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## Analog Electronic Instrumentation

4 to 20 mA analog current signals

The most popular form of signal transmission used in modern industrial instrumentation systems (as of this writing) is the 4 to 20 milliamp DC standard. 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, and any current value in between 4 and 20 milliamps represents a commensurate percentage in between 0% and 100%.

For example, if we were to calibrate a 4-20 mA temperature transmitter for a measurement range of 50 to 250 degrees C, we could relate the current and measured temperature values on a graph like this:

This is not unlike the pneumatic instrument signal standard or 3 to 15 pounds per square inch (PSI), where a varying air pressure signal represents some process measurement in an analog (proportional) fashion. Both signal standards are referred to as live zero because their ranges begin with a non-zero value (3 PSI in the case of the 3-15 PSI standard, and 4 milliamps in the case of the 4-20 mA standard). This “live” zero provides a simple means of discriminating between a legitimate 0% signal value and a failed signal (e.g. leaking tube or severed cable)1.

DC current signals are also used in control systems to command the positioning of a final control element, such as a control valve or a variable-speed motor drive (VSD). In these cases, the milliamp value does not directly represent a process measurement, but rather how the degree to which the final control element influences the process. Typically (but not always!), 4 milliamps commands a closed (shut) control valve or a stopped motor, while 20 milliamps commands a wide-open valve or a motor running at full speed.

Thus, most industrial control systems use at least two different 4-20 mA signals: one to represent the process variable (PV) and one to represent the command signal to the final control element (the “manipulated variable” or MV):

The relationship between these two signals depends entirely on the response of the controller. There is no reason to ever expect the two current signals to be equal, for they represent entirely different things. In fact, if the controller is reverse-acting, it is entirely normal for the two current signals to be inversely related: as the PV signal increases going to a reverse-acting controller, the output signal will decrease. If the controller is placed into “manual” mode by a human operator, the output signal will have no automatic relation to the PV signal at all, instead being entirely determined by the operator’s whim.

Relating 4 to 20 mA signals to instrument variables

Calculating the equivalent milliamp value for any given percentage of signal range is quite easy. Given the linear relationship between signal percentage and milliamps, the equation takes the form of the standard slope-intercept line equation y = mx + b. Here, y is the equivalent current in milliamps, x is the desired percentage of signal, m is the span of the 4-20 mA range (16 mA), and b is the offset value, or the “live zero” of 4 mA:

This equation form is identical to the one used to calculate pneumatic instrument signal pressures (the 3 to 15 PSI standard):

The same mathematical relationship holds for any linear measurement range. Given a percentage of range x, the measured variable is equal to:

Some practical examples of calculations between milliamp current values and process variable values follow:

**Example calculation: controller output to valve**

An electronic loop controller outputs a signal of 8.55 mA to a direct-responding control valve (where 4 mA is shut and 20 mA is wide open). How far open should the control valve be at this MV signal level?

We must convert the milliamp signal value into a percentage of valve travel. This means determining the percentage value of the 8.55 mA signal on the 4-20 mA range. First, we need to manipulate the percentage-milliamp formula to solve for percentage (x):

Next, we plug in the 8.55 mA signal value and solve for x:

Therefore, the control valve should be 28.4 % open when the MV signal is at a value of 8.55 mA.

**Example calculation: 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, we convert the flow value of 204 GPM into a percentage of range. This is a simple matter of division, since the flow measurement range is zero-based:

Next, we take this percentage value and translate it into a milliamp value using the formula previously shown:

Therefore, the transmitter should output a PV signal of 13.3 mA at a flow rate of 204 GPM.

**Example calculation: 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.

First, we convert the temperature value of 79 degrees into a percentage of range based on the knowledge of the temperature range span (140 degrees − 50 degrees = 90 degrees) and lower-range value (LRV = 50 degrees). We may do so by manipulating the general formula for any linear measurement to solve for x:

Next, we take this percentage value and translate it into a 4-20 mA current value using the formula previously shown:

Therefore, the transmitter should output a PV signal of 9.16 at a temperature of 79o F.

**Example calculation: pH transmitter**

A pH transmitter has a calibrated range of 4 pH to 10 pH, with a 4-20 mA output signal. Calculate the pH sensed by the transmitter if its output signal is 11.3 mA.

First, we must convert the milliamp value into a percentage. Following the same technique we used for the control valve problem:

Next, we take this percentage value and translate it into a pH value, given the transmitter’s measurement span of 6 pH (10 pH − 4 pH) and offset of 4 pH:

Therefore, the transmitter’s 11.3 mA output signal reflects a measured pH value of 6.74 pH.

**Example calculation: reverse-acting I/P transducer signal**

A current-to-pressure transducer is used to convert a 4-20 mA electronic signal into a 3-15 PSI pneumatic signal. This particular transducer is configured for reverse action instead of direct, meaning that its pressure output at 4 mA should be 15 PSI and its pressure output at 20 mA should be 3 PSI. Calculate the necessary current signal value to produce an output pressure of 12.7 PSI. Reverse-acting instruments are still linear, and therefore still follow the slope-intercept line formula y = mx + b. The only differences are a negative slope and a different intercept value. Instead of y = 16x + 4 as is the case for direct-acting instruments, this reverse-acting instrument follows the linear equation y = −16x + 20:

First, we need to to convert the pressure signal value of 12.7 PSI into a percentage of 3-15 PSI range. We will manipulate the percentage-pressure formula to solve for x:

Next, we plug in the 12.7 PSI signal value and solve for x:

This tells us that 12.7 PSI represents 80.8 % of the 3-15 PSI signal range. Plugging this percentage value into our modified (negative-slope) percentage-current formula will tell us how much current is necessary to generate this 12.7 PSI pneumatic output:

Therefore, a current signal of 7.07 mA is necessary to drive the output of this reverse-acting I/P transducer to a pressure of 12.7 PSI.

**Graphical interpretation of signal ranges**

A helpful illustration for students in understanding analog signal ranges is to consider the signal range to be expressed as a length on a number line. For example, the common 4-20 mA analog current signal range would appear as such:

If one were to ask the percentage corresponding to a 14.4 mA signal on a 4-20 mA range, it would be as simple as determining the length of a line segment stretching from the 4 mA mark to the 14.4 mA mark:

As a percentage, this thick line is 10.4 mA long (the distance between 14.4 mA and 4 mA) over a total (possible) length of 16 mA (the total span between 20 mA and 4 mA). Thus:

This same “number line” approach may be used to visualize any conversion from one analog scale to another. Consider the case of an electronic pressure transmitter calibrated to a pressure range of -5 to +25 PSI, having an (obsolete) current signal output range of 10 to 50 mA. The appropriate current signal value for an applied pressure of +12 PSI would be represented on the number line as such:

*Proportion:*

Finding the “length” of this line segment in units of milliamps is as simple as setting up a proportion between the length of the line in units of PSI over the total (span) in PSI, to the length of the line in units of mA over the total (span) in mA:

Solving for the unknown (?) current by cross-multiplication and division yields a value of 22.67 mA. Of course, this value of 22.67 mA only tells us the length of the line segment on the number line; it does not directly tell us the current signal value. To find that, we must add the “live zero” offset of 10 mA, for a final result of 32.67 mA.

Thus, an applied pressure of +12 PSI to this transmitter should result in a 32.67 mA output signal.

**Continue reading to the next page, Current Loops**

**Go Back to Lessons in Instrumentation Table of Contents**

written by Eric Kesinger, December 13, 2015

If I Purchase the T590X I/P transducer and input is 12 ma signal, what would be the output in (PSI)

How do I calculate for other (ma) inputs

Thanks

Eric kesinger

KADANWARI FLOW LINE TEMPERATURE TRANSMITTER ; ADDITIONAL DETAIL: TRANSMITTER, TEMPERATURE, ELECTRONIC, 4 - 20 MA, 0 - 300 DE

QYT 4