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This circuit is a high voltage generator using flyback transformer.

It works well, but i thought it shouldn't working by this polarity of the feedback winding !

I mean when the voltage on the primary side increasing(before saturation) the voltage of the feedback coil should be minus to force the transistor switch OFF.

But by the dot convention that shown in the circuit the transistor should be stay ON for ever (because there are no 180 degree phase shift between primary and feedback winding).

So how this circuit can works by this feedback winding polarity?

enter image description here

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But by the dot convention that shown in the circuit the transistor should be stay ON for ever (because there are no 180 degree phase shift between primary and feedback winding).

The idea behind this is to turn on and off the transistor harder. For example, when the transistor starts to turn off, the feedback and the series capacitor guarantees a harder and complete turn-off.

So how this circuit can works by this feedback winding polarity?

When the power is first applied, the base voltage will rise in a time interval defined by the Thevenin equivalent of the divider/delay circuit formed by 47k/3k9/47n. The feedback winding will accelerate the turn-on of the transistor. Since the base current is limited by the resistors, the collector current cannot rise above a certain level. If it rises more then the transistor will be out of saturation. This causes the collector voltage to increase, and thus the primary voltage to decrease. By transformer action, the feedback winding voltage will decrease and this will accelerate the turn-off of the transistor. You probably know the rest of the operation.

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