Modified Clapp Oscillator

This is a wireless electricity transfer circuit utilizing the following components.

1 Transistor – BD139

C1 = 470pFC2 = 1nFC3 and C5 = 4.7nF

C4 and C6 = 470nF

( and the coil is made out of a loop of wires )

I just don’t understand how the DC supply voltage becomes AC voltage through the coil.

Also what are some calculations which can be applied to this circuit

Answer

There are 3 main type of oscillators that can be constructed with one transistor: The Hartley oscillator, the Colpitts oscillator and the Clapp oscillator. The oscillator in your figure is a Clapp oscillator.

The image below is the simplest Clapp oscillator taken from wiki

The capacitor $C_0$ above is $C_1$ parallel $R_1$ in your figure, the capacitor $C_1$ above is $C_2$ in your figure and the capacitor $C_2$ above is ($C_3$ series $C_4$) in your figure.

Description of the above figure way of oscillating:

There is a back reaction from the collector (drain) to the base (gate) introduced by the coil (series C0) through the voltage source. A part of this feedback is also taken into the emitter (by the midway point of $C_2$ series $C_1$). This is done for stability of oscillations. When a small perturbation occurs in the base (gate) it is amplified by the transistor, and found in the collector. By the feedback circuit ($L$ series $C_0$) this amplified perturbation is retransmitted back into the base and further amplified. The process repeats itself. Thus oscillations occur.

The frequency of oscillations for the above figure is

$F_0 =1/(2*pi)*sqrt{(1/L*(1/C_0 +1/C_1+1/C_2))}$

(see http://en.wikipedia.org/wiki/Clapp_oscillator)

At the reception an AC voltage is induced in the coil $L_2$ and this is turned into DC voltage (rectified) by the diode $D_1$. $C_5$ is for making an LC oscillator parallel circuit tune on the emission frequency of the first oscillator (from the left figure) and $C_6$ is for filtering the DC voltage.