Indeed the power is low but only because the output tubes have been triode strapped. This can be undone and 12 Watt is available. Matching tubes is always a good idea even with seperate bias for each tube.
The preamp tube maybe overkill but has been done because it just sounded better (according to the author).
The preamp tube maybe overkill but has been done because it just sounded better (according to the author).
One of the reasons this is not viable in this situation is this: "Met een versterking van 22 kom je prima uit vanaf lijn-nivau: Plm. 1 Volt RMS in voor maximaal vermogen (potmeter tussen kwart-voor-12 en -over-12)."
The power supply is interesting with all those filters. Also demonstrates how minimal the Macintosh version of the program is, which doesn't allow adding to the supplied circuits.
I was going by my understanding of your statement. Thank you for correcting the misunderstanding. 🙂
Fine, looked into it and no data on the amplifiers characteristics are supplied on the website. Will Glassware provided this data? Inputting the three designs into the program, ignoring supplied data, would provide equal evaluation and equal comparison, correct?
Yet another question. Probably doesn't apply, just a curiosity. Reading the Morgan Jone's book as recommended and states a low pass filter starts with a resistor. However, a LC filter, with the intention of filtering out the high frequency has a capacitor first so per the book it be classified and work as a high pass filter. Which is correct?
Having the resister first would hide the condenser from the power tube allowing the capacitor to be of a higher value (leave the advantages to another discussion).
Having the resister first would hide the condenser from the power tube allowing the capacitor to be of a higher value (leave the advantages to another discussion).
Both are! High pass starts with a C, low pass with a R or L…
In case no R is used then an L has a higher resistance for the frequency it has been designed to react to.
It works then as a variable resistor having a higher value for higher frequencies. Actually wirewound resistors can behave the same way and are not always suitable in singnalpath’s unless induction free versions are used.
In case no R is used then an L has a higher resistance for the frequency it has been designed to react to.
It works then as a variable resistor having a higher value for higher frequencies. Actually wirewound resistors can behave the same way and are not always suitable in singnalpath’s unless induction free versions are used.
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Don't see how can be, then both be wrong.
So it does not matter if the inductor is followed by a capacitor or resistor? Seems then the book is incomplete.
To confirm, an inductor does not hide the capacitor from the tube?
You need Kirchhoff's Laws for AC circuits, and other basic electronic theory stuff if you want to really understand.
Also, what does it mean for an inductor to 'hide' a capacitor? The capacitor and inductor work together in a filter to pass (or make a more 'visible' connection, if you want to at it that way) depending on the frequency that's there.
Also, what does it mean for an inductor to 'hide' a capacitor? The capacitor and inductor work together in a filter to pass (or make a more 'visible' connection, if you want to at it that way) depending on the frequency that's there.
An inductor can “hide” the tube from the capacitor as it has a reactance (resistance at resonance). The resonance can be calculated but you could also copy from proven schematics ( best to use text book schematics as these use official values instead of “audiophile” ones that kill your rectifier tubes)
Kirchhoff's Laws have nothing to do with filtering as you have taught me. Additionally know this because the book does not use it in explaining filtering.
The tube can handle no more than 60μF before the capacitor becomes detrimental to the tube, one method for resolving this is to hide the capacitor behind a resistor, then the tube only sees the resistor.
Ah, so then it would not hide it in the power supply filter, as after the voltage has been converted from AC to DC by the tube.
So then with the inductor followed by resistor, this be a low pass filter, correct?
If so, what happens if the power supply has both a low pass and high pass filters?
Not so, the dc out of a rectifier is far from dc! It is part dc part ac as there always is a ripple. This ripple is key here. The load (amplifier in your case), the voltage and the first c determain the ripple before the L. To get the lowest ripple after the L it needs to be in resonance. Again, this is in most cases an estimate and need not to be exact.
I forgot... Been unwell this morning, unable to eat, so blood sugar is super low. Maybe based it on the PSDU results.
I just remembered saw in this morning's emails that USPS has delivered a book on power supply by same author as the recommended pre amplifier book. First, see if can eat.
A high pass filter in a power supply makes no sence as dc is the lowest frequency (and that is what you want). Removing dc from a dc power supply means no power at all …
To make things complicated :
We coul use ac power supplies instead but then this needs to be far away from the audio spectrum. In this case no filtering is required at all.
To make things complicated :
We coul use ac power supplies instead but then this needs to be far away from the audio spectrum. In this case no filtering is required at all.
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Then what is the purpose of the rectifier tube or bridge rectifier supplying B+? I have not seen an amplifier design without it.
There is an article I am thinking of, however, now realizing my text file with links is severely deficient in an ability to quickly find a link and will take hours and hours to find it again. No time now, got to fix lunch. Wouldn't it be nice if we didn't have to eat? We could save so much time. A compromise be a spouse. 😛 (Being silly.)
In an AC power supply there is no need for capacitors but a rectifier of some sort is still required for efficiency as most amps need a positive and/or negative voltage to work properly. Now a tube doesn’t care if the positive voltage is ac as long as the pulses are positive it will conduct the moment the voltage is high. If this is often enough a “steady” current flow is obtained. The audio signal will be chopped naturally because the current only flows in case the voltages are high. But this is of no consequence if the frequency is in the radio frequency range.
But all this is totally out if topic i guess.
More can found in this thread if you’re interested:
https://www.diyaudio.com/community/threads/zotl-and-ferrites.73135/
But all this is totally out if topic i guess.
More can found in this thread if you’re interested:
https://www.diyaudio.com/community/threads/zotl-and-ferrites.73135/
This does not make sense designers using a rectifier (bridge or tube) solely for voltage drop, be better to select a transformer with the desired B+ voltage.
I swear read that the input and output tubes' cathodes needed to be supplied with DC, not AC. Interesting they do not and does not affect the AC signal, would have thought the 50 or 60Hz of the mains would add into the signal. To take it further, then makes no sense why some amplifiers supply the filaments with DC, since the mains frequency isn't affecting the signal. Oh and EMF from the transformers then have no impact. I am clearly stunned by your statement.
Turns out the book has not arrived.
Not for voltage drop. Ac voltage without a rectifier will feed the amp also with negative voltages. This can get an issue in case the limit of the tube has reached ( max negative anode voltage rating).
Therefore providing only positive halves of the ac will be needed, hence the rectifier.
Therefore providing only positive halves of the ac will be needed, hence the rectifier.
So since ripple does not affect the signal, per your previous statement, then why are chokes and filters used?