QB_PH_SEC8_027: Operational Amplifiers & Analog Circuits

A Wien-bridge oscillator uses matched resistors $R = 10\text{ k}\Omega$ and matched capacitors $C = 15.9\text{ nF}$ in its lead-lag feedback network. What is the oscillation frequency $f_0$ in kHz? (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Wien Bridge Lead-Lag Network Transfer Function**: $\beta(s) = \frac{Z_p}{Z_s + Z_p}$ where $Z_s = R + \frac{1}{sC}$ and $Z_p = \frac{R}{1 + sRC}$. 2. **Zero Phase Shift Condition**: The imaginary part of the denominator vanishes at frequency: $\omega_0 = \frac{1}{RC} \implies f_0 = \frac{1}{2\pi RC}$ At this frequency, $\beta(\omega_0) = 1/3$ with zero phase shift. 3. **Numerical Computation**: - $R = 10\text{ k}\Omega = 10^4\,\Omega$ - $C = 15.9\text{ nF} = 15.9 \times 10^{-9}\text{ F}$ $f_0 = \frac{1}{2 \times 3.14159 \times 10^4 \times (15.9 \times 10^{-9})} = \frac{1}{9.99 \times 10^{-4}} \approx 1000\text{ Hz} = 1\text{ kHz}$ 4. **Official Answer**: 1 kHz.