L1, C1 and C2 are the tank. Capacitors in series add as resistors in parallel so I have a 500pfd cap a 100uh coil forming my tank. This will give a resonant circuit at about 712Khz.

I take a signal from the collector and feed the tank. Then send feedback to the base. That is C3 and C4s jobs.

Now I power the circuit and it sings. It is producing a 8 volt pk-pk signal. Lowering the value of C3 and C4 will produce a lower output. Lowering C3 will reduce the feed to the tank. Lowering C4 will produce less feedback to the amplifier input. In order to have oscillation we need a gain higher than 1. It simply has to overcome the circuit losses. If we had a load on the tank we would need more feedback to replace the signal drawn by the load.

The capacitor divider divides the AC voltage the same as resistors will DC.

You should have no trouble seeing the resistors as a voltage divider (for a DC circuit)

The capacitors provide the same function in an AC circuit. The reactance is set by the frequency. The series circuit has the same frequency so we can use the ratio of capacitance and do the division. Xc = 1 / (2 * pi * f * c) so the larger c has the smaller drop.

Vc1 will be greater than Vc2. With resistors we use a formula such as Vr1 = Ea * (R1 / (R1 + R2)).

With parallel resistors we use:

Rt = 1 / (1/r1 + 1/r2 .......) When we do this we are applying Ohm's law and assuming a voltage of 1 volt.

I=E/R Ir1=1volt/r1 Ir2=1volt/r2 It=Ir1+Ir2 R=E/I

Rt=1volt/It. See it is Ohm's law assuming 1 volt applied.

You can solve the capacitor circuit using the same approach BUT remember capacitor in series add as resistor in parallel so use the conductance to solve it.

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