I have build a DAC-Output with the 801A as LTP with an OPT to convert the differential signal to SE and load the 801A. At the bottom we have a cascoded current source.
I have read a couple of whte papers on biasing the 801a like this one
801A Tube
and this one
VinylSavor: The search for that 'magical' operating point
Now, my question is a bit different...as I am using this tube as a linestage, I run it currently with 265V, 28mA, -6 V Ug.
As the DAC outputs less than 1V into the grid you would say: No issue, right ?
NOW....my thoughts:
- The heater voltage is 7.5V...
- so, given that the CCS is connected to the positive side of the heater this mean we have at the plus side -6V while we have on the negative side -13,5V, correct ?
- now, we have a small Signal of 1V on the Grid.
So...We have a Cathode-Anode Voltage of 265+6V= 271V and on the other side 271V-7,5V=263,5V
Hmmm...I guess this will lead to a kind of grid distortion, right ?
So, how can this influenced in a positive way ?
I understand from Steve Bench, that starving the filament is one option...Steve's Tube Pages
...How far down could I go in case of the 801A without damaging the tube ?
Ok, I tried to do some homework and found this article, but it is beyond me, so please help me to understand if raising the Anode Voltage or current could have as well a positive effect:
From John Harper:
"Filamentary Tubes
Most of the physics behind filamentary tubes is the same as for indirectly-heated tubes, but there are some differences. First, there is the question of the effective area of the cathode. An accepted formula for this is to use the length of the filament times twice the filament-grid distance [Spang48, p189].
The most significant difference arises because the voltage along the filament is not constant, but varies from one end to the other by the applied filament potential. Although this potential is small, it must be remembered that the effective plate voltage as seen at the cathode (filament) is also small. For example, a 300B operating under quiescent conditions of 350V and 90mA, with –60V on the grid, has a potential as seen at the cathode of around 15V, against a filament voltage of 5V. At the negative extreme of grid voltage, modulated by the signal, this effective voltage will drop close to or even below 5V.
When the effective plate voltage is less than the filament voltage, only part of the filament contributes to the plate current, i.e. the part which is still more negative than the filament. Furthermore, the current varies along the filament. The effect of this is that the current becomes dependent on the 5/2 power of the effective plate voltage, rather than the 3/2 power. As the plate voltage increases beyond the filament voltage, there is a gradual transition between the 5/2 power and the 3/2 power, which is approximately given by the formula [Dow37]:
where: P = perveance
Veff = effective plate voltage
Vfil = filament voltage
It is this shift from a 3/2 law to a 5/2 law which explains the distinctive “tuck under” observed in the plate curves for filamentary tubes at high negative grid voltages and low currents. It has been observed [Bench99] that distortion can be measurably reduced with filamentary tubes by lowering the filament voltage to the lowest possible value consistent with avoiding saturation. In fact, in a filamentary tube with close cathode-grid spacing, at low currents the law will follow an even higher power, in theory 7/2."
Source: Tubes 201 - How Vacuum Tubes Really Work
I have read a couple of whte papers on biasing the 801a like this one
801A Tube
and this one
VinylSavor: The search for that 'magical' operating point
Now, my question is a bit different...as I am using this tube as a linestage, I run it currently with 265V, 28mA, -6 V Ug.
As the DAC outputs less than 1V into the grid you would say: No issue, right ?
NOW....my thoughts:
- The heater voltage is 7.5V...
- so, given that the CCS is connected to the positive side of the heater this mean we have at the plus side -6V while we have on the negative side -13,5V, correct ?
- now, we have a small Signal of 1V on the Grid.
So...We have a Cathode-Anode Voltage of 265+6V= 271V and on the other side 271V-7,5V=263,5V
Hmmm...I guess this will lead to a kind of grid distortion, right ?
So, how can this influenced in a positive way ?
I understand from Steve Bench, that starving the filament is one option...Steve's Tube Pages
...How far down could I go in case of the 801A without damaging the tube ?
Ok, I tried to do some homework and found this article, but it is beyond me, so please help me to understand if raising the Anode Voltage or current could have as well a positive effect:
From John Harper:
"Filamentary Tubes
Most of the physics behind filamentary tubes is the same as for indirectly-heated tubes, but there are some differences. First, there is the question of the effective area of the cathode. An accepted formula for this is to use the length of the filament times twice the filament-grid distance [Spang48, p189].
The most significant difference arises because the voltage along the filament is not constant, but varies from one end to the other by the applied filament potential. Although this potential is small, it must be remembered that the effective plate voltage as seen at the cathode (filament) is also small. For example, a 300B operating under quiescent conditions of 350V and 90mA, with –60V on the grid, has a potential as seen at the cathode of around 15V, against a filament voltage of 5V. At the negative extreme of grid voltage, modulated by the signal, this effective voltage will drop close to or even below 5V.
When the effective plate voltage is less than the filament voltage, only part of the filament contributes to the plate current, i.e. the part which is still more negative than the filament. Furthermore, the current varies along the filament. The effect of this is that the current becomes dependent on the 5/2 power of the effective plate voltage, rather than the 3/2 power. As the plate voltage increases beyond the filament voltage, there is a gradual transition between the 5/2 power and the 3/2 power, which is approximately given by the formula [Dow37]:
where: P = perveance
Veff = effective plate voltage
Vfil = filament voltage
It is this shift from a 3/2 law to a 5/2 law which explains the distinctive “tuck under” observed in the plate curves for filamentary tubes at high negative grid voltages and low currents. It has been observed [Bench99] that distortion can be measurably reduced with filamentary tubes by lowering the filament voltage to the lowest possible value consistent with avoiding saturation. In fact, in a filamentary tube with close cathode-grid spacing, at low currents the law will follow an even higher power, in theory 7/2."
Source: Tubes 201 - How Vacuum Tubes Really Work