Equivalent circuit diagram of current transformer.
i1: primary current.
i2: The value of the primary current divided by the rated current ratio converted to the secondary current.
ie: excitation current.
ib: The secondary current flowing into the secondary load.
ES: Secondary induction electromotive force.
RCT: Transformer secondary winding resistance.
XCT: Leakage reactance of secondary winding of transformer.
zb: secondary load impedance.
ze: excitation impedance.
kn: rated current ratio.
10% error test principle
From the above equivalent circuit diagram of the current transformer, it can be seen that the primary current I1 of the current transformer is divided into two parts after being converted to the secondary, a small part is used for internal loss excitation, magnetic loss, and eddy loss, and the other part flows into the secondary circuit, and this current is the secondary current IB that we measured externally, so the excitation current is the reason for the error of CT.
The excitation impedance Ze is not a constant constant, but Ze is deterministic under the same induced electromotive force, so we can obtain the relationship between you and IE by measuring the excitation characteristic curve, and the composite error of CT in various operating states can be obtained by calculation.
Since the error has been set to 10%, i.e., i2=10ie, there is the following relationship:
From the equivalent circuit diagram, it can be seen that i2=ie+ib and i2=10 ie can obtain: ib=9 ie
Define m as the maximum short-circuit current multiple on the primary side, and kn as the conversion ratio of the current transformer, then there is:
Where:
is the maximum short-circuit current on the primary side.
It is the rated current on the primary side.
It is the rated current on the secondary side (size is 1A or 5A).
The maximum load impedance allowed for the secondary circuit when the ratio is 10 bad
The formula for calculation is:
where:
It is the secondary winding impedance of the current transformer.
It is the secondary induction of electromotive force for the current transformer.
According to the formula
The relationship curve between IE and ES can be obtained, from which the ES value can be obtained, and the voltage value U corresponding to IE can be found on the excitation curve.
According to Equation 11 and 12. Finally, the maximum short-circuit current multiple m and the maximum load impedance in the allowable secondary circuit can be obtained
10 error curves described.
CTP test error curve
The above process is implemented by the tester software, and the error curve is drawn to reflect the characteristics of the CT itself. Based on the 10% error curve of the CT, we must also look at the maximum short-circuit current flowing through the CT and the impedance z of the loop on the secondary side of the CT. The maximum short-circuit current is usually obtained during the setting calculation, and is the short-circuit current at the most serious short-circuit under the maximum operation mode of the CT channel. The secondary loop impedance z can be measured with a CTP tester setup. Through the nameplate or the maximum short-circuit current multiple m and the measured loop impedance z, check the 10% error curve drawn by the tester, if the point (m, z) is below the measured curve, it meets the requirements, indicating that the current conversion error of CT is less than 10% in the most serious short-circuit case. If the point (m,z) is above the measured curve, it means that the current conversion error of CT is greater than 10% in the case of the most severe short circuit, which does not meet the requirements.
Here's an example
For example, if the parameters of a CT are: the rated current ratio is 200A 5A, 10P10 level, the rated load is 30VA, and the actual load is 10VA.
Answer: According to the nameplate, CT passes 2000A at one time, and when 10 errors need to be met, the maximum allowable load of the second time is 11446, less than 12 (=rated load 5A*5A), from which it can be concluded that the CT cannot carry the rated load.
In practice, when CT passes 2000A at one time and has to meet 10 errors, the maximum allowable load of the second time is 11446, greater than 04 (= actual load 5a*5a), from which the CT can be obtained with the actual load.