# How to Use High-Voltage and High-Current-Drive Op Amps in ±20mA or 4–20mA Current-Loop Systems

Abstract: This application note shows how to use a high-voltage, high-current-drive operational amplifier to convert a voltage signal into a ±20mA or 4–20mA current signal for use in process-control industrial applications. The MAX9943 op amp serves as the example device. Experiments are described and test results presented.

## Introduction

## Basics of a Current Loop

**Figure 1**). The sensor measures a physical parameter (such as pressure or temperature) and provides a corresponding output voltage. The transmitter converts the sensor's output into a proportional 4mA-to-20mA current signal. The receiver then converts the 4–20mA current into a voltage. An ADC or a microcontroller digitizes the receiver's voltage output.

*Figure 1. These major components form a simple current loop.*

**Figure 2**). A DAC converts the digital information into an analog voltage signal. A current-loop transmitter converts the DAC's output voltage into a 4–20mA or ±20mA current signal that drives the actuator. An example of such a system can be found in power-grid monitoring systems where sophisticated algorithms determine the current state of the system, predict the direction of changes in the system, and implement a control loop to dynamically adjust the system.

*Figure 2. A more complex system uses a second current loop for controlling an actuator.*

## Use the Op Amp as a VI Converter with a High-Current Drive

**Figure 3**shows how a simple VI (voltage-to-current) converter can be designed with two op amps and a few external resistors. When powered with ±15V, the op amp (here, the MAX9943) delivers more than ±20mA output current to small impedance loads.

*Figure 3. A VI converter transforms the DAC output to load current. The circuit uses two MAX9943 op amps.*

_{IN}, and the load current is given by Equation 1:

V_{IN} = (R2/R1) × R_{SENSE} × I_{LOAD} + V_{REF} |
(Eq. 1) |

R2 = 750kΩ

R

_{SENSE}= 12.5Ω

R

_{LOAD}= 600Ω

_{REF}can be synchronized with the reference voltage used by the DAC. In that case all voltages (V

_{IN}) are ratiometric with V

_{REF}, and errors from variation in V

_{REF}can be eliminated.

## Create a ±20mA Current Drive from a ±2.5V Range

_{REF}= 0V, the input range between -2.5V and +2.5V produces a nominal ±20mA current output, as shown in

**Figure 4**.

_{IN}) and the output voltage (V1) of the "forward" op amp is given by:

V_{IN} = (R2/R1) × (1 - α/β) × V1 + V_{REF} × (1 – (R2/R1) × 1/(β × (R2 + R1))) |
(Eq. 2) |

α = (1/R_{SENSE}) + R2/(R1 × (R1 + R2)) |
(Eq. 3) |

β = 1/R_{SENSE} + (1/R1) + 1/R_{LOAD} |
(Eq. 4) |

V1 = 4.897 × V_{IN} - 4.896 × V_{REF} |
(Eq. 5) |

_{IN}= +2.5V, the output of the lower op amp (V1) reaches approximately 12.2V. If the input voltage increases beyond 2.5V, eventually the output device reaches its saturation point and the output voltage can no longer increase. The Figure 4 curves flatten and no longer follow the ideal profile. A similar process happens when the negative input is lowered below -2.5V.

*Figure 4. A ±20mA output current range is produced by a ±2.5V input voltage range. The blue line is the ideal gain curve; the red line is the measured data. V*

_{CC}= +15V; V_{EE}= -15V.**Figure 5**.

*Figure 5. A ±24mA output current range is produced by a ±3V input voltage range. The blue line is the ideal gain curve; the red line is the measured data. V*

_{CC}= +18V; V_{EE}= -18V.## Create a 4–20mA Current Drive from a 0 to +2.5V Range

_{REF}= -0.25V the input range between 0 and +2.5V produces a 2mA-to-22mA current output (

**Figure 6**). Normally in 4–20mA current loops, designers want extra "room" in the dynamic range (e.g., from 2mA to 22mA) to allow for software calibration. If more current is required, then the MAX9943 circuit can be powered with a dual ±18V supply voltage, as explained earlier.

*Figure 6. A 4–20mA output current range is produced by a 0 to 2.5V input voltage range. The blue line is the ideal gain curve; the red line is the measured data. V*

_{CC}= +15V; V_{EE}= -15V.