Strain gauge connected to op-amp: finding its range of measurement

Question

Given here (copied below) is a quarter strain gauge with R1 = R2 = R3 = R4 = R = 120 Ω and R4 = R + ΔR as the variable resistance, where the change ΔR is caused due to mechanical forces applied to the test structure.

Given V1 = v_s = 1.35 V, R5 = R7 = ℜ = 10 kΩ and R6 = R8 = 10ℜ = 100 kΩ; the op-amp is supposed to be ideal with input impedance Z_i → ∞ and output impedance Z_o = 0, open loop gain A = 10⁵, and I = [i1, i2, i3, i4, i5, i6] represents mesh current vector, taken clockwise. Then the mesh equations are: $$ \begin{pmatrix}2R+\Delta R & 0 & 0 & -R-ΔR & 0 & 0\\\ 0 & 2R & -R & 0 & 0 & 0\\\ 0 & -R & Z_i+11ℜ+R & -10ℜ & -Z_i-10ℜ & 0\\\ -R-ΔR & 0 & -10ℜ & 11ℜ+R+\Delta R & 10ℜ & 0\\\ 0 & 0 & (A-1)Z_i-10ℜ & 10ℜ & Z_o+20ℜ+(A-1)Z_i & -Z_o\\\ 0 & 0& -AZ_i & 0 & AZ_i-Z_o & R_L+Z_o\end{pmatrix}I=\begin{pmatrix} -v_s\\\ v_s \\\ 0\\\ 0 \\\ 0 \\\ 0 \end{pmatrix} $$ Retaining the dominating terms only in the impedance matrix, and neglecting the term Z_o, it becomes: $$ \begin{pmatrix}2R+\Delta R & 0 & 0 & -R-ΔR & 0 & 0\\\ 0 & 2R & -R & 0 & 0 & 0\\\ 0 & -R & Z_i & -10ℜ & -Z_i & 0\\\ -R-ΔR & 0 & -10ℜ & 11ℜ & 10ℜ & 0\\\ 0 & 0 & (A-1)Z_i & 10ℜ & (A-1)Z_i & 0\\\ 0 & 0& -AZ_i & 0 & AZ_i & R_L\end{pmatrix}I=\begin{pmatrix} -v_s\\\ v_s \\\ 0\\\ 0 \\\ 0 \\\ 0 \end{pmatrix} $$ Solving in maxima for i6 in the limit Z_i → ∞, $$ \text{o/p voltage}, v_o=i_6R_L|_{Z_i \to \infty} = Av_s \frac {(R+ΔR)^2 + 9ℜ\Delta R-2ℜR}{(R+ΔR)^2 -11ℜ\Delta R-22ℜR} $$ Now it has been asked if the smallest o/p voltage detected is 100 μV, what is the minimum value of R that can be measured? Putting $$v_o = 100 \mu\text{V} \rightarrow \Delta R = 24.3 \Omega,R_{\text{min}} = R-\Delta R =95.7 \Omega $$

This is absurd, because the actual simulation gives a value of ΔR = 3.6 mΩ for v_o = 100 μV. Where did I go wrong?

Answer 1

Calculated with Maple sheet. Confirmed your sensitivity 3.6 mOhm for 100uV.