SRPP - Mu vs. Beta Follower
dang...@earthlink.net
Some posts have discussed the SRPP variants mu-follower and
beta-follower.
In a prior post I stated the beta-follower has 100x less distortion
than the mu-follower.
This is only correct for a mu-follower using triode as top plate,
where impedance on lower plate is only increased by a factor of 2 or
3.
Alan Kimmel's pentode version of the mu-follower is a different
animal. It presents a high plate impedance to the lower tube, thus
the lower tube plate distortion may be much improved, perhaps
approaching that of the beta-follower, but I can't present numbers, as
Alan didn't publish distortion figures.
I looked a little more closely at Alan's mu-follower using the
pentode, and here are some disadvantages relative to the beta-follower
SRPP.
The mu-follower has (2) caps affecting the frequency response and
linearity of the top tube. A coupling cap between bottom tube plate
and top tube grid. And, a second cap between cathode and screen grid.
The design is very complex and quite an achievement, allowing the
pentode to be used in lieu of a triode as top SRPP tube, however, the
network around the pentode is extremely complex, involving 7 resistors
and 2 capacitors, and all 5 pentode elements.
The transfer function must be much more complex than that of a
beta-follower.
The beta-follower is just a simple SRPP, w/ a simple BJT current
source, so the order of the transfer function must be lower, and the
stage complexity less.
Regards,
Dangerdave
dnot...@my-dejanews.com
<963069D49FF0F67C.36E7A0C7...@library-proxy.airnews.net
>, dang...@earthlink.net wrote:
How do they sound?
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/rg_mkgrp.xp Create Your Own Free Member Forum
dang...@earthlink.net
Excellent question. I will breadboard some and listen. Hopefully
some other RATs will do the same.
Regards,
Dangerdave
>In article
><963069D49FF0F67C.36E7A0C7...@library-proxy.airnews.net
>>, dang...@earthlink.net wrote:
>
>> Greetings RAT,
>>
>> Some posts have discussed the SRPP variants mu-follower and
>> beta-follower.
>>
Manfred Huber
>source, so the order of the transfer function must be lower, and the
>stage complexity less.
(use a fixed pitch font in your reader to read the ASCII-diagrams)
This is my view of the beta-follower:
The beta-folower looks like this:
B+
|
plate
+--- grid
| cathode
| |
| +--------------o
| |
| floating
| current
| source
| |
+-----+
|
plate
o------- grid
cathode
|
gnd
Let愀 assume that the current is perfect. This means
that the tubes operate at constant current all the time and that
the plate of the lower tube is totally decoupled from the
cathode of the upper tube. In this case the following circuit is
equivalent to the beta follower:
B+ B+
| |
current |
source |
| plate
+----------------- grid
| cathode
plate |
o------- grid +----------o
cathode |
| current
| source
| |
gnd gnd
This is a common cathode stage followed by a direct coupled
cathode follower. Nothing special to it.
The later circuit needs more parts but it has several advantages:
* The current sources have large voltage drops and for this
reason they can operate better.
* You can individualy choose the plate current for the tubes.
* You have more freedom of choice when selecting tubes.
The beta follower needs two tubes that can operate at the same
plate current. The upper tube of the beta-follower needs to
allow several volts of bias at it愀 operating point in order
to provide some headroom for the current souce.
Any errors in my logic?
Regards
Manfred
dang...@earthlink.net
Nice equivillant circuit, Manfred.
Thanks.
Regards,
Dangerdave
>>The beta-follower is just a simple SRPP, w/ a simple BJT current
>>source, so the order of the transfer function must be lower, and the
>>stage complexity less.
>
>(use a fixed pitch font in your reader to read the ASCII-diagrams)
>
>This is my view of the beta-follower:
>
>The beta-folower looks like this:
>
> B+
> |
> plate
> +--- grid
> | cathode
> | |
> | +--------------o
> | |
> | floating
> | current
> | source
> | |
> +-----+
> |
> plate
>o------- grid
> cathode
> |
> gnd
>
>Let╢s assume that the current is perfect. This means
> allow several volts of bias at it╢s operating point in order
Pascal Sternis
Since I haven't read glass audio, I used the schematics posted by
dangerdave to exercise myself , and calculated the voltage amplification
of this beta follower. maybe I did it for nothing and it's already
published in the article, but, all in all, it's a good exercise.
So, for this I took the following parameters :
2 identical tubes, with mu and Rp
beta =gain of the transistor
ro = collector resistance
RE = the resistance in the emmiter of the transistor
h11 = the base input resistance
Rp = the plate resistance of one tube
this gives a rather complicated formulae, which can be simplified if you
consider beta big enough and ro >> RE (which is the case in the
reality)
Av = -mu * (Rp + mu *( ro + Rgen)) / ( 2*Rp +(mu+1)*(ro +Rgen) )
where Rgen = (RE * beta * ro) / ( h11 + RE))
(for those interested, I could publish all the details somewhere on a
home page).
All in all, the term mu * (Rgen + ro) is way superior to Rp, and the
formulae can be simplificated more :
Av = -mu * ( mu *(ro + Rgen) ) ((mu+1) * (ro + Rgen) )
The beauty in this is you can simplify by (ro + Rgen) to obtain
Av = -mu * mu / (mu + 1) , roughly = -mu.
1) as the term Rgen dissapeared, and it contains beta, the influence of
beta is reduced.
2) you can consider also that variations in beta will cancel at the
numerator and denominator.
What I want to express, is, in the topology you described, with 2
current generators, you have to deal with the linearity of 2 current
generators.
In the "beta follower", which, imho, is not a well chosen name ;-), the
linearity problem of the transistor parameters are minimized cause they
cancel in the numerator and denominator.
Note 1: I neglected the influence of the return grid resistor to
simplify the calc.
Note 2: from these formulae, two different tubes can also be considered,
different mu (mu1, mu2) and different Rp (Rp1, Rp2). anyway, as Rp1 or
Rp2 stay way below mu * (Rgen + ro),
the formulae becomes simply :
Av = -mu1 * (mu2) / (mu2 +1). again is mu2 is big enough, Av = -mu1
I wonder, because of this property of beta cancellation, if this is the
reason why the designer of this afforded to choose a rather common
transistor , the 2N2222 ;)
Anayway, if anybody can comment, or point errors on my calculations, I
would be very happy. to one condition : it has to be mathematically
demonstrated :)
Best Regards
Pascal
--
--------
please, remove ".NOSPAM" from adress before replying
dang...@earthlink.net
<pascal....@art.NOSPAM.alcatel.fr> wrote:
Hi Pascal,
Here's something simpler to demonstrate.
Thorsten calculated the plate impedance at the bottom plate of 40 M
Ohm, in an EMail.
Using this figure, the Fo HF pole for the top 6SN7 would occur at 995
Hz ! Using the book value of 4pF for the Cgp, and no Miller effect,
because the top tube plate is common w/ the PS.
Boy, talk about a poor trade-off !
0.001% THD is unbelievable for an open loop tube stage, but at the
expense of 1kHz BW?
If Thorsten's plate impedance figure stands up to bench test, I would
assume the BW figure will as well. In any event, the HF performance
of the beta-follower version of the SRPP may well be it's fatal flaw.
Seems there's plenty of reason for some RAT to bench test this design,
before anyone actually pops one into their dream system. I will try
to, and then post the results to RAT. At the moment I have some
uncertainties ahead of me, so I might not be able to do it. If
someone else could bench test the beta-follower, it would probably be
a valuable contribution to RAT.
Regards,
Dangerdave
John Byrns
<4224E28B8F91B736.BD16FB6D...@library-proxy.airnews.net>,
dang...@earthlink.net wrote:
> On Mon, 10 Aug 1998 13:08:55 +0200, Pascal Sternis
> <pascal....@art.NOSPAM.alcatel.fr> wrote:
>
> Hi Pascal,
>
> Here's something simpler to demonstrate.
>
> Thorsten calculated the plate impedance at the bottom plate of 40 M
> Ohm, in an EMail.
>
> Using this figure, the Fo HF pole for the top 6SN7 would occur at 995
> Hz ! Using the book value of 4pF for the Cgp, and no Miller effect,
> because the top tube plate is common w/ the PS.
Come again?
I haven't been closely following this thread, and am not privy to
Thorsten's private email, but this HF cutoff is way out of line for the
"beta follower" circuit as I saw it. Now I could have easily missed
something, but isn't the bottom tube opearing basically as a common
cathode stage? If that is true, then you need to include the plate
resistance of the bottom tube in the calculation of the high frequecy
cutoff, which will move the cutoff way, way up in frequency.
Regards,
John Byrns
John Byrns
Hi Dangerdave,
I see someone else has also responded to your post, pointing out that
there is a 120k resistor shunting the current source, although he missed
the most important resistance, the plate resistance of the lower tube, or
the effective plate resistance, if there is no cathode bypass.
I don't think you need to go to iterative equations to solve it, I think
about 3 loop equations will lead to a straight forward solution, assuming
a perfect current source between the tubes. A somewhat imperfect current
source would add a slight bit of extra complication, but iteration still
should not be required.
Regards,
John Byrns
<96E659D752301965.624D30B4...@library-proxy.airnews.net>,
dang...@earthlink.net wrote:
> The drive signal to the top plate grid comes from the output of the
> current source.
>
> Thorsten calculated the load impedance seen at that point, as 40 M
> Ohm. He may have just made a rough calculation. I don't to put words
> in his mouth, and claim this is a definitive impedance calculation.
>
> But I think I see your point as well. This impedance would be in
> parallel, with the plate impedance of an 6SN7? ca. 10 or 20K?
>
> That would move the HF roll-off up to ca. 2 MHz, a much more promising
> figure.
>
> I haven't analyzed the impedance of the beta-follower myself, and I'm
> not sure I will. The problem w/ SRPP from an analytical viewpoint, to
> me, is that it's in a loop, so you have to set up recursive equations
> to solve for parameters, and I just don't want to mess with recursive
> equations and MCAD. It's always a bummer to de-bug MCAD stuff (for
> me) and it takes a long time when you write a new MCAD "spreadsheet".
>
> I think the best thing now is to do a good bench test, using spec.
> analysis, THD analysis, transient response, Zout, Fo, Gain, and so on.
>
> See what the beta-follower can really do using measurements.
>
> Regards,
> Dangerdave
Pascal Sternis
> Here's something simpler to demonstrate.
>
> Thorsten calculated the plate impedance at the bottom plate of 40 M
> Ohm, in an EMail.
>
> Using this figure, the Fo HF pole for the top 6SN7 would occur at 995
> Hz ! Using the book value of 4pF for the Cgp, and no Miller effect,
> because the top tube plate is common w/ the PS.
this figure of 995 Hz (obtained I suppose by 1/(4pf * 4Omeg * 2 * Pi) )
is fortunately erroneous...
- the current generator does not present an impedance of 40 meghom since
it is bypassed by the 120Kohm resistor thru the cap between the base of
the transitor and the anode of the bottom triode.
- This is also confirmed by some AC sweep Pspice simulations I did.
Since I am not gifted for ascii art, I can't reasonnably show it there .
As posting binary files here is not allowed, I 'll put that on a web
page soon.
- about the distorsion figures, I'm sure the tubes models I use would
not render real world figures. it is the simplified triode model based
on (Vp exp(3/2)), coming from Duncan Munro's site.
- All in all, it is already sufficient to at least have an idea of the
frequency response which is flat between 30Hz and 40Khz (assuming a 10uF
cap on the base, and no decoupling on the cathode of the bottom tube -
just for comfort in transient analysis)
- Only experimentation with a bench of 2 tubes will permit further
comparison. (if someone has done an extensive analysis of this with
accurate triode models, I would be pleased to see it posted in the
newsgroup).
- Dave, can you just remind me how much uF is the value of the
decoupling cap on the transistor, I do not recall it exactly and your
drawings have been canceled.
- I'm not pretty sure, but I begin to consider that this design is more
interesting for its way of setting the DC bias point rather than its
possible AC performances. Why I say that ? because the benefit of the
constant source is really on the DC point of view. on the AC side, the
bottom tube is loaded by .. a 120K resistor in parallel with a 40Mohm
resistance. Am I wrong somewhere ?
P.S : I could consider making a bench test to 2 conditions :
1) I will keep the batch of 20 NOS 6SN7 the requester would have given
to me,
2) He will have to loan me a distorsion analyser and a decent scope,
just remember I am located in France and i'm a poor hobbyist :)
Best Regards
Pascal
--- just remove the nospam in the return adress to email ---
dang...@earthlink.net
dang...@earthlink.net
<pste...@clubnospaminternet.fr> wrote:
>dang...@earthlink.net wrote:
>> Here's something simpler to demonstrate.
>>
>> Thorsten calculated the plate impedance at the bottom plate of 40 M
>> Ohm, in an EMail.
>>
>> Using this figure, the Fo HF pole for the top 6SN7 would occur at 995
>> Hz ! Using the book value of 4pF for the Cgp, and no Miller effect,
>> because the top tube plate is common w/ the PS.
>Hi Dave,
>this figure of 995 Hz (obtained I suppose by 1/(4pf * 4Omeg * 2 * Pi) )
>is fortunately erroneous...
>- the current generator does not present an impedance of 40 meghom since
>it is bypassed by the 120Kohm resistor thru the cap between the base of
>the transitor and the anode of the bottom triode.
I think the 120 K sets the current source value. The cap to the base
of the BJT is AC FB to the current source. I think Thorsten's
calculation of a very high impedance, seen as a plate impedance is
probably correct. Although I do not know how he calculated 40 M Ohm.
At any rate, John Byrns sent me some EMail, and I think it may be
possible to consider the plate impedance seen at the bottom tube, in
parallel w/ the plate resistance (10 or 20K) of a 6SN7. This would
give an upper tube Fo around 1 or 2 MHz. Much better. I posted the
dialogue w/ John it a prior post.
Good analysis. I love your handle. Some RAT needs to bench test this
circuit.
Regards,
Dangerdave
dang...@earthlink.net
I just found my RCA book. The book value for plate resistance of 6SN7
is around 7K. It varies w/ Ip and Vp.
Regards,
Dangerdave
Ned Carlson
This is a good point. Every once in awhile I get someone calling me
claiming they want X tube instead of Y tube, because in the book,
the listed Rp for X tube is lower...but they neglect to note that
Rp isn't a constant, and perhaps the book numbers for tube Y
were under different conditions than tube X.
Another source of confusion is the massive differences in Rp
between pentodes/tetrodes & triodes.
Ned Carlson Triode Electronics "where da tubes are!"
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
Your Start Page for Tube and Tube Amp info on the net...
http://www.triodeel.com/tlinks.htm
Thor...@tnt-audio.com
Dave wrote:
> Here's something simpler to demonstrate.
>
> Thorsten calculated the plate impedance at the bottom plate of 40 M
> Ohm, in an EMail.
I should note that this is based on a monolithic FET Current-source (about 2
Meg Ohm if memeory serves) and then multiplied by the Mu - 1 of the upper
valve....
This is the LOAD RESISTANCE (not Impedance) seen by the lower Valve. What it
is not is the Source-Impedance for the upper Grid....
The Upper Grid will be fed from a Impedance equal to the Valves Anode
Resistance (6SN7 - let's say 8kOhm) in Parallel to the Load (40 Meg estimated
- can ignore this)....
So the Source Impedance for the Upper Grid is about 8kOhm.... Giving us
something like 5 MHz as Bandwidth....
Hope that clears it....
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
======================================
> Using this figure, the Fo HF pole for the top 6SN7 would occur at 995
> Hz ! Using the book value of 4pF for the Cgp, and no Miller effect,
> because the top tube plate is common w/ the PS.
>
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
dang...@earthlink.net
Manfred Huber
please understand that I don´t want to invest too much time into
the mathematical analysis of this circuit. The hardest part of
it would be to prove the rightness of the simplifications that you
use.
>What I want to express, is, in the topology you described, with 2
>current generators, you have to deal with the linearity of 2 current
>generators.
This is true. On the other hand a current source that can drop say
150V of supply voltage can give much better result than one that has
only 4V to burn.
>the linearity problem of the transistor parameters are minimized cause they
>cancel in the numerator and denominator.
Onother way to say it is that the transistor is helped by the upper
valve which keeps the variation of the collector-emitter voltage low.
Linearity of transistor parameters is something I worry about if the
transistor is used as an amplifier. What I ask of a current source is
a high output impedance. A current source that can drop 150V can be
implemented with output impedances in the order of several megaohms
and it can maintain that high value way beyond the audio frequency
range. If one uses such a source to feed the plate of a valve with
say 10kOhms then the influence of the beta-variations of the used
transtistors is very low. I think that ne has to be more concerned
about the parasitic capacitances and the noise parameters.
>it has to be mathematically demonstrated :)
Not my cup of tea, unfortunately.
Please let me clarify that I don´t want to imply that the
beta-follower has to sound bad. After all I didn´t listen to it.
Regards
Manfred
Pascal Sternis
Thanks for your interest :) Be sure I don't want to have a struggle with
anybody, ok ;-)
Well, I have dug more deeply into this beta-follower, and I have studied
it, simulated it, built it...and not listen it yet. (as I make a lot of
this work in the evening, it's often to late to enlighten the
neigbourhood with music).
First, I apologize if i "sounded" agressive or "Mister I know
everything" that was not intended. I am here to learn things, and if I
speak about maths, it's because I am not myself very giften in this
domain, and it is in order to have enlightment by others, and better
understanding of things.
Well, after having working harder on this beta-follower, I have to admit
that my demo is pretty dumb. The transistor in the beta-follower is NOT
a current source, at least , it is not intended to works as a constant
current source. I let you discover what I think it is.
For this, The best I can do is to invite you to have look at my little
web page where I have put everything I have been able to compile on this
design. I hope you'll enjoy it. it is at
http://perso.club-internet.fr/psternis/betafol.htm
Kind Regards
Pascal
--
Visitez notre site nounours :
http://www.mygale.org/05/ourvotre/
dang...@earthlink.net
<pste...@club-int.ernet.fr> wrote:
Hi Pascal, and Manfred, and Steve.
Great work on the bench test, Pascal. Thanks.
I'm not quite sure I follow your website derivations, Pascal, but I
think you did a great job on the simulation as well.
I think the main attraction of the beta-follower, vs. normal SRPP is
decreased THD (about 280 times less) and reduced Zout (about 4 times
less), and a little more gain..
However, these are just magazine (Glass Audio) numbers, from SPICE
simulations.
I hope I will soon have the time to do bench test and listening eval.
Steve and another RAT, sorry can't remember the name, brought up
issuses of thermal stability.
I do, fortunately, have spec. analyzer and THD analyzer, and some
other lab equipment, to test some of the distortion claims. And even
a two little lab ovens to run temperature tests.
Thanks again, to all the contributors so far on this thread. I think
it's great that RAT can be used to verify the performance claims of
"new" tube circuits, and share the information w/ interested hobbyists
and designers.
Nice work Pascal. You really put a lot of effort into this, and it
shows. Beautiful web graphics.
BTW: although I think Pascal's calculation of Zout for the current
source may be slightly low, interested RATs could substitue FET or
"true" BJT-FET current sources, if desired, and as suggested by other
contributors. I think a limiting constraint here would be the voltage
drops across the current sources, and the range of operation
limitations given the need to sustain grid voltage of the top tube
within limits.
Regards,
Dangerdave
Manfred Huber
>Be sure I don't want to have a struggle with anybody, ok ;-)
I second that.
>First, I apologize if i "sounded" agressive.
You did NOT. No need to apologize.
>The transistor in the beta-follower is NOT a current source, at least,
>it is not intended to works as a constant current source.
Well, the inventor of the beta-follower states clearly that he intends
it to be a CS.
>I let you discover what I think it is.
I think it is a CS, but not a good one.
>For this, The best I can do is to invite you to have look at my little
>web page...
Impressive work. You obviously did invest a great deal of time into
this analysis.
Now I wait for the first listening tests to appear here on RAT.
Regards
Manfred