Draw FET based Hartley and Colpitt Oscillator. What is the frequency of oscillation if - (i) L1= 10 mH, L2 =10 mH and C=0.1pF for Hartley tank circuit - (ii) L=10 mil, Cl = 0.1pF and C2 = 0.1 pF. for Colpitt tank circuit. -

Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals

Marks: 10 Marks

Year: May 2014

Hartley Oscillator

If FET is used as an active device in an amplifier stage, then the circuit is called as FET Hartley oscillator. The practical circuit is shown in the figure.

FET Hartley Oscillator

The resistance $R_1, R_2$ bias the FET along with $R_s$. The $C_s$ in the source bypass capacitor. To maintain Q point stable, coupling capacitors $C_c1, C_{c2}$ are used. These have very large values compared to capacitor C. the tank circuit is shown in a box.

We know,

$$X_1+X_2+X_3=0$$

And

$$X_1=j\omega L_1,X_1=j \omega L_2 \ \ and \ \ X_3=\dfrac{1}{j\omega C}$$

Solving for ω, we get the same expression for the frequency

$$f=\dfrac{1}{2 \pi \sqrt{CL_{eq}}} \\ L_{eq}=L_1+L_2 \ \ or L_1+L_2+2M$$

This is dependent on whether $L_1, L_2$ are wound on the same core or not.

If $L_1=L_2=L$ then the frequency of oscillation is given by,

$$f=\dfrac{1}{2 \pi \sqrt{2} \sqrt{LC}}$$

Colpitt Oscillator

If in the basic circuit of Colpitts oscillator, the FET is used as an active device in the amplifier stage, the circuit is called as FET Colpitts oscillator. The tank circuit remains same as before.

The working of the circuit and oscillating frequency also remains the same.

The practical circuit of FET Colpitts oscillator is shown in the figure.

Problem:

1. Hartley tank circuit

   Here L1=10 mH and L2=10 mH

   So, L=L1+L2=10+10=20mH

   Hence, Oscillator frequency is given by,

   $f=\dfrac{1}{(2π\sqrt{LC}} \\ f=\dfrac{1}{2π\sqrt{20 x 0.1}}=0.11Hz$

2. Colpitt tank circuit

   Here C1=0.1 pF and C2=0.1 pF

   So, $C=\dfrac{C1 C2}{C1+C2}=\dfrac{0.1 x 0.1}{0.1+0.1}=\dfrac{0.01}{0.2}=0.05$

   Hence, Oscillator frequency is given by,

   $f=\dfrac{1}{2π\sqrt{LC}} \\ f=\dfrac{1}{2π\sqrt{10 x 0.05}}=14.04Hz$