Yes. I guess the first step is to recognise the problem for cascode stages. But I really think the opportunity lies mostly in the simple triode with a CCS anode load + buffer stage.
For an output stage driving real current, screen grid drive is reasonably linear. Just check the screen grid curves for a typical tube (bottom page 3):
http://frank.pocnet.net/sheets/123/6/6HE5.pdf
But that requires a lot of drive voltage and is low impedance besides. Tough on the driver stage instead.
So instead, the next step is to use a low internal Mu beam Sweep tube (Mu like 3 or 4) and split the drive between screen grid and grid1 with a 1/Mu resistive divider (down to some negative bias V). So each grid is doing equal work. With Mu 4, it only takes 2X the usual grid 1 drive to operate the screen grid. So Mu scaled g2/g1 drive is easier on the driver and less hazardous for the driven tube as far as runaway.
Then, one can go further. Since the screen grid tends to intercept a constant fraction of the plate current (like 5 or 10%) when the plate V is well above the screen V. It can be seen as a bipolar transister with a current gain Beta (10 or 20). Now since Beta will droop off when the plate V gets low, we again combine it with a resistive divider (down to neg. bias V) 1/Mu drive of grid1. So a current drive drives the screen grid, and the AC voltage appearing on the screen gets divided by Mu to operate grid1 (in neg. territory). Now grid1 has an expansive power law, so will tend to compensate the screen grid Beta droop. One can maybe tweak the divider ratio to get best compensation for overall linearity. And the driver impedance running the screen grid can also be optimised for best linearity compensation. (screen grid current drive is compressive, screen grid voltage drive is expansive, so we find the optimum in between.)
(screen grid will do 2/3 power I to V conversion, before Beta droop. g1 will do more like 2.0 power V to I so will be expansive, 2/3 * 4/2 => 4/3 power) (so two parameters to play with to get best linearity, driver impedance, and R divider ratio for grid1)
http://frank.pocnet.net/sheets/123/6/6HE5.pdf
But that requires a lot of drive voltage and is low impedance besides. Tough on the driver stage instead.
So instead, the next step is to use a low internal Mu beam Sweep tube (Mu like 3 or 4) and split the drive between screen grid and grid1 with a 1/Mu resistive divider (down to some negative bias V). So each grid is doing equal work. With Mu 4, it only takes 2X the usual grid 1 drive to operate the screen grid. So Mu scaled g2/g1 drive is easier on the driver and less hazardous for the driven tube as far as runaway.
Then, one can go further. Since the screen grid tends to intercept a constant fraction of the plate current (like 5 or 10%) when the plate V is well above the screen V. It can be seen as a bipolar transister with a current gain Beta (10 or 20). Now since Beta will droop off when the plate V gets low, we again combine it with a resistive divider (down to neg. bias V) 1/Mu drive of grid1. So a current drive drives the screen grid, and the AC voltage appearing on the screen gets divided by Mu to operate grid1 (in neg. territory). Now grid1 has an expansive power law, so will tend to compensate the screen grid Beta droop. One can maybe tweak the divider ratio to get best compensation for overall linearity. And the driver impedance running the screen grid can also be optimised for best linearity compensation. (screen grid current drive is compressive, screen grid voltage drive is expansive, so we find the optimum in between.)
(screen grid will do 2/3 power I to V conversion, before Beta droop. g1 will do more like 2.0 power V to I so will be expansive, 2/3 * 4/2 => 4/3 power) (so two parameters to play with to get best linearity, driver impedance, and R divider ratio for grid1)
Last edited:
Hmmm, looking at the 6HE2 g2 drive curves I get n is about 1.42, pretty close to 3/2 which is normal for g1 drive in most valves. The point I make in the OP is that this non-linearity appears avoidable no matter what n is, or which grid is driven ...............
Last edited:
I get 2.63 power (V to I) for the 6HE7 g1 (using -39V as cutoff for grid1 V), and 1.23 power for g2. Leaving 1.4 power (of V) difference between the grids. So Mu would be changing almost linearly with current. Most TV sweep tubes are in this range for g1 gm, 2.0 to 2.8 power.
The 2.63 power law for g1 could then be degenerated back to approx. 2.0 power (Vg to I) by a small tail resistor on a class A, LTP differential stage. Then gm varies as 1.0 power (of Vg) for each device, and sums to a constant for differential mode, giving linear gain directly from the plates (for pentodes). (opposite gm ramps sum to a constant)
But as you say, a constant current load will give near constant V gain too. Buffers needed to drive a load then.
What I was saying above earlier, was that by using compressive current screen grid drive and expansive grid1 V drive together, one could get a linear output tube directly. Even a 300B is not linear directly, with a load on it.
The 2.63 power law for g1 could then be degenerated back to approx. 2.0 power (Vg to I) by a small tail resistor on a class A, LTP differential stage. Then gm varies as 1.0 power (of Vg) for each device, and sums to a constant for differential mode, giving linear gain directly from the plates (for pentodes). (opposite gm ramps sum to a constant)
But as you say, a constant current load will give near constant V gain too. Buffers needed to drive a load then.
What I was saying above earlier, was that by using compressive current screen grid drive and expansive grid1 V drive together, one could get a linear output tube directly. Even a 300B is not linear directly, with a load on it.
Last edited:
Is that your tubelington PP OPS formed as SEPP power bridge( Single Ended Push Pull ) configuration ? or ...,
Which type of BJT`s you use for B+ = 200VDC ?
Maybe some basic schematic ?
Regards
Cascodes are not very linear. That is why they are only used for input circuits or where nonlinearity is a requirement. Big snag is that cascode has very poor PSRR, so difficult to use for input. Cascode also has high output impedance. However, in these respects they are no worse than a pentode and have the advantage of no partition noise.
Took a fresh look at the tublington thread, and curves for it posted there. Transfer characteristic appears to be exponential Ia versus Vg, ie most like a BJT, which is less linear than a regular triode with a 3/2 power law. No surprise I suppose, but doesn't look like that helps out here FWIW.........
Yes. I suppose cascodes have niche applications, in that the transfer characteristic is from the bottom triode, but beneficial features of the top valve such as power rating or voltage rating can be applied. Plus cascodes avoid miller C. But linear they ain't, at least not naturally.
Last edited:
Generally true, but not necessarily. I remember seeing a cascode circuit from Tim deParavicini that looked like it would be awful: ECC88s running at starved-current levels and a very high plate resistor. I built and measured it just for curiosity (Tim couldn't be THAT wrong, could he?) and was astonished at how low the distortion was (it's probably posted somewhere here on the forum).
In retrospect, I can at least get a hint about why this was true: the high value resistor on the top tube's plate provided a reasonable load for the bottom tube. There's a tradeoff between Miller capacitance reduction and linearity that he saw and I (initially) didn't.
"Took a fresh look at the tublington thread, and curves for it posted there. Transfer characteristic appears to be exponential Ia versus Vg, ie most like a BJT, which is less linear than a regular triode with a 3/2 power law. "
Some link to this tublington or tubelington thread?
All I could find with a search here was a comment by ThorstenL.
Some link to this tublington or tubelington thread?
All I could find with a search here was a comment by ThorstenL.
Feed the cascode or pentode from a CCS. Put the load resistor from the plate to ground. Have the CCS provide current for both the load resistor and the amplifying device.
You now have your high gain, great PSRR, and a very short signal current loop.
The resistor makes the DC operating point stable. You can pull the signal from the MU output of the CCS if you want or need the buffering.
You now have your high gain, great PSRR, and a very short signal current loop.
The resistor makes the DC operating point stable. You can pull the signal from the MU output of the CCS if you want or need the buffering.
Re: stocktrader
The 82 Ohm emitter resistors are changing the transistor drive sensitivity to voltage input somewhat. That might account for some bipolar exponential character intruding that Luckythedog mentioned. If the BU407 are near constant Beta, it could help to remove the emitter resistors, or at least lower them to a low R Isense value, for some idle current control servo to operate from.
Luckythedog:
"Took a fresh look at the tublington thread, and curves for it posted there. Transfer characteristic appears to be exponential Ia versus Vg, ie most like a BJT, which is less linear than a regular triode with a 3/2 power law. "
"Transiodarius" "triodemultipler, valvelington, tubelington, triodelington"
-----------------------------------------------
Re: Gary P
"Feed the cascode or pentode from a CCS. Put the load resistor from the plate to ground. Have the CCS provide current for both the load resistor and the amplifying device. "
Maybe take an R divider tap off the load resistor to control the cascode (top) grid/gate. Giving it finite Mu controlled gain.
The 82 Ohm emitter resistors are changing the transistor drive sensitivity to voltage input somewhat. That might account for some bipolar exponential character intruding that Luckythedog mentioned. If the BU407 are near constant Beta, it could help to remove the emitter resistors, or at least lower them to a low R Isense value, for some idle current control servo to operate from.
Luckythedog:
"Took a fresh look at the tublington thread, and curves for it posted there. Transfer characteristic appears to be exponential Ia versus Vg, ie most like a BJT, which is less linear than a regular triode with a 3/2 power law. "
"Transiodarius" "triodemultipler, valvelington, tubelington, triodelington"
-----------------------------------------------
Re: Gary P
"Feed the cascode or pentode from a CCS. Put the load resistor from the plate to ground. Have the CCS provide current for both the load resistor and the amplifying device. "
Maybe take an R divider tap off the load resistor to control the cascode (top) grid/gate. Giving it finite Mu controlled gain.
Last edited:
OldEurope's (Darius) design without the emitter resistor looks like it preserves the tube characteristic well (he just used 12AU7 that looks bad)
http://www.vintage-radio.net/forum/attachment.php?attachmentid=7529&d=1167819193
Putting an emitter resistor in, changes the circuit to a V to I converter, which will not sound or curve trace the same.
http://www.vintage-radio.net/forum/attachment.php?attachmentid=7529&d=1167819193
Putting an emitter resistor in, changes the circuit to a V to I converter, which will not sound or curve trace the same.
the emitter resistor linearizes the transistor while also feeding the tube transfer characteristic at manageable ( 75 ma to 150 ma ) current levels.
a ground biased 6922 will drive the resistor-less BU407 or MJE13007 to over an amp.
I have been thinking of trying an OTL 6922 amp running this way.
a ground biased 6922 will drive the resistor-less BU407 or MJE13007 to over an amp.
I have been thinking of trying an OTL 6922 amp running this way.
You can't have both high gain and great PSRR in this way, as the resistor value pulls these in opposite directions.Gary P said:
Low resistor value: low gain, high distortion, good PSRR.
High resistor value: high gain, low distortion, poor PSRR.
By using a CCS and putting the load resistor to ground you swap the PSRR behaviour. To get gain and PSRR pulling in the same direction you need the resistor to go to the supply rail, as in the non-CCS circuit.
From the AC equivalent circuit point of view there is no difference in having the plate resistor connected to B+ or ground. As such, the distortion will be determined by the active device (cascode, pentode) and the load resistor.
In the case of the plate resistor connected to B+ you have the power supply in the signal loop. Any noise on the power supply is added in series with the signal. With a cascode or high plate resistance pentode the PSRR approaches zero.
In the case of the plate resistor connected to ground the power supply is not in the signal loop. Any noise on the power supply is attenuated by the ratio of the CCS impedance divided by the plate resistor value. With a good performing CCS the PSRR ratio of the circuit can be very high, unlike the standard circuit with the plate resistor to B+.
The AC performance of the 2 circuits below is equivalent. The difference is in the PSRR.
An externally hosted image should be here but it was not working when we last tested it.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.