**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:

**Hartley tank circuit**

Here L1=10 mH and L2=10 mH

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

Hence, Oscillator frequency is given by,

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

**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,

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