If we let R1 = R2 = R and C1 = C2 = C which is usually the case for this type of oscillator (same values). We can see that C1, C2, R1 and R2 form a potential diver.

The impedance of the top part R2 in series with C2 is:

$$
Z_{top} = R + \frac{1}{j \omega C} = \frac{1+j\omega C R}{j\omega C}
$$

The impedance of the bottom part R1 in parallel with C1 is:

$$
Z_{bot} = \frac{R \frac{1}{j \omega C}}{R + \frac{1}{j \omega C}} = \frac{R}{1 + j \omega C R}
$$

So the gain of this potential divider is:

$$\begin{align}
G & = \frac{Z_{bot}}{Z_{top}+Z_{bot}} \\
& = \frac{\frac{R}{1 + j \omega C R}}{\frac{1+j\omega C R}{j\omega C}+\frac{R}{1 + j \omega C R}} \\
& = \frac{\frac{R}{1 + j \omega C R}}{\frac{(1 + j \omega C R)^2 + j \omega C R}{j \omega C(1+j\omega C R)}} \\
& = \frac{j \omega C R}{1 + 3j \omega C R -\omega^2 C^2 R^2} \\
& = \frac{\omega C R}{-j(1 + 3j \omega C R -\omega^2 C^2 R^2)}\\
& = \frac{\omega C R}{(\omega^2 C^2 R^2 -1)j + 3 \omega C R}
\end{align}$$

If we chose a frequency such that the imaginary (j) part is zero then the Gain G is 1/3. Now for an oscillator to work it needs a gain of exactly one so the gain of the amplifier needs to be 3 since 3 times 1/3 = 1. Typically there is a non linear element such as a PTC or bulb in the feedback path R6, R3 to provide a simple auto gain control to ensure this.

The frequency when this will oscillate is:

$$
\omega^2 C^2 R^2 - 1 = 0 \Rightarrow \omega = \frac{1}{C R} \Rightarrow f = \frac{1}{2 \pi C R}
$$

Finally consider what would happen if the output were exactly zero volt. All inputs to the opamp would be zero and the circuit would never start. However there is always some noise in the system to start it going.

**Edit:**

If you are simulating this Remove R4 and change R3 to 20k. This should make the gain exactly 3 as required. You will also need to add an initial charge to either C1 or C2 to get it started otherwise the inputs will be at 0V and it wont start.

If you are testing with real hardware you will need to play with the setting of the pot to much or too little gain and it wont work.

## Best Answer

Ignore (for now) the MPSA18 and the two zener diodes and 10M resistor - they are used to control the amplitude of the DC output once the circuit is producing high voltage: -

Upon power being applied the 10uF capacitor has a charging voltage (in red) rising from 0V and after a short while this voltage will cause the base-emitter of the 2N4403 to conduct which rapidly causes the MPSA42 to switch on via the 1k8 resistor.

The MPSA42 will switch on and immediately start dischrging the 10uF capacitor via the 1k and the 1N4007. Shortly afterwards the MPSA42 will turn off because the 2N4403 turns off due to the cap discharging.

The current that was flowing in the transformer primary has stored energy in its magnetic field and this is harvested by the secondary circuit which presumably has a higher turns ratio than the primary.

And the process starts again - MPSA42 turns on - energy is stored in magnetic field and discharges to sencondary when MPSA42 switches off.

The MPSA18 will start to conduct when approximately 240V is on the output and this will start to turn-off the 2N4403 more making the 10uF take longer to charge thus the duty cycle is reduced.

It looks to me like there will be a fixed period of a few microseconds during which the MPSA42 will conduct and an ever-increasing period where it is switched off as the output DC level gets to about 240V DC. That makes sense.