Is there any way to have a diode that have very small resistance from reverse path?
simulate this circuit &ndash; Schematic created using CircuitLab
Figure 1. 4 - 20 mA to 0 - 3.3 V conversion.

Input span: \$5 - 1 = 4\ \mathrm V\$.

Output span: \$ 3.3 - 0  = 3.3 \ \mathrm V \$.

Attenuation: \$ \frac {3.3}4 \$ so let's set R3 = 330 kΩ and that makes R2 = 70 kΩ.

Now fix the offset. The negative voltage has to be low enough that the \$ \frac {3.3}4 \$ divider pulls 1 V to 0 V. That means that
\$ (1 + V_1)\frac {3.3}4  - V_1  = 0 \$
so
\$ 0.175 V_1 = 0.825 \$ and
\$ V_1 = 4.71428 \ \mathrm V \$ (since you need such precision).

Note that R2 and R3 are loading R1 and introducing an error which you will need to correct.

You now have the problem of generating a precision -4.7142 V source.

D1 will only conduct when the input current drops below 4 mA but there may be some reverse leakage current in normal operation so you will need to check the datasheets for that.


By now it should be clear to you why a direct 20% to 100% conversion of the 4 - 20 mA signal just using a 165 Ω resistor is such a good idea. All this complexity is going to add to your calibration nightmare.
You can check the errors by running the simulation on Figure 1 and hover over the various points on the schematic for various values of I1.