MSP430F5659: ADC problem

Hello,

In the attached schematic,  pin 6.0, 6.1 and 6.2 is connected to adc pin p6.0 p6.1 and p6.2 of microcontroller.

These are plain transimpedance amplifiers, where V_out = −I_sensor × 47 Ω. So for currents between −4 mA and −20 mA, you get between 0.188 V and 0.94 V.

If the P5.0/P5.1 pins are available, you can apply something like 0.12 V and 1 V to them and use them as reference voltages for the ADC.

But it's probably easier to change the opamp configuration to get the correct output.
To increase the gain, simply increase the resistors.
To subtract an offset from the output voltage, add resistors to V_CC from pin 2 of the opamps; the offset is V_CC × R_1,2,3 / R_offset. (See, e.g., I-to-V Amp with Offset for details.)

Alternatively, check how the RCV420 does this:

THEORY OF OPERATION

For 0 to 5V output with 4–20mA input, the required transimpedance of the circuit is:

V_OUT/I_IN = 5V/16mA = 0.3125V/mA.

To achieve the desired output (0V for 4mA and 5V for 20mA), the output of the amplifier must be offset by an amount:

V_OS = –(4mA)(0.3125V/mA) = –1.25V.

The input current signal is connected to either +In or –In, depending on the polarity of the signal, and returned to ground through the center tap, C_T. The balanced input—two matched 75Ω sense resistors, R_S—provides maximum rejection of common-mode voltage signals on C_T and true differential current-to-voltage conversion. The sense resistors convert the input current signal into a proportional voltage, which is amplified by the differential amplifier. The voltage gain of the amplifier is:

A_D = 5V/(16mA)(75Ω) = 4.1667V/V.

The tee network in the feedback path of the amplifier provides a summing junction used to generate the required –1.25V offset voltage. The input resistor network provides high-input impedance and attenuates common-mode input voltages to levels suitable for the operational amplifier’s common-mode signal capabilities