Oh no, i did recalculate everything around the ECC84 but i believe you, "DF96" and "petertub" are all correct that this tube is not the correct tube for a LTP phase splitter, it might be made to work but still is not a suitable tube for this job.
I also noted that increasing plate current and lowering plate load resistors made the current distribution difference smaller and that's why i said that maybe a better CCS would work better.
I now also believe that and the extra effort that is needed does not worth it because the Concertina works so much better for the output amplitude i need and it is so simple...
I can't thank you all for your answer, your information is so valuable...
Chris
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The 2N3904 could be a good enough current source, but its base was driven by 750 Ohms (at least in the 12AX7 circuit shown).
It probably should have been driven by something closer to 100 Ohms. That would swamp out the base resistance and capacitance.
Years ago, I used a 2N3053, a resistor to pump current into an LED for a voltage reference, 100 Ohm base resistor, and an emitter resistor for current sense.
It worked very well. I selected the 2N3053 because it had a very high collector impedance, a nice metal case to get rid of heat (easy to add a heat sink on it if you wish), and because it was readily available locally. I still have several in case I can not find a newer suitable NPN locally.
It probably should have been driven by something closer to 100 Ohms. That would swamp out the base resistance and capacitance.
Years ago, I used a 2N3053, a resistor to pump current into an LED for a voltage reference, 100 Ohm base resistor, and an emitter resistor for current sense.
It worked very well. I selected the 2N3053 because it had a very high collector impedance, a nice metal case to get rid of heat (easy to add a heat sink on it if you wish), and because it was readily available locally. I still have several in case I can not find a newer suitable NPN locally.
Yes i took this in to account, the second half of the ECC84 has a 200v cathode to filament negative rating (the first half is only rated at 100V.
I understand this as i always buy more than i need just in case, i many times find my self panicked that replacement parts will run out
I finally think that the simple solution with the LED and a transistor is the most reliable solution but i will try and some other CCS topologies like the ones listed here:
Cascoded Ring of Two for Plate Load
Chris
Myself, I think the simple solution is a LED-as-voltage-source, an emitter follower NPN, and a "programmable" emitter-side load for it. You determine VBE by measuring it at 10 mA or so. Easy to do, outside the circuit. The LED is your beautifully glowing constant voltage supply. A nice high-value resistor connects it to B+. And to the NPN's base. Choosing any old yellow LED kind of 'does the trick'. Delivers somewhere from 1.8 to 2.2 volts to the base.
VLED - VBE = (1.8–2.2) - (0.7–0.8)
… = 1.13 V by gaussian mean
… = 1.13 V by gaussian mean
So our RE now becomes predictive of how much current is going to flow.
IC = 1.13 / RE
So if you prefer to choose the IC,
RE ← 1.13 / IC
And life is good. The VC is surprisingly adaptive in real life. “it becomes what the Tube wants” to achieve the IK ≡ IC
GoatGuy
It is caused by low anode-cathode voltage. In order to maintain the anode current (set by the tail) the grid-cathode voltage becomes less negative. Once it gets less than around -1V you start to get grid current. If the DC paths to the two grids have very different resistances then you get imbalance.
Really? I kind of thought that physics would have prevented that. That relative to the cathode, the grid has to be positive in order for there to be grid current.
I must have missed that in physics school. LOL.
GoatGuy
You're forgetting that electrons can leave the cathode with an initial velocity that gives them enough kinetic energy to hit a slightly negative control grid. Diodes show some small plate current at VPK= 0V. If the electrons can get there, there will be current.
Also don't forget that stray ions also contribute to grid current.
Also don't forget that stray ions also contribute to grid current.
Ah, you're right. I overlooked the kinetic energy of the drift electrons. Makes sense. In fact one's grid doesn't even need to be close to zero, right? If the distance from K → G is '1' unit, and G → A is 9 units (total 10), and if the curl of the electric field determines relative acceleration of the electrons, then by ∇×B or something close, you get a 1/d² relationship.
So when VA relative to VK is (e.g.) 200 V, then the electrons near the grid are only winging along at 2 V or so. Therefore, VG < –2 V is the cutoff.
OK, I'm happy now.
GoatGuy
VA-VK (1²/(1+9)²) = ¹/₁₀₀ VA-K or so.
So when VA relative to VK is (e.g.) 200 V, then the electrons near the grid are only winging along at 2 V or so. Therefore, VG < –2 V is the cutoff.
OK, I'm happy now.
GoatGuy
GoatGuy,
Thanks for that treatise, and a way to find what limits are reasonable to keep away from grid current. I had not thought of that method. I love field theory, so why did I not think of it earlier?
Makes me want to examine spacing of parts in tubes; measure the voltages (I already always do that part); then go through the calculations. But that has 2 sets of voltage measurements and calculations (1. quiescent; 2. under signal conditions, with the grid going less negative while the plate swings toward the cathode).
Two other methods (you already know) to measure if there is grid current:
1. Quiescent: measure the voltage across the grid leak resistor.
2. Under signal: My longtime friend came up with the following method. Oh, did I forget to say that was when we both worked at a major Test and Measurement Company?
A. Start with a low impedance generator, i.e. 50 Ohms (allways terminate generators with their characteristic output impedance, tube circuits can not normally do that). With that low impedance drive (25 Ohms), measure the harmonic distortion of the amp. B. Then increase the generator impedance (terminate the generator, and put a series resistor from the generator terminated output to the tube input circuit). You use a resistor the same value as the grid leak resistor. Increase the generator voltage output by 2x (+6dB) to make up the voltage loss of the 2:1 divider of the series resistor and the grid leak resistor. Now re-measure the harmonic distortion. If it went up versus driving from low impedance, you are drawing grid current (at the same signal level as before).
Note: You may need to use a series resistor that is only 0.11 x the grid leak resistance. Then increase the generator voltage output by 10% (+0.8 dB). Then re-measure the harmonic distortion. If it went up versus driving from low impedance, you are driving grid current (at the same signal level as before). The reason for the lower value series resistor is to ensure that the miller capacitance does not decrease the drive voltage to the grid.
I keep my tube grids quiescent state more negative than -1V relative to the cathode (-2V, -3V, -4V . . . etc.). Otherwise when the signal comes along, it would cause grid current and load the signal source. It all depends on the tube that is used, and the DC voltages and signal voltages of that stage. I thought that was a given criteria for amp classes other than A2, AB2, C(2), etc.
My LTP phase invertor circuits do not draw grid current during signal levels that I am listening to. If I get carried away listening loud, and get clipping, then I turn it down. I find that saves my ears, and my psyche.
The balance pot only works if the whole design is correct, the tubes and parts are good, etc. Using a couple of new old stock or used tubes that were built in different years, by different manufacturers, and that use plates, cathodes, control grids, beam formers, etc. that look totally different are not likely going to balance at DC, let alone have AC balance with medium to large signal levels.
In recent years I have only used current production tubes, heavily re-tested, matched pairs, and purchased from only one vendor. I am lucky because he is local, and I can call and have a matched pair waiting for me when I drive there. I know that limits me from using certain tube types. I do have some future plans for some tubes that may require me to use new old stock. 807, 4X150A, etc. But I will have to make that tradeoff, or try and find reliable current manufacturers of such tubes.
Thanks for that treatise, and a way to find what limits are reasonable to keep away from grid current. I had not thought of that method. I love field theory, so why did I not think of it earlier?
Makes me want to examine spacing of parts in tubes; measure the voltages (I already always do that part); then go through the calculations. But that has 2 sets of voltage measurements and calculations (1. quiescent; 2. under signal conditions, with the grid going less negative while the plate swings toward the cathode).
Two other methods (you already know) to measure if there is grid current:
1. Quiescent: measure the voltage across the grid leak resistor.
2. Under signal: My longtime friend came up with the following method. Oh, did I forget to say that was when we both worked at a major Test and Measurement Company?
A. Start with a low impedance generator, i.e. 50 Ohms (allways terminate generators with their characteristic output impedance, tube circuits can not normally do that). With that low impedance drive (25 Ohms), measure the harmonic distortion of the amp. B. Then increase the generator impedance (terminate the generator, and put a series resistor from the generator terminated output to the tube input circuit). You use a resistor the same value as the grid leak resistor. Increase the generator voltage output by 2x (+6dB) to make up the voltage loss of the 2:1 divider of the series resistor and the grid leak resistor. Now re-measure the harmonic distortion. If it went up versus driving from low impedance, you are drawing grid current (at the same signal level as before).
Note: You may need to use a series resistor that is only 0.11 x the grid leak resistance. Then increase the generator voltage output by 10% (+0.8 dB). Then re-measure the harmonic distortion. If it went up versus driving from low impedance, you are driving grid current (at the same signal level as before). The reason for the lower value series resistor is to ensure that the miller capacitance does not decrease the drive voltage to the grid.
I keep my tube grids quiescent state more negative than -1V relative to the cathode (-2V, -3V, -4V . . . etc.). Otherwise when the signal comes along, it would cause grid current and load the signal source. It all depends on the tube that is used, and the DC voltages and signal voltages of that stage. I thought that was a given criteria for amp classes other than A2, AB2, C(2), etc.
My LTP phase invertor circuits do not draw grid current during signal levels that I am listening to. If I get carried away listening loud, and get clipping, then I turn it down. I find that saves my ears, and my psyche.
The balance pot only works if the whole design is correct, the tubes and parts are good, etc. Using a couple of new old stock or used tubes that were built in different years, by different manufacturers, and that use plates, cathodes, control grids, beam formers, etc. that look totally different are not likely going to balance at DC, let alone have AC balance with medium to large signal levels.
In recent years I have only used current production tubes, heavily re-tested, matched pairs, and purchased from only one vendor. I am lucky because he is local, and I can call and have a matched pair waiting for me when I drive there. I know that limits me from using certain tube types. I do have some future plans for some tubes that may require me to use new old stock. 807, 4X150A, etc. But I will have to make that tradeoff, or try and find reliable current manufacturers of such tubes.
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People are often surprised to find that grid current starts from around -1V bias, because of the hot cathode. The actual value depends on the valve, and varies from sample to sample. In most cases you will not notice it, because it needs a high impedance grid circuit and low bias (e.g. low anode voltage). Most commonly seen in high mu valves, such as ECC83/12AX7, because for them there is only a narrow window between grid cutoff and grid current.
Pardon my heavy paraphrasing. I like the technique. It encourages thinking whilst testing, an increasingly lost ideal. I'll have to haul out some of my test equipment and give 'er a go.
Thanks for taking the time to write all that up. Tho' its like your Mother saying, “now don't you boys get in trouble!!!”, I advise extraordinary caution when fiddling with the 4X150's kilovolts of nominal plate tension. Be careful!
GoatGuy
Very interesting finding. Does this effect also set the 12AX7 grid to about –1.0 to –1.5 volts on true "grid leak" self biasing topologies? As rare as they are… I would imagine practically it takes the likes of a Wheatstone Bridge to perform a zero-current measure of a self-biased floating grid. Even gigaohms would load the thing!
GoatGuy
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