EE3901/EE5901 Sensor Technologies
Week 6 Notes
Signal conditioning circuits for resistive sensors

College of Science and Engineering, James Cook University

Last updated: 21 March 2023

How do you measure resistance? In the simplest case, Ohm's law may be all that you need. However, if a sensor experiences only small changes in resistance, then careful circuit designs are needed to maximise sensitivity and signal-to-noise ratio.

Interfacing with resistive sensors
- The two wire measurement circuit
- The four wire measurement circuit
Measuring small changes in resistance
- The Wheatstone Bridge Circuit
- Using the Wheatstone Bridge with strain gauges
Differential amplifiers
Instrumentation Amplifiers
References

Interfacing with resistive sensors

There are two requirements for all resistive sensor interface circuits:

The circuit must drive the resistive sensor with a voltage or current. Resistance can only be measured when there is a current flowing.
The voltage and current must be low enough to avoid self-heating of the sensor (unless self-heating is part of the operational mechanism of the sensor, in which the interface circuit must ensure appropriate levels of power dissipation).

The two wire measurement circuit

The simplest approach is to measure resistance with Ohm’s law. Apply a current and measure the voltage, or vice versa. A typical circuit is shown in Figure 1.

Figure 1: The two wire measurement circuit. Here

R

is the resistive sensor, and

R_{w_{1}}

and

R_{w_{2}}

are the resistance of the cabling used to connect to the sensor. Power is supplied by the current source

i

and the voltage is measured by the voltmeter

V

. Zoom:

The weakness of this approach is that the resistance of the cabling will affect the measurement. The cable resistance will vary with temperature so it not always easy to correct for. Hence, the two wire approach is generally unsuitable for precision measurements.

Answer: (c)

The four wire measurement circuit

A better design separates the voltage measurement from the excitation current, as shown in Figure 2.

Figure 2: The four wire measurement circuit, where

R_{w_1}

R_{w_2}

R_{w_3}

, and

R_{w_4}

are the resistances of the wires used to connect to the sensor

R

. Zoom:

If the voltmeter has a large input impedance, then the sensing current $I_{sense}$ is very small in comparison to the excitation current. Consequently the voltage drop in the wires is reduced.

Answer: (a)

Measuring small changes in resistance

Suppose that we want to measure small changes in a large resistance. For example, imagine we have a sensor whose resistance changes from 10,000 Ω to 10,001 Ω. How can we design circuits to interface with such a sensor?

To formalise the problem, consider a sensor with a transfer function of the form

\begin{equation} R=R_{0}\left(1+x\right) \end{equation}

where $x\ll1$ . We are specifically interested in $x$ rather than the large constant $R_{0}$ . If we would measure $R$ directly, then we’d use much of our measurement precision on $R_{0}$ and make it hard to detect the interesting change in $x$ . Hence, we often need to design interface circuits that are sensitive to small changes in resistance.

An example of such a circuit is the Wheatstone Bridge.

The Wheatstone Bridge Circuit

A Wheatstone Bridge is two voltage dividers in parallel. It is often drawn in a diamond shape, like Figure 3.

Figure 3: The concept of a Wheatstone bridge. Zoom:

There are two main ways to use the Wheatstone Bridge. These are called “balance mode” and “deflection mode”.

Balance mode

In balance mode, a current meter is connected across the bridge, as shown in Figure 4.

Figure 4: A Wheatstone bridge wired in balance mode. Zoom:

Resistor $R_{3}$ is the unknown resistance (e.g. the sensor). Resistance $R_{4}$ is a variable resistor (shown here as a potentiometer with its wiper shorted to one of the terminals). The idea is to adjust $R_{4}$ until a “balance” is achieved and the current meter reads $I=0$ . The balance condition is

\begin{equation} \frac{R_{1}}{R_{4}}=\frac{R_{2}}{R_{3}}. \end{equation}

The user adjusts $R_{4}$ until the bridge is balanced, then calculates the unknown resistance $R_{3}$ based upon the values of the other resistors.

Deflection mode

The second way to use the Wheatstone Bridge is in “deflection mode” where a voltage meter is connected across the bridge:

Figure 5: A Wheatstone bridge in deflection mode. Zoom:

Analysing this circuit, the output voltage is

\begin{equation} V_{0}=V_{i}\left(\frac{R_{3}}{R_{3}+R_{2}}-\frac{R_{4}}{R_{4}+R_{1}}\right). \end{equation}

The Wheatstone Bridge is typically used with strain gauges or RTDs where a small change in resistance must be detected. Let $R_{3}$ be the sensor with transfer function

\begin{equation} R_{3}=R_{0}(1+x), \end{equation}

where $x$ is the parameter of interest. To ensure that the bridge is balanced when $x=0$ , we must choose resistor values that satisfy

\begin{equation} \frac{R_{1}}{R_{4}}=\frac{R_{2}}{R_{0}}=k. \end{equation}

Designing a bridge circuit is then a matter of choosing $k$ . In the tutorial questions you will prove that maximum sensitivity is achieved when $k=1$ . However, larger values of $k$ give increased linearity of $V_{0}$ at the expense of reduced sensitivity.

Answer: (a), (b)

Using the Wheatstone Bridge with strain gauges

A common configuration is to place multiple strain gauges in a single bridge circuit. The configurations are called “quarter bridge” (one sensor), “half bridge” (two sensors) and “full bridge” (four sensors).

The half bridge

If the gauges measure equal but opposite strains, connect each sensor to the same side of the half bridge, as per Figure 6.

Figure 6: A half bridge circuit. The strain gauges on opposite sides of the beam are connected on the same arm of the Wheatstone bridge. Zoom:

Analysing this circuit, we find

\begin{equation} V_{0}=V_{i}\left(\frac{R_{0}(1+x)}{R_{0}(1+x)+R_{0}(1-x)}-\frac{R_{0}}{2R_{0}}\right). \end{equation}

Simplifying,

\begin{equation} V_{0}=\frac{V_{i}x}{2}. \end{equation}

Notice how the output voltage does not depend upon $R_{0}$ .

The full bridge

The full bridge circuit is shown in Figure 7.

Figure 7: A full bridge circuit. Strain gauges with equal responses are placed at diagonally opposite sides of the bridge. Zoom:

Here the output voltage is

\begin{equation} V_{0}=V_{i}x. \end{equation}

Notice that this is twice as sensitive as the half bridge.

Answer: 2,1

Differential amplifiers

The voltages obtained from a Wheatstone Bridge are measured between two nodes in the circuit. For many applications we need to amplify the signal and convert it to a “single ended” voltage measured with respect to ground (for example, to connect to a microcontroller’s ADC). In other words, we want something of the form

\begin{equation} V_{out} \propto V_1 - V_2, \end{equation}

where $V_1$ and $V_2$ are the nodes inside the Wheatstone bridge.

Your first year circuits textbook probably shows you an op-amp circuit that appears to solve this problem, most likely with something like Figure 8. However, there are some practical issues with this circuit that need to be explained.

Figure 8: Don’t do this. It is a circuit design that works in theory but will not succeed in practice, as discussed in the text. Use an instrumentation amplifier instead. Zoom:

Let’s first analyse this circuit to see how it’s supposed to work. Assuming the op-amp is operating within its linear regime, some elementary circuit analysis will show you that the output voltage is given by

\begin{equation} V_{0}=\frac{R_{4}}{R_{3}+R_{4}}\left(1+\frac{R_{2}}{R_{1}}\right)V_{2}-\frac{R_{2}}{R_{1}}V_{1}. \end{equation}

Since we want an output proportional to $V_1 - V_2$ , we split the output into common mode and differential mode components:

\begin{align*} V_{c} & =\frac{V_{1}+V_{2}}{2}\\ V_{d} & =V_{1}-V_{2}. \end{align*}

Solving for the input voltages:

\begin{align*} V_{1} & =V_{c}+\frac{V_{d}}{2}\\ V_{2} & =V_{c}-\frac{V_{d}}{2}. \end{align*}

Substituting into $V_{0}$ :

\begin{equation} V_{0}=A_{c}V_{c}+A_{d}V_{d} \end{equation}

where the common mode gain is

\begin{equation} A_{c}=\frac{R_{4}}{R_{3}+R_{4}}\left(1+\frac{R_{2}}{R_{1}}\right)-\frac{R_{2}}{R_{1}} \end{equation}

and the differential mode gain is

\begin{equation} A_{d}=\frac{-A_{c}}{2}-\frac{R_{2}}{R_{1}}. \end{equation}

To use this circuit as a differential amplifier we require $A_{c}=0$ . This places constraints on the values of the resistors. If we had exact resistors and an ideal op-amp then $A_{c}=0$ would be achievable. However in real life our resistors have some tolerance so we cannot exactly satisfy $A_{c}=0$ . Therefore, resistor tolerance makes this circuit impractical.

A desirable parameter for a differential amplifier is to have a high common mode rejection ratio (CMRR), defined as

\text{CMRR}=\frac{A_{d}}{A_{c}}.

There will some CMRR due to the tolerance of the resistors, as well as some CMRR due to the op-amp itself.

A further problem with this circuit is the input impedance. The resistors need to be in the range of 10s to 100s of kΩ for good op-amp performance. However, the resulting input impedance looking into $V_{1}$ and $V_{2}$ is far too small for some applications. For example, in tutorial questions you will analyse a resistance measurement where gigaohms of input impedance are required. This design is not suitable for many precision measurements. The solution is to use an instrumentation amplifier.

Instrumentation Amplifiers

An instrumentation amplifier (sometimes called an “in-amp”) is an IC that solves the problems of a basic op-amp differential amplifier. It has excellent CMRR and extremely high input impedance.

The basic design of an instrumentation amplifier is shown in Figure 9.

Figure 9: The simplified equivalent circuit of an instrumentation amplifier. The highlighted region is manufactured in a single integrated circuit, allowing for precise trimming of the matching resistors so that the tolerate problems mentioned in the previous section can be avoided. Zoom:

You should be able to recognise two voltage followers and the differential amplifier stage from earlier. Notice how the input impedance looking into $V_{1}$ and $V_{2}$ is extremely high because these signals are terminated directly into an op-amp.

The output voltage is

\begin{align} V_{0} & =\left(1+\frac{2R_{2}}{R_{1}}\right)\frac{R_{4}}{R_{3}}\left(V_{2}-V_{1}\right)+V_{ref}\\ & =G\left(V_{2}-V_{1}\right)+V_{ref}, \end{align}

where $G$ is the gain of the amplifier. In the case of AC signals and dual power supplies, $V_{ref}$ can be tied to ground. For single-supply operation, $V_{ref}$ can be set to some nonzero value to shift the output. For example you might have a microcontroller ADC and you could set $V_{ref}=V_{DD}/2$ to shift the output to the middle of the ADC’s range.

Often all components except for $R_1$ are manufactured in a single IC. $R_1$ is typically external and is used to set the gain.

In next week’s practical, you will design and build a circuit using an instrumentation amplifier. Specifically, you will use the Texas Instruments INA826. Please refer to that device’s datasheet to familiarise yourself with its properties.

References

Ramon Pallas-Areny and John G. Webster, Sensors and Signal Conditioning, 2nd edition, Wiley, 2001.

Winncy Y. Du, Resistive, Capacitive, Inductive, and Magnetic Sensor Technologies, CRC Press, 2015.