For IC 741 circuit shown below, the maximum input frequency above which the output will be distorted is ____________. 
To determine the maximum input frequency above which the output of the IC 741 will be distorted, we need to consider the slew rate of the operational amplifier. The slew rate is the maximum rate of change of the output voltage and is critical in determining the maximum input frequency to avoid distortion.
The slew rate (\(SR\)) is given by:
\(SR = \frac{dV_o}{dt}\)
For the IC 741, the typical slew rate is \(0.5 \, \text{V/μs}\).
Let \(V_m\) be the amplitude of the output voltage. The maximum frequency (\(f_{max}\)) without distortion can be found using the formula:
\(f_{max} = \frac{SR}{2\pi V_m}\)
Assuming a maximum output voltage swing of \(V_m = 10 \, \text{V}\), the maximum frequency is calculated as follows:
\(f_{max} = \frac{0.5 \times 10^6}{2 \pi \times 10}\)
Solving the above equation:
\(f_{max} = \frac{0.5 \times 10^6}{62.83} \approx 8.33 \, \text{kHz}\)
Thus, the maximum input frequency above which the output will be distorted is 8.33 kHz.
Important Questions from Op-Amp and Its Applications
In the circuit shown below, $R$ is $1\text{ k}\Omega$ and $R_f$ is $10\text{ k}\Omega$. If $V_{\text{in}}$ is $100\text{ mV}$ and the op-amps have supply voltages of $\pm 15\text{ V}$, then $V_{\text{out}}$ is _______ V.
Assume that an op-amp 741 connected as a unit gain inverting amplifier is applied with an input change of 10 V. Determine the time taken for the output to change by 10 V. [Slew rate for op-amp 741 = 0.5 V/µs]
What is the value of output voltage $V_o$ in the given circuit? Assume op-amp to be ideal and MOSFET operating in saturation region.
In an ideal op-amp, the common mode gain is _____________.
Assertion (A) : An Op-Amp is a direct coupled high gain amplifier.
Reason (R) : It consists of one or more differential amplifiers and usually followed by a level translator and push pull stage.
