Solution:

An opamp operates as a unity gain butter with 3VPP square wave input. If opamp is ideal with slew rate $0.5 \mathrm{~V} /$ microseconds, find the maximum frequency of operation.

$\text { slew rate }=2 \times \pi \times f \times v\\ $

where,

$f=$ The highest signal frequency, $H_2$

$V=$ The maximum peak voltage of the signal.

$\begin{aligned}\\ 0.5 \mathrm{v} / \mu \mathrm{s} &=2 \times \pi \times f \times 1.5 \\\\ 0.5 \times 10^{+6} &=9.42 \mathrm{~F} \\\\ F &=53078.55 \mathrm{H}_2\\ \end{aligned}\\ $