CURRENT TRANSFORMER (CT) BURDEN SAMPLE CALCULATION

Higher ohmic burdens in the ct secondary circuit will tend to result in greater saturation of the core, and therefore, larger errors in the secondary current waveform. The reason for this is that a given secondary current requires more voltage from the ct for a higher burden, and the core flux density is proportional to the time-integral of this voltage.

When the core becomes saturated, significant current is diverted through the cts magnetizing branch, and the desired secondary current is reduced and distorted. Burden calculations are, therefore, necessary to ensure that ct accuracy limits are not exceeded.

The total ohmic burden on the ct is the vector sum of the ct winding resistance, the connecting lead resistance, the impedance of any auxiliary cts, and the impedance of the connected relays and meters. Impedances of devices connected in the secondary of an auxiliary ct should be reflected (multiplied by the square of the auxiliary ct ratio) to the primary side, when calculating the burden on the main ct. This is only accurate if the auxiliary ct is not saturated.

As a first check in making the burden calculation, it is common practice to add the individual burdens arithmetically rather than vectorially. In many cases, this approach is very accurate, particularly if the ct winding resistance and the connecting lead resistance comprise the bulk of the secondary burden.

However, if this method predicts poor ct performance, and if information on burden power factor is available, the less conservative, but more complicated, vectorial method should be used. Electromechanical relays are usually subject to saturation themselves, at high currents.

Coil impedances at the currents of interest (as opposed to rated current) should be used in the burden calculation. A table of burdens vs, current (burdens may be expressed either in ohms or volt-amperes) is usually provided in the relay instruction book, but information on the power factor is often incomplete. In this case, it is customary to assume a purely resistive burden.

With the ohmic burden determined, the next step in predicting ct performance is to determine the required ct excitation voltage by multiplying the calculated total ohmic burden (using the magnitude, in the case of vectorquantities) by the maximum expected secondary fault current.

The ct excitation characteristic is then used to determine the excitation current. The higher the excitation current, as a proportion of the expected secondary current, the worse will be the actual replication of the primary current waveform. If errors greater than 10% are indicated (or more conservatively, if the calculated excitation voltage is above the knee-point), then the application is suspect and measures to reduce the burden are advised.

Sample burden calculation
Consider the 1200/5 ct of figure 1 below applied under conditions of a 24 000 A maximum fault current as illustrated in below. First consider the circuit without the auxiliary ct and then with the auxiliary ct.

The relay time-overcurrent unit is to be set for 5 A, and the instantaneous unit for 40 A. The secondary current under maximum fault conditions is expected to be 24 000/240 = 100 A.


1200/5 ct: From figure 1, the winding resistance is 0.61 W
1500 ft of #10 wire: 1500 ft ´ 1.0 W/1000 ft ´ 2 = 3.0 W.
TOC unit: The relay instruction book indicates a burden of 490 VA at 20 times tap-value (5 A), or 0.049 W.
IOC unit: The instruction book indicates a burden of 0.007 W for this unit.
Total burden: The scalar addition of all burdens (a fairly accurate approach which also simpliÞes calculations), results
in 0.61 + 3.0 + .049 + 0.007, or about 3.7 W.
The required excitation voltage is, therefore, 3.7 ´ 100 = 370 V. This is well above the knee-point voltage of the ct, and
is at best a marginal application.
Consider now the same application with the addition of a 5:1, T200 auxiliary ct.
Auxiliary ct: According to the manufacturer, the internal burden of the auxiliary ct is 1.11 VA at 5 A. The ohmic burden
is, therefore, 1.11 W on the secondary side.
TOC unit: The reduced current requires that the TOC unit now be set on the 1 A tap. The burden at 20 times tap-value
current is given as 265 VA, or 0.66 W
IOC unit: The IOC unit burden at the new tap setting is given as 0.125 W.
Total burden on the auxiliary ct: Again using a scalar addition, the secondary burden on the auxiliary ct is 1.11 + 3.0
+ 0.66 + 0.125, or 4.9 W. The required excitation voltage from the auxiliary ct is 4.9 ´ 20, or 98 V, well within the
capability of a T200 ct.


Total burden on the main ct: Reflected to the primary the auxiliary ct secondary burden is 4.9/25, or 0.196 ohm. The total burden on the main ct is, therefore, 0.61 + 0.196, or 0.81 ohm. The required excitation voltage on the main ct is now 81 V, representing a dramatic reduction compared with the previous example.

It should be pointed out that in general, other factors such as dc offset in the primary current waveform, ct remanence, the operating characteristics of the connected relays etc., should also be considered. This may result in a requirement for better cts (or smaller connected burdens) than calculations of the above type would indicate.