Question 6.FOM.1: Wheatstone Bridge Filter The Wheatstone bridge circuit of Ex...
Wheatstone Bridge Filter
The Wheatstone bridge circuit of Examples 2.10 and Focus on Measurements: Wheatstone Bridge in Chapter 2 is used in a number of instrumentation applications, including the measurement of force (see Example 2.13, describing the strain gauge bridge). Figure 6.14 depicts the appearance of the bridge circuit. When undesired noise and interference are present in a measurement, it is often appropriate to use a low-pass filter to reduce the effect of the noise. The capacitor that is connected to the output terminals of the bridge in Figure 6.14 constitutes an effective and simple low-pass filter, in conjunction with the bridge resistance. Assume that the average resistance of each leg of the bridge is 350 \Omega (a standard value for strain gauges) and that we desire to measure a sinusoidal force at a frequency of 30 \mathrm{~Hz}. From prior measurements, it has been determined that a filter with a cutoff frequency of 300 \mathrm{~Hz} is sufficient to reduce the effects of noise. Choose a capacitor that matches this filtering requirement.
By evaluating the Thévenin equivalent circuit for the Wheatstone bridge, calculating the desired value for the filter capacitor becomes relatively simple, as illustrated at the bottom of Figure 6.14. The Thévenin resistance for the bridge circuit may be computed by short-circuiting the two voltage sources and removing the capacitor placed across the load terminals:
R_{T}=R_{1}\left\|R_{2}+R_{3}\right\| R_{4}=350\|350+350\| 350=350 \Omega
Since the required cutoff frequency is 300 \mathrm{~Hz}, the capacitor value can be computed from the expression
\omega_{0}=\frac{1}{R_{T} C}=2 \pi \times 300
or
C=\frac{1}{R_{T} \omega_{0}}=\frac{1}{350 \times 2 \pi \times 300}=1.51 \mu \mathrm{F}
The frequency response of the bridge circuit is of the same form as equation 6.16:
\frac{\mathbf{V}_{\text {o }}}{\mathbf{V}_{i}}(j \omega)=\frac{1}{1+j \omega C R_{T}} (6.16)
\frac{\mathbf{V}_{\text {out }}}{\mathbf{V}_{T}}(j \omega)=\frac{1}{1+j \omega C R_{T}}
This response can be evaluated at the frequency of 30 \mathrm{~Hz} to verify that the attenuation and phase shift at the desired signal frequency are minimal:
\begin{aligned}\frac{\mathbf{V}_{\text {out }}}{\mathbf{V}_{T}}(j \omega=j 2 \pi \times 30) &=\frac{1}{1+j 2 \pi \times 30 \times 1.51 \times 10^{-6} \times 350} \\&=0.9951 \angle-5.7^{\circ}\end{aligned}
Figure 6.15 depicts the appearance of a 30-\mathrm{Hz} sinusoidal signal before and after the addition of the capacitor to the circuit.
Focus on Computer-Aided Tools- An EWB simulation of this circuit may be found in the accompanying CD-ROM.
