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What is the use of R2 in this oscillation circuit?

Discussion in 'Electronic Basics' started by [email protected], Apr 29, 2005.

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  1. Guest

    I have electronics hobby and just start learning how oscillator works,
    like the loop gain must be >= 1 and the phase between the input and
    output should be 0 degree. The problem is, when I see the actual (even
    simple) oscillation circuit, I am at a loss how it exactly works.

    "http://img13.imagevenue.com/loc70/a2a_osc123.jpg"

    For example, in Fig.1, what's the use of resistor R2, from oscillation
    point of view (not from transistor DC basing point of view)? I figure
    that, since C3 is basically to short AC (i.e. RF in this case), it
    seems to be equivalent to place one end of the crystal directly to the
    emitter, as shown in Fig.2. Unfortunately, it does not oscillate this
    way. Likewise, since C3 is to short the AC, then it should also be OK
    to remove R2 and C3 completely, and place the crystal in a way shown in
    Fig.3. However, it does not oscillate, either.

    (1) So what is the function of R2 here?
    (2) If it is a must for the oscillator to work, how to choose it?
    (3) How to choose the value of C3?

    Your help is appreciated!
     
  2. colin

    colin Guest

    the circuit in fig1 works by feeding back the signal at the emiter to the
    base through the emiter base capacitance, a tune circuit is formed between
    the crystal wich will be slightly inductive and the emiter capacitance in
    series with c3 wich should be slightly higher than the load capacitance
    specified for the crystal, the resultant signal at the base gives positive
    feedback.

    if you short out the emiter to ground you remove the signal, also the dc
    curent would no longer be controled exept by the hfe.

    Colin =^.^=
     
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