An amplifier using an Op-Amp with a slew rate SR = 1 V/μ sec has a gain of 40 dB. If this amplifier has to faithfully amplify sinusoidal signals from 10 to 20 kHz, without introducing any slew rate induced distortion, then the input signal level must not exceed. 

Mathematical Analysis: 

The maximum rate of change of the output voltage in response to a step input voltage is defined as the slew rate of the op-Amp.

Mathematically, the slew rate is defined as:

\(S.R = {\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}}\)

When the input is a sinusoid given as:

V_i(t) = Vm sin2πf t

And the voltage gain of the op-amp is A_v, the output voltage will be:

V_o = A_V V_i

Vo = AV Vm sin2πf t

Now, the rate of change in output will be:

\(\frac{{d{V_0}}}{{dt}} =A_VV_m2\pi f~cos2\pi ft\)

The maximum rate of change (Slew rate) will be:

\({\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}}=A_VV_m2πf_{max}\)   ---(1)

Calculation:

Gain = 40 dB, i.e.

20 log₁₀ A_V = 40

A_V = 100

Putting on the respective values in Equation (1), we can write:

\(\frac{1}{10^{-6}}=100\times V_m\times 2π\times20k\)

\(V_m=\frac{1}{2\pi \times 2}=79.5~mV\)

∴ The input level must not exceed above 79.5 mV.

• The output of an Op-Amp can only change by a certain amount in a given time. This limit is called the slew rate of the Op-Amp.
• Op-Amp slew rate can limit the performance of a circuit if the slew rate requirement is exceeded.
• This is best explained with the help of the given waveform: