Hi Gyula ,
The data of the tests you ask . In the impedance test i use the value i have available , i don't have many parts now 
but i try find something to try aproach to 100V
Negative probe in Emitter and positive prove in the base
Shorted DC = -0,96V AC =3,21V
Unload AC = -0,99V AC =3,4V
With bulb connected DC = -1,2V AC =3,27V
With a Resistor 33K =92V in output
Hi Nelson,
Thank you for doing these tests. The loading effect of the 33 kOhm resistor that dropped the 203 V DC output to 92 V indicates the possible output resistance of your oscillator is likely in the 35-40 kOhm range.
So the output is not really a 200 V voltage source but rather a current source with that much kOhm internal resistance. This can influence (slow down) for instance the charge up of an electrolytic capacitor from the 200V oscillator output.
Regarding the DC and AC voltages measured between the base and emitter by a DMM indicates the oscillator transistor operates as a switch rather than a linear amplifier I think, more exact deduction can be made by evaluating the waveform across the base-emitter when it will be available.
Here I attempt to give a possible operation principle for this oscillator as I see it.
Case 1: the output is unloaded, there is a certain amount of oscillating power in the circuit, established by the DC input power. Once the circuit oscillates, the impedances the transistor is embedded into are established by voltage and current levels
In the followings I refer to component labels used in your original schematic what I attached below. I edited your original schematic too and attached it also below in which I erased the transistor and the push button to show the coils and capacitors that can mainly influence the operating frequency.
It is possible that the on-off switching of the transistor (if it works as a switch) also changes the frequency when the output is loaded, this can be checked by a scope or a frequency meter, albeit this change may be small.
Case 2: the output is short circuited, the AC impedance of transformer T1 between its pins 2 and 3 decreases significantly to a low value (the short is transformed backwards of course). This low impedance surely changes the operating point of the transistor, so its collector current too. Without seeing the changing waveforms on a scope, I can only assume that this impedance change may reduce the on time of the transistor and the effect of this manifests in a lower input current. Reduced on time involves higher off time hence higher output impedance. And the conditions for oscillation is still maintained with the low impedance transformed back into the circuit.
Your oscillator includes the D1 and D2 diodes and (as I suspected and now your base-emitter voltage data confirmed) these participate in controlling the base-emitter bias voltage i.e. establish the operating point of Q1 after the start up.
The diodes partially rectify the AC oscillating voltage for the base-emitter.
When you push the start button, the 24V positive voltage drives forward current into the base-emitter via the E1 bulb and this current is limited mainly by the 300 Ohm base resistor and the cold resistance of the bulb only. So the initial base-emitter current is 24V / 300 Ohm = 80 mA or
so, neglecting the 13 Ohm bulb and the 2.6 Ohm L1 resistances.
The average DC voltage across the base-emitter is a negative value (by your measurements) which may keep the transistor off but whenever a positive oscillating wave appears on the base, it surely overdrives the negative bias and switches the transistor on again, till the positive waveform lasts.
Case 3: the output is loaded by say the 12 V, 1 W rated bulb, then its (nonlinear) resistance is transformed back to the oscillator via T1, and surely establishes yet another operating point for the transistor.
Let me notice that the changing load conditions (the short or the 12V bulb) across the output forces the oscillator to work less 'actively', on this I mean: the ability to operate under restricted conditions becomes narrower.
I am sure I have not covered all the small operational details in this oscillator, for instance the positive flyback pulse of L1 is steered to the base via C4 to defeat the negative bias, this process substitues the start-up push button.
You have surely noticed in this post that I did not refer to negative resistance in the transistor as it was addressed already above in this thread. The reason is if the operation is really of switching nature, then this is out of question.
And if the transistor is full on all the time and never off, then it works as linear or nearly linear amplifier (many active devices in oscillators work like that), so the base-emitter or the collector-emitter 'junctions' cannot receive excessive reverse bias that would be needed to create a negative resistance region. This question can also be answered after evaluating the waveforms between said electrodes.
Greetings,
Gyula
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