Impedance Matching Question

Its my First Sound preamp. Its a 6922 tube. The internet says its 1200 ohms.

The amp is my Found Music Blade. KT88 PP. About 35k input. Was considering a Schiit Ragnarok 2 paralleled on the woofers. The Ragnarok 2 is 10k input.

In parallel I think the pre would see about 6.7k. Kind of low? I don't know the math so much.
ok. cathode follower circuit then you get 100ohm. the problem is there is a rule of thumb for this circuit the load resistance has to be 5 times higher than the cathode resistance. to reach 100ohm it will be between ~8k - 10k ohm. but your load is also~ 8k ohm y(35k parallel to 10k. that won't work well because not enough current will flow to drive load+cable capacitance.
There is only one way out of this dilemma. You have to change the input resistance of the power amps to 100k ohms. But there is always a downside to this. By doing this you increase the possibility that interference can easier in the amp much more easily. This can be done with short, well-shielded cables, but I cannot give a guarantee.

P.S ok 1.2kohm other circut equal same procedure;)47k or 50k ohm volume controll in ragnarok2.
 
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Sorry, concerning Ron question formulated in post #1 any one not knowing about Thevenin theorem is an ignorant. This apparently simple question is complex in reality.

BTW, you can ask ChatGPT about this theorem. ;)
The Professor is back ! :) God to hear from you again Francisco, you were missed !
 
The easiest way to biamp with different amps is to adjust the volume of one amp. In this case, equip the low-impedance power amplifier with a potentiometer( 50kohm )at the input.
This solves two problems with one thing. You reduce the load on the preamp and can adjust the level precisely so that all drivers in the end are equally loud.
 
I was thinking another route. I have a Schiit Valhalla II. That has a 100k input. I could parallel the Blade and the Valhalla, then use the Schiit Aiger amp from the Valhalla to the speaker. That gives me gain control and a much higher impedance..
Just read post 24 and was wondering as such. Which amp would need the volume control. I can put the Valhalla to either amp. Would prefer it on the woofers.

Its just a preliminary experiment with biamping. Data point.
 
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I’m curious would a resistive L pad work in this situation?
If I really liked where a simple biamp worked, and I needed to attenuate one of the amps, I would think an L pad would work well. As long as I one could get the impedance where they wanted. What sucks about it all is having to add a set of interconnects.
 
I haven't been following this thread, but a quick read of the ChatGPT answer shows why humans are still around. It is generally OK but misleading in that a lot of details and worse potential pitfalls are left out. My very limited experience with ChatGPT is that it all too often provides an answer that sounds reasonable, but is incorrect in many details, and can lead the reader astray. I find similar things with Wikipedia when looking up things in my area of expertise -- answers incomplete, and sometimes misleading (wrong), but by and large Wikipedia seems more self-regulating and correct than ChatGPT. I suppose eventually AI will replace me, but since I'm retired now it will be the next generation's problem (Lord help them!)
 
The easiest way to biamp with different amps is to adjust the volume of one amp. In this case, equip the low-impedance power amplifier with a potentiometer( 50kohm )at the input.
I don't understand this. We want the attenuator on the net higher output amplifier.

This solves two problems with one thing. You reduce the load on the preamp and can adjust the level precisely so that all drivers in the end are equally loud.
I don't understand this. How does attenuating the level going into the amplifier reduce the load on the preamp? All we're doing is burning off some of the full range signal voltage; it's not reducing the load on the preamp.
 
Fortunately in this particular subject chatbots will give us an acceptable, but reductionist and misleading answer, in my opinion acceptable in an objectivist driven audio forum but unacceptable in an high-end forum where perfectionists want to extract the best individual subjective performance of their systems.

Just to point that in my opinion the accepted use of current AI chatbots in a subjective hobby such as high-end stereo will drive the hobby in mediocrity.

I suggest people ask AI chatbots what they think about "audiophile bias" :oops:

Happy New Year!
Happy New Year, Microstrip!

When it comes to calculations, I always recommend asking the chatbot to show or explain how it arrived at the results. This allows you to verify its output and ensure accuracy.

I’ve tested it on several occasions with my audio equipment—such as determining the output impedance of my MSB DAC or adjusting cartridge and phono stage settings. Each time, the chatbot provided excellent answers with clear explanations. In some cases, I even cross-checked the information with the manufacturer or dealer, and it was consistently accurate.

That said, your experience may differ from mine, so it’s always good to approach it with curiosity and a bit of caution.


.
 
I don't understand this. We want the attenuator on the net higher output amplifier.


I don't understand this. How does attenuating the level going into the amplifier reduce the load on the preamp? All we're doing is burning off some of the full range signal voltage; it's not reducing the load on the preamp.
In the case of Kingrex, both amps have a load of 8kohm on the preamp out (1.2kohm), which is a bad ratio. You solder the original 10kohm input resistor and replace it with a 50kohm potentiometer. Then these two amps create a 20.5k load for the preamp. The ratio of almost 1:20 works almost always. Then you have, for example, one power amplifier whose level you can adjust in the bass and another that supplies the high and mid-range. If you use a studio power amplifier from Yamaha, you even have 12db xover that can be adjusted from 25hz-150hz, you could even drive the bass actively. They cost used 350€ sound really good for bass( p 3500s).
20231130_140430.jpg
 
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I don't understand this. How does attenuating the level going into the amplifier reduce the load on the preamp? All we're doing is burning off some of the full range signal voltage; it's not reducing the load on the preamp.
If the IC had no capacitance and no inductance, then the only load on the preamp is the amplifier’s input impedance.
And the other DC load is the amp’s input impedance.

The L and C of the IC is generally pretty low, so the IC usually feels from the perspective of the preamp, that it is a feather of a load.
Unless those values get really large, or the cable really long… then the preamp is trying to drive a more significant load.

If you put a 1M ohm resistor on the end of the IC, then the load would be mostly the IC’s response to the AC signal (The L and the C).

It is probably worthwhile to consider doing an active XO in this case.
And one could just use a Y connector and some DSP as an active XO on the bass side. Then go from the DSP into the bass amp.
 
If the IC had no capacitance and no inductance, then the only load on the preamp is the amplifier’s input impedance.
And the other DC load is the amp’s input impedance.

The L and C of the IC is generally pretty low, so the IC usually feels from the perspective of the preamp, that it is a feather of a load.
Unless those values get really large, or the cable really long… then the preamp is trying to drive a more significant load.

If you put a 1M ohm resistor on the end of the IC, then the load would be mostly the IC’s response to the AC signal (The L and the C).
Sorry, but how is this answering the question: How does attenuating the level going into the amplifier reduce the load on the preamp?
 
Sorry, but how is this answering the question: How does attenuating the level going into the amplifier reduce the load on the preamp?
From https://www.electronics-tutorials.ws/attenuators/l-pad-attenuator.html

1735663201676.png

With a potentiometer, R1+R2 is constant, but the output tap varies along the resistor.

R1 + R2 = Rtotal
R1 = Rtotal - R2
R2 = Rtotal - R1

An L-pad uses fixed resistor values.

The output load is the amplifier input Rin and is in parallel with R2, so the effective R2 is always less than the attenuator's R2 value. If the control is set so R1 is not zero (i.e. maximum gain), then R1 is in series with and thus adds to the impedance the preamplifier (Vin) sees. You could also add another fixed resistor in series with R1 so R1 effective becomes R1 + R1series, further raising the impedance, but that will also increase the attenuation.

To use an example, consider a 50k attenuator (potentiometer, pot) so R1 + R2 = 50k-ohms. Assume the amplifier has 50k-ohm input impedance (there will be capacitance as well, dropping the impedance at higher frequencies, but ignore that for now). Now set the pot to the center, attenuating by about 6 dB (1/2 in voltage to the power amp).

That means R1 = R2 = 25k-ohms, but the amp's input puts 50k in parallel with R2 yielding 16.7k-ohms* and overall impedance of (25 + 16.7) = 41.7k-ohms, less than the power amp alone. Note the attenuation is also larger, now about 8 dB, so turning pot to get back to 6 dB will change the values a bit.

Now use a power amp with 10k-ohm input impedance (more like a SS amp) and the same 50k pot, again set to the half-way position. Then R1 = 25k as before, and R2 effective is 25k in parallel with 10k or 7.14k. Now the load the preamplifier sees is 25k + 7.14k = 32.14k-ohms, higher than the power amp alone. But the attenuation is also much greater (-13 dB) so you need to adjust the control to reach 6 dB of attenuation. After some algebra (hopefully correct, no promises), R1 = 8.1k and R2 = 41.9k (remember R1 + R2 = 50k, the overall value of the pot), providing R2 effective = 30.9k // 10k = 8.1k so we get our 6 dB of attenuation. The preamp load is now 16.2k-ohms, still higher than the amp alone.

This is why a simple answer is difficult; the impedances are all interrelated so you must know the amp's input impedance, attenuator's resistance (overall, end-to-end), and setting (attenuation, division range) to calculate the result. I have also ignored the preamp's output impedance as small enough to not significantly change the result.

HTH - Don

* For Ra in parallel with Rb the resulting resistance Rc = Ra * Rb / (Ra + Rb)
 
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Thank you very much, Don, for this extremely comprehensive explanation!

Happy New Year!
 
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Happy New Year, Microstrip!

When it comes to calculations, I always recommend asking the chatbot to show or explain how it arrived at the results. This allows you to verify its output and ensure accuracy.

I’ve tested it on several occasions with my audio equipment—such as determining the output impedance of my MSB DAC or adjusting cartridge and phono stage settings. Each time, the chatbot provided excellent answers with clear explanations. In some cases, I even cross-checked the information with the manufacturer or dealer, and it was consistently accurate.

That said, your experience may differ from mine, so it’s always good to approach it with curiosity and a bit of caution.


.

Surely if you just want to use chatbots to perform basic calculations any first year EE student can easily carry the AI chatbot will do it correctly - it is why the teaching and student evaluation paradigm is adapting to their use.

Ron question as formulated in post #1 is much more complex - it needs capacitive impedance analysis and knowledge of the Thevenin theorem, that must be considered not just from the signal point of view, but also from the audiophile perspective. Technical argumentation needs all the output and input proper details (measured, not internet guesses) and probably at best can exclude some of the propositions, not tell us what sounds better in his system. The suggested simple approach that was posted can, in effect, rule out the best sounding choice.

It is curious that although the high end is essentially a subjective hobby, influenced and supported by biases, audiophiles many times like to have the comfort of simple, basic measurements to support their biases.

In reality, behind a sometimes simple aspect, high end equipment design and system assembly can be very complex.

A simple question - would anyone put a 5 kohm resistor in series in the output of his preamplifier?

Edit - "output of his preamplfier", surely not amplifier. We are discussing preamplfier load, not amplifier load.
 
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Surely if you just want to use chatbots to perform basic calculations any first year EE student can easily carry the AI chatbot will do it correctly - it is why the teaching and student evaluation paradigm is adapting to their use.

Ron question as formulated in post #1 is much more complex - it needs capacitive impedance analysis and knowledge of the Thevenin theorem, that must be considered not just from the signal point of view, but also from the audiophile perspective. Technical argumentation needs all the output and input proper details (measured, not internet guesses) and probably at best can exclude some of the propositions, not tell us what sounds better in his system. The suggested simple approach that was posted can, in effect, rule out the best sounding choice.

It is curious that although the high end is essentially a subjective hobby, influenced and supported by biases, audiophiles many times like to have the comfort of simple, basic measurements to support their biases.

In reality, behind a sometimes simple aspect, high end equipment design and system assembly can be very complex.

A simple question - would anyone put a 5 kohm resistor in series in the output of his amplifier?
Happy New Year, Microstrip!

Thank you for your explanation—it’s clear that Ron’s question involves complex analysis and technical considerations. Your points about Thevenin’s theorem and measured details are well taken.

That said, not everyone has the expertise or interest in diving deep into such topics. For many audiophiles, tools like chatbots provide accessible insights and a good starting point, especially for simpler calculations or initial guidance. While they can’t replace hands-on experience or detailed analysis, they’re helpful for those without a technical background.

Regarding your question about placing a 5 kΩ resistor in series with an amplifier's output—this would typically be uncommon due to several reasons:
  1. Power Loss: A high-value resistor would dissipate significant power, reducing efficiency.
  2. Damping Factor Reduction: It would lower the damping factor, potentially affecting the control over speaker drivers.
  3. Signal Attenuation: The resistor would significantly attenuate the signal, impacting performance.
If someone were to consider using a 5 kΩ resistor in an amplifier's output, it would likely be for a highly specific and unconventional application.

Thanks again for your thoughtful input!
 
Happy New Year, Microstrip!

Thank you for your explanation—it’s clear that Ron’s question involves complex analysis and technical considerations. Your points about Thevenin’s theorem and measured details are well taken.

That said, not everyone has the expertise or interest in diving deep into such topics. For many audiophiles, tools like chatbots provide accessible insights and a good starting point, especially for simpler calculations or initial guidance. While they can’t replace hands-on experience or detailed analysis, they’re helpful for those without a technical background.

Regarding your question about placing a 5 kΩ resistor in series with an amplifier's output—this would typically be uncommon due to several reasons:
  1. Power Loss: A high-value resistor would dissipate significant power, reducing efficiency.
  2. Damping Factor Reduction: It would lower the damping factor, potentially affecting the control over speaker drivers.
  3. Signal Attenuation: The resistor would significantly attenuate the signal, impacting performance.
If someone were to consider using a 5 kΩ resistor in an amplifier's output, it would likely be for a highly specific and unconventional application.

Thanks again for your thoughtful input!

First, my apologies for my typo - I misspelled "preamplifier" in my post and accepted the speller correction to "amplifier" ... The question is trickier with a preamplifier.

The problem with the "initial guidance" of AI chatbots in high end matters is that they either present established science in audio - e.g. AES (Audio Engineering Society), that does not care or consider high-end - or the marketing of of the high-end industry, that we all know to systematically misrepresent science and make abusive claims.

In my opinion the too often praised Occam's razor principle in general does not apply to this hobby. And yes, I know some people disagree with my belief. :)
 
In the case of Kingrex, both amps have a load of 8kohm on the preamp out (1.2kohm), which is a bad ratio. You solder the original 10kohm input resistor and replace it with a 50kohm potentiometer. Then these two amps create a 20.5k load for the preamp. The ratio of almost 1:20 works almost always. Then you have, for example, one power amplifier whose level you can adjust in the bass and another that supplies the high and mid-range. If you use a studio power amplifier from Yamaha, you even have 12db xover that can be adjusted from 25hz-150hz, you could even drive the bass actively. They cost used 350€ sound really good for bass( p 3500s).
View attachment 142581
Yamaha P3500 is not longer available. Can you recommend another? Are all the Yamaha similar in quality.

I need to be able to cross over up to 250 to 450 on the low end. My woofers do a lot more work than most. The analog crossover with the speaker supposedly crosses up high. When you disconnect the 10" coax, you hear a lot of music from the woofers.

I can put the amp in full range and push the amp into the low end of the passive crossover. But it would be interesting to bypass the analog crossover and use the active to see how I can shape the sound.
 
Yamaha P3500 is not longer available. Can you recommend another? Are all the Yamaha similar in quality.

I need to be able to cross over up to 250 to 450 on the low end. My woofers do a lot more work than most. The analog crossover with the speaker supposedly crosses up high. When you disconnect the 10" coax, you hear a lot of music from the woofers.

I can put the amp in full range and push the amp into the low end of the passive crossover. But it would be interesting to bypass the analog crossover and use the active to see how I can shape the sound.
The newer one from yamaha have switching power supply, they have a different signature in the sound, I don't like it anymore. try to find the p3500s second-hand on ebay or other portals. the woofer can play so high, the crossover only has a 6db slope, then you can hear signals up to that range. please take a photo of the crossover. the coax plays in the baffle down to 100hz with no problem.
Exsample 1 6db/ 2/12db etc..Butterworth_Filter_Orders.svg.png
 
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There are so many things being wrong in this post that it is not possible to write an helpful corrections to it in a reasonable time. Mainly Ron's problem was wrongly formulated to ChatGPT from start .

It is known that we can't discuss with AI bots in such matters - they will omit the critical audiophile aspects and dominate with long extensive verbiage.

Properly used AI tools are great to make us more efficient and reliable - we should not expect them to turn ignorant people in experts.

Exactly, that's why I ask my grandsons ...............;)
 

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