Larkin's Oscillator Explained

Jim Thompson · 2010-06-13T23:19:23+00:00

Larkin's Oscillator ExplainedASCII drawing courtesy of .... Vcc = +5v --+-------+------+-- | | | | | |_ || | | _)|| .-. --- L1a _)||Rb | | C1 --- 1mH _)||100k | | 1uF | _)|| '-' | _)|| | | *| || | .-+------' || | |/ || +---| Q1 || | |>. 2n3904 || | | * || C2 --- +---------. || 1uF --- | L1b _)|| | | 25uH _)|| | | | === | === | '-------------> outputQuiet lurker kevin93 was close to the actualsolution. Contact me via my web page mailer to claim your bottle ofwine!It's really quite simple to understand.Most ALL _high_Q_ sinusoidal oscillators behave this way.Assume growing amplitude oscillation (not always a good assumption ;-)C-B junction of Q1 forward biases pulling voltage at top of C2 lower,pulling _voltage_ bias point of Q1 base lower.Since one end of L1b is tied to ground, there is less_conduction_angle_ to forward bias the Q1 emitter,When energy added (current) by the shrinking _conduction_angle_ at theemitter equals the losses in the LC (assume finite Q, otherwise you'llbe forever confused) equilibrium is reached. (So the transistor isacting as a duty cycle _switch_, NOT as AGC.)I experimented with the effects of turns ratio here:At high turns ratios some might claim class-A, but clearly the emitterCURRENT never becomes sinusoidal... it can't with the simple-mindedlow-impedance drive winding driving a non-linear bipolar junctioncharacteristic.At low turns ratios (