Question

The slew rate of an op-amp is 0.5 $V/\mu S$ when used as an inverting amplifier with a gain of 50. The voltage gain vs frequency curve is flat upto 20 kHz. The maximum peak to peak input signal can be applied without distorting the output is

• A. 142 mV ( p- p)
• B. 210 mV (p -p)
• C. 159 mV (p -p)
• D. 118 mV ( p- p)

Answer 1

The correct option is C 159 mV (p -p)

The maximum output voltage at 20 kHz, is

$0.5=\frac{2\pi \times 20\times {10}^3\times V_m}{{10}^6}$

$V_m=3.98V_{\left(peak\right)}$

$V_0=7.96V\left(peaktopeak\right)$

Hence, for the output to be undistorted sine wave, the maximum input signal should be less than
$\frac{7.96}{50}=159mV$ (peak to peak)