In factory instrumentation, process parameters such as pressure and flowrate are measured, converted to electrical signals, and sent some distance to an electronic controller. The controller then decides what actions should be taken. One of the main concerns in these systems is the physical

Question

In factory instrumentation, process parameters such as pressure and flowrate are measured, converted to electrical signals, and sent some distance to an electronic controller. The controller then decides what actions should be taken. One of the main concerns in these systems is the physical distance between the sensor and the controller. An industry standard format for encoding the measurement value is called the 4–20 mA standard, where the parameter range is linearly distributed from 4 to 20 mA. For example, a 100 psi pressure sensor would output 4 mA if the pressure were 0 psi, 20 mA at 100 psi, and 12 mA at 50 psi. But most instrumentation is based on voltages between 0 and 5 V, not on currents. Therefore, let us design a current-to-voltage converter that will output 5 V when the current signal is 20 mA.

Accepted Answer

The circuit in Fig. 2.51a is a very accurate model of our situation. The wiring from the sensor unit to the controller has some resistance, [latex]R_{wire}.[/latex] If the sensor output were a voltage proportional to pressure, the voltage drop in the line would cause measurement error even if the sensor output were an ideal source of voltage. But, since the data are contained in the current value, [latex]R_{wire}[/latex] does not affect the accuracy at the controller as long as the sensor acts as an ideal current source.

As for the current-to-voltage converter, it is extremely simple—a resistor. For 5 V at 20 mA, we employ Ohm’s law to find

R = [latex]\frac{5}{0.02}[/latex] = 250 Ω

The resulting converter is added to the system in Fig. 2.51b, where we tacitly assume that the controller does not load the remaining portion of the circuit.

Note that the model indicates that the distance between the sensor and controller could be infinite. Intuitively, this situation would appear to be unreasonable, and it is. Losses that would take place over distance can be accounted for by using a more accurate model of the sensor, as shown in Fig. 2.52. The effect of this new sensor model can be seen from the equations that describe this new network. The model equations are

[latex]I_{S}\,=\,\frac{V_{S}}{R_{S}}+\frac{V_{S}}{R_{wire}+250}[/latex]

and

[latex]I_{signal}\,=\,\frac{V_{S}}{R_{wire}+250}[/latex]

Combining these equations yields

[latex]\frac{I_{signal}}{I_{S}}\,=\,\frac{1}{1+\frac{R_{wire}+250}{R_{S}}}[/latex]

Thus, we see that it is the size of [latex]R_{S}[/latex] relative to ([latex]R_{wire}[/latex] + 250 Ω) that determines the accuracy of the signal at the controller. Therefore, we want [latex]R_{S}[/latex] as large as possible. Both the maximum sensor output voltage and output resistance, [latex]R_{S}[/latex], are specified by the sensor manufacturer.
We will revisit this current-to-voltage converter in Chapter 4.

Question 2.36: In factory instrumentation, process parameters such as press...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions