Considering the circuit below.:
What is the purpose of capacitor C5? I know it provides positive feedback to sustain the LC oscillations, but how does this work ?
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The C5 capacitor across the transistor serves to keep the tank circuit vibrating. In theory, as long as there is a supply voltage across the parallel inductor and variable capacitor, it should vibrate at the resonant frequency indefinitely. In reality however, the frequency decays due to heating losses. C5 is used to prevent decay.
For a circuit to oscillate it is necessary to have a LOOP GAIN of unity (Barkhausen criterion). Hence, the amount of output voltage that is fed back to the input of the amplifying unit depends on the gain (magnitude and phase!). In the present case, only a part of the output voltage is created across the impedance seen at the emitter node because of the voltage division rule. This emitter node impedance is \$re=r6||(1/gm)\$ with \$gm=transconductance\$.
For the circuit shown, the exact calculation of loop gain is rather involved since the feedback circuit (C5-re) acts as an additional load for the transistor and, hence, must be considered for gain calculation (in parallel to the LC tank). Something similar applies to the phase. Because the feedback path (high-pass C5-re) creates a certain phase shift (the capacitor C5 CANNOT be regarded as a short for "high frequencies") the resulting oscillation frequency will NOT be identical to the resonant frequency of the LC tank. Instead, there will be a small frequency shift.
If the base is kept fixed and the emitter voltage is reduced, the transistor sees a higher voltage between the base and emitter and it is turned ON harder. If the voltage on the emitter increases, the transistor turns OFF as the difference between the two is reduced. This is exactly what happens in this circuit. The 5pF capacitor between the collector and emitter influences the voltage on the emitter to turn the transistor on and off. It does this by constantly monitoring the voltage on the tuned circuit and passing the change to the emitter. The above fact combined with the fact that the capacitor has an impedance of about 600 ohms at 50 MHz, and so only a part of the output is fed back to the input. I think this answer gives me closure. Is my understanding right, even if it is really basic ?