Is there any way to have a diode that have very small resistance from reverse path?

Question

I'm trying to find a way so I will always have close to 0V at the measurement point (above the diode), when I have 0V as input. The -3.3V is a fixed voltage source.

At 0.64V at input, I have -5.7mV at the measurement point. That's OK to me. At 3.3V at input, I have 2.2V at the measurement point. That's OK to me.

But when I have 0V at input, then I have -271.2mV at the measurement point. That's to much negative!

If the diode have at least 10 Ohm in resistance, then the measurement point would be very close to zero. I what it to happen. But I cannot use a diode. Is there any better way to solve this issue without using a diode, or using a better diode?

Update:

Try this with a 3.6V zener diode. Change the current source from 4-20mA and look at the output. At 4mA, then we have close to zero volt. At 20mA, then we have about 2.2V. OK! But at 0mA, then we have -225mV. That's not OK! It should be low as zero volt as well.

Update:

Here is my solution. Is it good? At 4mA, we have some voltage for the ADC and therefore, we can see that there is a sensor connected.

Answer 1

^{simulate this circuit – 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.