[Alex Kitic]

Zitat:Original geschrieben von kahlo

Das ist nicht richtig. Ein Stromstoss auf der Sekundärseite, zum Beispiel durch einen Gleichrichter+Kondensator, stoppt den Oszillator auch. Dann entlädt sich die vergrösserte Primärkapazität über einen Transistor. Der Transistor brennt durch.

Die Schaltung ist nicht sicher.

Under which condition is this going to happen?
Are you referring to the secondary of the output transformer of the electronic transformer unit?

Anyway, the large capacity cap (large is a disputable term, maybe larger) will discharge very slowly if you do not provide some resistor or other means: this once the oscillation has stopped (for instance, turned off).

Oscillation can stop only if there is a fault condition (no-load, short circuit). If no load, the transistors are protected by flyback diodes. If short circuit, the short circuit protection wil sense the high current and cut everything down.

Part of this is explained in the text, and you can trace what happens by looking at the schematics. The rest comes from my own experience and experimenting with several units. Unless you stick your fingers inside, the circuit is safe as anything can be.

The cap, which is crucial in transforming a trivial circuit into innovative possibility, does just two potentially dangerous things:
1) Raises DC RMS voltage by eliminating 100Hz ripple - which is not a problem in general because the peak DC value remains unchanged.
Since the DC voltage defines the peak oscillation voltage by its peaks, nothing is changed in the operation of the transistors, nor the peak voltage across the primary of the output transformer of the electronic transformer unit (Volts per turn).

The increase in voltage is in RMS terms, because the dips at 100Hz intervals are "filled". What changes is the duty cycle, or rather the stability of the duty cycle (more stable since more regular).

2) Of course the capacitor may prove too large for the diodes to fill at startup, but you can add the NTC to solve that, or change the diodes for more powerful ones.

I accept that the circuit although very simple might be tough to understand, particularly the difference in voltage. The problem is thinking in terms of duty cycle instead of wave magnitude and frequency. Because, the output is not a sine wave, rather an (imperfect) square wave.