I have made a number of amps, PP and SE operating in Class A1. Now I am interested to push to Class A2. The reason would be to either get more power, or be able to operate at a slightly lower B+ (or a bit of both).
My target for now a type 50 or a 2A3. I have not found plots of grid current for either anywhere, nor for type 45. Is this because they are unsuitable for Class A2 or simply a common omission?
I know that A2 means you need a driver that can deliver current and also that the fixed bias source must be shunt regulated so that the bias voltage is not affected due to the rectifying effect of the grid current (G1 becomes a second 'anode'). What I don't know is, even approximately, how much current is required. For a 211 or 845 it can be a lot, maybe 100's of milliamps in the extreme.
My driver is a 45 or a VT25 (10Y) in SE into a 5k:5k interstage with a 30mA DC standing current rating. For bias I propose this:
I believe that most bias circuits such as by Pete Millet, Rod Coleman or Guido Tent can only sink a few milliamps. Is that right? If the driver has to deliver the meat and potatoes of grid current, what is the effect at the bias source of the rectified grid current?
BTW: I would like to use fixed bias so as to limit power loss and heat generation in a cathode resistor, however, with cathode bias this 'rectified G1 current' problem should go away since the grid would have a direct route to the low impedance 0V line through the interstage secondary winding. Yes, the interstage tolerates standing DC, although the rectified G1 current would in reality be a pulsating DC signal.
Also: The type 50 recommends no more than 10k ohms in the grid circuit, a low value.
Any comments gratefully received.
My target for now a type 50 or a 2A3. I have not found plots of grid current for either anywhere, nor for type 45. Is this because they are unsuitable for Class A2 or simply a common omission?
I know that A2 means you need a driver that can deliver current and also that the fixed bias source must be shunt regulated so that the bias voltage is not affected due to the rectifying effect of the grid current (G1 becomes a second 'anode'). What I don't know is, even approximately, how much current is required. For a 211 or 845 it can be a lot, maybe 100's of milliamps in the extreme.
My driver is a 45 or a VT25 (10Y) in SE into a 5k:5k interstage with a 30mA DC standing current rating. For bias I propose this:
I believe that most bias circuits such as by Pete Millet, Rod Coleman or Guido Tent can only sink a few milliamps. Is that right? If the driver has to deliver the meat and potatoes of grid current, what is the effect at the bias source of the rectified grid current?
BTW: I would like to use fixed bias so as to limit power loss and heat generation in a cathode resistor, however, with cathode bias this 'rectified G1 current' problem should go away since the grid would have a direct route to the low impedance 0V line through the interstage secondary winding. Yes, the interstage tolerates standing DC, although the rectified G1 current would in reality be a pulsating DC signal.
Also: The type 50 recommends no more than 10k ohms in the grid circuit, a low value.
Any comments gratefully received.
The grid current is going to be supplied by the driver, not by the bias circuit. You will have a resistor (typically 100K) between the bias voltage source and the grid.
Edit: Just noticed the 10K recommended resistor value. That will change things a bit.
Edit: Just noticed the 10K recommended resistor value. That will change things a bit.
1.) Both driver tube has low gain to drive #50, or 2A3, even to A1.
2.) Interstage is unsuitable to drive power tube grid to A2.
A2 capable drivers use cathode follower or FET source follower to drive several mA current to grid.
My circuit is running in A1 now and sounds great. I get around 3.75W from a 2A3 and some 4.75W from a '50. I am happy with A1, which is not to say that it is perfect, or even right.
I made the mistake of asking too many questions in one post, what I really want to know above all is the likely magnitude of grid current for modest positive grid voltages, and even better if there is actual data on this somewhere for the 45, 50 and 2A3.
I made the mistake of asking too many questions in one post, what I really want to know above all is the likely magnitude of grid current for modest positive grid voltages, and even better if there is actual data on this somewhere for the 45, 50 and 2A3.
Tubes designed for transmission (cable drivers, radio stations, radar etc.) even working in positive grid voltage range (with more or less grid current).
I'm afraid, that #50 or 2A3 grid structure not suitable for this.
For example I use 300B in A2 capable layout, but only few type can this without grid damaging.
In my collection e.g. 841, 10Y/801/801a, 845, 211 accepts this mode.
I'm afraid, that #50 or 2A3 grid structure not suitable for this.
For example I use 300B in A2 capable layout, but only few type can this without grid damaging.
In my collection e.g. 841, 10Y/801/801a, 845, 211 accepts this mode.
A2 operation was not common back when the 50, 2A3 and 45 were introduced. AB2 operation in push pull was sometimes used with transformer coupling. There have been several different specs for the type 45 issued over the years. All but the oldest mention AB2 operation with a pair of 45's generating nearly 20 watts in AB2. The US specs all discuss "driving power" needed, but do not mention grid current directly. I found an Australian 45 spec that states 1 watt of driving power is needed for 19.1 watts of power output with the peak grid current listed as 8.36 mA. I have tested lots of 45's over the years and many old and well used tubes that work fine in a SE amp running 30 mA or so of plate current are not up to the task of AB2 operation where the peak plate current is over 100 mA. There just isn't enough peak emission capability left in the old filaments.
Push Pull experiments with new Chinese 2A3's got almost 30 watts on 300 volts and over 35 watts into 3300 ohms on 375 volts of B+.
My Tubelab SE and TSE-II amps all incorporate A2 capability and use the 45, 2A3 or 300B all work fine in A2. Peak grid currents in the 20 to 25 mA range were seen on Chinese 2A3's at 300 volts for over 5 watts into a 3000 ohm OPT. The grid current is lower on 375 volts for almost 6 watts at 65 mA. Yes, this is a bit over spec. Ever see a 15 year old 2A3 with (benign) blue glow on the glass?
I measured grid currents of over 500 mA on an 833A making 200 watts. I have also seen over 50 mA on 6L6GA's in AB2 at over 100 watts per pair.
The "PowerDrive" circuit that I developed for feeding hungry tube grids is discussed here:
http://tubelab.com/articles/circuits/power-drive/
The complete SE amp schematic is in post #1 here:
https://www.diyaudio.com/community/threads/after-a-14-year-run-the-tse-must-die.331038/
The push pull driver schematic is in post #2 here:
https://www.diyaudio.com/community/threads/tubelab-universal-driver-board-2015-version.316225/
Attachments
Driving your triode output stage into a2 requires a strong low z driver tube because when in a2 the output grid z drops massively to something like 2k with grid current. This is why you see those japanese amps with 300b driving 300b aka marantz. They also use it so the secondary bias is suppliedthrough the it sec and can be of low z to supply grid current. A2 is comlpex and takes big power from your driver gd luck.
Class A2 or AB2 operation has asymmetrical drive requirements. As mentioned, on the rising edge of a sine or other wave the grid impedance will go from very high at highly negative voltages with respect to the cathode to very low as the grid is forced positive. The impedance remains high and mostly capacitive (miller) as the grid to cathode voltage approaches zero. Many tubes, especially DHT's will go to a relatively low impedance, and begin to draw current while the voltage on the grid is still a volt or more negative. It will transition to a very low impedance as the grid goes positive, and continue to drop as the grid goes more positive. The instantaneous impedance can be below 1 K ohms on some tubes. The driver MUST be capable of dealing with this load change WITHOUT any distortion. No capacitor coupled circuits will work as the load change will upset the state of charge on the cap leading to inevitable blocking distortion or worse.
In the early days a driver transformer fed by a tube capable of meeting the "drive power" requirements was the design choice. A decent high Gm cathode follower directly connected to the grid can also work. Today, we have mosfets. I use a follower with the cathode or source tied directly to the grid to be driven. There must be a resistor or active CCS circuit from this grid to a negative voltage source with enough voltage to completely cut off the driven tube. This resistor or CCS must be capable of discharging the miller capacitance of the driven tube quickly enough to avoid slew rate limiting on fast negative going transients. The plate or drain of the follower goes to a low impedance voltage source capable of meeting the drive voltage requirements. Some current limiting on this source can be useful on amps that may see continuous over drive (guitar amps, or bass thumpers) to avoid blowing a grid (possible).
Done properly A2 or AB2 can work well. Some tubes like the 6L6 family respond well and some don't. Most TV sweep tubes don't need or want A2 or AB2 operation.
In the early days a driver transformer fed by a tube capable of meeting the "drive power" requirements was the design choice. A decent high Gm cathode follower directly connected to the grid can also work. Today, we have mosfets. I use a follower with the cathode or source tied directly to the grid to be driven. There must be a resistor or active CCS circuit from this grid to a negative voltage source with enough voltage to completely cut off the driven tube. This resistor or CCS must be capable of discharging the miller capacitance of the driven tube quickly enough to avoid slew rate limiting on fast negative going transients. The plate or drain of the follower goes to a low impedance voltage source capable of meeting the drive voltage requirements. Some current limiting on this source can be useful on amps that may see continuous over drive (guitar amps, or bass thumpers) to avoid blowing a grid (possible).
Done properly A2 or AB2 can work well. Some tubes like the 6L6 family respond well and some don't. Most TV sweep tubes don't need or want A2 or AB2 operation.
As a side note there are still transformer manufactures making It transformers. Tango hashimoto audionote uk monolith magnetics etc. Generally they are 1:1 ratio devices. The bandwidth of some of these tranformers is amazing. The tango nc2o is rated 4hz to 160khz. Its power rating is 160rms @10hz.yes thats correct 10hz. So the resources are there.
The trouble with inductors in the grid circuit is that they ring when grid current cuts on and off (di/dt).
All good fortune,
Chris
All good fortune,
Chris
These tubes were designed to give useful power without the extra cost of high-power drivers. They were designed to give high plate current at zero grid and low plate voltage. For reasonable load impedance, you won't get a lot more out even if you extend the loadline from zero grid all the way to the left side of the plot (you can't actually get there anyway).
As a rule of thumb, I am not going to buy your replacement tubes: you can use a general purpose triode as cathode-follower and drive grids positive, and power grids probably won't melt, especially on unclipped speech/music. There's no specific limit on 6L6 grid, but Mac sure made enough AB6 6L6 amps, even in lab-work (steady tone all day). IIRC 6F6 also got flogged AB2.
Thank you for this advice and information. I see now that positive grid voltage makes the grid look more like the base of a transistor, current controlled. I hadn't appreciated this would mean they tend to operate as voltage clamps as well. It makes the voltage curves showing positive grid potentially (sorry for the pun) misleading.
I like the idea of the Power Drive circuit (and have no problem with silicon in my amps), however, I own a pair of Tamura A342 IT's which are quite substantial with big cores, gapped for 30mA DC and in heavy iron cases, so I am inclined to try to use them. I accept that the 45 or 10Y driver will run out of juice if the output valve grid current is too high, but it seems to me that in principle an IT could deliver grid current if it is 'powerful' enough. Now I also realise what 'driving power' means in old data sheets. I had always wondered about that.
Thank you also for actual values of grid current. I had guessed that if the anode draws, say, 50mA the grid would be less but not a lot less, being in such close proximity to a seething fog of electrons looking for the easiest route back home. Brings to mind the rather unusual mode of operation where the grid is the anode and the anode is the grid (can't remember what that is for, but I think it is used in transmitter circuits sometimes)
I am a retired electronics engineer with a long career in a different electronics realm and new to valves, so I thank you for your patience with my questions.
So here are two more:
1 When the output valve rectifies the grid current what effect does that have on the bias generation circuit? I know it tends to drive it more negative thereby re-biasing the valve 'colder' and cancelling some of the extra drive benefit, but by how much? If, say, the grid is 10V positive does that inversely impress the 10V on the bias voltage, e.g. if bias is at -50V does it momentarily change to -60V? (assuming a high impedance bias source). That's half a question, because I also need to know if the rectified portion dumps any current onto the bias circuit or does the cathode deal with it exclusively? Put more simply: how low a source impedance does the bias generator need to be if the grid is drawing current? Yes, I see that in the Power Drive circuit the bias is applied via the source follower so the bias circuit is not exposed to the operation of the output valve grid.
2 What about grid stopper resistance? In Class A1 I am using 1k8 at the moment. I presume that has to be eliminated or greatly reduced when in Class A2. Anyone care to suggest a value? 0R? 10R? 100R?
Finally, an observation: at or around 0V Vgk there may be grid current into the more negative end of a DHT filament while the more positive end is still operating electrostatically with the grid. Does this potentially impair valve filaments or have other consequences?
Thank you.
I like the idea of the Power Drive circuit (and have no problem with silicon in my amps), however, I own a pair of Tamura A342 IT's which are quite substantial with big cores, gapped for 30mA DC and in heavy iron cases, so I am inclined to try to use them. I accept that the 45 or 10Y driver will run out of juice if the output valve grid current is too high, but it seems to me that in principle an IT could deliver grid current if it is 'powerful' enough. Now I also realise what 'driving power' means in old data sheets. I had always wondered about that.
Thank you also for actual values of grid current. I had guessed that if the anode draws, say, 50mA the grid would be less but not a lot less, being in such close proximity to a seething fog of electrons looking for the easiest route back home. Brings to mind the rather unusual mode of operation where the grid is the anode and the anode is the grid (can't remember what that is for, but I think it is used in transmitter circuits sometimes)
I am a retired electronics engineer with a long career in a different electronics realm and new to valves, so I thank you for your patience with my questions.
So here are two more:
1 When the output valve rectifies the grid current what effect does that have on the bias generation circuit? I know it tends to drive it more negative thereby re-biasing the valve 'colder' and cancelling some of the extra drive benefit, but by how much? If, say, the grid is 10V positive does that inversely impress the 10V on the bias voltage, e.g. if bias is at -50V does it momentarily change to -60V? (assuming a high impedance bias source). That's half a question, because I also need to know if the rectified portion dumps any current onto the bias circuit or does the cathode deal with it exclusively? Put more simply: how low a source impedance does the bias generator need to be if the grid is drawing current? Yes, I see that in the Power Drive circuit the bias is applied via the source follower so the bias circuit is not exposed to the operation of the output valve grid.
2 What about grid stopper resistance? In Class A1 I am using 1k8 at the moment. I presume that has to be eliminated or greatly reduced when in Class A2. Anyone care to suggest a value? 0R? 10R? 100R?
Finally, an observation: at or around 0V Vgk there may be grid current into the more negative end of a DHT filament while the more positive end is still operating electrostatically with the grid. Does this potentially impair valve filaments or have other consequences?
Thank you.
For input Z calculation, you need IV characteristic curve which includes positive grid range. There is no such curve for 300b or 2a3 etc, you can "build" or extend one yourself like described in this article and cover most of the issues you raised:
https://www.tubecad.com/2016/12/blog0365.htm
You can build A2 curve with Paint Tool and LTspice, by simply extended positive grid range on the IV curve plot.
The grid stopper if any should be carrying equal or larger current than the maximum A2 grid current as not to interfere with the operation.
https://www.tubecad.com/2016/12/blog0365.htm
You can build A2 curve with Paint Tool and LTspice, by simply extended positive grid range on the IV curve plot.
The grid stopper if any should be carrying equal or larger current than the maximum A2 grid current as not to interfere with the operation.
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The positive grid equivalent resistance tends to be asymtotic to a value similar to 1/Gm. Infinity for negative grid and a few hundred ohms for a power tube at positive grid. This is almost logical. The off-cathode space passes electrons with some resistance. It does not matter a lot if they are going past grid to plate or right to grid. Actually they go both ways but Gm rises as current increases so the negative-grid Gm is a fair guide to the higher current positive grid area.
As a rational engineer, you can see that if you must buy a beefy driver and a beefy interstage transformer, you are half-way to just using two power tubes in push-pull, which is more power at less distortion and less iron. That's why you find few A2 circuits (Bogen had a push-pull A2 triode amp) and little A2 data on the makers' sheets.
As a rational engineer, you can see that if you must buy a beefy driver and a beefy interstage transformer, you are half-way to just using two power tubes in push-pull, which is more power at less distortion and less iron. That's why you find few A2 circuits (Bogen had a push-pull A2 triode amp) and little A2 data on the makers' sheets.
Parallel outputs also cuts the output transformer primary impedance in half, which you could argue is worth doing anyway.
All good fortune,
Chris
All good fortune,
Chris
The asymptotic grid current curve shown clearly here for 211:
The Western Electric Type 43 uses 211 in PP with a rather powerful 205D PP driving amplifier, but I'll bet it's miles away from 100's mA grid current.
I fully take the point that a PP would be simpler and more powerful. But maybe also less fun to design 🙂
My SE prototype sounds 'better' than my PP designs. Maybe I just like the sparkle of even harmonic distortion. It sure makes female vocals and brass instruments very 'crunchy' and exciting. Who knows if that makes it HiFi, but I like it. I have read theories about the OPT never reversing the field, always sitting at the best part of the B-H curve, being a reason for SE sound, but I don't know if that is just armchair theorisation...
The Western Electric Type 43 uses 211 in PP with a rather powerful 205D PP driving amplifier, but I'll bet it's miles away from 100's mA grid current.
I fully take the point that a PP would be simpler and more powerful. But maybe also less fun to design 🙂
My SE prototype sounds 'better' than my PP designs. Maybe I just like the sparkle of even harmonic distortion. It sure makes female vocals and brass instruments very 'crunchy' and exciting. Who knows if that makes it HiFi, but I like it. I have read theories about the OPT never reversing the field, always sitting at the best part of the B-H curve, being a reason for SE sound, but I don't know if that is just armchair theorisation...
The grid - cathode interface is in essence a diode. As the driver tries to pull the grid positive with respect to the cathode it will draw current that increases with an increase in positive grid voltage and decreases with an increase in positive plate voltage as the plate attracts more electrons. The effect of this current depends hugely on the driving circuit and its dynamic impedance. The PowerDrive with a low RDSon mosfet tied to a low impedance power supply can have an impedance of an ohm or less when sourcing grid current.
A driver transformer coupled circuit should reflect this dynamic impedance onto the plate of the driver tube according to its impedance ratio, minus any losses which are primarily from its DCR. Any DC bias shift from these losses will become a source of distortion.
Back in the early days of vacuum tube ham radio the diode clamp effect was often used with a high impedance RF source to generate enough negative voltage to bias the tube into class C (cutoff for most of the RF drive cycle). Pulling the crystal out of the Heathkit transmitter guaranteed a melted 6DQ6 tube due to loss of bias.
In either case the stopper resistor is a source of distortion whenever the grid goes positive. I use a 100 ohm 1/4 W metal film resistor as a fuse in case of bias mosfet failure, and a convenient place to hang a pair of scope probes to see the grid current (scope in differential mode).
For most DHT's the current from filament to plate is far higher than from filament to grid, so the effect of grid current is minimal.
This reminded me of some testing to failure I did with a bunch of really old 6L6G's and 6L6GA's driven hard in AB2 push pull. Yes, you can get 110 watts from a pair of old 6L6GA's, but a tube arc will blow the cathode sense resistor apart when attempting to get 120 watts. There are some AB2 experiments scattered throughout the universal driver thread, but the fun with 6L6's stares in post #56.
https://www.diyaudio.com/community/threads/tubelab-universal-driver-board-2015-version.316225/page-3
