I'm looking into building a 4-20mA receiver and found a reference schematic. In the description, they state that it uses a resistance value of 150 Ohms to produce a 3 V when the sensor's output is 20mA. enter image description here

When checking the schematic, the 150 Ohms are split into 2 resistors with a node that goes to the ADC input between them, so I would think that the ADC will measure 2 V when the current is 20 mA, instead of 3V.

  1. Is my assumption correct (voltage will be 2 V when current is 20mA)?

  2. The 3.3k resistor is there for high current protection and it doesn't have any influence on the voltage measurement?

enter image description here

  • 1
    \$\begingroup\$ 3K3 and 100nF are a low pass filter, to remove most of the interference that can be picked up on the cable. (The resistor does provide some protection too) At 480Hz they would add about 1 millisecond delay to the voltage reading. \$\endgroup\$
    – user16324
    Sep 30, 2021 at 15:01

1 Answer 1


Referring to the topmost unit (A0 input), R1 and R2 form a potential divider, which applies \$\frac{100\Omega}{100\Omega+50\Omega}=\frac{2}{3}\$ of the voltage at A0 to the actual ADC input of U1.

This means, for example, if you apply 3V at A0, then a current of \$I=\frac{V}{R}=\frac{3V}{100\Omega+50\Omega} = 20mA \$ will flow into A0 to ground. The voltage appearing at the ADC input on the IC will be \$\frac{2}{3}\times 3V=2V\$.

R3 and C3 form a low-pass filter, with a cut-off frequency of about \$\frac{1}{2 \pi R_3 C_3} = 480Hz\$, presumably to mitigate noise.

ScottSeidman commented about the sensors being 4mA to 20mA current sources, which is a good observation. You may wonder what the difference is, between a voltage source that supplies some voltage that results in a current flow through a fixed resistance connected across it, and a current source that "pushes" some current through that same resistance, resulting in voltage developing across it.

The answer is if that resistance is constant, then there is no functional difference. Compare these four scenarios:


simulate this circuit – Schematic created using CircuitLab

In all four scenarios the resistors are kept the same. In the left column (A and C), we use current sources, where you could say that current is being "pushed" down the resistor chain, causing the resistors to develop some voltage.

On the right (B and D) we use voltage sources to impose some voltage directly across the resistors, with exactly the same result.

In fact, from the perspective of the resistances, they have no idea what's happening. All they know is that there's some voltage across them, and some current through them, and they could not care less how that comes about, or even find out.

It's possible and valid to consider a current source to be a special kind of adjustable voltage source, that modifies its own voltage until a certain current is achieved. In scenario A, if that 10V voltage supply were to rise, the current source would simply drop a larger voltage across itself to compensate, to maintain current at 4mA.

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    \$\begingroup\$ Isn't that backward. The sensor produces 20mA into A0, and when all of that goes through 100R, there will be 2V at AIN0 \$\endgroup\$ Sep 30, 2021 at 15:06
  • \$\begingroup\$ I could consider the sensor to be a source of 20mA or a source of whatever voltage will cause 20mA to flow. Both interpertations are equivalent from the perspective of a fixed 150Ω to ground. \$\endgroup\$ Sep 30, 2021 at 15:12
  • \$\begingroup\$ @ScottSeidman Actually, the context of the question is clearer to me now, thanks to your comment. I shall edit my answer to address it. \$\endgroup\$ Sep 30, 2021 at 15:31

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