$An op-amp has slew rate = 0.5 V/ μ s. The amplitude of input signal is 0.02 V and the close loop voltage gain is 25. If the input signal is a sinusoidal signal, then the maximum frequency that may be used is equal to$

170.30 kHz

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159.15 kHz

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210.35 kHz

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190.45 kHz

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Solution

The correct option is B 159.15 kHz
$S R \geq {(\frac{d V_{o}}{d t})}_{max}$

$S R \geq 2 π f V_{on}$

$S R \geq ω V_{on}$

$f_{max} = \frac{0.5 \times 10^{6}}{25 \times 2 π \times 0.02} = 159.15 kHz$

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