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Question: With the help of a circuit diagram, explain the working of a Colpitt’s oscillator....

With the help of a circuit diagram, explain the working of a Colpitt’s oscillator.

Explanation

Solution

An oscillatory circuit, also called a tank circuit is an electronic circuit that produces a repetitive waveform with a constant amplitude and frequency and the oscillator does not require an external input signal. The tank circuit of a Colpitt’s oscillator consists of two capacitors in series and an inductance coil connected parallel to them.

Formulas used:
The frequency of oscillations of a tank circuit is given by, f=12πLCf = \dfrac{1}{{2\pi \sqrt {LC} }} where LL is the inductance of the inductor coil and CC is the capacitance of the tank circuit.
The effective capacitance of two capacitors C1C_1 and C2C_2 connected in series is given by, 1C=1C1+1C2\dfrac{1}{C} = \dfrac{1}{{C_1}} + \dfrac{1}{{C_2}}.

Complete Step by Step Answer:
Step 1: Sketching the circuit diagram of the Colpitt’s oscillator.

Step 2: Listing the elements of the circuit.
The tank circuit is made up of the two capacitors C1C_{1} and C2C_{2} in series and connected parallel to the inductor L1L_{1}.
Between the collector and the +Vcc + V_{cc} , a radio frequency choke RFC1RFC_{1} is connected. The RFC1RFC_{1} acts as a collector load and ensures proper bias to the collector. All ac signals that could be present in the power supply are also decoupled by the radio frequency choke.
The resistors R1R_{1} and R2R_{2} form the voltage divider component of the circuit.
The collector capacitor C5C_{5} couples the oscillatory circuit to the collector, the base capacitor C3C_{3} blocks the dc currents from entering the base and the emitter capacitor C4C_{4} bypasses the emitter resistance R2R_{2}.
Step 3: Explanation of the working of the circuit.
The capacitors C1C_{1} and C2C_{2} get charged when we switch on the circuit. They discharge through the inductance coil L1L_{1}. An oscillatory current is then produced. The oscillations across C2C_{2} are feedback to the input of the circuit. The input given to the transistor gets amplified. The output of the transistor will be received across the capacitor C1C_{1}. The process then repeats and leads to the generation of a sustained oscillating output across the secondary coil of the transformer. The primary coil of the transformer is the inductance coil L1L_1.
The frequency of the oscillations of the output will match the natural frequency of the oscillatory circuit (or tank circuit).
The frequency of the tank circuit is given by, f=12πL1×Cf = \dfrac{1}{{2\pi \sqrt {L_1 \times C} }}
where L1L_1 is the inductance of the inductor coil and CC is the effective capacitance of the capacitors connected in series.
We have the effective capacitance of the tank circuit as 1C=1C1+1C2\dfrac{1}{C} = \dfrac{1}{{C_1}} + \dfrac{1}{{C_2}}.
On simplifying the above expression we get, C=C1C2C1+C2C = \dfrac{{C_1C_2}}{{C_1 + C_2}}
Thus the frequency of the oscillations will be f=12πL1×(C1C2C1+C2)f = \dfrac{1}{{2\pi \sqrt {L_1 \times \left( {\dfrac{{C_1C_2}}{{C_1 + C_2}}} \right)}}}

Note:
For an oscillator, the phase difference between the output signal and the input signal must be 0° or 360°. This is referred to as positive feedback. There exists a phase difference of 180° between the output and the input as the ends of the tank circuit are oppositely charged. Also, the transistor used as an amplifier again produces a phase shift of 180° between the output and the input. Thus the total phase difference between the output and input will be 360°.