4-20 mA Current loop - Calculation of process variable


#1

What is 4-20mA signal?

Analog instrumentation in an industry is based on the signal transmission of the process variable in a standard range of 4-20mA, 1-5 VDC. A process variable such as pressure, temperature, flow, level, displacement, in any range of measurement is converted to 4-20mA current signal for transmission to the control panel.

This is an analog signal standard, meaning that the electric current is used to proportionately represent measurements or command signals. Typically, a 4 milliamp current value represents 0% of scale, a 20 milliamp current value represents 100% of scale.

Controllers such as PLC/DCS uses digital to analog converter to convert the voltage signal to digital numbers which are understandably by CPU. The measuring device has 250-ohm resistor. As for current signals, they are converted to voltage signals like this

4 ma * 250 ohm = 1 v
20ms * 250 ohm = 5v

Every instrument technician tasked with maintaining 4-20 mA instruments commits these values to memory, because they are referenced so often:

Why we use 4-20mA, not 0-20mA?

This an important question, should be answered by every instrumentation engineers.

When the signal starts from 0mA instead of 4mA, we cannot distinguish between system OFF and loop failure. Even if the system is in OFF state the signal shows 0mA and if the system is ON and there is loop failure, it shows the 0mA. That makes difficult to identify the system per cent.

And the range above 20mA can affect human and can be affected by nearby electrical lines in the industries.

How to relate 4-20mA with process variable?

A 4 to 20 mA current signal represents some signal along a 0 to 100 percent scale. Usually, this scale is linear as shown by this graph:

Being a linear function, we may use the standard slope-intercept linear equation to relate signal percentage to current values:

y = mx + b

Where,
y = Output from instrument
x = Input to instrument
m = Slope
b = y-intercept point (i.e. the live zero of the instrument’s range)

The slope is Rise/Run

m = Rise/Run = (20 − 4)/(100 − 0) = 16/100

y = (16/100)x + b

b is y-intercept point, from the graph it is 4

Then b = 4

y = (16/100)x + 4

Calculation of percentage of opening: controller output to valve:

We have to calculate how much percentage the valve stem moves, let calculate for 9mA.

y = (16/100)x + 4

9 = (16/100)x + 4

5 = (16/100)x ;

x = 31.25%; the valve is 31.25% open at an applied MV signal of 9 milliamps.

Calculation for flow transmitter:

A flow transmitter is ranged 0 to 350 gallons per minute, 4-20 mA output, direct-responding. Calculate the current signal value at a flow rate of 204 GPM.

First converting the flow rate into a percentage of flow:

which is: (204GPM/350GPM)* 100= 58.3%

Next, we plug this percentage value into the formula:

y = (16/100)58 + 4

y = 9.33 + 4

y = 13.33

Therefore, the transmitter should output a PV signal of 13.3 mA at a flow rate of 204 GPM. An alternative approach is to set up a linear equation specifically for this flowmeter given its measurement range (0 to 350 GPM) and output signal range (4 to 20 mA).

The slope (m) for this equation is rise over run, in this case 16 milliamps of rise for 350 GPM of
run:

y = ((20 − 4)/(350 − 0))x + b = (16/350)x + b

Now that the linear equation is set up for this particular flowmeter, we may plug in the 204 GPM
value for x and solve for current:

y = (16/350)204 + 4

solve: y = 13.33mA

Calculation for temperature transmitter:

An electronic temperature transmitter is ranged 50 to 140 degrees Fahrenheit and has a 4-20 mA
output signal. Calculate the current output by this transmitter if the measured temperature is 79
degrees Fahrenheit.

The graph changes and we should find the b intersection point.

y = ((20 − 4)/(140 − 50))x + b = (16/90)x + b

We know it shows 4mA volts for 50°F

thus; 4 = (16/90)50 + b

then; b = -4.89

the equation becomes; 4 = (16/90)x - 4.89

Thus for 79°F, it is;

y = (16/90)79 - 4.89

then y = 9.16

easy way:

Converting 79 oF into a percentage of a 50-to-140 oF range requires that we first subtract the live-zero value, then divide by the span:

Per unit ratio = (79 − 50)/(140 − 50) = 0.3222

Percentage = 0.3222 per unit × 100% = 32.22%

y = (16/100)32.22 + 4

y = 5.16 + 4

y = 9.16


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