Ballast circuit for a strain gauge

A ballast circuit is a simple potential-divider circuit in which the strain gauge $R_g$ is in series with a fixed ballast resistor $R_b$ across a supply $V$. The output is taken across the gauge. It is used mostly for dynamic strain, since any slow drift appears as a false signal.

   V (+) o---- R_b ----+-------o
                       |
                      R_g      V_o (across R_g)
                       |
   V (-) o-------------+-------o

Output change

$$V_o = V\frac{R_g}{R_g + R_b}$$

Differentiate with respect to $R_g$:

$$\frac{dV_o}{dR_g} = V\frac{R_b}{(R_g + R_b)^2}$$

For a small change $\Delta R_g$:

$$\Delta V_o = V\frac{R_bR_g}{(R_g + R_b)^2}\cdot\frac{\Delta R_g}{R_g}$$

Maximum sensitivity

Let $m = R_b/R_g$. Then $\Delta V_o = V\dfrac{m}{(1+m)^2}\dfrac{\Delta R_g}{R_g}$. Setting $d/dm[m/(1+m)^2] = 0$ gives $m = 1$, i.e. $R_b = R_g$. The maximum output change is

$$\Delta V_o = \frac{V}{4}\frac{\Delta R_g}{R_g}$$

(This is the same as a quarter bridge, but the output sits on a dc level of $V/2$, which must be blocked by a capacitor before amplification.)

Numerical

$\Delta R_g/R_g = G\varepsilon = 2.0 \times 1000\times10^{-6} = 2\times10^{-3}$, $R_b = R_g$.

$$\Delta V_o = \frac{6}{4}(2\times10^{-3}) = 3\times10^{-3}\ \text{V}$$

Answer: $R_b = R_g$ for maximum sensitivity; $\Delta V_o = 3$ mV.