Q.Calculate the frequency and feedback factor of the circuit shown below.

The network is a Colpitts tank: inductor $L = 100\,\mu\text{H}$ with series capacitors $C_1 = 0.1\,\mu\text{F}$ and $C_2 = 0.22\,\mu\text{F}$, so $f = \dfrac{1}{2\pi\sqrt{L\,C_{eq}}}$ with $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}$ and $\beta = \dfrac{C_1}{C_2}\ …

[!TLDR]

The series capacitors give $C_{eq} = 6.875\times10^{-8}\,\text{F}$, so with $L = 100\,\mu\text{H}$ the frequency is $f \approx 60.691\,\text{kHz}$; the feedback factor is $\beta = C_1/C_2 \approx 0.454$.

This is the Colpitts feedback tank: a single inductor across two series capacitors whose common point is grounded. The feedback voltage is developed across $C_2$ and the output across $C_1$, so the feedback factor is the capacitance ratio $\beta = \dfrac{C_1}{C_2}$. The frequency is the tank resonance with the series equivalent $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}$.

Step 1 — equivalent capacitance: $C_{eq} = \dfrac{(0.1\times10^{-6})(0.22\times10^{-6})}{(0.1\times10^{-6}) + (0.22\times10^{-6})} = \dfrac{0.022\times10^{-6}}{0.32} = 6.875\times10^{-8}\,\text{F}$.

Step 2 — frequency: $f = \dfrac{1}{2\pi\sqrt{(100\times10^{-6})(6.875\times10^{-8})}} = \dfrac{1}{2\pi\sqrt{6.875\times10^{-12}}} \approx 60.691\times10^{3}\,\text{Hz}$. …