How to Select Current Transformer
For a long time, as a standard device in instruments and equipment, current transformer has been used to measure precision current. Even in harsh environment and high temperature conditions, this device is very accurate, convenient and reliable.In applications such as switching power supply, motor current load detection, lighting and instruments, current transformer is generally used as control, circuit protection and monitoring devices. With the increasing stock of current transformer, how to select a suitable current transformer needs to consider many factors. This paper introduces a simple selection method, which is very helpful to select suitable cost-effective devices in many applications. Although spot devices are cheap and desirable, they have some functional limitations in use. Some applications may require special products or even complete customization.Many factors should be considered in the selection of current transformer, such as size, frequency, function and current range.
Input currentFirstly, the selection of current transformer must clarify and verify many indicators, such as size, frequency, function and the range of sampling current. Its accuracy and efficiency actually depend on these parameters. In addition to the possible compromise on the accuracy of the current transformer, if the current of the current transformer exceeds the rated current specification specified by the manufacturer, its operating temperature will continue to rise and cannot be controlled, resulting in circuit failure.In addition, if the rating of a current transformer is much higher than its "sampling current", the size of this device will inevitably be large, which is too expensive for its application. Generally speaking, it is a wise choice that the rated value of the selected current transformer is about 30% higher than the maximum expected value of its "sampling current".
Turns ratioThe turns ratio of common current transformer ranges from 1:10 to 1:1000. The higher the turn ratio (r = nsec / NPRI), the higher the resolution of current measurement.However, it should be noted that too high turn ratio will lead to the increase of distributed capacitance and leakage inductance, which will reduce the accuracy of current transformer and the working performance at high frequency (caused by self resonance). However, if the turn ratio is too low (low induction coefficient), the output signal may be distorted or "decreased" (the single-stage input signal must be skewed), resulting in unstable control circuit and inaccurate measurement results.
Induction coefficient and excitation currentThe secondary inductance of current transformer determines the fidelity of output signal. The value of the induction coefficient is inversely proportional to the excitation current, which is commonly known as "induced current".In order to ensure the maximum fault-tolerant performance of current transformer, the excitation current should be several times smaller than the amplitude of sampling current. For most applications such as switching power supply, it is ideal to take 10% of the sampling current as the maximum excitation current. For example, if a circuit must guarantee a maximum loss of 10% for a sampling current of 1 20A at 100kHz, the maximum value of the excitation current must be set to 100mA (i.e. 10% of the minimum sampling current value).
The sampling current of 1A will produce an error of 10%, and the sampling current of 20a will produce an error of 0.5%. If the excitation current is not indicated in the data book provided by the manufacturer, it can be calculated by the following formula:e=CLdI/dt|dI/dt|=e/L
Where e is the device output voltage (unit: V), l is the induction coefficient (unit: H), and|di / dt|is the ratio of excitation current to time (unit: A / s).Output voltage and load resistanceThe output voltage (VO) should be set as low as possible to reduce the intervention loss. Assuming that the optimal secondary output voltage of a circuit is 0.5V and the output current is 20a, the current transformer with a turn ratio of 1:100 will produce a secondary current of about 200mA. The load resistance should be: RO = VO / is = 0.5 / 0.2 = 2.5 .
