Electronic – Understanding a high voltage generator circuit

Two notes:

• the impedance transformation is the square of its turns ratio, the voltage transformation is the turns ratio itself. Hence your 1k/8r transformer will have a voltage transformation of sqrt(125) = ~ 11. But ...

• the voltage transformation calculated this way applies to a sine wave, not to the on-off switching done in this case. Theoretically there is no limit to the voltage you can generate with this method, even on the primary (8r) side. For that reason circuits like this often use a zener diode at the primary side, to prevent damage to the transistor.

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.