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

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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

There are two requirements for all resistive sensor interface circuits:

  1. The circuit must drive the resistive sensor with a voltage or current. Resistance can only be measured when there is a current flowing.
  2. 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 RR is the resistive sensor, and Rw1R_{w_{1}} and Rw2R_{w_{2}} are the resistance of the cabling used to connect to the sensor. Power is supplied by the current source ii and the voltage is measured by the voltmeter VV. 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.

Self check quiz 6.1

Referring to the circuit shown in Figure 1, what is the equation that you would use to calculate the sensor resistance RR?

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 Rw1R_{w_1}, Rw2R_{w_2}, Rw3R_{w_3}, and Rw4R_{w_4} are the resistances of the wires used to connect to the sensor RR. Zoom:

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

Self check quiz 6.2

Referring to the circuit shown in Figure 2, what is the equation that you would use to calculate the sensor resistance RR? You may assume that

Isense0.I_{sense} \approx 0.
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

R=R0(1+x)\begin{equation} R=R_{0}\left(1+x\right) \end{equation}

where x1x\ll1. We are specifically interested in xx rather than the large constant R0R_{0}. If we would measure RR directly, then we’d use much of our measurement precision on R0R_{0} and make it hard to detect the interesting change in xx. 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 R3R_{3} is the unknown resistance (e.g. the sensor). Resistance R4R_{4} is a variable resistor (shown here as a potentiometer with its wiper shorted to one of the terminals). The idea is to adjust R4R_{4} until a “balance” is achieved and the current meter reads I=0I=0. The balance condition is

R1R4=R2R3.\begin{equation} \frac{R_{1}}{R_{4}}=\frac{R_{2}}{R_{3}}. \end{equation}

The user adjusts R4R_{4} until the bridge is balanced, then calculates the unknown resistance R3R_{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

V0=Vi(R3R3+R2R4R4+R1).\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 R3R_{3} be the sensor with transfer function

R3=R0(1+x),\begin{equation} R_{3}=R_{0}(1+x), \end{equation}

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

R1R4=R2R0=k.\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 kk. In the tutorial questions you will prove that maximum sensitivity is achieved when k=1k=1. However, larger values of kk give increased linearity of V0V_{0} at the expense of reduced sensitivity.

Self check quiz 6.3

Select all the true statements.

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

V0=Vi(R0(1+x)R0(1+x)+R0(1x)R02R0).\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}


V0=Vix2.\begin{equation} V_{0}=\frac{V_{i}x}{2}. \end{equation}

Notice how the output voltage does not depend upon R0R_{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

V0=Vix.\begin{equation} V_{0}=V_{i}x. \end{equation}

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

Self check quiz 6.4

A set of strain gauges have the transfer function

R=120(1+200ϵ),R = 120(1 + 200\epsilon),

where RR is resistance in ohms and ϵ\epsilon is the strain.

These strain gauges are connected in either a full bridge or a half bridge layout. The bridge’s power supply is Vi=10V_i = 10 V and the applied strain is ϵ=106\epsilon = 10^{-6}.

What is the voltage that will appear across the bridge circuit?

Full bridge case:   V0=V_0 = mV.
Half bridge case:   V0=V_0 = mV.
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

VoutV1V2,\begin{equation} V_{out} \propto V_1 - V_2, \end{equation}

where V1V_1 and V2V_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

V0=R4R3+R4(1+R2R1)V2R2R1V1.\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 V1V2V_1 - V_2, we split the output into common mode and differential mode components:

Vc=V1+V22Vd=V1V2.\begin{align*} V_{c} & =\frac{V_{1}+V_{2}}{2}\\ V_{d} & =V_{1}-V_{2}. \end{align*}

Solving for the input voltages:

V1=Vc+Vd2V2=VcVd2.\begin{align*} V_{1} & =V_{c}+\frac{V_{d}}{2}\\ V_{2} & =V_{c}-\frac{V_{d}}{2}. \end{align*}

Substituting into V0V_{0}:

V0=AcVc+AdVd\begin{equation} V_{0}=A_{c}V_{c}+A_{d}V_{d} \end{equation}

where the common mode gain is

Ac=R4R3+R4(1+R2R1)R2R1\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

Ad=Ac2R2R1.\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 Ac=0A_{c}=0. This places constraints on the values of the resistors. If we had exact resistors and an ideal op-amp then Ac=0A_{c}=0 would be achievable. However in real life our resistors have some tolerance so we cannot exactly satisfy Ac=0A_{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


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 V1V_{1} and V2V_{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 V1V_{1} and V2V_{2} is extremely high because these signals are terminated directly into an op-amp.

The output voltage is

V0=(1+2R2R1)R4R3(V2V1)+Vref=G(V2V1)+Vref,\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 GG is the gain of the amplifier. In the case of AC signals and dual power supplies, VrefV_{ref} can be tied to ground. For single-supply operation, VrefV_{ref} can be set to some nonzero value to shift the output. For example you might have a microcontroller ADC and you could set Vref=VDD/2V_{ref}=V_{DD}/2 to shift the output to the middle of the ADC’s range.

Often all components except for R1R_1 are manufactured in a single IC. R1R_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.


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.