Question 4 of 7: Load Cell Strain Gauges

Question 4: Load Cell Strain Gauges (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) Under an axial compressive load the cylinder shortens along its own axis (negative axial strain), so the axial gage (G1, bonded parallel to the cylinder's long axis) physically shortens with it and its resistance decreases ($\Delta R/R=GF\cdot\varepsilon_{axial}<0$). At the same time, because the material is (very nearly) incompressible on a Poisson's-ratio basis, the shortened cylinder must bulge outward radially/circumferentially to conserve volume — the transverse gage (G2, bonded around the circumference) is stretched by that bulging and its resistance increases. The two strains are linked by Poisson's ratio, $\varepsilon_{transverse}=-\nu\,\varepsilon_{axial}$, so the transverse strain is always opposite in sign to (and smaller in magnitude than) the axial strain.

b) Sensitivity can be raised by reducing the cylinder's cross-sectional area (thinner wall or smaller diameter, staying within the elastic range) so the same applied weight produces a larger axial stress $\sigma=F/A$ and hence larger strain $\varepsilon=\sigma/E$; by choosing a lower-modulus material ($E$ smaller gives more strain per unit stress, again within its elastic limit); by using gages with a higher gage factor $GF$ (more $\Delta R/R$ per unit strain); by wiring all four active/compensating positions with gages (a full 4-active-arm bridge) instead of the 2 shown, which roughly doubles the bridge output for the same strain; and/or by increasing the excitation voltage $V_{exc}$ (bounded by the self-heating limit of Q3e).

c) Put G1 in place of R1 and G2 in place of R3 — i.e. the two active gages occupy the two arms of the same series branch (Out 1's divider), while R2 and R4 remain ordinary fixed, matched resistors. Both gages sit right next to each other on the same load cell, so any change in ambient/self-heating temperature changes both of their resistances by the same amount, $\Delta R_{T}$. Writing $R1=R_0+\Delta R_{1,strain}+\Delta R_T$ and $R3=R_0+\Delta R_{2,strain}+\Delta R_T$, the divider output $Out_1=V_{exc}\,R3/(R1+R3)$ depends (for small changes) on the difference $\Delta R_{2,strain}-\Delta R_{1,strain}$ — the common $\Delta R_T$ term cancels out of that difference entirely. Because the axial and transverse strains are opposite in sign (part a), $\Delta R_{2,strain}-\Delta R_{1,strain}$ does not cancel — the strain signal survives while the temperature signal is rejected.

d) A differential (instrumentation) amplifier should be used to take Out 1 and Out 2 as its two inputs, since both outputs ride on a large common signal (nominally $V_{exc}/2$) with only a small difference between them carrying the actual weight information. The critical performance characteristic is a high common mode rejection ratio (CMRR, Q2h) — the amplifier must reject the large shared $V_{exc}/2$ component (and any common-mode interference/ground offset riding on it) while faithfully amplifying only the small Out 1−Out 2 difference signal; it should also present a high input impedance so it does not load (and unbalance) the bridge itself.

e) Given. Excitation $V_{exc}$, fixed reference resistor $R_{ref}$, single active gage G1 with nominal (zero-strain) resistance equal to $R_{ref}$ and a load-induced change $\delta$ (so $G1=R_{ref}+\delta$), two matched fixed resistors $R$ forming a divider between $V_{exc}$ and the op-amp's own output, ideal op-amp with its non-inverting ($+$) input grounded.

Find. $V_{out}$ as a function of $\delta$ and the circuit values.

Approach. Recognize the left-hand loop as a standard inverting op-amp stage (input resistor $R_{ref}$ from $V_{exc}$, feedback element G1, feeding back from the op-amp's output to its inverting input) to get the op-amp's own output as a function of G1, then apply voltage-divider analysis to the right-hand $R$-$R$ pair.

1. Virtual ground. With the ($+$) input grounded and negative feedback present (G1 connects the op-amp's output back to the ($-$) input), the op-amp forces $V_-=V_+=0\,\text{V}$ at the node shared by $R_{ref}$ and G1, and its input draws no current.
2. Constant current through G1. The current from $V_{exc}$ through $R_{ref}$ into the virtual-ground node is $I=V_{exc}/R_{ref}$, fixed and independent of G1. With no current into the op-amp input, that same current $I$ must flow on through G1 to the op-amp's output node $V_o$: $$0-V_o=I\cdot G1\quad\Longrightarrow\quad V_o=-\dfrac{V_{exc}}{R_{ref}}\,(R_{ref}+\delta).$$
3. Output divider. The two matched $R$ resistors form a plain divider between $V_{exc}$ and $V_o$, tapped at their midpoint: $$V_{out}=\dfrac{V_{exc}+V_o}{2}=\dfrac{V_{exc}}{2}\left[1-\dfrac{R_{ref}+\delta}{R_{ref}}\right] =\dfrac{V_{exc}}{2}\left(-\dfrac{\delta}{R_{ref}}\right).$$ $$\boxed{V_{out}=-\dfrac{V_{exc}\,\delta}{2\,R_{ref}}}$$

$V_{out}$ is exactly proportional to $\delta$ — at $\delta=0$ (no load) the circuit is balanced and $V_{out}=0$, exactly as a bridge output should be, and the same constant-current linearization trick used for the RTD/bridge amplifier removes any non-linearity from the single-gage measurement.