Instrument amplifier circuit composed of operational amplifier and photocoupler

4mA ~ 20mA current is the current signal uniformly used by industrial instruments. Its advantage is that it does not require power supply. The maximum current available for the circuit part is 4mA, and exceeding 4mA is not allowed, so the circuit must work hard to reduce power consumption.

The attached figure is an amplifier circuit using a low-power op amp and a photocoupler, which uses the photocoupler to achieve insulation between the front and rear circuits. The input current Iin flows through the light-emitting diode (LED) of the photocoupler PC1, generating a voltage drop of 1V to 2V across the two ends of the light-emitting diode. If the op amp can work normally with such a small voltage as the power supply voltage, an isolation amplifier can be designed. In this circuit, the single-power-supply OP90 (AD company product) is selected.

The optocoupler is CNR201, which contains a light emitting diode and a pair of photodiodes PD1 and PD2 with exactly the same characteristics.

The current IPD1 flowing through the photodiode PD1 is proportional to the current flowing through the LED. When IPD1=linRl/R2[A]H, the circuit is in a stable working state. So the operational amplifier Al is used to control the current flowing through the LED. Because R2=l0kΩ, R1=25Ω, IPD1=lin/400. Since the efficiency of the light-emitting diode PC1 is about 0.5% (1/200), the transistor Trl is used to shunt the input current, dividing half of the current of 4mA to 20mA.

Photoelectric current IPD2 also flows through photodiode PD2. Since IPD1=IPD2, after amplification by operational amplifier A2, an output voltage Vout proportional to input current Iin can be obtained.