Question

1. Wheatstone Bridge Sensor: The circuit shown below is called a "Wheatstone Bridge". It was invented in 1833 by Samuel Christie and improved by Charles Wheatstone in 1843. It remains today one of the most useful circuits for measuring small changes in physical quantities to very high precision. By inspection, the symmetrical circuit produces no output voltage ($v_o = 0$) when $\Delta R = 0$. The bridge is then said to be balanced. In practice some environmental or physical influence changes the resistance in one arm of the bridge ($\Delta R \ne 0$), thus unbalancing it and a voltage will appear at the output. The resistance change $\Delta R$ can be produced by, e.g., a pressure transducer or a thermistor (temperature-dependent resistor). $V_s \pm$ $R$ $R$ $v_o$ $R$ $R + \Delta R$ a. Find an expression for the output voltage and express it in terms of the ratio $\Delta R/R$. b. Your result in part (a.) should be nonlinear but only weakly for small $\Delta R/R$. Make a plot of your function in part (a.) for $v_o(\Delta R/R)$ over the range $-1 \le \Delta R/R \le 1$. Specifically, plot $v_o/V_s$ versus $\Delta R/R$. c. Linearize your function $v_o(\Delta R/R)$ by eliminating terms for which $\Delta R/R < 1$. Plot this on the same graph you prepared in part (b.). [Hint: By linearizing you should obtain a straight line.] d. What is the maximum value of $\Delta R/R$ that will cause the fractional error between the linear approximation $v_{lin}$ you obtained in part (c.) and the true output voltage $v_o$, $\delta = \frac{v_o - v_{lin}}{v_o}$ (1) to be less than $\pm 1\%$?

          1. Wheatstone Bridge Sensor: The circuit shown below is called a "Wheatstone Bridge". It was
invented in 1833 by Samuel Christie and improved by Charles Wheatstone in 1843. It remains
today one of the most useful circuits for measuring small changes in physical quantities to very
high precision. By inspection, the symmetrical circuit produces no output voltage ($v_o = 0$) when
$\Delta R = 0$. The bridge is then said to be balanced. In practice some environmental or physical
influence changes the resistance in one arm of the bridge ($\Delta R \ne 0$), thus unbalancing it and a
voltage will appear at the output. The resistance change $\Delta R$ can be produced by, e.g., a pressure
transducer or a thermistor (temperature-dependent resistor).
$V_s \pm$
$R$
$R$
$v_o$
$R$
$R + \Delta R$
a. Find an expression for the output voltage and express it in terms of the ratio $\Delta R/R$.
b. Your result in part (a.) should be nonlinear but only weakly for small $\Delta R/R$. Make a plot of
your function in part (a.) for $v_o(\Delta R/R)$ over the range $-1 \le \Delta R/R \le 1$. Specifically, plot
$v_o/V_s$ versus $\Delta R/R$.
c. Linearize your function $v_o(\Delta R/R)$ by eliminating terms for which $\Delta R/R < 1$. Plot this on
the same graph you prepared in part (b.). [Hint: By linearizing you should obtain a straight
line.]
d. What is the maximum value of $\Delta R/R$ that will cause the fractional error between the linear
approximation $v_{lin}$ you obtained in part (c.) and the true output voltage $v_o$,
$\delta = \frac{v_o - v_{lin}}{v_o}$ (1)
to be less than $\pm 1\%$?

$1. Wheatstone Bridge Sensor: The circuit shown below is called a "Wheatstone Bridge". It was invented in 1833 by Samuel Christie and improved by Charles Wheatstone in 1843. It remains today one of the most useful circuits for measuring small changes in physical quantities to very high precision. By inspection, the symmetrical circuit produces no output voltage (vo = 0) when Δ R = 0. The bridge is then said to be balanced. In practice some environmental or physical influence changes the resistance in one arm of the bridge (Δ R 0), thus unbalancing it and a voltage will appear at the output. The resistance change Δ R can be produced by, e.g., a pressure transducer or a thermistor (temperature-dependent resistor). Vs ± R R vo R R + Δ R a. Find an expression for the output voltage and express it in terms of the ratio Δ R/R. b. Your result in part (a.) should be nonlinear but only weakly for small Δ R/R. Make a plot of your function in part (a.) for vo(Δ R/R) over the range -1 ≤Δ R/R ≤ 1. Specifically, plot vo/Vs versus Δ R/R. c. Linearize your function vo(Δ R/R) by eliminating terms for which Δ R/R < 1. Plot this on the same graph you prepared in part (b.). [Hint: By linearizing you should obtain a straight line.] d. What is the maximum value of Δ R/R that will cause the fractional error between the linear approximation vlin you obtained in part (c.) and the true output voltage vo, δ = (vo - vlin)/(vo) (1) to be less than ± 1%?$

Added by Joseph M.

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