Hello everyone, long time lurker with a simple question that doesn't get asked or phrased explicitly for two reasons, one being that most often the folks discussing circuits already understand the basics, and second because there are often hardware/device limitations keeping the rp value within a limited range....
So, how is the ideal or intended signal impedance to an output stage (or really between any two stages) determined? I see that often alot of effort is taken to reduce a drive stage's rp value, to a certain degree, including through very expensive means (interstage transformers). But is lower always better? How low, exactly? Are there negative consequences for dropping below a certain value, or is it just a matter of diminishing returns?
If it seems like I'm being obtuse - yes, that is the case. I've got more of an "English major" brain than an "engineering major" brain.
Let's take some specific examples? I don't know if that is necessary but.... I see some very expensive devices that will use interstage transformers to reduce a drive stage's rp from say, 3k ohm, to just 750 ohms, to drive let's say a single-ended 2a3. I see high-dollar line-stage seperates bragging about a 250 ohm output. I see that tubes like 2a3's and 300b's in a single ended arrangement seem to be more demanding in this regard, and with certain pentodes it is treated as nearly irrelevant in the circuit. So:
How is this knowledge, this number or range of values, derived from a data sheet?
Is lower always better? I.E. if it were possible to achieve let's say, 40 ohms in a way similar to OTL output stages, would there be any advantage in doing so? Would there be detriment?
Basically, how do we know that, for a 2a3, x is better than y when x=(y/2), but z is no better than x when z=(x/2)? Caveman math, I know. I can't get through any of these engineering books & I swear I'm not low iq....
My reason for asking is, in plunking around somewhat blindly with circuits, I got some "good sound" from switching driver tubes for those with lower rp while playing around on breadboard. Basically, how far should I chase that rabbit? For a single ended 2a3 or 45 or 300b (the three I really want to spend time listening to) what is about the lowest necessary? I know it's somewhere in-between 2k ohms and 250 (about the lowest from a commercial high-dollar, "cost as no object" design that I'm aware of)... Thinking about constructing a line-stage seperate to achieve crazy low rp... it's not only possible, but easy, in design, it just takes serous current for a line stage, and so, quite a massive power supply... probably twice the weight of a 300b power stage unit.... is that just silly? It's silly... right? Otherwise people would do it...? What I'm thinking about doing is the "cathode follower OTL amp" thing. There's a great tubecad page on this:
https://tubecad.com/2020/05/blog0504.htm
But half of it is over my head. But would that be a waste of time, for a drive stage? I only have so much time...
So, how is the ideal or intended signal impedance to an output stage (or really between any two stages) determined? I see that often alot of effort is taken to reduce a drive stage's rp value, to a certain degree, including through very expensive means (interstage transformers). But is lower always better? How low, exactly? Are there negative consequences for dropping below a certain value, or is it just a matter of diminishing returns?
If it seems like I'm being obtuse - yes, that is the case. I've got more of an "English major" brain than an "engineering major" brain.
Let's take some specific examples? I don't know if that is necessary but.... I see some very expensive devices that will use interstage transformers to reduce a drive stage's rp from say, 3k ohm, to just 750 ohms, to drive let's say a single-ended 2a3. I see high-dollar line-stage seperates bragging about a 250 ohm output. I see that tubes like 2a3's and 300b's in a single ended arrangement seem to be more demanding in this regard, and with certain pentodes it is treated as nearly irrelevant in the circuit. So:
How is this knowledge, this number or range of values, derived from a data sheet?
Is lower always better? I.E. if it were possible to achieve let's say, 40 ohms in a way similar to OTL output stages, would there be any advantage in doing so? Would there be detriment?
Basically, how do we know that, for a 2a3, x is better than y when x=(y/2), but z is no better than x when z=(x/2)? Caveman math, I know. I can't get through any of these engineering books & I swear I'm not low iq....
My reason for asking is, in plunking around somewhat blindly with circuits, I got some "good sound" from switching driver tubes for those with lower rp while playing around on breadboard. Basically, how far should I chase that rabbit? For a single ended 2a3 or 45 or 300b (the three I really want to spend time listening to) what is about the lowest necessary? I know it's somewhere in-between 2k ohms and 250 (about the lowest from a commercial high-dollar, "cost as no object" design that I'm aware of)... Thinking about constructing a line-stage seperate to achieve crazy low rp... it's not only possible, but easy, in design, it just takes serous current for a line stage, and so, quite a massive power supply... probably twice the weight of a 300b power stage unit.... is that just silly? It's silly... right? Otherwise people would do it...? What I'm thinking about doing is the "cathode follower OTL amp" thing. There's a great tubecad page on this:
https://tubecad.com/2020/05/blog0504.htm
But half of it is over my head. But would that be a waste of time, for a drive stage? I only have so much time...
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Ah, see I knew it was some simple thing that I was just ignorant of. So, okay - ohm's law is V=I×R. For a 2a3 operating at 250v and 60mA, R=(250v)÷(0.06a)=4166.7 ohms... um or is it grid bias voltage? Or...
If I understood that correctly, which I'm not confident about and would almost assume I did not, the value "necessary" is about 4,167... and anything below... but, anything anything? What is the advantage of $20k amps running 750 ohms into a 2a3? I mean, I'm not trying to be... I can hear the difference so I'm wondering if I should shoot for 40 ohms... take the concept and run to the limit.
I can see a pretty simple way to get in the ballpark of 40 ohm with about 200mA in stereo for the line-stage seperate, with an input sensitivity around -2v depending on the op point I settle on for the single amp stage. I have almost everything except the tubes in particular, maybe a couple of high wattage resistors at the value I settle on, so, about $40 to spend to make it happen. I have the iron for the current - including chokes - it's valuable to me but I got it all cheap by being a craigslister and going to estate sales etc, ham equipment mostly. I am used to breadboarding first and won't waste time on a layout and enclosure if it doesn't sound good.... I'm very very close to just pulling the trigger on the tubes and trying this out... shy to say exactly what I'm thinking before knowing it sounds okay... it involves very high gm compooter toobs...
What I don't have, and can't get for cheap, is quality interstage iron. Stuff is expensive. But we can play around with weird ideas, if it's cheap!
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Ran out of edit time...
If it's grid bias voltage, for that 2a3 example, let's say grid bias is 44. So, (44v)÷(0.06a)=733 and suddenly a well designed machine at 750 ohms makes sense. However, regardless of how wrong I got that, now or then, the question remains - is that therfore the ideal, or is below it better in some or any way? What happens as it approaches zero? Is there eventually a problem? Is the minimum distortion number the number, 733 ohm or whatever it would actually be if I had grasped the math part?
I don't understand the physics, and while I have read "Valve Amplifiers" (Jones) I would say I got about 1/5th of it, or less. Other books I've tried - even worse. Once it's in equations I am just... reading Greek... well actually, Greek would be much better - I do read Attic and Koine, with an amateur proficiency... which is above my skill in this domain. Amps is harder than Aristotle.
If it's grid bias voltage, for that 2a3 example, let's say grid bias is 44. So, (44v)÷(0.06a)=733 and suddenly a well designed machine at 750 ohms makes sense. However, regardless of how wrong I got that, now or then, the question remains - is that therfore the ideal, or is below it better in some or any way? What happens as it approaches zero? Is there eventually a problem? Is the minimum distortion number the number, 733 ohm or whatever it would actually be if I had grasped the math part?
I don't understand the physics, and while I have read "Valve Amplifiers" (Jones) I would say I got about 1/5th of it, or less. Other books I've tried - even worse. Once it's in equations I am just... reading Greek... well actually, Greek would be much better - I do read Attic and Koine, with an amateur proficiency... which is above my skill in this domain. Amps is harder than Aristotle.
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Posting again because I think I've found my answer - and got some of my terms straightened out, allowing me to effectively search for that answer:
https://forums.stevehoffman.tv/threads/preamp-amp-impedance-matching-question.669180/
Quote:
"General rule of thumb is load impedance 10X or more the source impedance, to minimize voltage loss at the load input. A 14 ohm preamp output impedance is extremely low. You should have no problem with pretty much any amp on Earth with that one..."
Alright then, so I'll embark on my line-stage project... I'm sorry for the thread over something so simple...
https://forums.stevehoffman.tv/threads/preamp-amp-impedance-matching-question.669180/
Quote:
"General rule of thumb is load impedance 10X or more the source impedance, to minimize voltage loss at the load input. A 14 ohm preamp output impedance is extremely low. You should have no problem with pretty much any amp on Earth with that one..."
Alright then, so I'll embark on my line-stage project... I'm sorry for the thread over something so simple...
1. Triode Driver Stage:
Suppose a triode has an rp of 5k, a u of 20. u = Gm x rp, so the Transconductance, Gm would be 4,000 microMhos.
Let the plate load be 20k, and Rg that follows the coupling cap to the next stages grid be 60k. The total plate load is 20k in parallel with 60k = 15k.
The gain is u x ((15k / (15k + 5k)) = 15.
For a ratio of 3:1 (Total RL to rp), we get a total gain of 3/4 x u.
Suppose we use a plate load of 5,455 Ohms. The total plate load is 5,455 in parallel with 60,000 = 5000 Ohms.
Gain is u x ((5k / (5k + 5k)), 20 x 1/2 = 10.
With rp = total plate load, we get gain of 1/2 x u.
You can see that when the load is not much more than rp, the gain is reduced.
Suppose the bias voltage is fixed at 12V.
If gain is 15, the output can swing 12 x 15 = 180V peak (360V peak to peak)
If gain is 10, the output can only swing 12 x 10 = 120V peak (240V peak to peak).
One reason to pick a particular rp, is to get a particular gain, versus the total load impedance.
Also, the stage with total load of 15k, at 180V peak, will have lower harmonic distortion . . .
Versus the stage with a total load of 5k at 120V peak.
Another reason to pick rp versus the total load, is to reduce distortion.
2. 2A3 Output Stage
The 2A3 has an rp = 800 Ohms, and u = 4.2.
A. Pick an output transformer with 2400 Ohm primary.
Gain is 4.2 x ((2400 / (2400 + 800)) = 4.2 x 0.75 = 3.15. The Damping Factor is 2400 / 800 = 3 (it will be less than 3, because of the loss of the output transformer, primary DCR and Secondary DCR loose power, and reduce Damping Factor.
B. Pick an output transformer with 4800 Ohm primary.
Gain is 4.2 x ((4800 / (4800 + 800)) = 4.2 x 0.857 = 3.6. The Damping Factor is 4800 / 800 = 6 (it will be less than 6, because of the loss of the output transformer, primary DCR and Secondary DCR loose power, and reduce Damping Factor.
At 1 Watt output, B. will have less harmonic distortion than A.
There is lots more to make a design work well, B+ voltage, bias voltage, plate current, etc.
Interaction between quiescent parameters, distortion with signal applied, and maximum output is what it is all about . . . Tradeoffs
Suppose a triode has an rp of 5k, a u of 20. u = Gm x rp, so the Transconductance, Gm would be 4,000 microMhos.
Let the plate load be 20k, and Rg that follows the coupling cap to the next stages grid be 60k. The total plate load is 20k in parallel with 60k = 15k.
The gain is u x ((15k / (15k + 5k)) = 15.
For a ratio of 3:1 (Total RL to rp), we get a total gain of 3/4 x u.
Suppose we use a plate load of 5,455 Ohms. The total plate load is 5,455 in parallel with 60,000 = 5000 Ohms.
Gain is u x ((5k / (5k + 5k)), 20 x 1/2 = 10.
With rp = total plate load, we get gain of 1/2 x u.
You can see that when the load is not much more than rp, the gain is reduced.
Suppose the bias voltage is fixed at 12V.
If gain is 15, the output can swing 12 x 15 = 180V peak (360V peak to peak)
If gain is 10, the output can only swing 12 x 10 = 120V peak (240V peak to peak).
One reason to pick a particular rp, is to get a particular gain, versus the total load impedance.
Also, the stage with total load of 15k, at 180V peak, will have lower harmonic distortion . . .
Versus the stage with a total load of 5k at 120V peak.
Another reason to pick rp versus the total load, is to reduce distortion.
2. 2A3 Output Stage
The 2A3 has an rp = 800 Ohms, and u = 4.2.
A. Pick an output transformer with 2400 Ohm primary.
Gain is 4.2 x ((2400 / (2400 + 800)) = 4.2 x 0.75 = 3.15. The Damping Factor is 2400 / 800 = 3 (it will be less than 3, because of the loss of the output transformer, primary DCR and Secondary DCR loose power, and reduce Damping Factor.
B. Pick an output transformer with 4800 Ohm primary.
Gain is 4.2 x ((4800 / (4800 + 800)) = 4.2 x 0.857 = 3.6. The Damping Factor is 4800 / 800 = 6 (it will be less than 6, because of the loss of the output transformer, primary DCR and Secondary DCR loose power, and reduce Damping Factor.
At 1 Watt output, B. will have less harmonic distortion than A.
There is lots more to make a design work well, B+ voltage, bias voltage, plate current, etc.
Interaction between quiescent parameters, distortion with signal applied, and maximum output is what it is all about . . . Tradeoffs