Question

An op-amp having a slew rate of 62.8 V/μsec is connected in a voltage follower configuration. If the maximum amplitude of the input sinusoidal is 10V, then the minimum frequency a which the slew rate limited distortion would set in at the output is
1. 1.0 MHz
2. 6.28 MHz
3. 10 MHz
4. 62.8 MHz

Answer 1

Correct Answer - Option 1 : 1.0 MHz

Concept:

Slew rate is the maximum rate of change of output voltage with respect to time.

Slew rate limits the maximum frequency of operation of op-amp

The slew rate is usually measured in volts per microsecond.

Mathematically,

Vin = Vm sin 2πfmt and

V0 = AVin

then, Slew rate = \({\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}} = 2{\rm{\pi }}{{\rm{f}}_{{\rm{max}}}}{{\rm{V}}_0}_{{\rm{max}}}{\rm{\;}}\)

∴ We can say that the signal bandwidth fm is limited by the Slew rate.

Calculation:

Slew rate \(= \frac{{{\rm{\Delta }}{V_0}}}{{{\rm{\Delta }}{V_i}}} × \frac{{{\rm{\Delta }}{V_i}}}{{{\rm{\Delta }}t}} = {A_{CL}}\left( {2\pi {f_m}{V_m}} \right)\)

Given that, slew rate = 62.8 V/μsec

62.8 × 10⁶ = 6.28 × 10 × f_m 

f_m = 1 MHz