DHT filament bias

[Hierfi]

20uA on a 3A5 is nearly 0 current. As you can see, with no plate current and no plate voltage, the circuit's output impedance has become so high that it's not able to really act as a preamp.  I would go as far as to say that this qualifies as a class B preamp.

With one end of the filament grounded and the other end being supplied by a current source, one end of the filament is at zero bias, which will further degrade the performance. (Especially depending on what's feeding the grid of the 3A5)  You actually have a preamp that requires a source with low impedance to drive it (100 Ohms), yet the gain stage itself has incredibly high output impedance. 

I don't mean to sound overly negative, just trying to highlight all the trades you have made to get rid of one capacitor.

Thanks. No negativity is considered. Class B is exceeding generous considering it is not deserving of any class as a preamplifier     

Decreasing the plate voltage decreases the attraction of electrons. The literature indicates that starving the plate causes the input impedance to drop drastically on the grid, as you have mentioned. However, this is considered to occur as a result of the cloud of electronics ending up surrounding the grid and  overwhelming the effectiveness of input signal currents. By drastically reducing the heating of the cathode/filament the cloud is drastically reduced. To confirm this, and despite any reality to this form of thinking, I added resistance in series with the input signal to determine the source impedance. The output dropped in 1/2 with 100Kohm connected in series with the input, hence a 100Kohm input impedance. The conclusion is that the magnitude of electron emissions from the heater/cathode drastically affects the input impedance. 100Kohm appears more than adequate for a preamplifier.     

It is not clear what you meant by a zero bias point in relation to attaching the center of the heater to ground. A voltage gradient is created by heater current over the length of the tubes heater and goes from ground (as physically attached) to 0.6 volts set by the current source. With the grid being at ground potential the cathode is on average 0.3 volts positive with respect to the grid.  Despite this voltage drop the tube can be considered at 0 volts on the cathode and on the grid in examining the characteristic curves of the tube. It is only under higher plate voltage with the cathode and grid set equally that the plate dissipation can destroy the tube.

Grid voltages that swing positively and negatively in relation to the heater/cathode voltage is not indicated of any unnatural nonlinearity in the characteristic curves. Neither was this observed in the reality even when using a 100Kohm signal source. It should be noted however, that the measurement was done using a X10 probe that minimized both resistance and capacitance on the output. Hence to make this network usable it requires some form of impedance converter on the output.

 It should be noted that this network has more than twice the peak to peak swing capability of the quickie in using only a 20 volt supply. This is not to suggest that any implementation can remotely compete or challenge the quickie. It is simply an interesting comparison.

By the way, how did the 3A5 tube work out at the 90 volts?         [/font][/size][/size]

[Paul Birkeland]

Try driving the grid with a 10K source, then measure distortion. When you drive the grid positive the impedance drops like crazy and grid current is drawn. This will pop up immediately as high distortion that has poor symmetry.

[Paul Joppa]

Before I get into technical stuff - thanks for posting this! I always hope that our products will inspire others to experiment and play with tubes, and the Quickie is especially inviting because it's so cheap ...uhhh... inexpensive, yeah that's it inexpensive. It's fun, and cheaper that taking the family to a movie! OK, now some tech notes:

Max rated plate current (presumably at normal filament voltage of 1.4 volts) is listed as 5mA. Cathode emission goes as the 13th power of filament voltage*, so at 0.7 volts or half the normal filament voltage the max current would be 0.61 microamps (uA). So based on this, 20uA would greatly exceed the current necessary to maintain the protective space charge, exposing the cathode to possible poisoning from ion bombardment. On the other hand, at such a low plate voltage the ion energy is very small, and much less likely to do any damage. Hard to say if that's meaningful.  :^)

* This rule is only known for pure tungsten filaments, I've never seen any comparison with thoriated tungsten or oxide cathodes. So it may be quite inaccurate for them!

Assuming the filament is being starved, there should be a saturation current which is the maximum the cool cathode can deliver. You can short the 470K plate resistor with a current meter to see what the maximum is. If it's less than twice the 20uA you are using then the output voltage capability should be adequate. This saturation effect is what causes reduced distortion with starved filaments or heaters - when it's just right you get all third harmonic with no second. As above, at normal voltages this will kill the cathode fairly fast, but all bets are off at these very low voltages.

Grid current begins at a bias voltage that depends on the metallurgy of the cathode and grid as well as the size of the space charge. A few tubes can take up to a half volt or so of positive bias before grid current becomes measurable, while most start at around -0.6 volts. I've measured anywhere from -1.0 volt to -0.2 volts on various 6DJ8s and 6922s, operated at normal heater and plate voltages and currents. Some of the battery tubes are made so that they can operate at zero bias and maintain high grid impedance. For example, the Tung-Sol 3A5 curves show zero plate current at zero bias for grid voltages below about 1.2 volts, which the 12AX7 curves show a tiny current at -0.5v grid bias - all these at normal filament voltage of course.

[Hierfi]

Try driving the grid with a 10K source, then measure distortion. When you drive the grid positive the impedance drops like crazy and grid current is drawn. This will pop up immediately as high distortion that has poor symmetry.

There are several things to consider. The magnitude of grid current is a function of free electrons boiled from the heater/cathode. This is drastically reduced. Secondly the circuit has a natural gain of about 10x. Given a preamplifier with some realistic value of output, lets say a 2 volt peak to peak swing requires 0.2 volt peak to peak to cause that output. This means that the gradient across the heater, being on average 0.3 volts remains more positive than the grid under all circumstances.

The network is effectively a DC amplifier. In the case of positive grid currents the negative half of the output will start to limit in amplitude relative to the positive half. In other words there would be some form of square law distortion function that decreases amplitude on the negative side. 

In the testing there is no observable variance in amplitude of the positive and the negative half relative to the mean, even for a +/- 5 volt peak to peak output swing. This was using a 10Kohm or 100Kohm series input resistance. In other words no square law function was observable to indicate any grid current issues. In the case of the 100Kohm, as more seriously affected by grid current, was only observable starting to limit at around 7volts or 8volts negative relative to positive half. 

The testing suggests that grid current artifacts would not be the dominant source of distortion under any normal circumstances. By comparison, in the case of the quickie there is only 4 volts across the 4Kohm plate load resistance for a 1mA plate current. This means that in applying a negative grid voltage there is only a 4 volt swing to the positive rail before hard clipping occurs and a series of high order harmonic components results. IMO there seems no comparison to suggest of any necessary disadvantage as a result of grid currents in comparison to the quality of the quickie.             

[Hierfi]

Before I get into technical stuff - thanks for posting this! I always hope that our products will inspire others to experiment and play with tubes, and the Quickie is especially inviting because it's so cheap ...uhhh... inexpensive, yeah that's it inexpensive. It's fun, and cheaper that taking the family to a movie! OK, now some tech notes:

Max rated plate current (presumably at normal filament voltage of 1.4 volts) is listed as 5mA. Cathode emission goes as the 13th power of filament voltage*, so at 0.7 volts or half the normal filament voltage the max current would be 0.61 microamps (uA). So based on this, 20uA would greatly exceed the current necessary to maintain the protective space charge, exposing the cathode to possible poisoning from ion bombardment. On the other hand, at such a low plate voltage the ion energy is very small, and much less likely to do any damage. Hard to say if that's meaningful.  :^)

* This rule is only known for pure tungsten filaments, I've never seen any comparison with thoriated tungsten or oxide cathodes. So it may be quite inaccurate for them!

Assuming the filament is being starved, there should be a saturation current which is the maximum the cool cathode can deliver. You can short the 470K plate resistor with a current meter to see what the maximum is. If it's less than twice the 20uA you are using then the output voltage capability should be adequate. This saturation effect is what causes reduced distortion with starved filaments or heaters - when it's just right you get all third harmonic with no second. As above, at normal voltages this will kill the cathode fairly fast, but all bets are off at these very low voltages.

Grid current begins at a bias voltage that depends on the metallurgy of the cathode and grid as well as the size of the space charge. A few tubes can take up to a half volt or so of positive bias before grid current becomes measurable, while most start at around -0.6 volts. I've measured anywhere from -1.0 volt to -0.2 volts on various 6DJ8s and 6922s, operated at normal heater and plate voltages and currents. Some of the battery tubes are made so that they can operate at zero bias and maintain high grid impedance. For example, the Tung-Sol 3A5 curves show zero plate current at zero bias for grid voltages below about 1.2 volts, which the 12AX7 curves show a tiny current at -0.5v grid bias - all these at normal filament voltage of course.

The substance is interesting Paul. Thanks.

I still recall that in any circuit with a "device" connected through a fixed resistor to a power supply that drops half the voltage across the resistor means that the "device" is operating at it's maximum power dissipation. Hence the tube dissipation would drop if the voltage goes either more positive or negative from 10 volts. This means that power dissipation of the tube peaks at the quiescent operating point or 20uA x 10volts or 200uW. It seems more likelihood the tube will expire by losing its vacuum in the millennium than for any other cause.

The cathode isn't apparently starved. Shorting the 470Kohm resistor to the 20 volt supply resulted in 0.3mA, or 0.3mA x 20volts or 6mW dissipation. I suspect the current would still go up with a higher B+ supply. Of note is that the output goes into more symmetric clipping at the extremes, hence producing more odd order harmonics. 

As you suggest the 3A5 appears as a special breed. There were no substantial negative issues observed in the experimentation. It is all very interesting stuff. Thanks again.
         

[Paul Birkeland]

In the testing there is no observable variance in amplitude of the positive and the negative half relative to the mean, even for a +/- 5 volt peak to peak output swing. This was using a 10Kohm or 100Kohm series input resistance. In other words no square law function was observable to indicate any grid current issues. In the case of the 100Kohm, as more seriously affected by grid current, was only observable starting to limit at around 7volts or 8volts negative relative to positive half. 
   

Man, you're either really good, or you're really lucky!

Slap a fet follower on it and you'd have an interesting preamp!

[Paul Joppa]

The easy check for saturation current is to connect the grid to the plate  :^)

[Hierfi]

In the testing there is no observable variance in amplitude of the positive and the negative half relative to the mean, even for a +/- 5 volt peak to peak output swing. This was using a 10Kohm or 100Kohm series input resistance. In other words no square law function was observable to indicate any grid current issues. In the case of the 100Kohm, as more seriously affected by grid current, was only observable starting to limit at around 7volts or 8volts negative relative to positive half. 
   

Man, you're either really good, or you're really lucky!

Slap a fet follower on it and you'd have an interesting preamp!

 
There seems no end to what you can try with grounded heaters, or otherwise. To experiment, all that is necessary is to use an adjustable current source or a voltage source that can operate at low voltages. It seems overtly limiting for experimenters like us to restrict heater currents based upon criteria of normal operation that doesn't apply, particularly in the hunting in the bushes for the last bit of resolution we strive to obtain.

It is a simple matter to try the same thing with the 3S4 tube. Just select the plate load resistor and then adjust the heater current to cause a 10volt drop for the 20volt supply. Done. Either you exceed the heater current limits, end up with ion bombardment (sounds appealingly heavy and ominous), excessive grid current or low input impedance. What can be most critical is having low inherent voltage gain that requires higher input amplitudes to produce a desired output amplitude.       

In some ways it seems silly not to remove the resistor/capacitor pair in the cathode of the current quickie, short the heater to ground and diminish heater current to cause the same voltage drop across the 4Kohm plate load resistor. You end up using less heater current, no components in the capacitor to ground and the same output signal parameters. In other words, no fet follower or other buffer required. With that said I am not remotely suggesting that the removal of the capacitor/resistor pair in the cathode would improve and not seriously deteriorate the reproduction of music by the quickie as it stands. The reality rules. 

Logic is the systematic method of coming to the wrong conclusion with confidence. It seems that feeling confident, smart or lucky is a diversion to emotions from reasoned thought. It seems for another discussion on which is more appearling, the chicken or the egg. 

[Hierfi]

The easy check for saturation current is to connect the grid to the plate  :^)

Sure Paul. Saturation voltage occurs at 700mV from ground or 19.3Volts below the 20Volt supply. Hence saturation current is equal to 19.3 volts across the plate load resistance of 500 Kohm or 38.6uAmps. The plate saturation current cannot exceed 40uA as restricted by having all voltage drop across the 500Kohm plate resistor and none remaining on the plate. Under saturation conditions the plate dissipation was 27uW.

In an act of exceeding trepidation and brevity I connected the grid and plate together and attached it directly to the 20volt supply. The resulting saturation current was about 38mA or more than 7 times the 5mA maximum design limit. Plate dissipation was only about 760mW although more than 30,000 times that with a 500Kohm plate load resistor. I suspect some awkwardness developed by various elements in the tube and some consideration by the ions.   

[Hierfi]

The easy check for saturation current is to connect the grid to the plate  :^)

I hope everyone doesn't mind my input. A couple of errors to report. First, the saturation current with the grid and plate attached to 20volts appeared unbelievable high. In rechecking it was around 3mA not 38mA. As Gilda Radner would say "Oh, never mind."

Secondly, the input resistance of the circuit was far higher than reported. I inadvertently left a 100Kohm resistor to ground under testing, hence the input impedance was found to be 100Kohm. Wow, I have made this mistake before. Anyway, after removing the offending resistor the input is around 5Mohm.

To determine further what was happening I left a 1Mohm resistor in series with the input to observe grid currents. The findings were that at around positive 0.5 or 0.6 volts the input voltage began visibly limiting on the grid. This suggests that for an observable error of 0.1 volt using a 1Mohm resistor implies grid currents where in the 0.1uA region. This is not necessary low. However, under conditions of a circuit with 10x gain this would become observable at -5 or -6volts on the plate. It appears proper to consider that the output is a reflection of what is happening on the grid and judged accordingly.   

The findings appear correlating in a reasonable way to expectations in the discussions. This raises the question as to the interaction of the input potentiometer having a variant source impedance to the input. For a zero source impedance to the pot the maximum input impedance to the amplifier is 1/4 of the value of the pot. Even for 100Kohm this would be 25Kohm. It seems this could work.           

[Paul Joppa]

...
In an act of exceeding trepidation and brevity I connected the grid and plate together and attached it directly to the 20volt supply. The resulting saturation current was about 38mA 3mA... [corrected]

Wow - I had hoped you might try that, and I am stunned. Even 3mA is far, far higher than I expected. Clearly that 13th-power rule is NOT applicable to oxide cathodes! Thanks so much for this contribution to our knowledge.

Incidentally, the more important purpose of the cathode resistor is to adjust the bias and operating point as the high-voltage battery ages, to maintain reasonably good operation over the 24-36v range. With a fixed high voltage, I would have used an LED or something similar - because I totally agree, the resistor/capacitor combination does limit the ultimate audio performance.

[Hierfi]

Incidentally, the more important purpose of the cathode resistor is to adjust the bias and operating point as the high-voltage battery ages, to maintain reasonably good operation over the 24-36v range. With a fixed high voltage, I would have used an LED or something similar - because I totally agree, the resistor/capacitor combination does limit the ultimate audio performance.

It is nice when the preamplifier expires when the batteries expire and a circuit that uses the fewest  components to address a multiplicity of concurrent issues. It is a sign of admirable elegance. Anyway, LED's are simple and have a reasonable transfer curve without going unnecessarily blunt as a constant voltage drop. For better or worse it adds a semiconductor in the signal path and that I will need of add of such devices to create a buffer in the circuit I would like to try out.

As an aside, what are your thoughts on using constant current sources in the B+ supply? I have been considered this for awhile and have some thoughts on why this can have benefits.
               

[Paul Joppa]

Sorry for the late response - the "show unread posts" function has been very unreliable for me lately.

The PJCCS optional kit is a current source as plate load. All of our shunt regulators in the higher-end kits use a current source to feed the shunt regulator. Obviously I - and we - like them!

[Hierfi]

Sorry for the late response - the "show unread posts" function has been very unreliable for me lately.

The PJCCS optional kit is a current source as plate load. All of our shunt regulators in the higher-end kits use a current source to feed the shunt regulator. Obviously I - and we - like them!

Current sources seem as a clear advantage respecting noise on the B+ supply. I have a coupe of attachments to explain what might be going on. Given that B+ is in series with the plate load resistor feeding the output this can be considered a voltage source with an output impedance that is the load resistor. This is one of two networks feeding the output, the other is the tube with its inherent plate resistance. Both can be represented as indicated by A in the AC equivalent circuits sketches.

Both sources are connected to feed the same point, the output of the preamplifier for example. This is indicated in figure C. The second page indicates the actions of resistive division. Hence we can simply set 1 volt  of noise on the B+ and zero volts on the tube side. We can then find the output as a result of voltage division and determine the dB values of B+ noise attenuation.  For the analysis I have used 8 examples numbered on the second page from the top to bottom on the left side of the page. Note that the voltages appearing in the center can be considered as the output of the preamplifier. For a 1 volt output this translates into 0dB of attenuation of B+ noise, the worst case. For zero volts output this translates into infinite attenuation of B+ noise, the best case.

1. With a plate load resistor at 0 ohms and the plate resistance at 10K ohms the output is 1 volt as equal to the B+ This indicates 0dB of attenuation of B+ noise. This is also independent of any plate resistance that is non-zero. In other words no attenuation of B+ noise exists at the output.
2. Again 0dB of attenuation given that the tube resistance is infinite. In other words there is no voltage drop across the 10K ohm given that no current can flow through the infinite plate resistance 
3. Only  about 1dB of attenuation in cases when plate resistance is considerably higher than the plate load resistor. As an example for the quickie with a 4k ohm load resistor and perhaps 8K ohm plate resistance (don't really know) the attenuation might be around 3dB. Also of note is that pentodes can have inherently higher plate resistance and result in less B+ noise attenuation
4. There is 6dB of attenuation when plate resistance and plate load resistances are equal
5 When the plate load resistance is 9x higher than the plate resistance the attenuation of B+ noise is now about 20dB ( this is around the attenuation I am getting in the manner implementing the 3A5 tube)
6 This is the current source (if perfect) having infinite B+ noise attenuation.
7 This shows there is no dependency on plate resistance, still infinite B+ noise attenuation
8. This shows infinite B+ noise attenuation for plate resistances at zero ohms. For something like the 300B tube with 800 ohms plate resistance and the same plate load resistor of 4K ohm as the quickie the 300B can achieve perhaps 15dB of nose attenuation compared to the quickies at 3dB.

What is important to note, seemingly the most critical point, is that D in the AC equivalent circuits can be viewed the reality. For situations with 0 dB B+ noise attenuation the output can be considered fed by the B+ supply instead of ground and whatever signal happens to be on top. On the tube side the output is being fed from ground and whatever signal happens to be on top. In other words it seems far better to keep to the tube side having proportionately lower resistance. 

Clearly constant current sources never reach infinite attenuation yet yours must be more than adequate given the reality of your findings. .   

Hopefully my thinking isn't totally flawed. Anyway, back to the dungeon. Since my 3A5 couldn't drive anything useful I am trying to turn it into a little 2-3 watt "quickie integrated"  and  skipping the preamp or headphone amp, though I fear revealing of bipolars amongst the bottleheads!