I built a 6AS7 SETA yesterday
Kurt Strain
I quickly threw together a SETA using the ridiculously cheap 6AS7 tube
just to see if this tube has the purity of sound that it has in the OTL
configuration through a transformer. With a One Electron UBT-1 and
using the 16 ohm output tap, the primary impedance was 800 ohms for an
8 ohm speaker load. I used a single dual triode 6AS7 per channel with
both triodes driven. Power output was only 2W/ch, but the sound could
be described as big, authoratitive, and clear (within its limitations).
The low rp of the 6AS7 (280 ohms per triode section) gave me a damping
factor of 5.7.
I calculate the best I could do with this tranny using 6AS7's is to
use 2 per channel and I could get about 9W/ch out of it with a damping
factor of 11.4 with no feedback. I ought to try this.
I think the sound was more neutral and controlled than the 2A3 and the
Chinese 300B's I've tried. And the cost for the 6AS7 from New Sensor
is only $4.95 ea.
Why is this outstanding tube, one that can deliver all that current and
sound so good, with two triodes per package, so cheap? With the $95
OPT, the $10 surplus power transformers and $8 surplus chokes, this
could be the finest sounding el-cheapo SETA you could make.
Please don't ask for schematics for this circuit. It's all just hacked
in there right now. Try figuring it out yourself.
Kurt
Sander deWaal
On 18 May 1998 16:26:16 GMT, kst...@sr.hp.com (Kurt Strain) wrote
the following:
>Why is this outstanding tube, one that can deliver all that current and
>sound so good, with two triodes per package, so cheap? With the $95
>OPT, the $10 surplus power transformers and $8 surplus chokes, this
>could be the finest sounding el-cheapo SETA you could make.
While the tube may be cheap (due to enormous surplus),
the drive circuit isn't that simple, if done correctly.
You'll need at least as much as the Vpeaks at the plate
to drive it, and maybe, that's why you only manage to
get 2 watts out of a paralleled triode.
In theory, you should be able to draw 25 % . 2.13 watts
out of it, which is 6,5 watts.
Allowing for some losses in the transformer and circuit, 4 watt should
be possible.
I think you're having a problem with the driver clipping here.
BTW, when fully driven, there's grid current (clas A2).
The driver should be able to sink this current, so a DC coupled
driver can be made.
I did this several years ago in a OTL circuit, wasn't impressed
by the sound, certainly not after all the pain I went through
in designing the driver stage.
If you're getting into this, the 6336A double triode offers even more
than a 6AS7G/6080, with 30 watts each, and a mu of 2.7 .
This really is a nice tube to build an OTL with, even nicer
than a 6C33, IMO.
While the 6C33 is easier to drive, I found the 6336A to be better
sounding, especially in the bass region.
YMMV :-)
_
Sander deWaal
postm...@pegasus.demon.nl
www.pegasus.demon.nl
_______________________________________________
Russ Sadd
Hi,
Sounds like you've done a good job there! I tried designing for one of these
myself a while back - theoretically, a single double-triode with 250V HT in
Class A push-pull would give you 10W. Alas, the grid volt swing proved too
problematic - you needed some 125V peak-peak to do this. I suppose you could
easily get this in single-ended mode with a high-ratio coupling transformer.
Anyway, 5W should be more than enough for efficient speakers! Otherwise,
yep, damn cheap valves, damn good quality, damn easy circuit to design for -
just shame about the mu!
Best of luck!
Russ
Kenny MacKay, P. Eng.
Kurt:
Shhhh!
It's posts like yours that cause non-technical tube dealers to drive
the prices to ridiculously high levels!! The same thing happened in
vintage guitars... non-musician dealers driving the resale prices to a
level that no real musician in his right mind would pay!!!
Let's try to keep tube prices at reasonable levels for engineers and
experimenters in spite of the speculators.
kenny
Kurt Strain
Kenny MacKay, P. Eng. (cyc...@interlog.com) wrote:
: Kurt:
: Shhhh!
: It's posts like yours that cause non-technical tube dealers to drive
: the prices to ridiculously high levels!! The same thing happened in
: vintage guitars... non-musician dealers driving the resale prices to a
: level that no real musician in his right mind would pay!!!
: Let's try to keep tube prices at reasonable levels for engineers and
: experimenters in spite of the speculators.
: kenny
The good news is that for this 6AS7, both Sovtek and Svetlana are still
making them. I had a shipment of 20 of them sent to me for a total price
including shipping costs coast-to-coast for $106. They must have a lot
of them. I tried to add a Sovtek 5U4G, but they were currently out of
those.
Here's the part I could not believe. I tried to order 9 pin ceramic
sockets from Antique Electronic Supply awhile ago and they told me there
were no ceramic sockets left in stock. Nada. They are in too high a demand
for the Chinese (who are the only suppliers it seems for the good cheap
ones) to keep up with shipments. But you can get the molded kind.
Kurt
Thor...@tnt-audio.com
Hi all,
> If you're getting into this, the 6336A double triode offers even more
> than a 6AS7G/6080, with 30 watts each, and a mu of 2.7 .
> This really is a nice tube to build an OTL with, even nicer
> than a 6C33, IMO.
Speaking of which, there is a 6C33 SE Amp developed by an Italian. It also
uses a Parafeed Output Stage with some real Trick-schemes for a little
NFB.....
Driver is an EL34, probably you'd want to X that one in favour of any flavour
of DHT and slighlty redesign the Drivercircuit to run without NFB....
http://www.geocities.com/ResearchTriangle/8231/ba/msb_aloia_des.jpg
From the same Guy a "REAL" Speaker for SE Triodes:
Apocalypse Now - The Ultimate Loudspeaker
Woofer Beyma 18G400
Mid Audax PR17TX100
Tw Stage Accompany SA8520
NOTE: The Beyma is connected with a filter to reduce its efficiency and
to extend the very low frequency response.
http://www.geocities.com/ResearchTriangle/8231/ba/apocalypse_now_1.jpg
Sensitivity should make it close to 97db/W....
Later Thorsten
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading
André Jute
<Thor...@tnt-audio.com> wrote:
> a "REAL" Speaker for SE Triodes:
The REAL speaker for SE triodes and every other decent amp is a Quad
electrostat. The problem is not the speaker, it is the triodes not
putting out enough power, or the owner wanting too much volume. Get more
powerful triodes and you too can enjoy real speakers.
<G>
Andre
PS Thanks for the reference to the interesting 6C33 amp and accompanying
speaker.
--
Andre Jute an...@indigo.ie COMMUNICATION JUTE
--see our pages for music lovers, writers and audiophiles at
http://indigo.ie/~andre/ComJuteF1.html
http://www.foundmark.com/ComJute/ComJuteF1.html
Kurt Strain
Thor...@tnt-audio.com wrote:
: Hi all,
: > If you're getting into this, the 6336A double triode offers even more
: > than a 6AS7G/6080, with 30 watts each, and a mu of 2.7 .
: > This really is a nice tube to build an OTL with, even nicer
: > than a 6C33, IMO.
: Speaking of which, there is a 6C33 SE Amp developed by an Italian. It also
: uses a Parafeed Output Stage with some real Trick-schemes for a little
: NFB.....
Borbely was selling a 15W/ch monoblock SETA with the 6C33C. It had a
special 600 ohm primary operating at something like .25A bias current.
You don't find a lot of those OPTs around.
By the way, I did add another 6AS7 per channel to my amp. Now I'm up
to 4W/ch. Yep, the big problem is that the low mu (mu = 2) is so low
that you do have to drive the sucker class A2 to get all the power.
Still, not bad at 4W/ch, on par with a single 2A3 amp. I noticed there
are significant losses in this OPT, trying to use the 16 ohm tap for the
8 ohm speaker. That also cuts into damping factor for sure.
It sure sounded different at 4W/ch than 2W/ch. While there was
good big sound at only 2W, it was obviously running with noticeable
distortion, not harsh, but midrange "hyped" for some weird reason. At
double the power the timbre went more toward the expected, or should I
say accurate. Quite decent. It's amazing how much of certain types of
distortion you can tolerate, and even how much of certain types of
clipping. And then on the flipside, it's amazing how so little of another
type of distortion and clipping is totally annoying.
Kurt
Thor...@tnt-audio.com
Kurt,
> Borbely was selling a 15W/ch monoblock SETA with the 6C33C. It had a
> special 600 ohm primary operating at something like .25A bias current.
> You don't find a lot of those OPTs around.
Well, no problem. Have some wound. Better even, wind them yourself. I used to
wind OPT's for my Amp's back in East-Germany.....
> It's amazing how much of certain types of distortion you can tolerate,
> and even how much of certain types of clipping.
Well, to be honest, I can't. I find almost all 300B SET's hopelessly colored
and don't like them much for low level listening.....
> And then on the flipside, it's amazing how so little of another
> type of distortion and clipping is totally annoying.
Yes. X-Over distortion for Example. I do get my "Muscle" Solid State AMp into
clipping sometimes, but usually even my antisocial listening levels do not
require more than about 5 Watt Peak....
Hans Beijner
Hi
If someone is interested in building a SET with 6C33, Tango makes a transformer
especially for this, part no is XE-20-600S, primary 600 ohm, secondary 4,8 and
16 ohm, max 320 mA, price of this is JPY 19160 here in Japan. I have a diagram
of a SET using SRPP coupled 12AX7, anode coupled 12BH7, followed by cathode
follower with the other half of the 12BH7, driving the 6C33 at 200 mA. This
combination gives 13 W from 30 Hz to 50 kHz, no global feedback. I use this
combination myself and I think the sound of this is very nice, I have never
understood why people dont like 6C33, it is a very sweet sounding tube IMHO.
Hans
Kurt Strain wrote:
> Thor...@tnt-audio.com wrote:
> : Hi all,
>
> : > If you're getting into this, the 6336A double triode offers even more
> : > than a 6AS7G/6080, with 30 watts each, and a mu of 2.7 .
> : > This really is a nice tube to build an OTL with, even nicer
> : > than a 6C33, IMO.
>
> : Speaking of which, there is a 6C33 SE Amp developed by an Italian. It also
> : uses a Parafeed Output Stage with some real Trick-schemes for a little
> : NFB.....
>
> Borbely was selling a 15W/ch monoblock SETA with the 6C33C. It had a
> special 600 ohm primary operating at something like .25A bias current.
> You don't find a lot of those OPTs around.
>
> snip
> Kurt
Thor...@tnt-audio.com
Hi,
> If someone is interested in building a SET with 6C33, Tango makes a
> transformer especially for this, part no is XE-20-600S, primary 600 ohm,
> secondary 4,8 and 16 ohm, max 320 mA, price of this is JPY 19160 here
> in Japan.
Would you mind telling us just about how much Dollar the Yen for these
Day's...?
>I have a diagram of a SET using SRPP coupled 12AX7, anode coupled
>12BH7, followed by cathode follower with the other half of the 12BH7, driving
>the 6C33 at 200 mA. This combination gives 13 W from 30 Hz to 50 kHz, no
>global feedback.
Sounds interesting. Any Chance you could scan the Schematic....
>I use this combination myself and I think the sound of this is very nice, I
>have never understood why people dont like 6C33, it is a very sweet
>sounding tube IMHO.
I think it's not the Valve itself that is the Problem. The 600 Ohm X-Formers
are kinda hard to find as Kurt pointed out. I may try Sowter Transformers
here in England to make me a Pair....
Ross Lipman
Adam Stouffer wrote:
>
> Hans Beijner wrote:
> >
> > Hi
> >
> > If someone is interested in building a SET with 6C33, Tango makes a transformer
> > especially for this, part no is XE-20-600S, primary 600 ohm, secondary 4,8 and
> > 16 ohm, max 320 mA, price of this is JPY 19160 here in Japan.
>
>
> --
> -----------------------------------------
> | Anti GEKKO #2 |
> | Public enemy #10 |
> -----------------------------------------
At current exchange rates ~$175.00 US BEFORE shipping and other fees.
Kurt Strain
This is a follow-up about the 6AS7 amp. I A/B'ed this amp in comparison
with the Atma-Sphere M-60's, at the lower volume levels that the 6AS7
can handle. It was unfortunate for me I did this because I am now
reminded again about what that SETA thing is good for. It has a warm
"rightness" that is wonderful that the Atm-Sphere doesn't have. It is
unfortunately much less detailed, and will not power my 87 dB sensitive
speakers. The 6AS7's in the SETA sounded better overall than the 2A3's did
top to bottom and much better than the Chinese 300Bs, the best SE tube I
have tried. But I have not tried the SV572, or any of the high priced
300B versions out there.
I am frustrated that I can't seem to get it all. I love the sound of
my speakers, and I really don't have space for big horns and I'm not
sure I like those anyway. But I love the sound of the SETA for whatever
reasons it's doing it. But I need more power, and my budget has run out
at this moment to replace entire amps.
I think there might be something I don't like about the Atma-Sphere that
is not related to their topology. I think I don't like the way the
6SN7 sounds in there, at least the way it is implemented in this amp
(Kurt ducks the flying debris from this blasphemy). In my SETA I use
1J6 DHT's to drive the 6AS7 and those sound beautiful. I am considering
a major operation of the Atma-Sphere amp to replace the 6SN7 gain stage
with maybe the 1J6 in a differential amp - it is a dual triode with a
common heater/cathode. I hate to hack it up too much, but I want some
added warmth to this amp. The Atma-Sphere can be considered neutral,
but might be considered leaning toward the cold side, at least in my system.
But I am one that likes warm sounding amps. I also like it detailed at
the same time - kind of a tough trick to get right. I tried a few different
6SN7's but all were still a bit sterile to me. The best I found was an
RCA version.
If you know of an octal tube that can warm things up but not slow things
down, a dual triode that can operate in a 1mA 200V per plate diff amp
environment, let me know. I tried a Tung-Sol 6US7 (matched 6SL7 for
diff amp use) but it sounded just lousy - fat and slow and still somewhat
sterile.
Thanks.
Kurt
Thor...@tnt-audio.com
Hi there,
> I think there might be something I don't like about the Atma-Sphere that
> is not related to their topology.
I think it is related to their Topology....
From what I understand the Athmasphere uses Cathode Followers. I don't like
CF's. Not Even Allen Wrights "Superlinear CF". But this SLCF is allmost
acceptable....
Essentially, there is a Current-Source in the Cathode Line and a Cascode
Valve on Top of the CF, thus operating the CF at constant Voltage and
Constant Current.
My take would be to try a REALLY MEATY (say at least a 5687 or bigger run
close to max. Ia) Differential Amp as Drivers instead of the CF. But it nay
prove a Problem in the Athmasphere's Topology.
I have looked very carefully at OTL Amp's (the Transcendent, the Wiggins
Circlotron and the "Technics" Type plus that Japanese Job with the modulated
Current-source) in PSpice and read peoples experiences....
I had to decide in the End that no current OTL Topology was very much to my
liking.... And I don't feel like inventing P-Channel Valves.... ;-)
har...@q3-consulting.com
In article <6k3fj9$u8l$1...@nnrp1.dejanews.com>,
Thor...@tnt-audio.com wrote:
> I had to decide in the End that no current OTL Topology was very much to my
> liking.... And I don't feel like inventing P-Channel Valves.... ;-)
In fact, P-channel valves wouldn't help. The Circlotron design is MORE
symmetrical than a transistor amp - I wonder if anyone has tried it
as a transistor topology? Maybe it would sound better than complementary
designs. (But Ralph's patent covers solid state too).
The reason for using CFs is to keep the output impedance down. You COULD
design an OTL that didn't use CFs, but if you didn't use NFB either
you would need a LOT of tubes (e.g. 16 x 6C33C-B to get 10 ohms in
a bridge). The problem with tubes is their high impedance.
Of course, if you had enough money you could custom-build a Godzilla
valve with tiny electrode spacing and huge electrodes.
John
Thor...@tnt-audio.com
Hi there,
> I had to decide in the End that no current OTL Topology was very much to
> my liking.... And I don't feel like inventing P-Channel Valves.... ;-)
>
> In fact, P-channel valves wouldn't help.
Yes they would. They would allow a Direct-coupled Driver using an N-Channel
Input Diff Pair into a P-Channel Pair in the "folded cascode" Connection.
This stage could readily drive a pair of N-Channel powervalves in a
Single-ended Push Pull Output Stage (a'la Kaneda) giving a fully DC coupled
OTL with no Cathode Followers but the ability to drive the Powervalves Grid
positive...
It would be similar to the transcendent OTL, but with the CF XXXX'ed out....
>The Circlotron design is MORE symmetrical than a transistor amp -
There are Circlotron Transistor-Amp's and ANY current Jeff Rowland Amp is
fully symmetric.
>I wonder if anyone has tried it as a transistor topology?
I have. In the early 80's. Using a Cacoded Quad of Video-Transistors in the
Driverstage and a Pair of Industrial Darlington Transistors (something silly
like a few hundert Ampere's).
It did not sound half-bad, but I had to use some serious NFB as the Driver
was unable to source enough current lineary near the top of the load
Envelope.
It sounded fine on the Bench (without NFB driving a single 8 ohm Speaker) but
in the BIG soundsystem it run out of steam due to the Darlingtons lousy Beta
maintainance.
I think later versions also got another EF (Emitter Follower) in Front of the
Transistors.
This was used in large sound-systems as "Monster-Monoblock". It had something
silly like 700-900 W RMS into 8 Ohm and accepted 8 to 12 8 Ohm Speaker in
parallel before running out of steam.
It had two 3KVA X-Formers for the Output stage. As these Darlington Modules
had a complete on-board protection circuit (thermal and current) the Amp was
also Roadie proof....
I build quite a few of them back in East Germany, the big transitors,
heatsinks, x-formers and capacitors all out of Spec units from the big
Russian and East German plants making Industrial Electronics stuff.....
Believ it or not, for a while we thought about a Valved Frontend, but we
needed way more Current than these could do to drive these pig-darlingtons...
>Maybe it would sound better than complementary designs.
Doubtfull. Most problems with Transistor Amp's are in the Driverstage and in
X-Over Distortion. These remain problems in the Wiggins Circlotron. Solve
these in any Circuit and give the poor Amp a decent PSU and even Sand sounds
not bad....
>(But Ralph's patent covers solid state too).
I'm amazed he could patent a Circuit known and used since the early 50's....
> The reason for using CFs is to keep the output impedance down.
Actually, the reason is that there is no way to get decent output from a 6AS7
or 6C33 without driving the Grid positive. You need a meaty directcoupled CF
for that. I don't like CF's (I prefer an X-Former instead - even a OPT)....
>You COULD design an OTL that didn't use CFs,
Correct.
>but if you didn't use NFB either you would need a LOT of tubes
Correct.
Hence my rejection of the OTL when using Valves as final Devices. I do have a
version of my old Monster-Amp on the Drawingboard using a Valved Frontend and
a Fet/Bipolar Complementary Follower as the output Devices.
But somehow I don't "fancy" building something like that at the moment.
For those on a Budget and in need of a GOOD cheap Amp this could be the
ticket.
Later Thorsten
har...@q3-consulting.com
In article <6k435c$kol$1...@nnrp1.dejanews.com>,
Thor...@tnt-audio.com wrote:
>
> Hi there,
>
> > I had to decide in the End that no current OTL Topology was very much to
> > my liking.... And I don't feel like inventing P-Channel Valves.... ;-)
> >
> > In fact, P-channel valves wouldn't help.
>
> Yes they would.
Of course you are right in the more global scheme of things as you
describe. I was thinking specifically of the output stage.
> >(But Ralph's patent covers solid state too).
>
> I'm amazed he could patent a Circuit known and used since the early 50's....
>
Well, I'm surprised at some of the claims too. But that's the way
patents work, you write the claims as broadly as possible...
Claim 1. Anything that moves and also anything that doesn't.
Claim 2. Anything that moves.
Claim 3. Anything that moves on wheels.
.....
Claim 99. Any device built of grade B37 stainless steel mounted on seven
wheels shod with tyres inflated to a pressure of 30-33 psi
intended for dragging logs in the teak forests of Outer
Mongolia.
> > The reason for using CFs is to keep the output impedance down.
>
> Actually, the reason is that there is no way to get decent output from a
6AS7
> or 6C33 without driving the Grid positive.
I'm not sure about this. The AtmaSphere M60 doesn't seem to draw grid
current, although the driver could handle it. It's real limits are
20V into 10 ohms, i.e. 2A, between 8 tubes, i.e. 250mA per tube. With
90V on the plate the curves show that this will happen with the grid
still negative, though not very. But you still need a hefty driver to
deal with the input capacitance of all those parallel tubes.
John
Thor...@tnt-audio.com
Hi there,
> Claim 1. Anything that moves and also anything that doesn't.
> Claim 2. Anything that moves.
> Claim 3. Anything that moves on wheels.
> .....
> Claim 99. Any device built of grade B37 stainless steel mounted on seven
> wheels shod with tyres inflated to a pressure of 30-33 psi
> intended for dragging logs in the teak forests of Outer
> Mongolia.
Yes. Like the Story "The Onion" run a while back on Microsoft patenting
Zero's and Ones....
> I'm not sure about this. The AtmaSphere M60 doesn't seem to draw grid
> current, although the driver could handle it. It's real limits are
> 20V into 10 ohms, i.e. 2A, between 8 tubes, i.e. 250mA per tube.
Have you tried lower Impedances? Is the M-60 Feedback less?
>But you still need a hefty driver to deal with the input capacitance of all
>those parallel tubes.
Mea Culpa of course. That is another one....
Later Thorsten
Kurt Strain
har...@q3-consulting.com wrote:
: > > The reason for using CFs is to keep the output impedance down.
: >
: > Actually, the reason is that there is no way to get decent output from a
: 6AS7
: > or 6C33 without driving the Grid positive.
: I'm not sure about this. The AtmaSphere M60 doesn't seem to draw grid
: current, although the driver could handle it. It's real limits are
: 20V into 10 ohms, i.e. 2A, between 8 tubes, i.e. 250mA per tube. With
: 90V on the plate the curves show that this will happen with the grid
: still negative, though not very. But you still need a hefty driver to
: deal with the input capacitance of all those parallel tubes.
When one half of the M-60 is on and the other cut off, then there are 4 tubes
driving the output. It will drive 30Vpk into 8 ohms with those 4 tubes,
3.75A, which is 0.94A per tube. That is definitely driven well into
positive grid current.
Output impedance is a function of cathode degeneration, and in this case that
cathode resistor happens to be the speaker load. So one of the peculiarities
of this amp is that the amp's output impedance will be a function of the
speaker impedance.
I found a more exact equation for a straight cathode follower output
impedance to be:
Zo = rp / (u + 1 + rp/Rk)
rp for a 6AS7 triode section is rated at 280 ohms,
u is rated at 2
and take the Rk to be an 8 ohm speaker load.
For 4 6AS7s on a side in the circlotron, rp = 280/8 = 35 ohms.
So then for one side, Zo = 35 / (2 + 1 + 35/8) = 4.75 ohms.
But with two sides driving in parallel, out of phase, the output impedance
drops in half, to 2.37 ohms.
Putting Rk=16 ohms, Zo calculates to 6.75 ohms for one half.
Both halves driven makes it 3.37 ohms.
For Rk=4 ohms, final output impedance calculates to 1.7 ohms.
For an open, the impedance rises to 5.83 ohms total.
Approaching a short, the impedance approaches 0 ohms.
But if you ground one side of the circlotron with a scope ground you can
effectively make the parallel operation not work since the terminals need
to float. Measurements of this type might be one-sided, and if you measure
output impedance at an open to a load comparison in voltage, it is really
messed up. I did this and got 13 ohms output impedance for the zero
feedback configuration. Well, for one side at the open condition, the
formula above gives me 11.7 ohms, close to what I got. Adding feedback
would drop that impedance measurement to about 10 ohms, which is what
people believe this amp is doing. Or, if you just eliminate the rp/Rk term
above and ignore the other half operating, you calculate 11.7 ohms.
I don't think Zo could be 10 ohms and do what it does.
I asked Atma-Sphere what the output impedance of the amp is. They say
that it is 1.8 ohms for zero feedback, 1.6 ohms with feedback - no
load stated however.
I could and should more accurately measure this impedance. To do so you
need to measure it with two close known loads - say 8 and 10 ohms. Then you
need a good voltmeter. Then solve the equations:
V1 = Vs (8 / (8 + Zo))
V2 = Vs (10 / (10 + Zo))
Vs = constant source voltage into voltage divider.
Any corrections to this is welcome.
Kurt
har...@q3-consulting.com
In article <6k4s1l$9...@canyon.sr.hp.com>,
kst...@sr.hp.com (Kurt Strain) wrote:
> When one half of the M-60 is on and the other cut off, then there are 4
tubes
> driving the output. It will drive 30Vpk into 8 ohms with those 4 tubes,
> 3.75A, which is 0.94A per tube. That is definitely driven well into
> positive grid current.
Mine top out at 23V or so into 10 ohms, at which point the input
stage is clipping (at 60V or so on the plate). This is 290mA per half
tube. How do you get yours to go further?
> I found a more exact equation for a straight cathode follower output
> impedance to be:
>
> Zo = rp / (u + 1 + rp/Rk)
>
Check. RDH has this formula (in a different form), and it is readily
confirmed by a page or so of messy algebra.
> I don't think Zo could be 10 ohms and do what it does.
My reaction to your post was to rush off and check my calculations.
I found a silly mistake, and indeed the paper result is as you say.
"Oh s**t", I said, feeling embarrassed. So my next move was to
redo the measurements -- I have already been caught out by the
unbalancing problem, when I did distortion measurements. Unfortunately,
the result is simply confusing.
I used an unbalanced generator at 1kHz. To measure the output,
I used a battery powered true RMS meter (Thurlby Thandar) with
various power resistors. Here are my results, at two different
signal levels (output RMS in volts).
No load 6.32 28.0
9.9 ohms 3.20 14.27
8.2 ohms 2.88 12.90
5.0 2.07 9.90
Using the 9.9/8.2 measurments (9.9 is the actual value of my
nominal 10ohm resistor, by the way) gives Zout of 11.4 and 10.4
ohms. This is what I measured before; and it is no longer
consistent with the theory, as I had believed. The others
give various different values going as low as 8 ohms or so.
So, what to make of all this? I don't know. I'd be delighted
for someone to show me that I've forgotten how to solve
simultaneous equations (or rather that Mathcad has).
>
> I could and should more accurately measure this impedance.
Please do! And post, or at least tell me, the results.
> To do so you
> need to measure it with two close known loads - say 8 and 10 ohms. Then you
> need a good voltmeter. Then solve the equations:
>
> V1 = Vs (8 / (8 + Zo))
> V2 = Vs (10 / (10 + Zo))
>
> Vs = constant source voltage into voltage divider.
Yep. That's what I did (for various values of 8 and 10).
John
John Byrns
[Snip]
> When one half of the M-60 is on and the other cut off, then there are 4 tubes
> driving the output. It will drive 30Vpk into 8 ohms with those 4 tubes,
> 3.75A, which is 0.94A per tube. That is definitely driven well into
> positive grid current.
>
> Output impedance is a function of cathode degeneration, and in this case that
> cathode resistor happens to be the speaker load. So one of the peculiarities
> of this amp is that the amp's output impedance will be a function of the
> speaker impedance.
>
> I found a more exact equation for a straight cathode follower output
> impedance to be:
>
> Zo = rp / (u + 1 + rp/Rk)
>
> rp for a 6AS7 triode section is rated at 280 ohms,
> u is rated at 2
> and take the Rk to be an 8 ohm speaker load.
There are several problems with trying to apply this formula to the
Circlotron. First the Circlotron is not a cathode follower, it is exactly
the same circuit as the McIntosh amps, with the power supplies
rearranged. Half the load is in the plate circuit, and half is in the
cathode circuit, what really counts is where the grid drive is
referenced. As such the cathode follower formula doesn't apply, I will
leave it as an exercise for the reader to derive the correct formula,
which should get the old RAT juices flowing.
The second problem is that even if the Circlotron were a cathode follower,
you are not applying the formula correctly. The "speaker load" is not Rk
in the formula, the "speaker load" is the load, and as such does not enter
into the calculation of the output impedance. Rk in the formula is the
cathode resistor in a normal single ended cathode follower, which appears
in parallel with the output impedance of the tube itself. In the case of
the Circlotron which is a push pull design, there is no equivalent
resistor, so Rk is infinity. Well depending on the circuit, Rk is
probably finite, but it is a lot larger than the speaker load.
Applying the cathode forllower formula to 8 6SA7's, with rp = 280 Ohms,
and u = 2, and setting Rk = infinity, we would have:
Zo = (280/8) / (2 + 1) = 11.67 Ohms
In reality, since half the speaker load is in the plate circuit of the
Circlotron, the above formula doesn't apply, and Zo is higher.
Regards,
John Byrns
> For 4 6AS7s on a side in the circlotron, rp = 280/8 = 35 ohms.
> So then for one side, Zo = 35 / (2 + 1 + 35/8) = 4.75 ohms.
>
> But with two sides driving in parallel, out of phase, the output impedance
> drops in half, to 2.37 ohms.
>
> Putting Rk=16 ohms, Zo calculates to 6.75 ohms for one half.
> Both halves driven makes it 3.37 ohms.
> For Rk=4 ohms, final output impedance calculates to 1.7 ohms.
> For an open, the impedance rises to 5.83 ohms total.
> Approaching a short, the impedance approaches 0 ohms.
>
> But if you ground one side of the circlotron with a scope ground you can
> effectively make the parallel operation not work since the terminals need
> to float. Measurements of this type might be one-sided, and if you measure
> output impedance at an open to a load comparison in voltage, it is really
> messed up. I did this and got 13 ohms output impedance for the zero
> feedback configuration. Well, for one side at the open condition, the
> formula above gives me 11.7 ohms, close to what I got. Adding feedback
> would drop that impedance measurement to about 10 ohms, which is what
> people believe this amp is doing. Or, if you just eliminate the rp/Rk term
> above and ignore the other half operating, you calculate 11.7 ohms.
> I don't think Zo could be 10 ohms and do what it does.
>
> I asked Atma-Sphere what the output impedance of the amp is. They say
> that it is 1.8 ohms for zero feedback, 1.6 ohms with feedback - no
> load stated however.
>
> I could and should more accurately measure this impedance. To do so you
> need to measure it with two close known loads - say 8 and 10 ohms. Then you
> need a good voltmeter. Then solve the equations:
>
> V1 = Vs (8 / (8 + Zo))
> V2 = Vs (10 / (10 + Zo))
>
> Vs = constant source voltage into voltage divider.
>
har...@q3-consulting.com
In article <jbyrns-2305...@asahel-25.d.enteract.com>,
jby...@enteract.com (John Byrns) wrote:
>
> In article <6k4s1l$9...@canyon.sr.hp.com>, kst...@sr.hp.com (Kurt Strain)
wrote:
>
I thought some more about the analysis of the circuit. In the M60,
there is a ground reference established by a 600ohm resistor from
each cathode (or group of cathodes). In effect, this splits
the load in the middle - at least for a resistive load.
So probably the best analysis is to look at each phase
separately, with half the load in the cathode and half in the
plate (as one post suggests).
The analysis for a cathode follower with a plate load is
simple, you just add the plate resistor to the plate resistance
of the tube. (I can post/send the analysis if anyone is
interested). Thus instead of a plate resistance of 35 ohms,
you have 39 ohms (for an 8 ohm total load) - this makes little
difference to the result:
Rout = (Rp+Rl)Rk/((mu+1)Rk+Rp+Rl)
= (35+4).4/((2+1).4+35+4)
= 39.4/(12+35+4)
= 156/51
= 3 (near enough)
Unfortunately this is not at all what the measurements show,
as in my previous post.
I don't see why the bridge puts the two Routs in parallel.
Rather, each in effect sees half of the load.
It is of no consequence whether it is the load or Rk
that we are talking about. The tube(s) can't tell the
difference! Rout is defined as dVout/dIout (where d is
really curly d of course) -- so the model I use has a
constant current for Iout in parallel with Rk. But you
can also calculate dVout/dRk -- which is the same apart
from a constant -- it comes to:
(Eb + mu.Vin).Rp/((mu+1).Rk+Rp)^2
= Rout.(Eb+mu.Vin)/((mu+1).Rk+Rp).Rk)
By way of confirming my measurements, I took advantage of having
all the kit in place to measure the Z vs freq of my speakers.
They show the usual ups and downs, up as high as 42 ohms and down
to 8 ohms. And sure enough, the output from the M60 varies
according to a 10ohm or so impedance, and the power delivered
to the speakers varies by 3dB or so. I used the same test setup -
battery powered, otherwise unconnected, Thurlby-Thandar 1705
true RMS meter, in a series/parallel arrangement with the
speaker (allowing both voltage and current measurment). (The 1705
is a neat, if pricey, instrument, and I'm very glad I decided
to splash out on it).
(If anyone needs the Z vs freq for the Spendor BC1 - I'm the one
to ask!).
John Byrns
> In article <jbyrns-2305...@asahel-25.d.enteract.com>,
> jby...@enteract.com (John Byrns) wrote:
> >
> > In article <6k4s1l$9...@canyon.sr.hp.com>, kst...@sr.hp.com (Kurt Strain)
> wrote:
> >
> This posting is in response to a couple of others (as above)...
>
> I thought some more about the analysis of the circuit. In the M60,
> there is a ground reference established by a 600ohm resistor from
> each cathode (or group of cathodes). In effect, this splits
> the load in the middle - at least for a resistive load.
OK, I have worked out the formula for the output impedancce of the
Circlotron circuit. Here it is, for the small signal case:
Rout = (rp / n) / (1 + u / 2)
where
rp is the plate resistance
n is the number of tubes in parallel
if class A this is the total number of tubes on both sides of the bridge
if class B this would be the number of tubes on one side of the bridge
u is the amplification factor
for the case where
rp = 280
n = 8
u = 2
and assuming class A operation
Rout = 17.5 Ohms
Any cathode resistors such as the 600 Ohm reference resistors mentioned
above would be in parallel with Rout. In this case Rout would become
Rout = 17.5 || (600 + 600) = 17.25 Ohms
Regards,
John Byrns
Sportster Dave
John Byrns wrote:
>
> In article <6k4s1l$9...@canyon.sr.hp.com>, kst...@sr.hp.com (Kurt Strain) wrote:
>
>
> > When one half of the M-60 is on and the other cut off, then there are 4 tubes
> > driving the output. It will drive 30Vpk into 8 ohms with those 4 tubes,
> > 3.75A, which is 0.94A per tube. That is definitely driven well into
> > positive grid current.
> >
>
>
> Circlotron. First the Circlotron is not a cathode follower, it is exactly
> the same circuit as the McIntosh amps,
Sorry, but you are mistaken!!! These Atma-Spheres are OTL's!! Now, I
built this circuit from scratch and, I KNOW what I am talking about.
The basic circuit is a bridged SE CATHODE FOLLOWER pair.
>
> The second problem is that even if the Circlotron were a cathode follower,
> you are not applying the formula correctly. The "speaker load" is not Rk
> in the formula, the "speaker load" is the load,
The second problem is that there is no Rk in this circuit other than the
speaker!!!
You see BOTH of the cathodes are tied to the speaker (I used power FETS
driven by a longtail 6SN7 splitter in my favorite version). So as such
the speaker does very much enter into the picture!!!
> Applying the cathode forllower formula to 8 6SA7's, with rp = 280 Ohms,
> and u = 2, and setting Rk = infinity, we would have:
>
> Zo = (280/8) / (2 + 1) = 11.67 Ohms
Hmmm, I seem to remember my good old RCA Radiotron handbook formula as:
Zo = 1/Gm; with Rk infinite...(I think it works out the same as your
formula, but it is much more elegant...)
>
> In reality, since half the speaker load is in the plate circuit of the
> Circlotron, the above formula doesn't apply, and Zo is higher.
Again, unfortunately, the above reality exists only in your mind...
Eric at Svetlana Electron Devices
> The good news is that for this 6AS7, both Sovtek and Svetlana are still
> making them. I had a shipment of 20 of them sent to me for a total price
Uh, Kurt, sorry to tell you this......
...."there is no "Sovtek" factory.
And they do NOT make a 6AS7.
New Sensor's "Sovtek" 6AS7
is an old surplus Svetlana, bought
and marked in Russia.........
Just wanted to keep the confusion
to a tolerable level.......
Kamaruddin
Kurt Strain wrote:
>
>
> I think the sound was more neutral and controlled than the 2A3 and the
> Chinese 300B's I've tried. And the cost for the 6AS7 from New Sensor
> is only $4.95 ea.
>
>
>
> Kurt
The prototype amp using the 6AS7 that was supposed to be given to me by
the Loth-X owner was hijack on its way from WCES98. He told me that it
sound better than the 300B and similiar to the 2A3. That amp is built to
drive the 104dB Azimuth and 1 watt is more than enough. Try to get a
really high efficiency speakers to listen to them. It should be a good
tube. I'm waiting for the MKII prototype now doing its service in
Frankfurt HiFi show. I hope he bring it back for me to try it. It's
supposed to use TWO! OPT per channel. This I gotta see. The amp should be
here next week.
One Eye Jack
har...@q3-consulting.com
In article <jbyrns-2405...@asahel-16.d.enteract.com>,
jby...@enteract.com (John Byrns) wrote:
> OK, I have worked out the formula for the output impedancce of the
> Circlotron circuit. Here it is, for the small signal case:
>
> Rout = (rp / n) / (1 + u / 2)
>
Can you either post or e-mail to me how you get to this? What
is the corresponding gain figure? Another point I haven't mentioned
is that the CF analysis gives good results as far as gain is
concerned -- the output stage is looking at ~60V input for
~20V output.
Thanks,
Proteus
I've heard the Atma-Sphere OTLs described as "circlotron" circuits, but
they are presumaby not the same as the transformer-coupled E-V Circlotrons
of the '50s.
For those of us who are not familiar with the Atma-Spheres, is there a
schematic easily accessible (maybe on-line somewhere)? Or could someone who
knows the topology describe the output stage in a bit more detail? Dave
says the amp uses bridged cathode followers -- for starters, where does the
DC idle current flow?
BTW: if the Atma-Spheres are bridging amps the two halves appear in series
for AC signals, so the output impedance is twice that of each half.
Thanks,
Proteus
har...@q3-consulting.com
In article <199805250632...@nym.alias.net>,
Proteus <prot...@nym.alias.net> wrote:
>
> I've heard the Atma-Sphere OTLs described as "circlotron" circuits, but
> they are presumaby not the same as the transformer-coupled E-V Circlotrons
> of the '50s.
>
> For those of us who are not familiar with the Atma-Spheres, is there a
> schematic easily accessible (maybe on-line somewhere)?
Take a look at www.q3-consulting.com/m60.html . This has the general
organisation of the output stage and an explanation of how it works.
I have the complete M60 circuit scanned, but I think it is up to AtmaSphere
to make it available online if they want to do so.
Thor...@tnt-audio.com
Hi all,
Proteus wrote:
>
> I've heard the Atma-Sphere OTLs described as "circlotron" circuits, but
> they are presumaby not the same as the transformer-coupled E-V Circlotrons
> of the '50s.
They are. But minus the X-Formers.
> For those of us who are not familiar with the Atma-Spheres, is there a
> schematic easily accessible (maybe on-line somewhere)?
Okay. Here are my OTL Patent-links:
Ralph Karstens/Atma-Sphere:
Patent #: 4719431
http://www.patents.ibm.com/details?patent_number=4719431
Bruce Rozenblit/Trancendent:
Patent #: 5604461
http://www.patents.ibm.com/details?patent_number=5604461
Both include either detailed or simplified Schematics.
And here a few more OTL Links:
Andrea Ciufoli:
http://www.geocities.com/ResearchTriangle/8231/
Covi 100W OTL - Circlotron Circuit and OTL Theory:
http://members.aol.com/aria3/index.html
Earmax Headphone Amp - OTL with a White CF:
http://www.nttlabs.com/people/sumisu/golden-tube/earmax.gif
Vaccum Tube Schematic of the Month:
http://www.mbnet.mb.ca/~rohringe/schematic.html
Later Thorsten
John Byrns
> John Byrns wrote:
> >
> > In article <6k4s1l$9...@canyon.sr.hp.com>, kst...@sr.hp.com (Kurt
Strain) wrote:
> >
> > [Snip]
> >
> > > When one half of the M-60 is on and the other cut off, then there
are 4 tubes
> > > driving the output. It will drive 30Vpk into 8 ohms with those 4 tubes,
> > > 3.75A, which is 0.94A per tube. That is definitely driven well into
> > > positive grid current.
> > >
> >
> >
> > There are several problems with trying to apply this formula to the
> > Circlotron. First the Circlotron is not a cathode follower, it is exactly
> > the same circuit as the McIntosh amps,
>
> Sorry, but you are mistaken!!! These Atma-Spheres are OTL's!! Now, I
> built this circuit from scratch and, I KNOW what I am talking about.
> The basic circuit is a bridged SE CATHODE FOLLOWER pair.
Dave,
I am disappointed in you, I thought you knew better than this. I confess
I know little about the Atma-Spheres, but from what I have heard and seen
they are basically an OTL version of the old Circlotron, if this is
incorrect, then I am wrong and please accept my apologies. The fact is
the tubes in the Circlotron circuit do not operate as pure cathode
followers, part of the load is in the plate circuit, just as in the
McIntosh design. Basically, you can think of it as a cathode follower
with the reference for the grid drive tapped half way up the load, this is
not a cathode follower in the traditional sense, and modifies the output
impedance.
> >
> > The second problem is that even if the Circlotron were a cathode follower,
> > you are not applying the formula correctly. The "speaker load" is not Rk
> > in the formula, the "speaker load" is the load,
>
> The second problem is that there is no Rk in this circuit other than the
> speaker!!!
> You see BOTH of the cathodes are tied to the speaker (I used power FETS
> driven by a longtail 6SN7 splitter in my favorite version). So as such
> the speaker does very much enter into the picture!!!
Another poster indicated that there are 2 600 Ohms resistors to provide
the reference for the grid drive. The speaker does not enter into the
calculation of the source or output impedance, the speaker enters the
picture only as the load.
> > Applying the cathode forllower formula to 8 6SA7's, with rp = 280 Ohms,
> > and u = 2, and setting Rk = infinity, we would have:
> >
> > Zo = (280/8) / (2 + 1) = 11.67 Ohms
>
> Hmmm, I seem to remember my good old RCA Radiotron handbook formula as:
>
> Zo = 1/Gm; with Rk infinite...(I think it works out the same as your
> formula, but it is much more elegant...)
I have now derived the formula for the Circlotron which is:
Zo = rp / (1 + (u /2))
> >
> > In reality, since half the speaker load is in the plate circuit of the
> > Circlotron, the above formula doesn't apply, and Zo is higher.
>
> Again, unfortunately, the above reality exists only in your mind...
Dave I advise you to think again, and maybe try a little math before
making these statements, or if the Atma-Spheres do not use the traditional
Circlotron circuit, tell us what they do use.
Regards,
John Byrns
John Byrns
> I've heard the Atma-Sphere OTLs described as "circlotron" circuits, but
> they are presumaby not the same as the transformer-coupled E-V Circlotrons
> of the '50s.
>
> For those of us who are not familiar with the Atma-Spheres, is there a
> schematic easily accessible (maybe on-line somewhere)? Or could someone who
> knows the topology describe the output stage in a bit more detail? Dave
> says the amp uses bridged cathode followers -- for starters, where does the
> DC idle current flow?
I may be wrong, but I think there has been enough information made
available over the last couple of years, that we can conclude that the
Atma-Spheres use pretty much the same circuit as the E-V Circlotrons. The
idle current flows around the loop composed of the two tubes, and the two
power supplies, assuming of course that the idle currents in the two
halves of the bridge are ballanced, as they should be in a properly
functioning Push Pull amplifier.
Regards,
John Byrns
John Byrns
> In article <jbyrns-2405...@asahel-16.d.enteract.com>,
> jby...@enteract.com (John Byrns) wrote:
> > OK, I have worked out the formula for the output impedancce of the
> > Circlotron circuit. Here it is, for the small signal case:
> >
> > Rout = (rp / n) / (1 + u / 2)
> >
>
> Can you either post or e-mail to me how you get to this? What
> is the corresponding gain figure? Another point I haven't mentioned
> is that the CF analysis gives good results as far as gain is
> concerned -- the output stage is looking at ~60V input for
> ~20V output.
>
> Thanks,
>
> John
Hi John,
I just wrote two loop equations, one around the plate to catheode loop,
and a second around the grid to cathode loop, then I solved them. I am
not in the mood to try some ASCII graphics to illustrate this, but if
someone wants to post an anotated schematic of the output stage, I will
adapt the equations to it. I may be able to scan in my hand drawn
schematic and derivation of the equation later in the week, and make it
available by email, since graphics are not allowed in the news groups.
I will rework equation into a form giving the gain off line and post it
later. When u is large, the gain will tend toward 2x. When u is smaller,
as in this case, the gain will be less than 2x.
Regards,
John Byrns
John Byrns
In article <jbyrns-2505...@kebob-23.d.enteract.com>,
jby...@enteract.com (John Byrns) wrote:
I thought I should point out that my formula for the output impedance of a
Circlotron output stage
Ro = rp / (1 + u / 2)
assumes no feedback. In addition to negative feedback, this assumes that
there is no positive feedback to the driver stage, such as is used in the
Electro-Voice Circlotron.
Since I am posting this I will include my first cut at the gain equation
for the Circlotron output stage. There may be errors in this as I haven't
had a chance to check it out.
Gain = Eo / Ein = u / (1 + (u /2) + (rp / Rl))
where
u is the amplification factor of the tubes
rp is the total plate resistance of all the tubes on both sides of the "bridge"
Rl is the load resistance (including any cathode resistors)
As in the formula for Ro where any cathode resistors must be paralled with
the value of Ro from the formula, to arive at a final Ro, so to with the
gain, and cathode resistors must be considered as part of Rl.
If I find any errors in gain formula, on rechecking, I will post them.
Regards,
John Byrns
Proteus
Thanks for the many excellent replies re: Atma-Sphere circlotron OTLs.
Next time I fire up the browser I'll have lots to look at!
John Byrns writes:
>The idle current flows around the loop composed of the two tubes, and the
>two power supplies, assuming of course that the idle currents in the two
>halves of the bridge are ballanced, as they should be in a properly
>functioning Push Pull amplifier.
Quite right. I had been thinking that the idle current in a
transformer-coupled circlotron flows in the transformer primary, but you
are right -- it doesn't.
For those interested:
_________ _________
| | | |
| |+ |+ |
------- Bat Bat -------
----- - - - - | | - - - - -----
______ | | ______
| | \ / | |
| \ / |
| \ |
|_____/ \_____| "Bat" = battery or floating PS
| |
|___ LOAD ____|
Basic circlotron output stage
________ ________
| | | |
| |+ |+ |
| Bat Bat |
< | | >
> \ / <
< \ / >
| \ |
| / \ |
| / \ |
|______/ \______|
| |
|___ LOAD ____|
Simplified circlotron circuit
(Rs substituted for tubes)
_______ Bat _______ B
| + |
| |
| |
< >
> <
< >
| |
| |
| |
|______ Bat ______|
A +
Untwisted simplified circuit
(LOAD connects A to B)
As long as the bridge is balanced, no current flows in the load. When grid
drive unbalances the bridge, the unbalanced current flows in the load.
Proteus
Kurt Strain
John Byrns (jby...@enteract.com) wrote:
: In article <6k4s1l$9...@canyon.sr.hp.com>, kst...@sr.hp.com (Kurt Strain) wrote:
: [Snip]
: >
: > Output impedance is a function of cathode degeneration, and in this case that
: > cathode resistor happens to be the speaker load. So one of the peculiarities
: > of this amp is that the amp's output impedance will be a function of the
: > speaker impedance.
: >
: > I found a more exact equation for a straight cathode follower output
: > impedance to be:
: >
: > Zo = rp / (u + 1 + rp/Rk)
: >
: > rp for a 6AS7 triode section is rated at 280 ohms,
: > u is rated at 2
: > and take the Rk to be an 8 ohm speaker load.
: There are several problems with trying to apply this formula to the
: Circlotron. First the Circlotron is not a cathode follower, it is exactly
: the same circuit as the McIntosh amps, with the power supplies
: rearranged. Half the load is in the plate circuit, and half is in the
: cathode circuit, what really counts is where the grid drive is
: referenced. As such the cathode follower formula doesn't apply, I will
: leave it as an exercise for the reader to derive the correct formula,
: which should get the old RAT juices flowing.
etc.
John,
I'm impressed. First, I'm impressed that you took the time to look at
my analysis for an amp you don't even have experience in using, and second
I am especially impressed in that you so easily came to the correct answer
that my equations are indeed bogus. My formula for output impedance includes
a cathode resistor, Rk, but that can't be the load it goes into. So it's
more difficult to solve. I don't know how you solved it, and even if
it's correct, but your numbers more accurately represent my measurement
I made yesterday.
So what is the output impedance? One impedance taken at Rload = 10 ohms
and another at 6.7 ohms showed an output impedance of 15.7 ohms in my
current version of this amp (w/o NFB). Yikes!
So now I don't understand where the folks at Atma-Sphere come up with
a 1.8 ohms figure? I believed them and tried to see where it came from,
but I don't think so now. I should have trusted my frequency response
and impedance response plots that indicated a high output impedance exists -
you get free bass extension by virtue of the fact that the woofer impedance
rises toward its resonance point.
Another thing. I still don't know how this amp works so well. It controls
bass well with a damping factor of about 0.5? It does something different
I would guess.
Kurt
Kurt Strain
Eric at Svetlana Electron Devices (svet...@earthlink.net) wrote:
: > The good news is that for this 6AS7, both Sovtek and Svetlana are still
Oh, yeah, I forgot about the fact that there's this "Sovtek" label
that has no factory. But it confuses me. Who makes the Sovtek 6922,
or the Sovtek 300B if not Svetlana? Those are being made aren't they,
and not old stock?
Kurt
Kurt Strain
John Byrns (jby...@enteract.com) wrote:
: OK, I have worked out the formula for the output impedancce of the
: Circlotron circuit. Here it is, for the small signal case:
: Rout = (rp / n) / (1 + u / 2)
: where
: rp is the plate resistance
: n is the number of tubes in parallel
: if class A this is the total number of tubes on both sides of the bridge
: if class B this would be the number of tubes on one side of the bridge
: u is the amplification factor
: for the case where
: rp = 280
: n = 8
: u = 2
: and assuming class A operation
: Rout = 17.5 Ohms
: Any cathode resistors such as the 600 Ohm reference resistors mentioned
: above would be in parallel with Rout. In this case Rout would become
: Rout = 17.5 || (600 + 600) = 17.25 Ohms
: Regards,
: John Byrns
John,
It looks like you are maybe off by a factor of 2 somewhere. I ran the
amp at low power to assure class A operation (1W), which involves all 8
6AS7s, or 16 triodes in parallel. That should have given an Rout of
8.75 ohms, which is 56% of my measured Rout. So however you did this,
it's not real accurate to my case.
Want to take another stab at it? This is non-trivial. Perhaps the
explanation can be found below. Are these tubes matched enough to
make that equation usable?
....
I was able to achieve 60W/ch at the onset of clipping
into 10 ohms instead of just under 50W/ch by adding a 0.15 ohm resistor
at the cathode of every 6AS7. I tried larger values but the sound was too
umm, indescribably overcontrolled in an artificial way - punch but not
undistortedly rounded. Very odd stuff going on here. Apparently there
is a current hogging phenomenon - the gm's probably vary a lot. Do the
mu's vary that much that it doesn't work well in your equation you
think? But if one is hogging, wouldn't mu be a little higher and Rout
just that much lower?
....
By the way, I improved the sound of these amps again over the weekend by
using an improved version of a constant current sourced CF driving the
outputs. Linearizing this stage really helps, but I realize CF's are
not what I want in a tube circuit. I have designed them into my projects,
but this tube really needs it to get the power out. My gratitude to the
person (I forgot who) who suggested this circuit. I agree with that
man's opinion on cathode followers. But how good would a White follower
be - with a little added feedback in the stage to linearize it further and
lower output impedance. Worse or better?
Geez, this is fun. Real tube amp talk that I'm learning from. Is something
wrong out there. :-)
Kurt
Kurt Strain
har...@q3-consulting.com wrote:
: In article <6k4s1l$9...@canyon.sr.hp.com>,
: kst...@sr.hp.com (Kurt Strain) wrote:
: tubes
: > driving the output. It will drive 30Vpk into 8 ohms with those 4 tubes,
: > 3.75A, which is 0.94A per tube. That is definitely driven well into
: > positive grid current.
: Mine top out at 23V or so into 10 ohms, at which point the input
: stage is clipping (at 60V or so on the plate). This is 290mA per half
: tube. How do you get yours to go further?
I meant 30V peak, well beyond clipping. The driver stage will pull up
the grid to get it into positive grid current.
: > I found a more exact equation for a straight cathode follower output
: > impedance to be:
: >
: > Zo = rp / (u + 1 + rp/Rk)
: >
: Check. RDH has this formula (in a different form), and it is readily
: confirmed by a page or so of messy algebra.
Yes, but I misused the formula. You can't place the load in parallel
with Ro to compute final Ro into the load. So the formula does not
apply and it is more compicated a situation.
: > I don't think Zo could be 10 ohms and do what it does.
: My reaction to your post was to rush off and check my calculations.
: I found a silly mistake, and indeed the paper result is as you say.
: "Oh s**t", I said, feeling embarrassed. So my next move was to
: redo the measurements -- I have already been caught out by the
: unbalancing problem, when I did distortion measurements. Unfortunately,
: the result is simply confusing.
: I used an unbalanced generator at 1kHz. To measure the output,
: I used a battery powered true RMS meter (Thurlby Thandar) with
: various power resistors. Here are my results, at two different
: signal levels (output RMS in volts).
: No load 6.32 28.0
: 9.9 ohms 3.20 14.27
: 8.2 ohms 2.88 12.90
: 5.0 2.07 9.90
: Using the 9.9/8.2 measurments (9.9 is the actual value of my
: nominal 10ohm resistor, by the way) gives Zout of 11.4 and 10.4
: ohms. This is what I measured before; and it is no longer
: consistent with the theory, as I had believed. The others
: give various different values going as low as 8 ohms or so.
I think this is the correct output impedance. I got higher with
no feedback, 15.7 ohms. But then my amp has a different driver
circuit than the original as well.
Kurt
John Byrns
I am familiar with the Circlotron circuit, but not specifically with the
atma-sphere m60. I received an email which pointed out that there are
really 16 triodes in the output stage of the atma-sphere m60.
I we apply my formula for calculating the Rout of a Circlotron amplifier:
Rout = rp / (1 + (u / 2))
Using rp = 280 Ohms (per triode)
and u = 2
We get the following result, assuming the negative feedback was
disconnected, as Kurt indicated it was.
Rout = ((280 / 16) / (1 + 2 / 2)) || (600 + 600) = 8.69 Ohms
This also assumes that there is no bootstrapping, or positive feedback
used in the atma-sphere driver stage. If there is positive feedback in
the drivers, then Rout would be somewhat higher, exactly how much higher
would depend on the amount of positive feedback involved, and more
calculations would be needed.
The 600 + 600 Ohms that I have added in parallel, are the driver reference
resistors that an earlier poster said were used.
Regards,
John Byrns
Thor...@tnt-audio.com
Hi all,
> So what is the output impedance? One impedance taken at Rload = 10
> ohms and another at 6.7 ohms showed an output impedance of 15.7 ohms
> in my current version of this amp (w/o NFB). Yikes!
Hey. I know a lot of SET's that measure better ;-).
> So now I don't understand where the folks at Atma-Sphere come up with
> a 1.8 ohms figure?
The same way Mullard arrived at Valve-ratings? Creative thinking with a good
deal of optimism? It seems fairly common amongst Manufacturers of SE and OTL
Amp's to be slightly Optimistic with their assessment of Poweroutput and
Output Impedance.
Not that that matters as such, it's the sound that matters.
> Another thing. I still don't know how this amp works so well. It controls
> bass well with a damping factor of about 0.5? It does something different
> I would guess.
The whole issue of Damping factor in itself being related to the Bass Quality
and Quantity is only partially true. Most Speakers (Woofers) are damped both
mechanically and magnetically.
Case-point. My Speakers are blessed with a fairly flat Impedance Curve (there
is a steep narrow drop at 2 kHz to under 2 Ohm and they go down to 2.3 Ohm at
about 90 Hz with a slight bump in the Frequency Response) and the Bass is
well tuned (Helmholz Resonance at 25-27 Hz - 2.7 Ohm Minimum).
Most of the time the Speakers Impedance stays around 5.5 Ohm, with a maximum
Impedance Rise from the dual Woofers being limited to about 12 Ohm.
I recently decided to see if a high Source impedance could degrade the Bass
performance and thus inserted increasing Resistances into the Speakerlead, up
to 8 Ohm.
With source-impedances of up to 4 Ohm there was VERY little change in
character, the Bass possibly becomming ever so slightly smoother and the
mid's being a bit recessed. Eight Ohm was definitly too much also because of
the powerloss, but a DF of about 0.5 does not imply bad bass....
Later Thorsten
Inserting
Kamaruddin
Eric at Svetlana Electron Devices wrote:
> > The good news is that for this 6AS7, both Sovtek and Svetlana are still
> > making them. I had a shipment of 20 of them sent to me for a total price
>
> Uh, Kurt, sorry to tell you this......
>
> ...."there is no "Sovtek" factory.
> And they do NOT make a 6AS7.
If there is no "Sovtek" factory, who make Sovtek now? Old Svetlana factory?
New Svetlana factory? This is news to me.
One Eye Jack
BTW. Loth-X Audio tried all the various brands of 6AS7 for their prototype amp
and found the Svetlana to be the best.
Ned Carlson
>> Uh, Kurt, sorry to tell you this......
>>
>> ...."there is no "Sovtek" factory.
>> And they do NOT make a 6AS7.
>If there is no "Sovtek" factory, who make Sovtek now? Old Svetlana factory?
>New Svetlana factory? This is news to me.
The "6550-WE" Sovtek tubes I have are labelled thusly:
6550-WE
SOVTEK (R)
REFLECTOR
MADE IN RUSSIA
>
>One Eye Jack
>
>BTW. Loth-X Audio tried all the various brands of 6AS7 for their prototype amp
>and found the Svetlana to be the best.
>
We have a mess of 6AS7-G with mid-80's datecodes, with the
big "C" logo same as appears on Svetlana 6550-C boxes.
They're in Russian boxes, and did not come from Sovtek.
Also a bunch of 811A with 1975 datecodes and "P"
logos, which I assume means the Reflektor factory.
I have no problem at all with "old surplus", Cold War era
Russian stock, BTW, so long as made & priced right.
Ned Carlson Triode Electronics,2225 W Roscoe Chicago, IL, 60618 USA
ph 773-871-7459 fax 773-871-7938
12:30 to 8 PM CT, (1830-0200 UTC) 12:30-5 Sat, Closed Wed & Sun
http://www.triodeel.com
Text file catalogs:Catalog 'Bot at cat...@triodeel.com
Kurt Strain
John Byrns (jby...@enteract.com) wrote:
: OK, I have worked out the formula for the output impedancce of the
: Circlotron circuit. Here it is, for the small signal case:
: Rout = (rp / n) / (1 + u / 2)
: where
: rp is the plate resistance
: n is the number of tubes in parallel
: if class A this is the total number of tubes on both sides of the bridge
: if class B this would be the number of tubes on one side of the bridge
: u is the amplification factor
: for the case where
: rp = 280
: n = 8
: u = 2
: and assuming class A operation
: Rout = 17.5 Ohms
: Any cathode resistors such as the 600 Ohm reference resistors mentioned
: above would be in parallel with Rout. In this case Rout would become
: Rout = 17.5 || (600 + 600) = 17.25 Ohms
I found a paper on the history of the OTL and lists the output impedance
formulas. See http://members.aol.com/aria3/otlpaper/otlhist.htm
I'm not sure of its correctness, but they show the OTL output impedance
formula as:
Rout = Rp / (2 + mu), where Rp = 2*rp/N, N = total number of triodes in the
circlotron.
And if you work that out, you have the same answer.
Why it doesn't quite measure the same is due to many possibilities, but
what I want to learn is how you solved this equation.
Kurt
SBench
Hi All,
Been monitoring this thread. Kudoos to John for his math
manipulation. Don't forget the gm/rp/mu data given in the
tube manuals are only valid at one point (namely, 135 volts and
125 mA). You'll need to get more accurate numbers by
finding the appropriate mu/rp numbers by tangenting the
tube curves at the operating point you're using.
Might explain the discrepancy in the measured numbers.
You can, of course, derive the tube curves FOR THE TYPE
of 6AS7s you're using by using the RAT gm/mu tester ;-)
Best Regards,
Steve
John Byrns
Kurt,
OK, here's how I calculated the output resistance for the Circlotron
amplifier. Please excuse my poor ASCII diagram.
+---------------+
plate | <--- Ip |
\ |
/ |
\ rp |
/ |
\ |
/ Loop #2 |
| + |
_____ |
grid / \ |
+-----------| -uEg | |
| \_____/ |
| + <--Eg--> - | cathode |
| + | |
_____ | - |
/ \ +---o <--+ |
| Ein | Loop #1 ( + | |
\_____/ ( | |
| - ( | |
+--------------( Eout |
| ( | |
| ( | |
| ( - | |
| +---o <--+ |
| | |
| +---------------+
|
v
(GND)
Assumptions
1. For Class A operation, both halves of the Circlotron Bridge can be
considered to be in parallel.
2. rp represents the equivalent rp of all the paralleled tubes.
3. The grid drive circuit is referenced to the center of the load, which
is also the ground reference. I have shown this as a tapped inductor
across the load.
4. The grid drive circuit can be represented as a Voltage source, with no
dependencies on the output Voltage or current. i.e. no bootstrapping or
positive feedback.
5. No current flows in the grid loop.
Equations:
(1.) Rout = - Eout / Ip (Identity for Rout)
(2.) Ein = 0.0 Volts (Set input to zero)
(3.) Eg - Ein + Eout / 2.0 = 0.0 Volts (Equation for Loop #1)
(4.) Ip * rp - u * Eg + Eout = 0.0 Volts (Equation for Loop #2)
(5.) Eg = - Eout / 2.0 (Substitute equation #2 for Ein in equation #3)
(6.) Ip * rp + u * Eout / 2.0 + Eout = 0.0 Volts (combine equations #4 and #5)
(7.) - u * Eout / 2.0 - Eout = Ip * rp (Rearrange equation #6)
(8.) - Eout * (1.0 + u / 2.0) = Ip * rp (Factor out Eout)
(9.) - Eout / Ip = rp / (1.0 + u / 2.0) (Move Terms around)
(10.) Rout = rp / (1.0 + u / 2.0) (combine equations #1 and #9)
There you have it. Questions?
Regards,
John Byrns
Kurt Strain
John Byrns (jby...@enteract.com) wrote:
: Kurt,
: OK, here's how I calculated the output resistance for the Circlotron
: amplifier. Please excuse my poor ASCII diagram.
...
Thanks for this nice work. I'm going to study it. Great job.
Kurt
Kurt Strain
John Byrns (jby...@enteract.com) wrote:
: In article <6kk3e6$4...@canyon.sr.hp.com>, kst...@sr.hp.com (Kurt Strain) wrote:
: +---------------+
I see the key to solving this is understanding that in the bridge,
the way to reference the input voltage is to the center of the load.
That I did not figure out. Pretty good, John.
Kurt
Chris Broadbent
Thor...@tnt-audio.com wrote in article
<6k3fj9$u8l$1...@nnrp1.dejanews.com>...
> <...snip>
>
> From what I understand the Athmasphere uses Cathode Followers. I don't
like
> CF's. Not Even Allen Wrights "Superlinear CF". But this SLCF is allmost
> acceptable....
>
> <snip...>
What's wrong with having a cathode follower in the signal path? In my
(admittedly limited) experience, it adds very little character of its own
to the signal. Certainly, an anode follower gain stage's characteristics
swamp those of a unity gain cathode follower.
Anyhow, sometimes there's no choice but to use one for impedance matching
purposes.
Of course, this is just my opinion based on what experience I have. Could
you could clarify your opposition to CF's?
Cheers,
Chris
NOTE: Remove the x in my reply address (xc...@bga.com) if you wish to email
me.
Thor...@tnt-audio.com
Hi there,
I wrote:
> > From what I understand the Athmasphere uses Cathode Followers. I don't
> > like CF's. Not Even Allen Wrights "Superlinear CF". But this SLCF is allmost
> > acceptable....
>
> What's wrong with having a cathode follower in the signal path?
Not much. They just don't sound good. But then, neither does SRPP. To my ears
they add unacceptable colorations and also reduce resolution.
If you don't mine them fine. But have you ever tried a Signalpath truely free
from followers?
>In my (admittedly limited) experience, it adds very little character of its own
> to the signal. Certainly, an anode follower gain stage's characteristics
> swamp those of a unity gain cathode follower.
No way. Cathode Follwers without a Current-Source and Bootstrap/Cascode
Valve/FET distrorts a lot more than a simple Gain Stage. Measure it.... (or
read the good old books - you know which ones).
> Anyhow, sometimes there's no choice but to use one for impedance
>matching purposes.
I agree that sometimes there is no way around a CF. But in my opinion, "like
war, using CF's is the last resort of the Incompetent" (paraphrased and not
meant too literally).... ;-)
But yes, if there is no other option to make the Design work and this design
is required, I too would use a CF (and count me incompetent)....
> Of course, this is just my opinion based on what experience I have. Could
> you could clarify your opposition to CF's?
A member of the Joe-list quite recently posted a few Web-Pages with a
theoretical and practical analysis and measurements from various CF-Shemes.
This included the normal CF, one with a Current-source and the White
Follower.
All three have a very non-linear Gain-Curve. Allen himself solved the Problem
by Cascoding the upper CF Valve, so the Follower Valve allways sees the same
Current through it and the same Voltage across it.
This solves the Linearity issue up to a point (depends on the Stability of
the Current-source and the load). Having tried this Circuit against a simple
Anode loaded Stage with paralleled Valves insead of a Gainstage/CF
arrangement including the additional Circuitry Alllen specifies, the "pure"
gainstage sounds notably less colored and more "alive" (untill you bump the
load capacitance - then the CF pulls ahead)....
I use in Development of my Circuits extensive Computer-Analysis, a copius
Amount of measurement and extended listening. All three tell me - I DON'T
LIKE CF's.....
Kind regards Thorsten.
======================================
e-mail:
Thor...@tnt-audio.com
Visit TNT-Audio on the Web - the only advertising
free audio web-zine.
http://www.tnt-audio.com
======================================
har...@q3-consulting.com
In article <6l8e0j$7nv$1...@nnrp1.dejanews.com>,
Thor...@tnt-audio.com wrote:
> A member of the Joe-list quite recently posted a few Web-Pages with a
> theoretical and practical analysis and measurements from various CF-Shemes.
> This included the normal CF, one with a Current-source and the White
> Follower.
>
Do you have a URL? I would like to understand this issue. I presume that
the non-linearity depends heavily on the valve you use.
John
Sheldon D. Stokes
> A member of the Joe-list quite recently posted a few Web-Pages with a
> theoretical and practical analysis and measurements from various CF-Shemes.
> This included the normal CF, one with a Current-source and the White
> Follower.
>
> All three have a very non-linear Gain-Curve. Allen himself solved the Problem
> by Cascoding the upper CF Valve, so the Follower Valve allways sees the same
> Current through it and the same Voltage across it.
>
> This solves the Linearity issue up to a point (depends on the Stability of
> the Current-source and the load). Having tried this Circuit against a simple
> Anode loaded Stage with paralleled Valves insead of a Gainstage/CF
> arrangement including the additional Circuitry Alllen specifies, the "pure"
> gainstage sounds notably less colored and more "alive" (untill you bump the
> load capacitance - then the CF pulls ahead)....
>
> I use in Development of my Circuits extensive Computer-Analysis, a copius
> Amount of measurement and extended listening. All three tell me - I DON'T
> LIKE CF's.....
>
That's interesting, I always assumed that a CF was a very low distiortion
stage due to the 100% feedback. I'll have to pull out RDH and spice and
play a bit.
I typically don't use them either, but I do use SRPP. There are a lot of
tubes with low plate reistances that really make cathode followers not
needed except when you really need a very low output impedance (like an
OTL headphnoe amp)
Take the above example of a headphone amp, it would be interesting to hear
the difference between a cathode follower OTL one and a transformer
coupled one, it's not clear in my mind which would be better. I have
heard an OTL one based on the 6080 tube that sounded really good.
Sheldon
--
"...from Gingus Kahn to the Fuller brush man, they're just a bunch of
loosers like me" -Dave VanRonk
Remove SPAM_BE_GONE from my e-mail to reply directly to me
Thor...@tnt-audio.com
Hi there,
> I typically don't use them either, but I do use SRPP.
Not quite as bad as a plain (resitor loaded) Cathode Follower, but almost....
>There are a lot of tubes with low plate reistances that really make cathode
>followers not needed except when you really need a very low output impedance
Correct. Hence I managed to avoid them so far.....
> Take the above example of a headphone amp, it would be interesting to hear
> the difference between a cathode follower OTL one and a transformer
> coupled one, it's not clear in my mind which would be better. I have
> heard an OTL one based on the 6080 tube that sounded really good.
That will heavily depend on the Quality of the transformer. With a really
excellent Transformer I'd say the OPT approach will win over OTL Cathode
Follower....
With the usual not so hot Output Transformers all bet's are off....
Later Thorsten