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22-Agric-A5 Principles of Instrumentation · December 2015

Question 3 of 7: Strain-Gage Wheatstone Bridge for a Cantilever Beam

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 04-Agric-A5 Principles of Instrumentation, National Exams December 2015 — a three-hour open-book exam; any non-communicating calculator is permitted. Questions 1 and 2 are compulsory (20 marks each); candidates then choose any three (3) of Questions 3-7 (20 marks each) for a 100-mark paper. All seven questions are worked here.

Reference texts. E.O. Doebelin, Measurement Systems: Application and Design, 5th ed. (calibration, standards, static/dynamic sensor characteristics, second-order step response); J.P. Bentley, Principles of Measurement Systems, 4th ed. (accuracy vs. precision, error propagation, signal conditioning); P. Horowitz and W. Hill, The Art of Electronics, 3rd ed. (bridge circuits, instrumentation amplifiers, CMRR, ADC architectures); J. Fraden, Handbook of Modern Sensors: Physics, Designs, and Applications, 5th ed. (thermistors, strain gages, photodetectors); R.W. Fox, A.T. McDonald and P.J. Pritchard, Introduction to Fluid Mechanics, 7th ed. (orifice and venturi metering).

Question 3: Strain-Gage Wheatstone Bridge for a Cantilever Beam (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.

a) Given. A Wheatstone bridge with four active strain gages, all of nominal resistance $R$ when unloaded: the two TOP gages ($R_1$, $R_3$) go into tension under the applied load (resistance $R+\Delta R$), the two BOTTOM gages ($R_2$, $R_4$) go into compression (resistance $R-\Delta R$), arranged with a top gage and a bottom gage in series on each bridge leg (leg $A$-$B$: top then bottom; leg $A$-$C$: bottom then top), excited by $V_{ex}$ between node $A$ and ground node $D$, and read differentially between nodes $B$ and $C$. Find. $V_{out}$ as a function of the four gage resistances, then in terms of $\Delta R$, $R$ and $V_{ex}$.

Vex (+)GND (−)R1 (top, +ΔR)R2 (bottom, −ΔR)R4 (bottom, −ΔR)R3 (top, +ΔR)BCDifferentialAmplifierVout
Fig. 1 — Full Wheatstone bridge with two top (tension) and two bottom (compression) strain gages; the differential amplifier reads $V_B-V_C$.

Approach. Write each bridge-node voltage from the divider rule (no current flows into either op-amp input of the differential amplifier), subtract to get $V_{out}$, then substitute the loaded resistances of the four gages.

  1. Node voltages from the divider rule. With no current drawn by the differential amplifier's inputs, the same current flows through each two-resistor leg, so $$\begin{aligned} V_B&=V_{ex}\cdot\dfrac{R_2}{R_1+R_2} \\ V_C&=V_{ex}\cdot\dfrac{R_3}{R_3+R_4} \end{aligned}$$
  2. Bridge output. $$V_{out}=V_B-V_C=V_{ex}\left(\dfrac{R_2}{R_1+R_2}-\dfrac{R_3}{R_3+R_4}\right).$$
  3. Substitute the loaded gage resistances. With $R_1=R_3=R+\Delta R$ (top, tension) and $R_2=R_4=R-\Delta R$ (bottom, compression), each leg's total resistance is exactly $R_1+R_2=R_3+R_4=2R$ regardless of $\Delta R$, so $$V_{out}=V_{ex}\left(\dfrac{R-\Delta R}{2R}-\dfrac{R+\Delta R}{2R}\right)=V_{ex}\cdot\dfrac{-2\Delta R}{2R}.$$
  4. Result. Writing $\Delta R/R=\mathrm{GF}\cdot\varepsilon$ (gage factor GF, strain $\varepsilon$), $$\boxed{|V_{out}|=V_{ex}\cdot\dfrac{\Delta R}{R}=V_{ex}\cdot\mathrm{GF}\cdot\varepsilon.}$$ This is an EXACT result (the $2R$ denominator never depended on $\Delta R$), not the small-signal approximation a quarter- or half-bridge needs.
QuantityResult
$V_B$, $V_C$ (bridge node voltages)$V_{ex}R_2/(R_1+R_2)$, $V_{ex}R_3/(R_3+R_4)$
$V_{out}$ in terms of gage resistances$V_{ex}\left(\dfrac{R_2}{R_1+R_2}-\dfrac{R_3}{R_3+R_4}\right)$
$V_{out}$, full-bridge loaded result$-V_{ex}\,\Delta R/R$ (magnitude $V_{ex}\,\mathrm{GF}\,\varepsilon$)

b) This full-bridge (all four arms active) configuration gives (1) FOUR TIMES the sensitivity of a single (quarter-bridge) gage — a quarter bridge gives $V_{out}=(V_{ex}/4)\cdot\Delta R/R$, while the full bridge derived above gives the full $V_{ex}\cdot\Delta R/R$, because the two tension arms and two compression arms all add constructively instead of one arm doing all the work; and (2) automatic first-order TEMPERATURE COMPENSATION — a uniform thermal resistance drift $\Delta R_T$ affects all four gages identically (same material, same ambient temperature), and because $R_1+R_2$ and $R_3+R_4$ both change by the same $2\Delta R_T$ regardless of load, that common shift cancels exactly out of the $V_B-V_C$ subtraction and never reaches $V_{out}$, unlike a single unbalanced gage whose thermal drift is indistinguishable from a real strain signal.

c) Over a long lead run, the actual excitation voltage delivered to the bridge itself is degraded by resistive $I\!R$ drop in the cable and by any pickup or supply ripple along the way — and because $V_{out}=V_{ex}\cdot\Delta R/R$ scales directly with $V_{ex}$, any uncontrolled variation in the excitation reaching the bridge is indistinguishable from a real strain signal. A Zener diode provides a stable, regulated reference/excitation voltage established right at (or very close to) the bridge itself, so the bridge's actual excitation is held to a known, constant value independent of supply fluctuations or cable losses upstream of the Zener.

d) Grounding a shield at BOTH ends, when those two ground points are physically separated by a long cable run, creates a closed conductive loop through the shield between two points that are rarely at exactly the same true ground potential; any potential difference drives a current around that loop (a ground loop), and that current induces additional noise voltage onto the signal conductors via mutual inductance — adding noise rather than removing it. Grounding the shield at only ONE end still lets it perform its intended electrostatic (capacitive) shielding function — intercepting and diverting external electric-field pickup to ground — while guaranteeing no current can ever flow along the shield itself.

e) Low: excitation power dissipated in each gage is $I^2R=V_{ex}^2/(4R)$ per arm; too high an excitation self-heats the gages, and that self-heating both drifts the gage's own resistance (a spurious "strain" signal with no mechanical cause) and can thermally strain the beam material it is bonded to, corrupting the zero point. High: the wanted output signal $V_{out}=V_{ex}\cdot\mathrm{GF}\cdot\varepsilon$ scales directly with $V_{ex}$, so a higher excitation gives a larger raw signal against the SAME fixed downstream noise floor and amplifier offset — improving signal-to-noise ratio and reducing the amplifier gain (and its own noise/offset contribution) needed to reach a usable output level. The excitation voltage is therefore a genuine design trade-off: the practical choice is the highest excitation that keeps self-heating-induced drift within the error budget the gage manufacturer specifies for the expected ambient/airflow conditions.