Question

$\text{An op-amp amplifier with slew rate of 10 V/}\mu \text{sec has a voltage gain of 66.0206 dB. If this}\text{amplifier has to faithfully amplify the sinusoidal signal from 0.6 to 10 kHz, without}\text{introducing any distortion, then the maximum value of input signal amplitude is\_\_\_\_\_mV}$

• A. 79.57

Answer 1

The correct option is A 79.57
$\text{Slew rate =}{\frac{d{V}_0}{dt}}_{max}\therefore \text{Slew rate = A.}\frac{d{V}_0}{dt}=A.\frac{d}{dt}{A}_{m}sin\left(\omega t\right)\text{Slew rate = AA}{}_m\omega \text{now,}A={10}^{\frac{66.0206}{20}}=2000\therefore 10\times {10}^6=2\times {10}^3\times A_m\times 2\pi \times 10\times {10}^3\therefore A_m=79.57mV$