[Alex Kitic]

Zitat:Original geschrieben von Rumgucker

Zitat:Original geschrieben von Alex Kitic
You have made mistakes in setting the simulation, 100uF leaves a lot of ripple, which is visible at the output... etc.

Mal ist der Kondensator zu groß. Dann ist er zu klein. Wie auch immer. Ich kann ihn dimensionieren wie ich will: es kommen keine 200Hz raus

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Erst die RMS-Geschichte. Und nun das 200Hz-Desaster. Das war offensichtlich zu viel.

Warum sind viele Röhrenfreaks nur so seltsam gnaddelige und unhumorige Menschen?

This time I am replying only because I am getting angry:
1) Do a simulation for yourself and see how much ripple will remain with a 100uF, 200uF, or 330uF cap after the rectifier - do it in PSUD2 if you cannot set your software to show you the ripple. Of course, set a fixed current sink, not a resistor which will change the current in the circuit changing the resulting ripple.
2) Now that you have done 1) you have probably understood that what ripple remains in the DC signal, will be passed to the output of the electronic transformer. More, or less. This is what you call "100Hz modulated high frequency AC", but a more precise definition is that the let's say 50kHz AC has lower or higher peak values due to the 100Hz ripple that changes the DC value and thus the output voltage. This is most prominent with the unmodified electronic transformer, where DC gets so low that the diac is needed to restart oscillation.
3) The true result of your simulation, besides the "100Hz" modulation which is visible in the result, should be 100kHz because the input before the full wave doubler is 50KHz. As a matter of fact, such high frequency is perfectly filtered with the caps you are using for the doubler, and thus it is hard to see it even if you were to increase very much the resolution of the displayed result.
4) This is why you are getting 100Hz - the initial "ripple". If you had filtered it enough, it would not be present, and what you would get is DC with 100kHz ripple. To get to that, the best way would be to replace the initial cap with a voltage regulator (as I said several times so far) that would add -60dB of ripple rejection to the already good result with a 200-330uF cap.
And yes, the 10000uF cap is surreal - too large, too high initial peak current, etc. The 100uF cap is too low, thus too much ripple remains and is visible in your simulation. Try a nice regulator, and to grasp the idea, try a DC source instead of the first rectification. To be able to see the 100kHz ripple you will need to keep a very low capacity in your doubler, otherwise it will be filtered out.

Now for the 200Hz you are mentioning:
If you use 100Hz AC with a full wave rectifier you will get 200Hz ripple. But since this is not 100Hz AC, rather "modulation" of HF AC at 100Hz, you will not get the 200Hz but the 100Hz modulation will be the only visible result, among other reasons because the higher the frequency, the less capacity is needed to filter it... thus leaving the 100Hz visible and effectively filtering out the 100kHz (in this case). If it looks the same to you, it is because you do not know where to look.

On the other hand, if we take a 100Hz signal, even a modulation, and use it to heat a direct heated cathode, and employ a so called hum-dinger pot, it will cause the main frequency (100Hz modulation) to null, while the remaining 2nd harmonics will be obviously 200Hz and will be audible in your system as a higher than usual frequency hum (more alike to buzz than hum, actually).

The above has nothing to do with tube-freaks or similar stuff. It has to do with whether you use your brain in advance, and whether you have some electronics traning or not (I am an economist, BTW) - and whether the only reason why you have a scope is being a repairs technician, or because you can afford it...

Enough said. If after this you still do not understand, feel free to make jokes about it: joking about stuff you do not understand is good for you, it will alleviate the stress of being challenged...