4-20mA Formulas and Examples

4-20 mA transmitter currentĀ output can be convertedĀ to the voltage within a specified range, from 0 to a maximal amplitude Vmax, as shown in the following picture:

4-20mA Formulas

4-20mA Formulas and Examples

Ā Calling I the output current and V the Voltage—or other physical variables (PV) —between Vmin (which is usually 0) and Vmax; the conversion formulas are as follows:

Ā 

  • I [mA] = ( ( (V āˆ’ Vmin) / ( Vmax āˆ’ Vmin) ) Ɨ 16 ) + 4

  • V [Voltage] = ( (I āˆ’ 4 ) / 16) Ɨ ( Vmax āˆ’ Vmin ) + Vmin

  • PV [units] = ( (I āˆ’ 4 ) / 16) Ɨ ( PVmax āˆ’ PVmin ) + PVmin

Where,

I is the current,

V is the voltage,

PV is the Process Variable

min is the Lower Range Value of the respective signal

max is the Upper Range Value of the respective signal

Note that the values and units of minĀ and maxĀ are units must be same.

Example 1 :

A Analog output card gives an output voltage of 2.5V and its overall range is 0-5V. calculate the equivalent 4-20mA current value.

I =Ā  ( ( (V āˆ’ Vmin) / ( Vmax āˆ’ Vmin) ) Ɨ 16 ) + 4

I =Ā  ( ( (2.5Ā āˆ’ 0) / ( 5Ā āˆ’ 0) ) Ɨ 16 ) + 4

I = 12 mA

Example 2 :

A transmitter output range is 13.5mA and its output range is 4-20mA. Calculate equivalent voltage. consider 0-10V is the output voltage range.

V =Ā ( (I āˆ’ 4 ) / 16) Ɨ ( Vmax āˆ’ Vmin ) + Vmin

V = ((13.5 – 4)/16 x (10 – 0 ) + 0

V = 5.9375 V

Example 3 :

A Flow transmitters output current is 16.5mA and its range is 0-1500 m3/hr. Calculate the equivalent flow rate value.

PV =Ā ( (I āˆ’ 4 ) / 16) Ɨ ( PVmax āˆ’ PVmin ) + PVmin

PV =Ā ( (16.5Ā āˆ’ 4 ) / 16) Ɨ ( 1500Ā āˆ’ 0Ā ) + 0

PV = 1171.87 m3/hr

Also Read : 4-20mA Transmitter Principle

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