Question

An accelerometer outputs 4.9 mV per g where $g=9.8m/s^2$ is connected to an ideal op-amp signal conditioning circuit as shown in the given figure. If the outputs of this $\left(V_0\right)$ provides a velocity signal sealed at $0.25V/\left(m{s}^{-1}\right)$ then the feedback resistance ratio $\left(\frac{R_2}{R_1}\right)$ is ________.

• A. 250
• B. 100
• C. 50
• D. 500

Answer 1

The correct option is D 500

$\because \text{Acceletrometer output =}\frac{4.9mV}{g}=\frac{4.9mV}{9.8m/s^2}=0.5mV/\left(m{s}^{-2}\right)$

Redraw the circuit,



output voltage of the accelerometer is given by,

$V_a=Ka$

Where $K=0.5mV/\left(m{s}^{-2}\right),a=inputacceleration\left(m/s^2\right)$

$\therefore V_{in}=-\frac{1}{RC}\int V_{a}dt=-\frac{K}{RC}\int a.dt=\frac{-K}{RC}v$

Where $V_{01}$ is the integrator output and v is the velocity in m/s.

$\therefore V_0=-\frac{R_2}{R_1}\left[V_{01}\right]$

$V_0=-\frac{R_2}{R_1}\left[\frac{-K}{RC}v\right]...\left(i\right)$

In the given problem when v = 1 m/sec, $V_0=0.25V$

$\therefore$ From equation (i)

$\therefore 0.25=\frac{R_2}{R_1}\left[\frac{0.5\times {10}^{-3}}{{10}^6\times {10}^{-6}}\right]\times 1$

$\frac{R_2}{R_1}=500$