How to choose the current transformer&transformer

For a long time, mutual inductance current transformer as standard equipment in a sophisticated device has been used to measure current. Even in harsh environments and high temperature conditions, this device is very accurate, easy to use and reliable.
 Such as switching power supplies, motor current load detection, lighting and instrumentation applications, the current transformer transformers are generally as control, circuit protection and monitoring devices to use. With the current increasing mutual inductance of the Electrical Transformers spot, how to choose a suitable current transformer transformers need to consider many factors, this paper introduces a simple selection method, this method for many applications to choose the right device is very cost-effective there to help. Although the spot devices are cheap, instant provision, but there are some functional limitations on use, some applications may require special products, or even need to fully customize.
 Choice of transformer current transformer to consider factors such as size, frequency, function and current range.
 Input current
 First, the current transformer and the transformer must be selected to verify the number of clear indicators, such as size, frequency, function and sampling current range. It really depends on the accuracy and efficiency indicators of these parameters. In addition to current transformer may be a compromise on the accuracy of the transformer, the transformer current transformer used if the current exceeds the manufacturer’s rated current specification, then the temperature will rise and can not control, leading to circuit failure.
 In addition, if the mutual inductance of a transformer rated current than the “sample current” much higher, then the device will inevitably large size, for its applications becomes too expensive. In general, the selected current transformer transformer rated higher than some “sample current” 30% of the maximum expected value is a wise choice.
 Turns ratio
 Common current transformer transformer turns ratio range from 1:10 to 1:1000 range. Turns ratio (r = Nsec / Npri) higher, the higher the resolution of current measurement.
 However, it is worth noting that high turns ratio will lead to distributed capacitance and leakage inductance increases, thereby reducing the current transformer accuracy and high frequency transformers operating performance (due to self-resonance caused). However, if the turns ratio is too low (low inductance), then the output signal may appear distorted or “down” (single-stage input signal skew will occur), causing the control circuit unstable, inaccurate results.
 Inductance and excitation current
 Current transformer secondary inductance of the Power Distribution Transformer determines the output signal fidelity. The value of inductance is inversely proportional to the excitation current, excitation current known as the “induced current.”
 In order to ensure the maximum transformer current transformer fault tolerance, the excitation current amplitude should be smaller than the sampling current several times. For most applications such as switching power supply, take the sample 10% of current maximum excitation current is ideal. For example, if a circuit under 100kHz sampling for 1 ~ 20A current to ensure maximum loss of 10%, then the maximum excitation current must be set to 100mA (the minimum value of 10% of the current sample).
 1A current will produce a sampling error of 10%, 20A current will produce a sampling error of 0.5%. If the manufacturer’s data sheet does not indicate the excitation current, then calculated by the following formula:
 e = CLdI / dt
 | DI / dt | = e / L
 Where e is the device output voltage (in units of V), L is the inductance (in units of H), | dI / dt | is the ratio of the excitation current and time (in units of A / s).
 Output voltage and load resistance
 Output voltage (Vo) should be set as low as possible, in order to reduce insertion loss. Assuming an optimal secondary circuit output voltage is 0.5V, the output current is 20A, then the turns ratio of 1:100 of the current transformer will produce about 200mA transformer secondary current. Load resistance should be: Ro = Vo / Is = 0.5/0.2 = 2.5Ω.