Larkin's Oscillator Explained
ASCII drawing courtesy of snipped-for-privacy@yahoo.com ....
Vcc = +5v --+-------+------+-- | | | | | |_ || | | _)|| .-. --- L1a _)|| Rb | | C1 --- 1mH _)||
100k | | 1uF | _)|| '-' | _)|| | | *| || | .-+------' || | |/ || +---| Q1 || | |>. 2n3904 || | | * || C2 --- +---------. || 1uF --- | L1b _)|| | | 25uH _)|| | | | === | === | '-------------> output
Quiet lurker kevin93 was close to the actual solution. Contact me via my web page mailer to claim your bottle of wine!
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 the emitter equals the losses in the LC (assume finite Q, otherwise you'll be forever confused) equilibrium is reached. (So the transistor is acting 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 emitter CURRENT never becomes sinusoidal... it can't with the simple-minded low-impedance drive winding driving a non-linear bipolar junction characteristic.
At low turns ratios (