Thanks for clarifying and for correcting me on which transformer you're using (LL1660 vs. LL1660s).
When you wire up the LL1660 according to Alt+S, the primary inductance is 130H, but the rated primary current is only 10mA.
According to the schematic, the 6C45P is drawing 12.9mA plate current (1.6V/124 ohms = 0.0129A).
If looking at it by the book, you're drawing too much current through the LL1660 primary in that configuration. That risks rapidly decreasing primary inductance as the limits of the transformer are reached and the core goes into saturation.
I may be wrong, but this looks pretty marginal.
Anybody else see that? Am I wrong?
47 uF or 68 uF could be interesting. Values of a few hundred microfarad lead to relatively poorly damped resonances.
Regarding analysing it as a damped LC parallel circuit: to get reasonably damped poles (say, Q not much above √2/2), you either have to make the decoupling capacitor so large that the LC circuit is damped adequately by the series resistance ri/(mu + 1) of the inductance you see at the cathode, or so small that it is damped adequately by the parallel resistor Rk. For the first option, 1000 uF is already on the low side, for the second 47 uF or 68 uF works well.
That is, I calculated the characteristic polynomial assuming that the primary inductance of the interstage transformer is 100 H due to the somewhat too large current (wild guess), that the voltage gain of the valve is 52 (datasheet value) and that the internal resistance is either 1684.91525 ohm (extrapolated from datasheet value using Child's law) or 2500 ohm (value from post #1). For 47 uF, I find either fn ~= 18.94389481 Hz and Q ~= 0.6316674341, or fn ~= 19.85691976 Hz and Q ~= 0.6346589069.
With 47 uF and 124 ohm, the second zero lies at -171.5854496 rad/s, corner frequency of the zero 27.30867246 Hz. If the corner frequency of the pole of the output transformer lies at roughly this frequency, you get a flatter response with 47 uF than with a large capacitance.
With 68 uF, the values are either fn ~= 15.74939031 Hz and Q ~= 0.7306017495, or fn ~= 16.33365341 Hz and Q ~= 0.7300749122. The corner frequency of the zero becomes 18.87511185 Hz.
With 1000 uF, I find either fn ~= 4.106938529 Hz and Q ~= 1.035762987, or fn ~= 4.30487762 Hz and Q ~= 0.8180474716. The second zero lies at -8.064516129 rad/s, corner frequency of the zero 1.283507606 Hz.
Regarding analysing it as a damped LC parallel circuit: to get reasonably damped poles (say, Q not much above √2/2), you either have to make the decoupling capacitor so large that the LC circuit is damped adequately by the series resistance ri/(mu + 1) of the inductance you see at the cathode, or so small that it is damped adequately by the parallel resistor Rk. For the first option, 1000 uF is already on the low side, for the second 47 uF or 68 uF works well.
That is, I calculated the characteristic polynomial assuming that the primary inductance of the interstage transformer is 100 H due to the somewhat too large current (wild guess), that the voltage gain of the valve is 52 (datasheet value) and that the internal resistance is either 1684.91525 ohm (extrapolated from datasheet value using Child's law) or 2500 ohm (value from post #1). For 47 uF, I find either fn ~= 18.94389481 Hz and Q ~= 0.6316674341, or fn ~= 19.85691976 Hz and Q ~= 0.6346589069.
With 47 uF and 124 ohm, the second zero lies at -171.5854496 rad/s, corner frequency of the zero 27.30867246 Hz. If the corner frequency of the pole of the output transformer lies at roughly this frequency, you get a flatter response with 47 uF than with a large capacitance.
With 68 uF, the values are either fn ~= 15.74939031 Hz and Q ~= 0.7306017495, or fn ~= 16.33365341 Hz and Q ~= 0.7300749122. The corner frequency of the zero becomes 18.87511185 Hz.
With 1000 uF, I find either fn ~= 4.106938529 Hz and Q ~= 1.035762987, or fn ~= 4.30487762 Hz and Q ~= 0.8180474716. The second zero lies at -8.064516129 rad/s, corner frequency of the zero 1.283507606 Hz.
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By the way, if it's of any use to anyone, I'll gladly write out the derivation in a pdf file with schematics and equations and write a corresponding Excel-compatible spreadsheet. If not, then I won't bother.
I would love to learn more about this, and as I said I can compare to a measurement when I get the chance to take down the amp off the shelf which is a little too heavy for 1 person to be totally confident it will not end in a regret, those output transformers are heavy!
Attachments
The spreadsheet doesn't exist yet, but attached is the derivation.
I've written the spreadsheet, extended the report with a couple of plots made with the spreadsheet and put them both in the attached zip file. The spreadsheet is in Excel format, but it was actually made with LibreOffice Calc.
Isn't that a large bump up in response between about 50Hz to 120Hz? Is that a desired effect?
yeah it does look a little, I will try with an 8R load resistor and see how this looks, not sure of the impact of the speakers. I was really only looking at the roll off at low frequencies. There is also somethign strange going on at higher freuencies.
Thanks
Thanks
This is excellent and very educational for me - thank you very much.
A few quick questions:
- For the Auxiliary high pass coming from the frequency response of the transformer itself, in ALT S. configuration the quote is 25Hz, and I wonder why the plots are with and without this, when would the without be possible - is this assuming a different model of transformer that has a much more extended frequency response?
- Can you explain more of the logic to use 100H instead of 130H, I am not sure I follow this.
Rich
Hi Rich,
I misinterpreted your 25 Hz. I thought there was a 25 Hz high-pass somewhere else, in the output stage for example. The interstage transformer is included in the response without auxiliary high-pass.
Regarding the 100 H, this relates to rongon's post #41, https://www.diyaudio.com/community/threads/cathode-bypass-calculator-help.421677/post-7883950 Rongon observed that the DC current through the transformer's primary is above spec. This will reduce its inductance, but I don't know by how much - so that 100 H is just a wild guess.
Best regards,
Marcel
Edit: in the LL1660 datasheet, there is a 25 Hz to 40 kHz +/- 1 dB frequency response specified for connection Alt S when driven from a 14 kohm resistive source. Your source impedance is much lower, which reduces the lower cut-off frequency - assuming the inductance doesn't drop too much due to the higher DC current.
I misinterpreted your 25 Hz. I thought there was a 25 Hz high-pass somewhere else, in the output stage for example. The interstage transformer is included in the response without auxiliary high-pass.
Regarding the 100 H, this relates to rongon's post #41, https://www.diyaudio.com/community/threads/cathode-bypass-calculator-help.421677/post-7883950 Rongon observed that the DC current through the transformer's primary is above spec. This will reduce its inductance, but I don't know by how much - so that 100 H is just a wild guess.
Best regards,
Marcel
Edit: in the LL1660 datasheet, there is a 25 Hz to 40 kHz +/- 1 dB frequency response specified for connection Alt S when driven from a 14 kohm resistive source. Your source impedance is much lower, which reduces the lower cut-off frequency - assuming the inductance doesn't drop too much due to the higher DC current.
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Marcel,
Thanks again - so I can just put 0 in the calculator for corner frequency for my application (or ignore the Red plot?)
Happy New Year!
Rich
Thanks again - so I can just put 0 in the calculator for corner frequency for my application (or ignore the Red plot?)
Happy New Year!
Rich
Hi Rich,
Indeed. You do get a bump in the response at 47 uF then, due to the zero that is at the wrong place.
When you fill in 0, the calculation for the red plot fails (division by 0 somewhere). In LibreOffice Calc, the red curve then disappears from the graph.
Happy New Year to you and everyone too!
Marcel
Indeed. You do get a bump in the response at 47 uF then, due to the zero that is at the wrong place.
When you fill in 0, the calculation for the red plot fails (division by 0 somewhere). In LibreOffice Calc, the red curve then disappears from the graph.
Happy New Year to you and everyone too!
Marcel
If you apply q.s. to tube audio design, you'll enjoy the hobby so much more. Tube audio is built for fudging with simple elements and still make beautiful music. The pursuit of perfection is never satisfied, tube tube audio still manages to surprise with imperfect design. Simplify, minimize and enjoy.
Quantum satis is Latin and a term used to quantify amounts that are just sufficient to do the job and no more. Approach designs with q.s. then add some headroom for reliability where that can make sense, like power consumption, and sleep well.
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The design was made for me by a friend who 'finished' my part build Ongaku copy, by completely changing the design to mimic/echo one of his commercial units. The circuit and choice of components were all his 30 years ago, with the use of my originally sourced Oil filled chokes. I had a busy life at that time and was studying at the Royal College of Art and my brother sympathised with my lack of time, and lack of cash and paid our friend to finally finish the build. It has stayed like this ever since and I am looking to change the component choices and quality but not to redesign the whole thing.
I really appreciate this forum for all the help and learning I have had over the years since I started way back with a preamplifier humm that no one I took it to could fix, DiyAudio coming to the rescue 🙂 THANKS
😀
I really appreciate this forum for all the help and learning I have had over the years since I started way back with a preamplifier humm that no one I took it to could fix, DiyAudio coming to the rescue 🙂 THANKS
😀