# Colpitts oscillator

Colpitt’s  Oscillator is an excellent circuit and is widely used in commercial signal generators upto 100MHz.

It consists of a single-stage inverting amplifier and an LC phase shift Network.

The two capacitors $C_{1}$ and $C_{2}$ provides potential divider used for providing $V_{f}$$C_{1}$ is the feedback element and which provides positive feedback required for sustained Oscillations.

The amplifier circuit is a self-Bias Circuit with $R_{1}$ , $R_{2}$ and parallel combination of $R_{E}$ with $C_{E}$.

$V_{CC}$ is applied through a resistor  $R_{C}$ (or) RFC choke some times. This RFC choke offers very high impedance to high frequency currents.

$R_{C}$ value has chosen in such a way that it offers high impedance. Two coupling Capacitors $C_{C1}$ and $C_{C2}$ are used to block d.c currents, that means they do not permit d.c currents into tank circuit.

These capacitors $C_{C1}$ and $C_{C2}$ provides a path from Collector to Base through LC Network.

when $V_{CC}$ is switched on , a transient current is produced in the tank circuit an consequently damped oscillations are setup in the circuit.

The oscillatory current in the tank circuit produces a.c voltages across $C_{1}$ and $C_{2}$ . If terminal 1 is more positive w.r. to 2 , then voltages across $C_{1}$ and $C_{2}$ are opposite thus providing a phase shift of $180^{o}$ between 1 and 2.

as the transistor is operating in CE mode , it provides a phase shift of $180^{o}$.

Therefore the over all phase shift provided by the circuit results $360^{o}$ which is an essential condition for developing oscillations.

If the feedback is adjusted so that the loop gain $A\beta&space;=1$ then then the  circuit acts as an Oscillator.

The frequency of oscillation depends on the tank circuit and is varied by gang (or) group tuning of $C_{1}$ and $C_{2}$ means $C_{1}=C_{2}$.

working:-

The capacitors $C_{1}$ and $C_{2}$ are charged by $V_{CC}$ and are discharged through the coil $L$ setting up of oscillations with frequency

$f_{o}=\frac{1}{2\pi&space;}\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$.

these oscillations across $C_{1}$ are applied to the Base-Emitter junction  and the amplified version of output is collected across Collector (the frequency of amplifier output is same as that of input of the amplifier) .

This amplified energy is given back to tank circuit to compensate losses.

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

chose $\left&space;|&space;A\beta&space;\right&space;|\geq&space;1$ for sustained oscillations.

Analysis(Qualitative):-

if $Z_{1}$ , $Z_{2}$  and $Z_{3}$  are pure reactive elements  such that $Z_{1}=\frac{1}{j\omega&space;C_{1}}&space;=\frac{-j}{\omega&space;C_{1}}$ ,  $Z_{2}=\frac{1}{j\omega&space;C_{2}}&space;=\frac{-j}{\omega&space;C_{2}}$   and  $Z_{3}=j\omega&space;L$.

from the general condition for an Oscillator

$\left&space;|&space;A\beta&space;\right&space;|&space;=1$  $\Rightarrow&space;h_{ie}(Z_{1}+Z_{2}+Z_{3})+Z_{1}Z_{2}(1+h_{fe})+Z_{1}Z_{3}=0$.

$h_{ie}(-\frac{j}{\omega&space;C_{1}}-\frac{j}{\omega&space;C_{2}}+j\omega&space;L)+\frac{j^{2}}{\omega&space;^{2}C_{1}C_{2}}(1+h_{fe})-\frac{j}{\omega&space;C_{1}}.j\omega&space;L=0$

find the real and imaginary parts,

$-j(\frac{1}{\omega&space;C_{1}}+\frac{1}{\omega&space;C_{2}}-\omega&space;L)h_{ie}-\frac{1}{\omega&space;^{2}C_{1}C_{2}}(1+h_{fe})+\frac{L}{C_{1}}=0$

equating imaginary part to zero  $(\frac{1}{\omega&space;C_{1}}+\frac{1}{\omega&space;C_{2}}-\omega&space;L)=0$  ,  since $h_{ie}\neq&space;0$ .

$\frac{\omega&space;C_{1}+\omega&space;C_{2}}{\omega^{2}&space;C_{1}C_{2}}=\omega&space;L$.

after simplification

$\omega&space;^{2}=\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$.

by substituting $\omega&space;=2\pi&space;f$    results $f_{o}=\frac{1}{2\pi&space;}\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$.

substituting the value of $\omega&space;^{2}$  in the real part gives $h_{fe}=\frac{C_{2}}{C_{1}}$  . this is the condition for sustained oscillations.

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## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems.

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