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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?

enter image description here

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.

enter image description here

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.

enter image description here

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  • 1
    \$\begingroup\$ What is that circuit supposed to do? Maybe there's a better way to reach your goal. \$\endgroup\$
    – JRE
    Mar 15, 2021 at 18:05
  • \$\begingroup\$ hm, why are expecting the voltgae across that voltage divider to ever be exactly 0 V here? As JRE says, what's the idea behind this? \$\endgroup\$ Mar 15, 2021 at 18:08
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    \$\begingroup\$ The OP is trying to get full-scale ADC range on a 4 - 20 mA input. This has been the subject of two previous questions. In one the OP has been challenged to do the calculations for accuracy on 20% - 100% range usage versus errors introduced by scaling and offset. There is no sign of these calculations yet. \$\endgroup\$
    – Transistor
    Mar 15, 2021 at 18:20
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    \$\begingroup\$ You don't have a negative voltage on the input so you don't need a clamp. You have an offset that you are trying to remove. \$\endgroup\$
    – Transistor
    Mar 15, 2021 at 18:24
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    \$\begingroup\$ Clarification: you don't have a negative ADC input as long as your sensor is connected and powered. If your signal < 4 mA then you will need a clamp to protect against negative input to the ADC. \$\endgroup\$
    – Transistor
    Mar 15, 2021 at 18:30

1 Answer 1

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schematic

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.

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  • \$\begingroup\$ Yes, I understand that is very much complex. But I have an idea. If I remove the regular diode and change the zener to a 2.7V zener and make sure that it has a very low forward voltage. Then I will be able to get about -100mA for the ADC. And the ADC have -300mA limit. \$\endgroup\$
    – Heretic
    Mar 15, 2021 at 23:03
  • \$\begingroup\$ Have a look at my solution in the updated question. What do you think? \$\endgroup\$
    – Heretic
    Mar 15, 2021 at 23:13
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    \$\begingroup\$ Ok. I think I will give up. It's much better to accept 20% lose, than working with external complexity. But how do they doing in real industrial applications? \$\endgroup\$
    – Heretic
    Mar 15, 2021 at 23:24
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    \$\begingroup\$ I told you the answer to that two questions ago. Read it again - it may be in the comments. \$\endgroup\$
    – Transistor
    Mar 15, 2021 at 23:26
  • \$\begingroup\$ "Have a look at my solution in the updated question. What do you think?" It looks a mess. There is no good reason for a Zener (note capital for the guy's name) diode. \$\endgroup\$
    – Transistor
    Mar 15, 2021 at 23:28

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