Design steps of two-wire AC current transmitter
Known high-current current transformers convert different currents into 0~5A alternating current for on-site display. For long-distance transmission, the current must be converted into a standard DC current signal of 4-20mA before transmission.
Most of these AC current transmitters on the market adopt the "four-wire system" method: two AC power lines and two DC current signal lines. And what we designed is the "two-wire AC current transmitter" that only uses two wires: that is, while providing DC power to the circuit in the transmitter, it will output standard DC according to the change of 0~5A AC current. The current signal 4-20mA is transmitted to the control room for display or enters the computer for processing and then displays on the monitor screen.
Design ideas 1. Choose low-power components and try to simplify the circuit on the premise of meeting the functional requirements to meet the requirements of two-wire meters.
2. Take effective measures to improve the anti-interference ability of the system and reduce temperature drift.
3. Improve system protection measures and increase the reliability of the instrument.
One, the choice of transformer
A current transformer is an alternating current/current converter. When an alternating current flows through the primary, the secondary coil generates an alternating current corresponding to its transformation ratio. It is then converted into an AC voltage signal through load resistance.
It is very important to choose the transformation ratio of the transformer reasonably.
Before selecting the transformation ratio, first determine whether the load voltage generated by the transformer meets the input signal voltage required by the transmission circuit. Usually we choose the maximum value of the input signal voltage around 2~3V/AC.
At the same time, select the transformer load resistance as the standard series resistance. Choose RL=1KΩ. (See Figure 1)
For example: the input signal voltage is 2.5V.
Known: AC current input is 0~5A,
The transformation ratio is: 5A/0.0025A=2000
Therefore, when the primary current of the current transformer changes from 0 to 5A, the voltage across the secondary load resistance is 0 to 2.5V.
Choose a 5A/2.5mA transformer.
If the maximum value of the input signal voltage is required to be selected at 3V, only the load resistance needs to be selected as RL=1.2KΩ.
Still choose 5A/2.5mA transformer.
Second, the choice of rectifier circuit
If the input signal is very weak, you need to amplify the signal before rectifying it. In order to simplify the circuit, we have chosen a relatively large input signal voltage range, 0~2.5V/AC. Therefore, it can be directly rectified without amplification.
If the commonly used crystal diodes are used directly for rectification, the forward voltage drop of the diodes will cause normal output at small currents, which will cause the transmitter to be unable to linearly output standard current signals when the transformer input current is ≤1A. The reason is that the forward voltage drop of the crystal diode is about 0.5~0.7V. When the input current of the transformer is ≤1A, the voltage across the secondary load resistance is ≤0.5V, and the crystal diode cannot be turned on!
We use the feedback circuit of the operational amplifier to achieve the characteristics of the ideal diode to obtain zero-crossing rectification, that is, the ideal rectification of small signals, so as to obtain the characteristics of high-precision linear rectification.
At the same time, in order to simplify the circuit and reduce the power consumption of the transmitter, a half-wave precision rectifier circuit is adopted. (See Figure 2)
The R2, R3, D1 and N1 operational amplifiers in the picture constitute an ideal diode rectifier circuit with positive output. D1 is connected in series with the output terminal of operational amplifier N1, and feedback starts from the cathode of D1. R2 is the input resistance Ri connected in series, and R3 is the feedback resistance Rf. Since there is no need to zoom in, choose R3=R2. Generally, the input impedance of a general-purpose operational amplifier is tens of kiloohms, so you can choose R2=R3=10KΩ~51KΩ. The relative error is required to be as small as possible, otherwise the output DC voltage will cause errors.
For the input negative half cycle signal, N1 is a typical inverting amplifier. The gain at this time is A=(—Vi)×(—R3/R2)=Vi
For the input positive half-cycle signal, the output of N1 becomes negative
At this time, D1 is reversely biased and cut off, and the input signal Vi is directly output to the subsequent circuit through the series circuit of R2 and R3