Week 6 Tutorial

Question 1

A four-wire resistance measurement is performed using the circuit shown in Figure Q1.

Figure Q1: A circuit intended for resistance measurement. The op-amp works as a constant current source, as you will prove in part (a). The component

V

is a voltage meter, measuring the voltage drop across the sensor

R

. The resistors

R_{w_i}

represent the unwanted resistances in the wires that connect to the sensor. Zoom:

The wires connecting the interface circuit to the sensor have resistance $R_{w_{1}}=R_{w_{2}}=R_{w_{3}}=R_{w_{4}}=1$ Ω. The device $V$ is a voltmeter, which has an input impedance of $Z_{meter}.$ The circuit is designed to measure resistances in the range $10\ \text{Ω}\leq R\leq500\ \text{kΩ}$ .

(a) Find the excitation current $i_{excite}$ . Assume that the op-amp is ideal.

(b) Notice that some current will be diverted through the voltmeter, so the measured voltage $V$ will not be the expected value of $V=Ri_{excite}$ . If the current flowing into the meter is $i_{sense}$ then the measured voltage is actually $V=i_{sense}Z_{meter}$ . Find the minimum value of $Z_{meter}$ such that the relative measurement error is no worse than $\pm$ 0.01%.

(c) In next week’s practical, you will use a Texas Instruments INA826 instrumentation amplifier. This device has an input impedance of approximately 20 GΩ. Would this device be a suitable buffer for the analog front end of the voltage meter?

Answer

(a) Recognise that the op-amp has negative feedback, and so it will drive its output in such a way that a voltage of $V_{ref}$ appears across $R_{ref}$ . Consequently, the excitation current is simply

i_{excite}=\frac{V_{ref}}{R_{ref}}.

(b) We need to analyse the circuit to find the voltage $V$ . Perhaps the easiest approach is to use a current divider formula to find how much of the current flows into the meter. Let the current entering the meter be denoted $i_{sense}$ . Since the wire resistance is a total of 2 Ω, we have

i_{sense}=i_{excite}\frac{R}{R+2+Z_{meter}}.

Hence the measured voltage is

V=i_{sense}Z_{meter}=\frac{Ri_{excite}Z_{meter}}{R+2+Z_{meter}}.

The ideal value (with no loading due to the meter) is $V_{ideal}=Ri_{excite}$ , so the relative error is

\begin{align*} E & =\frac{V-V_{ideal}}{V_{ideal}}\\ & =\frac{V}{V_{ideal}}-1\\ & =\frac{Z_{meter}}{\left(R+2+Z_{meter}\right)}-1. \end{align*}

We require $|E| < 0.0001$ to meet the specification given in the question. Hence, the question arises: will the worst case error occur for small $R$ or large $R$ ? Which end of the measurement range should we use when calculating the required $Z_{meter}$ ?

In the case of small $R$ :

\begin{align*} \lim_{R\to0}E & =\lim_{R\to0}\frac{Z_{meter}}{\left(R+2+Z_{meter}\right)}-1\\ & =\frac{Z_{meter}}{\left(2+Z_{meter}\right)}-1\\ & =\frac{Z_{meter}-\left(2+Z_{meter}\right)}{2+Z_{meter}}\\ & =\frac{-2}{2+Z_{meter}}\\ & \approx0\ \ \ \text{(because }Z_{meter}\ \text{is large)}. \end{align*}

Hence we conclude that small $R$ is the “easy case” to measure. Intuitively, we want current to flow through the sensor instead of the meter, and this is achieved when the sensor has lower resistance.

We can also consider the case of large $R$ :

\begin{align*} \lim_{R\to\infty}E & =\lim_{R\to\infty}\frac{Z_{meter}}{\left(R+2+Z_{meter}\right)}-1\\ & =-1. \end{align*}

This means there is a relative error of -100%. Hence we conclude that the most difficult case to measure accurately will be when the sensor resistance $R$ is at the top end of the measurement span ( $R=500\times10^{3}$ Ω).

Substituting this limiting case:

E=\frac{Z_{meter}}{500002+Z_{meter}}-1.

Given that the error is negative, substitute $E=-0.01\%=-0.0001$

\begin{align*} -0.0001 & =\frac{Z_{meter}}{500002+Z_{meter}}-1\\ Z_{meter} & =5\ \text{GΩ. } \end{align*}

(c) Yes, the larger the input impedance the better, so 20 GΩ is ideal in this scenario.

Question 2

The Wheatstone Bridge can be used to compensate for changes in temperature. Suppose that you have two identical strain gauges. The “active gauge” is glued to the stressed material, and the “dummy gauge” is kept away from any mechanical stress. If the gauges are at equal temperature, then the effects of temperature will be the same on both.

Suppose that the strain gauges have a nominal resistance of $R_{0}$ at a temperature $T_{0}$ . Also, they have a temperature coefficient of resistance (TCR) of $\alpha$ and a gauge factor of $G$ .

(a) Write down the transfer function for the gauges as a function of strain ( $\epsilon$ ) and temperature ( $T$ ).

Hint: a TCR represents a relative change per Kelvin. In other words, to account for temperature, multiply the entire resistance by a correction factor $\left(1+\alpha(T-T_{0})\right)$ .

(b) Suppose that these gauges are placed in a half-bridge configuration as shown below:

Figure Q2: The circuit for Question 2. Zoom:

Sensor $R_{1}$ is the dummy gauge (with zero strain), and sensor $R_{2}$ is the active gauge (with strain of $\epsilon$ ).

Analyse this circuit and show that $V_{0}$ is independent of temperature.

(c) Find the sensitivity of this circuit for small values of strain.

Question 3

A quarter-bridge circuit is used to interface with a resistive sensor as shown below:

Figure Q3: The circuit for Question 3. Zoom:

The sensor is placed at position $R_{3}$ and is described by a transfer function of the form

R_{3}=R_{0}(1+x),

where $R_{0}$ is a large fixed constant, and $x$ is the sensor response that we seek to measure.

(a) Analyse each side of the voltage divider and prove that the bridge is balanced (i.e. $V_{0}=0$ ) when

\frac{R_{1}}{R_{4}}=\frac{R_{2}}{R_{3}}.

(b) Suppose that the resistances satisfy a relationship

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

Find an expression for the output voltage in terms of $x$ and $k$ .

(c) Find the sensitivity of $V_{0}$ to changes in $x$ .

(d) Find the value of $k$ that maximises the sensitivity at $x=0$ .

Answer

Using voltage divider formulas,

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

(a) Let $V_{0}=0$ and solve to obtain the result.

(b) The definition of $k=R_{2}/R_{0}=R_{1}/R_{4}$ suggests that we should rearrange $V_{0}$ to this form:

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

Hence, we can substitute $R_{3}=R_{0}(1+x)$ and $k=R_{2}/R_{0}=R_{1}/R_{4}$

\begin{align*} V_{0} & =V_{i}\left(\frac{1}{1+k/(1+x)}-\frac{1}{1+k}\right)\\ & =V_{i}\left(\frac{1+x}{1+x+k}-\frac{1}{1+k}\right)\\ & =V_{i}\left(\frac{\left(1+x\right)\left(1+k\right)-\left(1+x+k\right)}{\left(1+x+k\right)\left(1+k\right)}\right)\\ & =V_{i}\left(\frac{kx}{\left(1+x+k\right)\left(1+k\right)}\right)\\ & =\frac{V_{i}kx}{(1+k)(1+x+k)}. \end{align*}

(c) The sensitivity of $V_{0}$ is given by

\begin{align*} \frac{\partial V_{0}}{\partial x} & =\frac{V_{i}k}{1+k}\frac{\partial}{\partial x}\left(\frac{x}{1+x+k}\right)\\ & =\frac{V_{i}k}{1+k}\ \frac{\left(1+x+k\right)-x}{\left(1+x+k\right)^{2}}\\ & =\frac{V_{i}k}{1+k}\ \frac{1+k}{\left(1+x+k\right)^{2}}\\ & =\frac{V_{i}k}{\left(1+x+k\right)^{2}}. \end{align*}

(d) At $x=0$ the maximum sensitivity is obtained by solving

\begin{align*} 0 & =\frac{d}{dk}\frac{V_{i}k}{\left(1+k\right)^{2}}\\ & =\frac{d}{dk}\frac{k}{k^{2}+2k+1}\\ & =\frac{\left(k^{2}+2k+1\right)-\left(2k+2\right)k}{\left(k^{2}+2k+1\right)^{2}}\\ & =k^{2}+2k+1-2k^{2}-2k\\ & =-k^{2}+1\\ k & =1. \end{align*}

We can exclude the negative square root because resistances must be positive.

Question 4

The circuit below is used to perform a resistance measurement. Balanced current sources $i_{1}=i_{2}$ inject power into the circuit. There is a sensor $R$ with transfer function

R=R_{0}(1+x)

where $x$ is a small relative change in resistance that we seek to measure. The circuit also has a fixed resistance $R_{z}$ .

Figure Q4: The circuit for Question 4. Zoom:

(a) Assuming perfectly balanced current sources ( $i_{1}=i_{2}$ ), find the value of $R_{z}$ such that $V_{0}=0$ when $x=0$ .

(b) Write the transfer function for this circuit (i.e. the relationship between $V_{0}$ and $x$ ).

(c) Discuss the practical issues that would be associated with this design.