Q.A Colpitts oscillator oscillates at $1.13\,\text{MHz}$. If the inductor in the feedback network has a value of $20\,\mu\text{H}$ and one of the capacitors value is $0.1\,\mu\text{F}$, calculate the value of the other capacitor.

The Colpitts oscillator is an LC oscillator that takes its feedback from a capacitive voltage divider — hence it uses capacitive feedback. Its tank circuit is a single inductor $L$ in parallel with two capacitors $C_1$ and $C_2$ connected in series, their common junction being grounded. The feedback voltage is developed across $C_2$.

The amplifier and biasing are the same as in other transistor LC oscillators: $R_1$, $R_2$ and $R_E$ set the DC bias, $C_E$ is the bypass capacitor, $C_C$ the coupling capacitor, and the RFC (radio-frequency choke) isolates the AC oscillations from the DC supply.

Working: at switch-on the rising collector current charges $C_1$ and $C_2$. Because their common terminal is earthed, the voltages across the two capacitors are in opposite phase; the voltage across $C_2$ is fed back to the input. When the capacitors are charged they discharge through $L$, sustaining the oscillation. The tank provides $180^\circ$ and the amplifier $180^\circ$, giving the required $360^\circ$.

Because the two tank capacitors are effectively in series across the inductor, the frequency of oscillation is

The feedback factor is $\beta = \dfrac{C_1}{C_2}$, and to meet $A\beta = 1$ the amplifier gain must be $A = \dfrac{C_2}{C_1}$. Colpitts oscillators produce stable high (radio) frequencies and are used in radio and TV transmitters and receivers and as local oscillators; their main drawback is that they are unsuited to low frequencies, where the required components would be bulky.