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 _)|| | | | === | === | '-------------> outputQuiet 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 (