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Saturday, October 23, 2010

OP90 4 mA to 20 mA Current Loop Transmitter Circuit


This is a circuit for current loop circuit. An output of 4 mA to 20 mA that is linearly proportional to the input voltage is provided by the current transmitter on figure below. Line rejection is 0.0005%/volt and linearity of the transmitter exceeds 0.004%. This is the figure of the circuit;
The REF-02EZ provides biasing for the current transmitter. The output current is regulated by OP90EZ to satisfy the current summation at the noninverting node:
Iout = 1/R6((Vin.R5)/R2 + (5V.R5)/R1)
Iout = (16/100 ohm)Vin + 4mA
That’s will give a full-scale output of 20 mA with a 100mV input. By adjusting R2, we can provide a offset trim and adjustment of R1 will provide a gain trim. Since the non-inverting input of the OP90 is at virtual ground, these trims do not interact. The input voltage spike will be prevented by D1 (the Schottky diode) from pulling the non-inverting input more than 300 mV below the inverting input. Without diode, such spikes could cause phase reversal of the OP90 and possible latch-up of the transmitter. This circuit has compliance from 10V to 40V. For transducer excitation, the voltage reference output can provide up to 2 mA. [Circuit diagram source: Analog Application Note]



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