Working principle of Hartley oscillator

Hartley oscillator:

The circuit diagram for Hartley oscillator is shown in figure. It consists of C-E amplifier and tank circuit. The tank circuit consist of a inductor coil with central tapping with mark values of inductance \(L_1\) and \(L_2\) as in figure . A variable capacitor connected in between in parallel with \(L_1\) and \(L_2\). The RFC is used as load resistor. The biasing resistor \(R_1\) and \(R_2\) are used for base biasing circuit . The bypass capacitor \(C_E\) used for stabilization of circuit whereas coupling capacitor \(C_1\) and \(C_2\) couples the feedback signal into the base and output signal to tank circuit. The DC power supply +\(V_{CC}\) is used as the source. The swamping resistance \(R_E\) is added for stabilization with temperature variation.

Working expression:

When the supply (\(V-{CC}\)) is given to the oscillator current raises to its Quiescent value. Due to collector current a transient current it initiates the natural oscillation in the tank circuit. As a result of oscillation the small emf is developed across \(L_2\) due to mutual inductance in inducting coil. This emf across \(L_2\) acts as feedback voltage which is given to the base of transistor and it is amplified by the CE amplifier and appears across RFC as output which small part is used up by the tank circuit to fulfill its loss and remaining energy is radiated in the form of electromagnetic wave. This process continues and generates the sine wave by Hartley oscillator.

The feedback input signal is amplified by transistor in CE mode. So that the total phase shift of \(180^{\circ}\) occurs. Since the inductor coil is central trapped in tank circuit, so that the two coil \(L_1\) and \(L_2\) acts as auto transformer, whose two ends are with opposite terminal. Hence the signal at tank circuit is again phase inverse by \(180^{\circ}\). Hence the total phase shift will be \(360^{\circ}\), which is necessary condition for sustained oscillation and generated sine wave continuously.

For ac analysis of Hartley oscillator, the resistance offered by RFC is infinitely large for the AC signal so its effect is neglected. Since, parallel combination of \(R_1\) and \(R_2\) i.e. \(R_1 \parallel R_2\) is in parallel combination with input resistance (\(r_{in}\)) of transistor so that the total input resistance is \(r_{in}\) of transistor so that the total input resistance is \(r_{in}\).Where,

\(r_{in}=\beta r_e’\) where,

\(\beta\)=amplification factor in CE mode

\(r_e’\)=ac input resistance

Also, capacitor \(C_1\) and \(C_2\) are shorted as the allow the ac signal then the ac equivalent circuit becomes,

Here,$$Z_l=jX_L=j\omega(L_1+M)$$

$$Z_2=j X L_2=j \omega(L_2+M)$$

$$Z_3=\frac{1}{j\omega_c}$$and $$r_{in}=\beta r_e’$$Now, the load impedance \(Z_L\) is given by

$$Z_L=Z_1\parallel[Z_3+(Z_2\parallel r_{in})]$$Now, the gain of amplifier in CE mode is given by,

$$A=\frac{-Z_1}{r_e’}=\frac{-Z_1\parallel [Z_3+(z_2\parallel r_{in})]}{r_e’}\dotsm(1)$$

The voltage across \(z_1\) is \(v_\circ\) .

Voltage across the \(z_2\) voltage \(v_f) then the feedback ratio /beta is given by,

$$\therefore \beta =\frac{v_f}{v_\circ}$$

$$\beta=\frac{(z_2\parallel r_{in})}{(z_3+z_2\parallel r_{in})}\dotsm(2)$$Now for the oscillation, we should have,

$$A\beta=1$$

$$\biggl[\frac{-z_1[z_3+(z_2\parallel r_{in})]}{r_e’}\biggr].\frac{(Z_2\parallel r_{in})}{(z_3+z_2\parallel r_{in})}=1$$

$$z_1z_2+z_2 z_3+z_1z_2\beta+(z_1+z-2+z_3)\beta r_e’=0$$In this equation (2), the real part and imaginary part of it should separately equal to zero.

Oscillation frequency :

Comparing imaginary part with zero,

$$(z_1+z_2+z_3)\beta r_e’=0$$

$$\Rightarrow z_1+z_2+z_3+0$$

$$j\omega (L_1+M)+j\omega(L_2+M)+\frac{1}{j\omega C}=0$$

$$\frac{j}{\omega}\biggl[\omega^2(L_1+M)+ \omega^2(L_2+M)-\frac{1}{C}\biggr]=0$$

$$\omega^2\biggl[L_1+M+L_2+M-\frac{1}{\omega^2C}\biggr]=0$$

$$L_1+L_2+2M=\frac{1}{\omega^2C}$$

$$\omega^2=\frac{1}{LC}$$ Where,$$\omega=\frac{1}{\sqrt{LC}}$$

$$f=\frac{1}{2\pi\sqrt{LC}}$$This is the frequencies generated by Hartley oscillator.

Oscillation criteria:

Equating real part of (2) with zero,

$$z_1z_2(1+\beta)+z_2z_3=0$$

$$z_1z_2(1+\beta)=-z_2z_3$$

$$\therefore 1+\beta=\frac{-z_3}{z_1}$$

We have \(z_1+z_2+z_3=0\)\(\Rightarrow z_3=-(z_1+z_2)\)

$$1+\beta=\frac{(z_1+z_2)}{z_1}$$

$$1+\beta=1+\frac{z_2}{z_1}$$

$$1+\beta=1+\frac{j\omega(L_2+M)}{j\omega(L_1+M)}$$

$$\beta=\frac{L_2+M}{L_1+M}$$This is condition for sustained oscillator.

Advantages:

• Instead of two separate coils as\(L_1\) and \(L_2\), a single coil of bare wire can be used and the coil grounded at any desired point along it.
• By using variable capacitor or by making core movable(varying the inductance), frequency of oscillation can be varied.
• The amplitude of the output remains constant over the working frequency range.
• Very few components are needed including either two fixed inductors or a tapped coil.

Disadvantages:

• Ti can’t be used as low frequency oscillator since the value of inductors become large and size of the inductors becomes bulky.
• The harmonic contain in the output of this oscillator is very high and hence it is not suitable for the application which require pure sine wave.

References:

(1)Theraja, B.L. Basic Electronics. N.p.: S.Chand, n.d. Print.

(2)C.L.Arora. Refresher Course in Physics. Vol. II and III. N.p.: S.Chand, 2006. Print.

(3)Malvino. Electronic Principles. N.p.: Tata McGraw-Hill, n.d. Print.

(4)N.Nelkon and P.Parker. Advanced Level Physics. 5th ed. N.p.: Arnold Heinemann, n.d. Print.

(5)Priti Bhakta Adhikari,Diya Nidhi Chaatkuli, Ishowr Prasad Koirala. A Textbook of Physics (2nd Year). N.p.: Sukunda Pustak Bhawan, 2070. Prin