QB_PH_SEC8_009: Operational Amplifiers & Analog Circuits

In a first-order active low-pass RC filter built using an operational amplifier in non-inverting configuration, if the resistor is $R = 10\text{ k}\Omega$ and capacitor is $C = 15.9\text{ nF}$, the cutoff frequency $f_c$ is approximately

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Transfer Function of First-Order Active LPF**: $H(s) = A_v \frac{1}{1 + sRC}$ 2. **Cutoff (-3 dB) Frequency Relation**: $\omega_c = \frac{1}{RC} \implies f_c = \frac{1}{2\pi RC}$ 3. **Numerical Computation**: - $R = 10\text{ k}\Omega = 10^4\,\Omega$ - $C = 15.9\text{ nF} = 15.9 \times 10^{-9}\text{ F} = \frac{100}{2\pi} \times 10^{-9}\text{ F} \approx \frac{1}{2\pi \times 10^4}\text{ F}$ $f_c = \frac{1}{2 \times 3.14159 \times 10^4 \times (15.9 \times 10^{-9})} = \frac{1}{9.99 \times 10^{-4}} \approx 1000\text{ Hz} = 1\text{ kHz}$ 4. **Conclusion**: Cutoff frequency is $1\text{ kHz}$. Option (A) is correct.