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    Tuesday, 18 April 2017

    DESIGN A HARTLEY OSCILLATOR

     DESIGN A HARTLEY OSCILLATOR, Fo=1MHZ, Vcc=18V(BJT TRANSISTOR)                         


    Aim:- To design a Hartley Oscillator using a transistor, to find out the frequency of oscillation 1MHz with 18volt supply.
    components:
    n-p-n transistor, carbon resistor, two inductors, capacitor and a 2x9volts Dc Batteries.

    Description:
    The Hartley oscillator is design for generation of sinusoidal oscillations in the R.F range (20 KHz-30MHz). It is very popular and used in radio receivers as a local oscillator.
    The circuit diagram of Hartley oscillator (parallel or shunt-fed) using BJT is shown in figure. It consists of an R-C coupled amplifier using an n-p-n transistor in CE configuration. R1 and R2 are two resistors which form a voltage divider bias to the transistor. A resistor RE is connected in the circuit which stabilizes the circuit against
    temperature variations. A capacitor CE is connected in parallel with RE, acts as a bypass capacitor and provides a low reactive path to the amplified ac signal. The coupling capacitor CC blocks dc and provides an ac path from the collector to the tank circuit. The L-C feedback network (tank circuit) consists of two inductors L1, and L2 (in series) which are placed across a common capacitor C and the center of the two inductors is tapped as shown in fig. The feedback network (L1, L2 and C) determines the frequency of oscillation of the oscillator.

    Theory:
    When the collector supply voltage Vcc is switched on, collector current starts rising and charges the capacitor C. When this capacitor is fully charged, it discharges through coils L1 and L2, setting up damped harmonic oscillations in the tank circuit. The oscillatory current in the tank circuit produces an AC voltage across L1 which is applied to the base emitter junction of the transistor and appears in the amplified form in the collector circuit. Feedback of energy from output (collector emitter circuit) to input (base-emitter circuit is) accomplished through auto transformer action. The output of the amplifier is applied across the inductor L1, and the voltage across L2 forms the feedback voltage. The coil L1 is inductively coupled to coil L2, and the combination acts as an auto-transformer. This energy supplied to the tank circuit overcomes the losses occurring in it. Consequently the oscillations are sustained in the circuit. The energy supplied to the tank circuit is in phase with the generated oscillations.
    The phase difference between the voltages across L1 and that across L2 is always 180° because the center of the two is grounded. A further phase of 180° is introduced between the input and output voltages by the transistor itself. Thus the total phase shift becomes 3600 (or zero), thereby making the feedback positive or regenerative which is essential for oscillations. So continuous undammed oscillations are obtained
                                                                                                                                                                     Precautions:
    I would have ensured that I do the following when building the circuit:
    1.     Check the continuity of the connecting terminals before going to connect the circuit.
    2.     Identify the emitter, base and the collector of the transistor properly before connecting it in the circuit.
    3.     Measure the horizontal length of the inductor between the two successive peaks should be accurately measured. 
    Design:
    Given specifications,

             Vcc=12v, Fo=2khz.
    Let stability, S=6, Vre=3v, hfe=300, Vbe(sat),
    ICq=1.6mA,L1=100uH, L1/L2=hfe        
    L2=?


          Vce Q=Vcc/2=6V
          Vre=ICq.Re=1.6x10-3xRe 
    Re   =1.857kΩ
    V          Vcc=VceQ+IcQ(Rc+Re)
    12      12=6+1.6x10-3(Rc+1.875 x 103)
    R     Rc=1.875kΩ∏∏
    R        R2=S X Re=11.25kΩ
           Vcc(R2/(R1+R2)=Vre+Vbesat
            R1=25.29k Ω
            L1/L2=hfe=300
             Assume L1=100µH

            L2=0.33µHAssume fc=1kF
    CE=630KF
    Zin=R1//R2//hie
          Assume hie=1K
    Zin=884.3 Ω
    Xc=Zin/10
    Cin=1.8µF

    kindly view the word document format for the full detail of this project. thanks
    click HERE





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