Question

For the active filter shown, the DC gain and the 3 dB cutoff frequency are nearly: \[ C_1 = 1.0\ \text{nF},\ R_1 = 15.9\ \text{k}\Omega,\ R_2 = 159\ \text{k}\Omega \]

• A. 40 dB, 3.14 kHz
• B. 40 dB, 1 kHz
• C. 20 dB, 6.28 kHz
• D. 20 dB, 1 kHz

Hint:

Use \( f_c = \frac{1}{2\pi RC} \), and \( A_v = 1 + \frac{R_2}{R_1} \) for active low-pass filters.

Answer 1

The Correct Option is D

DC Gain: \[ A_v = 1 + \dfrac{R_2}{R_1} = 1 + \dfrac{159}{15.9} = 1 + 10 = 11 \Rightarrow 20 \log(11) \approx 20.8\ \text{dB} \approx 20\ \text{dB} \] Cutoff frequency: \[ f_c = \dfrac{1}{2\pi R_1 C_1} = \dfrac{1}{2\pi \cdot 15.9 \times 10^3 \cdot 1 \times 10^{-9}} \approx 1\ \text{kHz} \]