# Converting 4 to 20mA Linear Signal to Square Root Extraction Output

**Some pressure measurements are used to indirectly derive another type of measurand. One of these is the Rate of Flow of a gas or liquid. The flow rate along a closed pipe is directly proportional to the square root of the pressure drop or differential pressure between two points. **

Since the relationship is non-linear, there is a greater change in flow at low pressures compared to higher ones. In order to optimise the resolution of flow measurement, the output on some differential pressure transmitters can be altered so that it is directly proportional to the flow rate rather than the differential pressure.

## Linear → Sq Rt 4-20mA Conversion

The following formula can be used for converting a linear 4-20mA current loop signal to a square root extraction type:

Output_{SqRt} = 4mA + (4 x √ (Output_{Linear} – 4mA))

The following table shows values for linear to square root extraction 4 to 20 milliamp current loop signal.

Linear (mA) | Square Root (mA) |
---|---|

4 | 4.00 |

5 | 8.00 |

6 | 9.66 |

7 | 10.93 |

8 | 12.00 |

9 | 12.94 |

10 | 13.80 |

11 | 14.58 |

12 | 15.31 |

13 | 16.00 |

14 | 16.65 |

15 | 17.27 |

16 | 17.86 |

17 | 18.42 |

18 | 18.97 |

19 | 19.49 |

20 | 20.00 |

Convert any linear measurement to sqrt extraction 4-20mA signal

## Sq Rt → Linear 4-20mA Conversion

The formula for converting a square root extraction 4-20mA signal to a linear one is:* *

Output_{Linear} = 4mA + ((Output_{SqRt} – 4mA)² / 16)

The following table shows values for square root extraction to linear 4 to 20 milliamp current loop signal.

Square Root (mA) | Linear (mA) |
---|---|

4 | 4.00 |

5 | 4.06 |

6 | 4.25 |

7 | 4.56 |

8 | 5.00 |

9 | 5.56 |

10 | 6.25 |

11 | 7.06 |

12 | 8.00 |

13 | 9.06 |

14 | 10.25 |

15 | 11.56 |

16 | 13.00 |

17 | 14.56 |

18 | 16.25 |

19 | 18.06 |

20 | 20.00 |

Convert sqrt extraction 4-20mA signal to any linear measurement