Data Description of Using Operational Amplifier to Form VI and IV Conversion Circuit

Fig. 1 is a V / I conversion circuit composed of operational amplifier, resistance capacitance and other elements, which can linearly convert a 0-5V DC voltage signal into a 0-10mA current signal. A1 is a comparator. A3 is a voltage follower to form a negative feedback loop. The input voltage VI is compared with the feedback voltage VF, and the output voltages VL and V1 are obtained at the output end of comparator A1 to control the output voltage V2 of operational amplifier A1, Thus, the output current IL of transistor T1 is changed, and the output current IL affects the feedback voltage VF to track the input voltage VI. The output current IL can be calculated by the following formula: Il = VF / (RW R7). Due to the effect of negative feedback, VI = VF, so il = VI / (RW R7). When RW R7 is 500 , the V / I conversion of 0-5V / 0-10mA can be realized. If the performance parameters of the selected device are relatively stable and the amplification factors of A1 and A2 are large, the conversion accuracy of this circuit, Generally, it can meet high requirements.

2. 0-10V / 0-10mA V / I conversion circuit

In Fig. 2, VF is the feedback voltage generated by the output current IL flowing through the resistor RF, that is, the voltage difference between V1 and V2. This signal is added to the two input terminals VP and VN of operational amplifier A1 through resistors R3 and R4. The feedback voltage VF = V1-V2. For operational amplifier A1, VN = VP; VpV1/R2R3 R2VNV2Vi-V2 R4 / (R1 R4), so V1 / (R2 R3) R2V2Vi-V2 R4 / (R1 R4), which can be deduced according to VF = V1-V2 and the above formula:

VF = R4 / R1 VI = 1 / 5vi. If the current flowing through the feedback circuits R3 and R4 is ignored, there are IL = VF / RF = VI / 5rf. It can be seen that when the open-loop gain of the operational amplifier is large enough, the output current IL and the input voltage Vi meet the linear relationship, and the relationship is only related to the resistance of the feedback resistance RF. Obviously, when RF = 200 , this circuit can realize the V / I conversion of 0-10V / 0-10mA.

3. 1-5V / 4-20mA V / I conversion circuit

In Fig. 3, the input voltage VI is superimposed on the reference voltage VB (VB = 10V) and input from the reverse input VN end of the operational amplifier A1. The transistors T1 and T2 form a composite tube as an emitter tracker to reduce the base current of T1 (i.e. ignore the feedback current I2), so that IL I1, and the operational amplifier A1 meets VN VP. If R1 = R2 = R, R4 = R5 = Kr in the circuit diagram, Then there is the following expression:

If RF = 62.5 , k = 0.25, VI = 1-5V, I1 = 4-20mA, and the actual conversion current IL is smaller than I1, with a difference of I2 (IL = i1-i2). I2 is a variable varying with the input voltage VI. when the input voltage is the smallest (VI = 1V), the error is the largest. In practical application, in order to minimize the error, the resistance values of R1, R2 and RF are generally 40.25k , 40K and 62.5 respectively.

4. 0-10mA / 0-5V I / V conversion circuit

In practical application, for the current input signal without common mode interference, a precision wire wound resistance can be directly used to realize the current / voltage conversion, as shown in Figure 4. If the precision resistance R1 RW = 500 , the I / V conversion of 0-10mA / 0-5V can be realized, and if the precision resistance R1 RW = 250 , the I / V conversion of 4-20mA / 1-5V can be realized. In the figure, R and C form a low-pass filter to suppress high-frequency interference. RW is used to adjust the output voltage range, and a zener diode is added at the current input.

For the current input signal with common mode interference, the isolation transformer coupling mode can be adopted to realize the I / V conversion of 0-10mA / 0-5V. Generally, the load capacity of the output end of the transformer is low. In practical application, a voltage follower should be connected at the output end as a buffer to improve the driving capacity.

5. 0-10mA / 0-5V I / V conversion circuit composed of operational amplifier

In Fig. 5, the magnification of op amp A1 is a = (R1 RF) / R1. If R1 = 100k and RF = 150k , a = 2.5; If R4 = 200 , for the current input signal of 0-10mA, a voltage signal of 0-2v will be generated on R4. It can be seen from a = 2.5 that the input current of 0-10mA corresponds to the output voltage signal of 0-5V.

In the figure, the current input signal II is input from the in-phase input terminal of operational amplifier A1, so it is required to select operational amplifiers with high common mode rejection ratio, such as op-07, op-27, etc.

6. 4-20mA / 0-5V I / V conversion circuit

Through the circuit analysis in Fig. 6, it can be seen that the current flowing through the feedback resistance RF is (VO VN) / RF, which is equal to VN / R1 (VN VF) / R5. Therefore, the expression of output voltage Vo can be deduced:

When 4-20mA current signal is input, 0-5V voltage signal is output accordingly.