Question

An op-amp amplifier with slew rate of $10V/\mu sec$ has a voltage gain of 66.0206 dB. If this amplifier has to faithfully amplify the sinusoidal signal from 6 to 10 kHz without introducing any distortion, then the maximum value of input signal amplitude is __ mV.

• A. 79.57

Answer 1

The correct option is A 79.57

$\text{Slow rate}=\frac{d{V}_0}{dt}{|}_{max}$

$\therefore \text{Slow rate}=A\frac{d{V}_0}{dt}=A.\frac{d}{dt}{A}_m\sin \left(\omega t\right)$

$\text{Slow rate}=A{A}_m\omega$

now, $A={10}^{\frac{66.0206}{20}}=2000$

$\therefore 10\times {10}^{-6}=2\times {10}^3\times A_m\times 2\pi \times 10\times {10}^3$

$\therefore A_m=79.57mV$