead resistance, relay burden, and a nonlinear exciting branch. The two excitation-curve points define the saturated log-log slope rather than a simple knee-point pass/fail check.

First it calculates the ratio N = IPR / ISR, where IPR is rated primary current and ISR is rated secondary current. From excitation points (Ie1, Ve1) and (Ie2, Ve2), it calculates:

S = ln(Ie2 / Ie1) / ln(Ve2 / Ve1)

Vs = Ve1 x (10 / Ie1)^(1 / S)

S is the inverse log-log saturation exponent, and Vs is the extrapolated RMS excitation voltage at 10 A exciting current. The relay VA is converted to impedance at rated secondary current. Round-trip lead resistance is 2 x length x resistance / 1000. The total resistive burden is winding resistance plus lead resistance plus the resistive part of device burden.

The ideal secondary current includes AC fault current and a decaying DC offset:

is(t) = (sqrt(2) Ip / N) x [cos(wt - phi0) - Off x exp(-t / tau)]

The CT flux linkage is then integrated with a fixed-step fourth-order Runge-Kutta method. At each step, exciting current is calculated from flux linkage as ie = A x sign(lambda) x abs(lambda)^S, actual secondary current is i2 = is - ie, and excitation voltage is d lambda / dt.

What the reported error means

First-cycle composite error is the RMS error current over the first complete cycle divided by the RMS ideal secondary current for that same cycle:

error = 100 x sqrt(integral(ie^2 dt) / integral(is^2 dt))

The 10% result is a visible engineering reference. It is not a universal acceptance rule for every relay element. Some schemes tolerate more transient error, while differential, high-impedance, distance, and sensitive overcurrent applications may need a more detailed relay-specific review.

What moves the answer most

Higher fault current, higher X/R ratio, reinforcing remanence, longer leads, higher winding resistance, and larger connected burden all push the CT toward saturation. A 1 A secondary usually handles long leads better than a 5 A secondary because the same VA burden corresponds to a larger impedance but much lower current through the leads.

What this calculator leaves out

This is a screening model. It simplifies below-knee excitation behavior and does not model hysteresis loops, eddy-current loss, leakage reactance, residual flux history, frequency-dependent material behavior, relay algorithm response, or current-limiter and breaker timing. Use actual excitation and winding-resistance data when possible, and confirm final protection decisions with relay-manufacturer guidance and a qualified protection engineer.

Worked example

For a 1200:5 CT with excitation points of 1 A at 360 V and 10 A at 400 V, the saturated-curve exponent is ln(10 / 1) / ln(400 / 360) = 21.85, and the extrapolated 10 A voltage is 400 V RMS. With 100 ft one-way No. 10 copper example leads at 1.21 ohm/1000 ft, lead resistance is 2 x 100 x 1.21 / 1000 = 0.242 ohm.

Add 0.3 ohm CT winding resistance and a 1 VA unity-power-factor device burden on a 5 A secondary, which is 1 / 5^2 = 0.04 ohm. The total resistive burden is 0.582 ohm, and the external rated-current burden is 7.05 VA. With a 12 kA symmetrical primary fault, X/R of 12, 100% DC offset, no remanence, and a 60 Hz system, the first-cycle composite error is effectively 0.00%, the actual first-cycle secondary current is about 74.47 A RMS, peak excitation voltage is about 72.91 V, and the reverse solver finds about 34,524 A primary for a 10% first-cycle composite-error reference.