Question 6 of 7: Wheatstone Bridge — Strain Gage / RTD Signal Conditioning

Question 6: Wheatstone Bridge — Strain Gage / RTD Signal Conditioning (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

[Figure not reproduced: Fig. 3 — Wheatstone bridge redrawn from the source figure: $R_1$ (top-left), $R_2$ (top-right), $R_3$ (bottom-left), and the sensing element $R+\Delta R$ (bottom-right, orange) fed by $V_{exc}$ through series resistor $R_z$ over long wires from the power source; a Zener diode clamps $V_{exc}$ . See the official exam paper.]

a) Given. Bridge excitation $V_{exc}$; left leg $R_1=R_3=R$ with Out 1 tapped between them; right leg $R_2=R$ in series with the sensing element $R+\Delta R$ with Out 2 tapped between them; both legs grounded at the bottom. (Node assignment read directly off the source figure's junction dots — the Out 1 lead crosses the sensing leg without a dot, so it belongs to the $R_1/R_3$ node.)

Find. $V_{out}=V_{Out1}-V_{Out2}$ as a function of $V_{exc}$, $R$, and $\Delta R$.

Approach. Treat each leg as an independent voltage divider from $V_{exc}$ to ground and subtract the two divider outputs.

1. Write each leg as a voltage divider. $$V_{Out1}=V_{exc}\cdot\frac{R_3}{R_1+R_3}=V_{exc}\cdot\frac{R}{R+R}=\frac{V_{exc}}{2},$$ $$V_{Out2}=V_{exc}\cdot\frac{R+\Delta R}{R_2+(R+\Delta R)}=V_{exc}\cdot\frac{R+\Delta R}{2R+\Delta R}.$$
2. Subtract to get the bridge output. $$V_{out}=V_{Out1}-V_{Out2}=V_{exc}\left(\frac12-\frac{R+\Delta R}{2R+\Delta R}\right) =\boxed{-\,\dfrac{V_{exc}\,\Delta R}{2\,(2R+\Delta R)}.}$$ The minus sign simply records that the ACTIVE leg is the one brought out on Out 2, so a positive $\Delta R$ pulls Out 2 above Out 1; the magnitude $|V_{out}|=V_{exc}\Delta R/[2(2R+\Delta R)]$ is what sets the measurement sensitivity, and in practice the sign is absorbed by which way round the two leads enter the differential amplifier. Taking the difference the other way ($V_{Out2}-V_{Out1}$) gives the same expression with a $+$ sign.
3. Linearize for the usual small-signal case ($\Delta R\ll R$). Dropping $\Delta R$ next to $2R$ in the denominator, $$|V_{out}|\approx\frac{V_{exc}\,\Delta R}{4R}\qquad(\Delta R\ll R),$$ the familiar quarter-bridge sensitivity result — e.g. for $V_{exc}=10$ V, $R=1000\ \Omega$, $\Delta R=5\ \Omega$ (a 0.5% change): exact $|V_{out}|=12.469$ mV vs. linear $|V_{out}|\approx12.500$ mV, a $\sim$0.25% linearization error, confirming the approximation is good whenever $\Delta R/R$ stays small.

b) Any resistance in the long wires themselves — and its own drift with temperature — adds directly in series with the bridge legs (or with the excitation line if unregulated), corrupting the tiny $\Delta R$ signal the bridge is built to resolve; long wires also enclose a larger loop area and pick up more capacitively/inductively coupled interference (the same physical mechanism as Question 2's electrical-interference question). Locating the bridge itself right at the sensing resistor minimizes both the added, temperature-sensitive series lead resistance and the interference pickup; where the bridge genuinely cannot be placed there, a 3- or 4-wire lead-compensation scheme is needed instead.

c) The Zener diode regulates/clamps $V_{exc}$ to a fixed breakdown voltage, holding the bridge's excitation constant despite fluctuations in the raw “Power” supply or in the drop across $R_z$ as bridge current varies. Since $V_{out}$ from part (a) is directly PROPORTIONAL to $V_{exc}$, any instability in the excitation voltage would appear directly as a spurious error in the measured output; the Zener removes that error source (and can also provide basic over-voltage/transient protection for the bridge).

d) Mount a second, matched gage on the same specimen material so it experiences the identical local temperature as the active gage, but oriented/positioned so it does NOT see the mechanical strain being measured (a strain-free “dummy” gage) — or, for double sensitivity, mounted to see an equal-and-opposite mechanical strain (a half-bridge). Both gages then undergo the SAME temperature-induced resistance shift, $\Delta R_{T}=\alpha_T\Delta T\cdot R$ (identical material, identical $\Delta T$), but the mechanical-strain-induced $\Delta R_{\varepsilon}$ appears in ONLY the active gage. Placing the active gage in one bridge arm (e.g. the $R+\Delta R$ position) and the dummy in the ADJACENT arm means the bridge output depends on the DIFFERENCE between the two arms: the identical, same-sign temperature term cancels out of that difference exactly as derived in part (a) (it shifts both dividers' relevant resistor equally and leaves the balance condition unchanged), while the strain-only term — present in only one arm — survives fully into $V_{out}$. The bridge is therefore temperature-compensated by construction, not by post-processing.

e) From part (a)'s result, $|V_{out}|\approx V_{exc}\Delta R/4R$ is directly proportional to $V_{exc}$: a HIGHER excitation gives a proportionally LARGER output signal for the same $\Delta R$, improving signal-to-noise ratio and resolution against whatever fixed noise floor the downstream amplifier/DAQ contributes. However, a higher $V_{exc}$ also drives more current through the bridge resistors, increasing $I^2R$ SELF-HEATING in the sensing element itself — the same self-heating error mechanism already identified in Question 2(h) — which can bias a strain gage or RTD reading and, at extremes, damage a delicate gage. The excitation voltage must therefore be chosen as high as possible for good SNR while staying low enough that self-heating error remains an acceptably small fraction of the measurement's required accuracy.