Question

An op-amp with a slew rate of $1V/msec$ has been used to build an amplifier of gain 10. If the input to the amplifier is sinusoidal voltage having peak amplitude of 1 V. The maximum allowable frequency of the input signal for undistorted output is

• A. 15.92 kHz
• B. 18.74 kHz
• C. 12.47 kHz
• D. 13.62 kHz

Answer 1

The correct option is A 15.92 kHz

$\text{Gain},A_V=\frac{V_0}{V_i}=10$

$\text{Slew rate}={\left(\frac{dV}{dt}\right)}_{max}={10}^{6}V/sec$

$V_0=10V_i=10\sin \omega t$

$\frac{d{V}_0}{dt}=10\omega \cos \omega t$

$\text{Slew rate}={\left(\frac{dV}{dt}\right)}_{max}=10\omega =10\times 2\pi \times f$

${10}^6=10\times 2\pi \times f$

$f=15.92kHz$