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4/20mA to 0/20mA loop current converter for grounded loads

Ground-referenced loads are common in industry. This circuit implements current loop conversions for them.

Recently, there have been several published Design Ideas on converting 0/20mA to 4/20mA current and 4/20mA to 0/20mA current as full-circle current loop conversions. However, these circuits have all focused on floating loads. It’s common to also come across loads that are ground-referenced. The circuit in Figure 1 addresses this alternative requirement, converting 4/20mA current to 0/20mA current for feeding grounded loads.

Wow the engineering world with your unique design: Design Ideas Submission Guide


Figure 1 In this 4/20mA to 0/20mA converter for grounded loads, R4 and RB can be replaced by multi-turn potentiometers for tuning purposes.

How does it work? Input current of 4-20mA feeds into R1 and is converted into 0.4V – 2.0V, which is buffered by U2A. U1 generates a reference current of 1mA which is is fed into R4, and which converts it to 0.4V. This converted current is buffered by U2B. U2C subtracts the two voltages.

Next, let’s look at the positive input of U2D. There are two currents:

  • One current going through Ra= frac{(Iinput*R1)-(Iref*R4)}{Ra}
  • Another current going through Rb= –frac{Iout.Rc}{Rb+Rc}

Since the negative input of U2D is grounded, these two currents must be same:

frac{(Iinput*R1)-(Iref*R4)}{Ra} = frac{Iout.Rc}{Rb+Rc} where I ref*R4 is 0.4V

Iout = (frac{Iinput.R1-0.4V}{Ra})*(1+frac{Rb}{Rc})

Select the values of Rb and Rc such that Rb/Rc =124. Substituting the values of Ra, Rb, Rc and R1 from Figure 1:

Iout = (frac{Iinput.R1-0.4V}{Ra})*125

Hence, Iout= (Iinput-4 mA)*1.25. For example, if the I input is 20 mA, I out= (20-4)*1.25=20mA. And if the I input is 4mA, I out= (4 – 4) * 1.25=0mA.

How do you tune the circuit? Implement R4 and Rb as multi-turn potentiometers. R1 conversely should be a precision 100 ohm resistor. Adjust R4 such that voltage across it is 0.4 V. Feed 20mA current from the precision current source as the Iinput and adjust Rb to get 20mA as Iout. Repeat this exercise by feeding the circuit with 4mA and 12mA.

Simulation test results follow:

Iinput (mA)

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

Iout (mA)

0.44

2.56

5.06

7.56

10.1

12.6

15.1

17.6

20.1

Calculated Iout (mA)

0

2.5

5.0

7.5

10.0

12.5

15.0

17.5

20.0

Error (%)

2.2

0.3

0.3

0.3

0.5

0.5

0.5

0.5

0.5

These results suggest that the circuit delivers high accuracy, with error no higher than 0.5%, except with a 4mA input. The error and associated accuracy can be improved by selecting high-end operational amplifiers such as instrumentation amplifiers with negligible offset and a high common-mode rejection ratio (CMRR).

Q2 prevents the output current from exceeding a few mA above the 20mA input threshold, as a safety measure. R5, R6, and R7 should be identical in value. And also implement R8 as a multi-turn potentiometer. The resultant tuning capability helps to reduce the output to near-zero for a 4mA input.

Jayapal Ramalingam has over three decades of experience in designing electronics systems for power & process industries and is presently a freelance automation consultant.

Related Content

  • Linearly variable two-wire loop current generator
  • Full circle current loops: 4mA-20mA to 0mA-20mA
  • Another silly simple precision 0/20mA to 4/20mA converter
  • Silly simple precision 0/20mA to 4/20mA converter
  • Combine two TL431 regulators to make versatile current mirror
  • A 0-20mA source current  to 4-20mA loop current converter

The post 4/20mA to 0/20mA loop current converter for grounded loads appeared first on EDN.

19 May 2026
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