SonicTrance
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Nice! Oblivion on the top shelf 
Oblivion | UltraSonic Studios
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MrCurwen
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There's been some talk here about tube choice for these UltraSonic amps. This thread isn't about tube theory, but I think it might be interesting for people interested in these amps to talk a little bit about how and why tubes are chosen for a specific job.
Now me personally I've tried out a bunch of tubes for these circuits, including 1626, 6V6GT, 71A, 47, 2C22 (output), 6SN7 (output), 6E5P, EL84, EL81, EL41, 6P31, 4P1L, 3C24, 6L6, 6N16P, 8025A, 8012A, 6BQ6, ECL82, 6S45P and some I've forgotten by now. My current everyday amp runs 6E5P finals.
First since not everyone is a DIY guy (you don't have to be to understand this tube selection thing), let's in short go thru the basics.
Aside from "trying out different tubes" how can one know in advance what tube type will sound best in a specific circuit?
Well, what is "best sound"? That's a whole discussion in itself. For the sake of brevity, best sound is lowest possible distortion in an open loop system (Sonic's amps do not employ gNFB at all). I would venture to say if one is specifically fond of coloured sound, perhaps one should look for SE amps with traditional series feed OT to suit their tastes.
How is minimum distortion achieved? Here we come to load lines. Curves and load lines are the heart and soul of any tube system. They tell you so much about the tube and it's operation in a specific circuit I cannot even begin to explain it in full here. The basics are very simple though.
Let's draw a load line. Now in an old fashioned amp, which utilizes the output transformer primary coil as the load for the output tube (primary is in series with the tube), the load is something between 1k to 10k, usually let's say 5k. There were old time rules of thumb for "good results", for example the 3x rule, OT primary impedance should be 3x the output tube anode resistance for "best results" i.e. an acceptable compromise between output power and amount of distortion generated. If you increase OT primary impedance, you lessen output power and also lessen distortion, and vice versa.
Now let's pick a tube everybody is familiar with, the 6V6. To draw a load line for 5k primary load, you take your B+, let's say 300V, and divide it by 5000. Ohm's law says the result is in amperes, converted to mA it's 60mA.
Now look at the curves. The right side endpoint of the load line is 300V / 0mA and the left side endpoint is 0V / 60mA:
All possible op points are on this red line. Op point or operation point is the static condition, the rest condition where the output tube is biased.
The curves denote grid voltage. The control grid is the 'input' of the tube. This is where the input signal is applied. Each curve has a grid voltage attached to it on the top.
Which point should you choose? Now some people like to just try things out and try to find by luck or accident a good sounding op point. That is just shooting in the dark blindfolded and with no ammo.
The 'point' of an op point is that once signal is applied, there is sufficient amount of headroom on BOTH SIDES of the load line (looking out from the op point).
Let's say we chose grid bias point at -25V. So that's the op point now, the resting starting position. Now apply a signal to the grid.
Voltage at grid starts going up, we move left on the load line. Plenty of room to go around.
Now the sine wave reaches it top, and starts going down. Grid voltage goes down, down, past the op point, then what.
There is only a tiny amount of room left to go on the right side of the load line.
Now a sine wave has equal amounts of positive and negative peak to it. If we select this op point, the tube is unlikely to be able to reproduce the sine wave even close to original.
So logically, one should choose an op point that is rightabouts halfway on the load line.
So, about -10V. This is not an uncommon op point for a 6V6 amp running at 300V B+.
Now let's also draw the biggest possible input signal swing to the load line. This is the hard clipping point for the amp, since the grid cannot go positive unless specifically driven by a certain kind of circuit (which was not available / used in traditional circuits). Sonic's amps can drive output tube grids well into the positive part, but we're not there yet.
Now with this load (5k) and chosen bias point (27mA bias current at -10V grid), with the grid going between 0V and -20V, we have the anode voltage varying from 110 to 230 V, at op point it's about 175V.
What does this mean? Well, for one, we have amplification, 20V input signal produces about 120V output signal for a gain of 6. But with regards to signal distortion?
The blue part is the positive input signal portion, the green part the negative. You can easily see with your eyes, that they are not of equal lenght, meaning the equal variance in input signal does not produce equal variance in output signal.
The output signal waveform doesn't look exactly like the input signal waveform. This is distortion.
Now it's mostly second harmonic and third harmonic, but we won't go into that here. We'll stay on the level of avoiding distortion alltogether.
Now you can also see, that if you chose the op point to be any more on the right, the distortion would be significantly higher, because the grid lines start to bunch up closer together on the right.
Similarly, if you chose the op point more on the left, your maximum available anode voltage swing would be lower. So with the design limitations (5k load) you have, you are now at the optimum place for distortion and output power.
But since we are not limited by having to use a 5k load, or the 6V6 tube, what would be the ultimate tube and load line that would produce NO DISTORTION AT ALL? (This is hyperbole, there is always some distortion. But in comparison to the presented example.)
Well, for one, you would want a tube that has curves that are equally spaced apart. Not like those right side curves on that 6V6 load line.
Second, you can see that the curves always tend to have a different slope on different 'heights' (on the Y axis). So closer to bottom they are more level, and closer to top they are more steep.
So, the load line should stay within the same vertical region. The logical ultimate load line is thus horizontal or close to it.
So in summary:
1. Curves should be as evenly spaced in the voltage swing area you plan to utilize as possible
2. The load line should go thru only one "vertical section" of the curves, so as close to horizontal as possible
3. The op point should be about halfway point between B+ and 0V gridcurve.
These are the ingredients that produce maximum fidelity, transparency and liveliness to a tube circuit.
Now me personally I've tried out a bunch of tubes for these circuits, including 1626, 6V6GT, 71A, 47, 2C22 (output), 6SN7 (output), 6E5P, EL84, EL81, EL41, 6P31, 4P1L, 3C24, 6L6, 6N16P, 8025A, 8012A, 6BQ6, ECL82, 6S45P and some I've forgotten by now. My current everyday amp runs 6E5P finals.
First since not everyone is a DIY guy (you don't have to be to understand this tube selection thing), let's in short go thru the basics.
Aside from "trying out different tubes" how can one know in advance what tube type will sound best in a specific circuit?
Well, what is "best sound"? That's a whole discussion in itself. For the sake of brevity, best sound is lowest possible distortion in an open loop system (Sonic's amps do not employ gNFB at all). I would venture to say if one is specifically fond of coloured sound, perhaps one should look for SE amps with traditional series feed OT to suit their tastes.
How is minimum distortion achieved? Here we come to load lines. Curves and load lines are the heart and soul of any tube system. They tell you so much about the tube and it's operation in a specific circuit I cannot even begin to explain it in full here. The basics are very simple though.
Let's draw a load line. Now in an old fashioned amp, which utilizes the output transformer primary coil as the load for the output tube (primary is in series with the tube), the load is something between 1k to 10k, usually let's say 5k. There were old time rules of thumb for "good results", for example the 3x rule, OT primary impedance should be 3x the output tube anode resistance for "best results" i.e. an acceptable compromise between output power and amount of distortion generated. If you increase OT primary impedance, you lessen output power and also lessen distortion, and vice versa.
Now let's pick a tube everybody is familiar with, the 6V6. To draw a load line for 5k primary load, you take your B+, let's say 300V, and divide it by 5000. Ohm's law says the result is in amperes, converted to mA it's 60mA.
Now look at the curves. The right side endpoint of the load line is 300V / 0mA and the left side endpoint is 0V / 60mA:
All possible op points are on this red line. Op point or operation point is the static condition, the rest condition where the output tube is biased.
The curves denote grid voltage. The control grid is the 'input' of the tube. This is where the input signal is applied. Each curve has a grid voltage attached to it on the top.
Which point should you choose? Now some people like to just try things out and try to find by luck or accident a good sounding op point. That is just shooting in the dark blindfolded and with no ammo.
The 'point' of an op point is that once signal is applied, there is sufficient amount of headroom on BOTH SIDES of the load line (looking out from the op point).
Let's say we chose grid bias point at -25V. So that's the op point now, the resting starting position. Now apply a signal to the grid.
Voltage at grid starts going up, we move left on the load line. Plenty of room to go around.
Now the sine wave reaches it top, and starts going down. Grid voltage goes down, down, past the op point, then what.
There is only a tiny amount of room left to go on the right side of the load line.
Now a sine wave has equal amounts of positive and negative peak to it. If we select this op point, the tube is unlikely to be able to reproduce the sine wave even close to original.
So logically, one should choose an op point that is rightabouts halfway on the load line.
So, about -10V. This is not an uncommon op point for a 6V6 amp running at 300V B+.
Now let's also draw the biggest possible input signal swing to the load line. This is the hard clipping point for the amp, since the grid cannot go positive unless specifically driven by a certain kind of circuit (which was not available / used in traditional circuits). Sonic's amps can drive output tube grids well into the positive part, but we're not there yet.
Now with this load (5k) and chosen bias point (27mA bias current at -10V grid), with the grid going between 0V and -20V, we have the anode voltage varying from 110 to 230 V, at op point it's about 175V.
What does this mean? Well, for one, we have amplification, 20V input signal produces about 120V output signal for a gain of 6. But with regards to signal distortion?
The blue part is the positive input signal portion, the green part the negative. You can easily see with your eyes, that they are not of equal lenght, meaning the equal variance in input signal does not produce equal variance in output signal.
The output signal waveform doesn't look exactly like the input signal waveform. This is distortion.
Now it's mostly second harmonic and third harmonic, but we won't go into that here. We'll stay on the level of avoiding distortion alltogether.
Now you can also see, that if you chose the op point to be any more on the right, the distortion would be significantly higher, because the grid lines start to bunch up closer together on the right.
Similarly, if you chose the op point more on the left, your maximum available anode voltage swing would be lower. So with the design limitations (5k load) you have, you are now at the optimum place for distortion and output power.
But since we are not limited by having to use a 5k load, or the 6V6 tube, what would be the ultimate tube and load line that would produce NO DISTORTION AT ALL? (This is hyperbole, there is always some distortion. But in comparison to the presented example.)
Well, for one, you would want a tube that has curves that are equally spaced apart. Not like those right side curves on that 6V6 load line.
Second, you can see that the curves always tend to have a different slope on different 'heights' (on the Y axis). So closer to bottom they are more level, and closer to top they are more steep.
So, the load line should stay within the same vertical region. The logical ultimate load line is thus horizontal or close to it.
So in summary:
1. Curves should be as evenly spaced in the voltage swing area you plan to utilize as possible
2. The load line should go thru only one "vertical section" of the curves, so as close to horizontal as possible
3. The op point should be about halfway point between B+ and 0V gridcurve.
These are the ingredients that produce maximum fidelity, transparency and liveliness to a tube circuit.
MrCurwen
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So now that the basics of tube load lines and operations are clear, let's directly compare two types; the 6V6 and EL36. Both are beam tetrodes, but we used them in triode mode.
So here are the curves, first the 6V6:
then the EL36:
Now different ways of drawing and presenting the curves can sometimes trick your eyes even if you're quite familiar with looking at curves. So it's best to draw the load lines and simply check the voltage swings.
So first, let's look at the 6V6 with a B+ of 300V and using a horizontal 500k impedance load line at 18mA bias current:
So op point is at 150V, the positive swing anode voltage is at about 55V and negative swing 235V.
So left side swings 95V while right side swings 85V. Is this 'good' or 'bad'? Depends. It's clearly much superior to the slanted old fashioned load line presented in the previous message.
Now let's look at the EL36:
Op point is again about 150V, grid at -26V. Positive swing goes to about 15V and negative swing to 285V.
So both sides swing 135V. Notice that in addition to being symmetric, the overall swing is considerably more. This is because the tubes 6V6 and EL36 have different anode resistances. The lower the anode resistance, the closer to 0V anode voltage you can get, so the larger the swing.
Now graphic solutions are always subject to accuracy problems, I wasn't super careful with the pixels here. If you look closely op point is not 150V exactly, it's something like 153V. 0V grid line is at 13V or so in reality.
So it is not 100.00% perfectly symmetrical, but it is several magnitudes more symmetrical than the 6V6 example. In technical solutions you are left to find the best solution, never the perfect solution.
If we were to do a serious comparison, we would look at both tubes swinging the same amount of anode voltage. In this comparison the EL36 would beat the 6V6 much more clearly.
So now using these simple tools anybody can find a nice tube to use to get the best possible (lowest possible distortion) results. Just look for a tube with a decently low internal resistance (Ra or plate resistance) and curves that are nice and evenly spaced.
No need to try the tube out first, this method produces the factual outcome when using this particular circuit. If the swings are more symmetrical on tube B than tube A, then tube B will always sound more transparent and more musical and lively in this circuit. Guaranteed.
So here are the curves, first the 6V6:
then the EL36:
Now different ways of drawing and presenting the curves can sometimes trick your eyes even if you're quite familiar with looking at curves. So it's best to draw the load lines and simply check the voltage swings.
So first, let's look at the 6V6 with a B+ of 300V and using a horizontal 500k impedance load line at 18mA bias current:
So op point is at 150V, the positive swing anode voltage is at about 55V and negative swing 235V.
So left side swings 95V while right side swings 85V. Is this 'good' or 'bad'? Depends. It's clearly much superior to the slanted old fashioned load line presented in the previous message.
Now let's look at the EL36:
Op point is again about 150V, grid at -26V. Positive swing goes to about 15V and negative swing to 285V.
So both sides swing 135V. Notice that in addition to being symmetric, the overall swing is considerably more. This is because the tubes 6V6 and EL36 have different anode resistances. The lower the anode resistance, the closer to 0V anode voltage you can get, so the larger the swing.
Now graphic solutions are always subject to accuracy problems, I wasn't super careful with the pixels here. If you look closely op point is not 150V exactly, it's something like 153V. 0V grid line is at 13V or so in reality.
So it is not 100.00% perfectly symmetrical, but it is several magnitudes more symmetrical than the 6V6 example. In technical solutions you are left to find the best solution, never the perfect solution.
If we were to do a serious comparison, we would look at both tubes swinging the same amount of anode voltage. In this comparison the EL36 would beat the 6V6 much more clearly.
So now using these simple tools anybody can find a nice tube to use to get the best possible (lowest possible distortion) results. Just look for a tube with a decently low internal resistance (Ra or plate resistance) and curves that are nice and evenly spaced.
No need to try the tube out first, this method produces the factual outcome when using this particular circuit. If the swings are more symmetrical on tube B than tube A, then tube B will always sound more transparent and more musical and lively in this circuit. Guaranteed.
MrCurwen
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6L6 and eqs were mentioned in this thread. So let us look at how the 6L6 would do in triode mode and utilizing the best possible load line:
So op point is about 160V / 15mA, (the 3mA difference is not important, the curves don't change that fast while going a little up or down in current).
Positive side swing is 120V, negative side swing is 100V. So it will not perform even close to as good as EL36 and eqs.
So op point is about 160V / 15mA, (the 3mA difference is not important, the curves don't change that fast while going a little up or down in current).
Positive side swing is 120V, negative side swing is 100V. So it will not perform even close to as good as EL36 and eqs.
baronbeehive
Headphoneus Supremus
Besides being more symmetrical, I believe I'm right in saying that the more the anode voltage swing the livelier the sound? Which means the EL36 is the livelier sounding, as well as being the more transparent and less distorted tube, at least for that operating point?
baronbeehive
Headphoneus Supremus
That would have saved me a lot of work in the Little Dot Supermod thread!
baronbeehive
Headphoneus Supremus
So what exactly is it that can drive the amp into the positive grid area, I'm guessing it is something to do with the gyrators, or the use of the OT... or something special in the hybrid design of the amp?
MrCurwen
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More voltage swing makes the sound livelier in the sense that it makes it louder, and louder sound is more engaging.
What I was after when writing that was simply lower distortion. Distortion masks details in the reproduced sound; lack of this masking makes the sound livelier.
Simply saying "lower distortion" over and over again doesn't often send the desired message of how much it really matters. This is of course because of the gNFB low distortion amps, which usually do not sound very lively or realistic. Sonic's amps are open loop AND low distortion. That's another animal in itself.
Yep. Always do your homework, you'll save effort in the long run.
Hint is in the 6L6 and 6V6 curves. It's the ability to provide current to the grid, so the grid drivers.
The curves have extra curve lines marked with + voltages. They denote current that the grid eats when the grid is at a specific voltage. You can see it's quite much for extreme positive grid conditions. It's quite much for even +1V or, in fact, for the 0V grid line itself!
Tubelab George is the progenitor of this grid driver technology, incredibly knowledgeable about tubes and especially this grid current phenomenom. Most tubes eat grid current to some extent even while the grid is still negative. This is in part due to different parasitic capacitances inside the tube, between the different elements.
Old time radio transmitters used a tube cathode follower to drive the transmitter final tube, to provide current for the control grid on big pulses.
The grid drivers are simply FETs operating as source followers (FET version of cathode follower). So they just 'follow' the voltage, do not do anything to it, and provide current as needed.
Sonic's amps are able to drive the output tube grid well into positive territory. This in part provides towards the first class transient response.
The gyrators are basically as well source followers.
MrCurwen
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Well the EL36 and eq don't really have room for positive grid operations, maybe a volt or two, but grid current is a real thing in low negative grid region already, as well as thruout the whole range to some extent.
The control grid is located very close to the cathode in comparison to the anode (or g2 in a triode strapped beam tetrode). It is physically near the electron cloud that just hangs around the space close to the cathode.
Cathode is heated, electrons boil off and form a cloud, anode positive voltage starts to draw them towards the anode.
Now what happens when grid is positive? Then the grid is the LOCAL anode, it's easier to get to than the far far away anode. So the electrons go to the grid. After that they have to go SOMEWHERE, and if there is nowhere to go, they stay at the grid. What does that mean? That means the voltage potential of the grid changes, and NOT BY THE AUDIO SIGNAL. That's distortion.
So the grid system must have ways of getting rid of the electrons that get lost there. That's what the grid driver system does. It sinks the current neatly away. (Remember current is holes, I'm mixing actual physical electrons and hole here but I hope you follow.)
The control grid is located very close to the cathode in comparison to the anode (or g2 in a triode strapped beam tetrode). It is physically near the electron cloud that just hangs around the space close to the cathode.
Cathode is heated, electrons boil off and form a cloud, anode positive voltage starts to draw them towards the anode.
Now what happens when grid is positive? Then the grid is the LOCAL anode, it's easier to get to than the far far away anode. So the electrons go to the grid. After that they have to go SOMEWHERE, and if there is nowhere to go, they stay at the grid. What does that mean? That means the voltage potential of the grid changes, and NOT BY THE AUDIO SIGNAL. That's distortion.
So the grid system must have ways of getting rid of the electrons that get lost there. That's what the grid driver system does. It sinks the current neatly away. (Remember current is holes, I'm mixing actual physical electrons and hole here but I hope you follow.)
baronbeehive
Headphoneus Supremus
Yes, reviewers should be very careful to control the volume when doing comparisons to avoid spurious results.
SonicTrance
Member of the Trade: Custom Amp Builder
A lot of tech talk, which is nice but I thought I'd spice things up with some pics!
This Oblivion is just complete. Burning in now.
Next I'll build @joseph69 Citadel. The chassis is supposed to be delivered next week.
This Oblivion is just complete. Burning in now.
Next I'll build @joseph69 Citadel. The chassis is supposed to be delivered next week.
UntilThen
Headphoneus Supremus
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2 knobs?
SonicTrance
Member of the Trade: Custom Amp Builder
Yes, it's just a selector switch for speaker/headphones on this one. No impedance switch.2 knobs?
UntilThen
Headphoneus Supremus
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I have been listening to music with hd800 and Oblivion these last 3 weeks because my Verite is with my son.
The level of details and clarity is outstanding. I’m not sure about HEDDphone and Oblivion - it could be more details than you can handle.
The level of details and clarity is outstanding. I’m not sure about HEDDphone and Oblivion - it could be more details than you can handle.
MrCurwen
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Try watching a well mixed horror movie with Sonic's amp and headphones. It's a unique experience.
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