Question

An ideal OPAMP circuit with a sinusoidal input is shown in the figure. The 3 dB frequency is the frequency at which the magnitude of the voltage gain decreases by 3 dB from the maximum value. Which of the options is/are correct?

• A. The circuit is a low pass filter.
• B. The circuit is a high pass filter.
• C. The 3 dB frequency is $1000\ \text{rad/s}$.
• D. The 3 dB frequency is $\dfrac{1000}{3}\ \text{rad/s}$.

Hint:

A capacitor in series with the input of an op-amp amplifier always indicates a high-pass filter. The cutoff is found from the input RC network, not from the feedback network.

Answer 1

The Correct Option is B, C

The circuit consists of a capacitor in series with the input, followed by a resistor before reaching the non-inverting terminal of the op-amp. A series capacitor at the input produces a high-pass filter behavior, because low-frequency signals (including DC) are blocked while higher frequencies pass through.
The op-amp configuration given is a non-inverting amplifier with feedback resistors \(R_f = 2\,\text{k}\Omega\) and \(R = 1\,\text{k}\Omega\). Thus, the mid-band (maximum) gain is: \[ A_0 = 1 + \frac{R_f}{R} = 1 + \frac{2000}{1000} = 3. \] Step 1: Determine the high-pass cutoff frequency.
The cutoff frequency \( \omega_c \) is determined by the series capacitor–resistor combination at the input: \[ C = 1\,\mu\text{F}, \qquad R = 1\,\text{k}\Omega. \] The high-pass cutoff angular frequency is: \[ \omega_c = \frac{1}{RC} = \frac{1}{(1000)(1\times 10^{-6})} = 1000\ \text{rad/s}. \] Step 2: Verify 3 dB point.
For a high-pass filter, the magnitude at the cutoff frequency satisfies: \[ |A(j\omega_c)| = \frac{A_0}{\sqrt{2}} \] which matches the 3 dB drop definition. Thus the calculated cutoff frequency is indeed the 3 dB frequency.
Therefore: - The circuit is a high-pass filter.
- The 3 dB frequency is 1000 rad/s.
Final Answer: (B) and (C)