4-20 mA analog output circuit for increased resolution

Question

I have a device that uses the 4-20 mA output. I want to convert this to a range of 0 or 0.48 V to 2.4 V. So I decided to use a precision resistor of 120 ohm resistor.

How can I adjust my circuit in order to get the maximum resolution possible? I mean, what would a good circuit be that can catch the change from 6 mA to 6.2 mA or 6.4 mA?

Maybe even a noise reduction add on my 120 ohm circuit to help me get better results?

Any advice welcome.

EDIT: I must have not explained well my situation and I am sorry for the misunderstanding.

I have an output of 4-20 mA and I want to convert to the range of my ADC 0-2.4 V.

So the best solution is a 120 ohm resistor that I use in order to have 4 mA * 120 = 0.48 V minimum voltage at my ADC and 20 mA * 120 = 2.4 V maximum voltage at my ADC.

My question is about the circuit. Now I use only the 120 ohm resistor connected at the 4-20 mA output pins and then I connect my resistor to the ADC.

How can I adjust my circuit, in order to catch the smallest change in milliamps possible? I mean that I want to catch even the smallest change. For example 6.0 mA to 6.2 mA. This change in the amps gives me an 0.2 mA * 120 = 24 mV of change in my voltage. How can I adjust my circuit to make these kind of changes better visible to my ADC?

This is why I spoke about noise reduction at my first post.

Searching I found that if my output signal is slow I can add a 100nF capacitor parallel to the 120 Ohms resistor.

I am looking for something like this but to help the ADC get better readings from the 120 ohms resistor.

Answer 1

You don't say what resolution your A/D is, but you probably don't have a problem to begin with. A 12 bit A/D (quite common and available built into a number of microcontrollers) has a resolution of 1/4095 of its whole input range. If that is 3.3 V, then the resolution is 810 µV.

You want to detect a 200 µA difference thru a 120 Ω resistor, which amounts to 24 mV. That's 30 times the resolution of the 12 bit A/D. Put another way, you will see a 30 count difference in two readings that are 200 µA apart. Even a 10 bit A/D would give you about 7.5 counts.

Where is the problem?

Answer 2

I (possibly incorrectly) read your question as

"I want to take a subset of the current range of a 4-20 mA output and use it to create a voltage that covers the whole of my ADC output.

Example: 0-20 mA full range.
Range of interest = eg 6.0 to 6.4 mA ADC max input = 5V say.
ADC min input = 0V say.

Object:

Convert 6.0 mA to 0V at ADC input.
Convert 6.4 mA to 5V at ADC input. With linear response in between.

One method:

• Convert whole current range to a Voltage range.

  Supply a differential amplifier which converts the range subset to the desired out put range.

  Deal with out of range signals.

Use a resistor to convert I to V.
Say use 100 ohms to make example easier. So:
4 mA = 0.4V 20 mA = 2.0V

Target start = 6.0 mA = 0.6V
Target end = 6.4 mA = 0.64V
Voltage range of interest = 0.60 - 0.64V = 0.04V

Output swing = 0-5V
Input swing = 0.04V (above)
Gain required = 6.0/0.04 = 150 x

Use differential amplifier.
Amplifier gain = 150
Amplifier non inverting input is set to 0.06V

When Iin < 6 mA, Vin < 0.06V and amoplifier output is negative or illegal. Ignore.

When Iin > 6.4 mA Vin >= 0.64V.
Vout of amplifier > 5V = illegal. Ignore.

When 6 mA <= Iin <= 6.4 mA then 0.60 <= Vin <= 0.64V and
amplifer output = 0-5V.

QED

In practice where I say "ignore" and "illegal" you need to arrange the circuitry to deal properly with the out of range signals. This is "just a matter of engineering" (which may be trivially easy or very hard but probably somewhere in between) but is outside th ebasic principles listed above.