An RLC op amp oscillator

Discussion in 'Homework Help' started by samy moenis, Feb 10, 2010.

  1. samy moenis

    Thread Starter New Member

    Feb 10, 2010
    [​IMG][​IMG] circuit.jpg

    given : (r2/r1 ) = (r4/r3 )

    c = 1nf , F= 2 Mhz

    Required : output voltage and oscillation frequency at 3 cases

    a- (r2/r1 )= (r4/r3 ) b- (r2/r1 )> (r4/r3 )

    c-(r2/r1 )< (r4/r3 )

    intial conditions : I (coil )= Io (i.e. doesnot equal 0 )

    Vcap = 0

    My attempts so far :

    1- i deduced the L that satisfies parallel resonance between the parallel R-L-C , such that L= 6.33uH and at resonance the Z ( total ) = L /C

    2-Since the RLC are at resonance , then the intial current Io travels in the parallel RC giving a rise to the cap voltage.

    3- since the cap voltage equals the V+ then V+ increases

    what i failed to understand is that :

    from the voltage division
    V- = Vo ( r1 / r1+ r2 )
    V+ = Vo ( Z / Z+ r4 ) where z is the parallel RLC ; Z= L/C

    so when Vc increases V+ increases and so does V- , this doesnt seem logical and i can't figure out what im missing out here

    Thx in advance for your help:)
  2. mik3

    Senior Member

    Feb 4, 2008
    Why do you think it is not logical?
  3. samy moenis

    Thread Starter New Member

    Feb 10, 2010
    hi mik3

    because its supposed to be an oscillator , Vo should change .

    If V+ and V- and Vo increase together , Vo would reach Vcc and then stabilzes
  4. mik3

    Senior Member

    Feb 4, 2008
    When Vo will be at its peak value, the inductor will be discharged and the capacitor charged. Then the capacitor will discharge and V+ will reduce again. The capacitor will discharge, V+ will be at its lower value and the charged inductor will discharge again chatging the capacitor and risisng V+. This happens continuously.
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    What you have here is a "negative impedance converter" (NIC).

    The impedance looking into the plus input of the opamp needs to be slightly smaller in magnitude (and negative in sign) so that the parallel combination of the impedance presented by the NIC and R3 is a net negative impedance. When it's sufficient to overcome the losses of the inductor (and capacitor), the circuit will oscillate.

    If the inductor is fairly good quailty, it won't take much to sustain oscillations. It would be reasonable to assume that when the negative input impedance of the NIC is equal to R3, that will be the boundary between oscillations and no oscillations.

    If the negative impedance of the NIC overwhelms the positive impedance of R3 by a large margin, the oscillations will be quite non-sinusoidal; the opamp output will be driven to the rails.