Question

An OPAMP has a slew rate of 100 V/msec. For a frequency of 10 MHz, the maximum value of the sine wave output voltage will be
1. 100 V
2. 1/200π V
3. 10 V
4. 5 V

Answer 1

Correct Answer - Option 2 : 1/200π V

Concept:

When the input is a sinusoid given as:

V_i (t) = A_m sin(2πf_mt)

Let, the gain of OPAMP is 'A_v', the output is given as:

V_o(t) = A_v A_m sin(2πf_mt)

The rate of change of output:

\(\frac{{d{V_o}}}{{dt}} = A_v A_m ~cos (2\pi f_m t)\)

The maximum rate of change = |A_v A 2πf_m cos(2πf_mt)|_max

\({\left. {\frac{{{d{V_o}}}}{{dt}}} \right|_{max}} = {A_v}A_m~2\pi {f_m} =Slew~Rate\)

Calculation:

Given, 

Slew Rate (i.e.) the maximum rate of change of the output = 100 V/ms

The frequency of sinusoidal signal = 10 MHz

\( \frac{{100~V}}{{{{10}^{ - 3}}}} = {A_v A_m}× {10 × 10^6}\times 2\pi\)

\(A_v A_m = \frac{10^5}{2\pi \times10^7} V\)

Since the value of gain is not given we can take A_v = 1

\(\therefore ~A_m = \frac{1}{200 \pi} V\)