| Author |
Message |
Jon Yaeger
Guest
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 Posted:
Tue Jan 11, 2005 10:54 pm Post subject:
Basic Question |
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For tubes used in single-ended mode, what is the relationship between R
plate and optimal R load?
TIA,
jon
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John Stewart
Guest
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| Posted:
Wed Jan 12, 2005 12:44 am Post subject:
Re: Basic Question |
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Jon Yaeger wrote:
| Quote: | For tubes used in single-ended mode, what is the relationship between R
plate and optimal R load?
TIA,
jon
|
For triodes Rl = 2 Rp gives max power transfer.
Rl = 3 Rp lowers the distortion quite a bit & the max power goes down a
little.
Rl = 3 Rp is probably optimum, although some would go farther.
If for a voltage amplifier, higher is always better. That is why some use a
mu stage.
For pentodes & beam tetrodes the relation varies a lot, but Rl is usually a
lot less than Rp. The harmonics produced are very much influenced by the
load chosen. Things get much worse in real applications since the load will
be complex ( a loudspeaker) & the program material (music & voice) is
complex.
In a PP stage since even order harmonics are canceled in a well balanced
amp, I usually use a load that is somewhat less than double the SE value
since that will reduce the 3rd harmonic.
I like to use lower values of load Rl for pentode voltage amplifier since
the pentode is a current source.
My 2 cents, anyway!!
Cheers, JLS |
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TubeGarden
Guest
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| Posted:
Wed Jan 12, 2005 12:54 am Post subject:
Re: Basic Question |
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Hi RATs!
If R load is equal to, or less than R plate, power is low and distortion is
high.
As R load gets bigger, power goes up and distortion goes down, for a while,
then power comes back down and distortion keeps going down.
How does it sound? try it and see. Only morons assume that most power or least
distortion must be the same as my favorite sound ;)
You hear what you like when you hear it.
Everything else is techno babble.
Somewhere between 2x and 10x is prolly the sound of your dreams ;)
Happy Ears!
Al
Alan J. Marcy
Phoenix, AZ
PWC/mystic/Earhead |
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Ian Iveson
Guest
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| Posted:
Wed Jan 12, 2005 1:29 am Post subject:
Re: Basic Question |
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"Jon Yaeger" <jono_1@bellsouth.net> wrote in message
news:BE097CF6.152DB%jono_1@bellsouth.net...
| Quote: | For tubes used in single-ended mode, what is the relationship
between R
plate and optimal R load?
|
Optimum for what?
For power, Rout should be equal to Rload. If Rout is equal to Rp,
then Rp should be equal to Rload, but Rp changes, typically by quite
a lot in drivers and power stages, which muddies the water
considerably.
Assuming a constant load, best linearity is at infinite RL, as with
bottom valve of mu-follower. Not so with top valve
though...obviously with a truly infinite load there is no power
output.
Best linearity of all would be with an active, varying and
sympathetic load, such as srpp etc.
For triodes, a common compromise between linearity and power seems
to be Rload = 2 to 3 times Rp for triodes. Pentodes are more of a
mystery...similar perhaps if you take Rp of triode-connected
pentode. Amps that switch between output modes don't seem to bother
changing the load.
I thought of producing a graph from simulation, hoping for a knee
somewhere. Restricting to SE makes it easier, but it is still
necessary to find the optimum operating point for each change in RL,
so the job is long and tedious, and would be easier in real time
with real instruments.
Perhaps someone has done the maths? Ignoring the foibles of
individual valves, it should be possible to derive the relationships
between load, distortion and power from the classic equations. Is it
in RDH? Patrick?
cheers, Ian |
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Fabio Berutti
Guest
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| Posted:
Wed Jan 12, 2005 3:01 am Post subject:
Re: Basic Question |
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I saw that many others (and more expert) RATs already answered, but I think
that one pont is missing, which is: at which operating point? Usually, the
closer You go to the maximum allowable voltage, the higher the reflected
load should be. Just take the anode curves and play a little with a clear
ruler: it is apparent that at high B+ a higher impedance is needed to avoid
reaching the "zero current" line for a symmetric driving voltage swing
(reaching the 0V on the grid "on the other side").
IMHO a higher impedance gives a more "equilibrated" sound even with
difficult loudspeakers, a lower one yields a more "dynamic" sound, but at
high volumes it seems to "yell" a bit. (PERSONAL impression, don't shoot me
I'm only the piano player...). I recently finished a 6S33 SE having a dual
output winding for 600 and 900 Ohms, and it confirmed this (to my ears..).
I first noticed it with a 6V6 PP-UL amp using Hammond OPTs at 5000 Ohms,
which is far lower than optimal (all D/S say 8000). Well, it's a great amp
for "energetic" music, say R&B.
OTOH, my beloved 2A3SE uses 3K5 OPTs to take care of impedance "dips".
What's better, a 10-yrs single-malt Scotch or a Jack Daniel's Reserve? A
good excuse to taste another small sip or to build another amp: the best is
always the next one.
Ciao
Fabio
"Jon Yaeger" <jono_1@bellsouth.net> ha scritto nel messaggio
news:BE097CF6.152DB%jono_1@bellsouth.net...
| Quote: | For tubes used in single-ended mode, what is the relationship between R
plate and optimal R load?
TIA,
jon
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Ian Iveson
Guest
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| Posted:
Wed Jan 12, 2005 3:48 am Post subject:
Re: Basic Question |
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"John Stewart" <jh.stewart@sympatico.ca> wrote in message
news:41E42C99.A544EC77@sympatico.ca...
| Quote: |
For triodes Rl = 2 Rp gives max power transfer.
|
I've seen this before and you are probably right in some sense, but
how?
If you regard the valve as a fixed voltage source with series output
resistance Rp, then maximum power output is when Rload = Rp. To get
2Rp you must be working with a different definition or assumption
somewhere.
Just tried with a simulated triode-connected EL84. Set up with -12V
on grid and 330V on anode via a 1kH perfect choke, with large cap
and resistor load to ground. Measured Rp as 1.7k by injecting 1V
signal at anode and measuring ensuing current variation. Then tried
a range of load resistor values. Compared power outputs from 1V grid
input signal by measuring voltage and current into load resistor and
multiplying together.
Holding constant 1V signal to grid, highest power was with load of
1.7k
So what does "max power transfer" mean?
| Quote: | I like to use lower values of load Rl for pentode voltage
amplifier since
the pentode is a current source.
|
lower than what, John? And why does lower follow from the fact that
it is a current source? Is it a more linear current source at low
Rload?
I have wondered for ages...can't remember resolving the matter...why
the common loading for 4 * EL84 pp amp (for example) is 4k no matter
whether it is triode, UL or pentode mode. Exceptions are for class A
"hi-fi" which use 5k.
cheers, Ian
| Quote: | Rl = 3 Rp lowers the distortion quite a bit & the max power goes
down a
little.
Rl = 3 Rp is probably optimum, although some would go farther.
If for a voltage amplifier, higher is always better. That is why
some use a
mu stage.
For pentodes & beam tetrodes the relation varies a lot, but Rl is
usually a
lot less than Rp. The harmonics produced are very much influenced
by the
load chosen. Things get much worse in real applications since the
load will
be complex ( a loudspeaker) & the program material (music & voice)
is
complex.
In a PP stage since even order harmonics are canceled in a well
balanced
amp, I usually use a load that is somewhat less than double the SE
value
since that will reduce the 3rd harmonic.
My 2 cents, anyway!!
Cheers, JLS
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Chris Hornbeck
Guest
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| Posted:
Wed Jan 12, 2005 5:34 am Post subject:
Re: Basic Question |
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On Tue, 11 Jan 2005 22:48:58 GMT, "Ian Iveson"
<IanIveson.home@blueyonder.co.uk> wrote:
| Quote: | I've seen this before and you are probably right in some sense, but
how?
|
The disparity comes from trying to avoid grid current. Given
unlimited grid current, load matching applies. But given zero
grid current, well, yadayada.
Not yet discussed is the issue of source impendance to the
loudspeaker for the tre, tre moderne zero-loop-feedback
output stages. Loading impedance ratio equals damping
ratio, (-ish).
Great to see the topic back, yet again,
Chris Hornbeck
"Conscious that we must have sprung from somewhere, we are
lured to the riddle of our origins." -Malcolm W. Browne |
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John Stewart
Guest
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| Posted:
Wed Jan 12, 2005 6:29 am Post subject:
Re: Basic Question |
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Ian Iveson wrote:
| Quote: | "John Stewart" <jh.stewart@sympatico.ca> wrote in message
news:41E42C99.A544EC77@sympatico.ca...
For triodes Rl = 2 Rp gives max power transfer.
I've seen this before and you are probably right in some sense, but
how?
|
Rather than me trying to explain why such a strange thing should happen,
I will leave it to the experts. See the exert from Samuel Seely's
'Electron Tube Circuits' over at ABSE. Basically, even though the triode
is a pretty good voltage source it is not perfect. So to get max power
we need to treat it a little differently.
| Quote: | If you regard the valve as a fixed voltage source with series output
resistance Rp, then maximum power output is when Rload = Rp. To get
2Rp you must be working with a different definition or assumption
somewhere.
Just tried with a simulated triode-connected EL84. Set up with -12V
on grid and 330V on anode via a 1kH perfect choke, with large cap
and resistor load to ground. Measured Rp as 1.7k by injecting 1V
signal at anode and measuring ensuing current variation. Then tried
a range of load resistor values. Compared power outputs from 1V grid
input signal by measuring voltage and current into load resistor and
multiplying together.
Holding constant 1V signal to grid, highest power was with load of
1.7k
So what does "max power transfer" mean?
I like to use lower values of load Rl for pentode voltage
amplifier since
the pentode is a current source.
lower than what, John?
|
Usually much lower than Rp. To get bandwidth, I like 100K or less.
Unlike a triode, the pentode becomes more non-linear as the load Rl
increases. I have something in mind to have a pentode drive into a
virtual short circuit. You often see that with Op-Amps. That way it
becomes a better voltage to current convertor. So we will see when I've
got some time!!
| Quote: | And why does lower follow from the fact that
it is a current source? Is it a more linear current source at low
Rload?
|
Yes.
| Quote: | I have wondered for ages...can't remember resolving the matter...why
the common loading for 4 * EL84 pp amp (for example) is 4k no matter
whether it is triode, UL or pentode mode. Exceptions are for class A
"hi-fi" which use 5k.
|
Most of these bottles will tolerate loads from half to twice of nominal
& give passable results. If you think of it a loudspeaker load does that
anyway. I usually aim for a load somewhat less than optimum required to
drive an R load since the loudspeaker will be all over
anyway. Cheers, John
> cheers, Ian |
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Chris Hornbeck
Guest
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| Posted:
Wed Jan 12, 2005 8:35 am Post subject:
Re: Basic Question about RL/Ra |
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On Wed, 12 Jan 2005 15:49:05 +1100, Patrick Turner
<info@turneraudio.com.au> wrote:
something, I guess. Dude, you're doubtless a sweet person,
but you seriously need an editor. "Life is short. Art is long.
Opportunity fleeting. Experiment treacherous. Judgement
difficult." -Archimedes
I'd personally love to read your comments, but(!) a good
ROT on Usenet is to keep it to a single page. If you don't
care about me reading your thoughts, you can, of course,
continue. It's a free world. But I'd like to be able to.
Please, please don't take this as some kinda harsh criticism.
Anything but.
Good fortune,
Chris Hornbeck
"Happiness isn't something you experience; it's something you remember."
-Oscar Levant |
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Patrick Turner
Guest
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| Posted:
Wed Jan 12, 2005 8:35 am Post subject:
Re: Basic Question about RL/Ra |
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Jon Yaeger wrote:
| Quote: | For tubes used in single-ended mode, what is the relationship between R
plate and optimal R load?
|
" R plate " as you call it is normally called the plate resistance, or
anode resistance, Ra, in the books.
Ppl may wonder WTF is Ra?
Well if you were given a triode set up in a mystery box and told there's
the 0V terminal,
and there is the anode terminal, and asked what the resistance or impedance
was between the two, you'd
be able to find out by connecting the anode to a low known impedance or
resistance output
and measuring Va change and Ia change with two values of series R.
You should be utterly familiar with such basic measurements, and in fact
the results you obtain will be the
Ra in parallel with whatever RL is inside the mystery box.
If RL is known, we can work out Ra. The measurements are done using small
signal voltages so
the distortion in the signals is below 1% and the results will then be
untainted by
major inaccuracies.
The best way to appreciate the relationship between Ra and power output
and distortion is to plot a graph for the maximum power output limited by
clipping
for various values of RL for a given tube at a given set of idle operating
conditions
of Ia and Ea.
Since so few ppl use pentodes, and because the rules about pentodes differ
to triodes,
which are used more often as SE configs at least in small signal circuits,
then let us consider a
triode such as a 6SN7.
You could set one up with B+ = 320v, RL = 50k, and Ia at 3.3 mA.
This will give you about Ea = 155v across the anode to cathode, with Ek at
about 6v
to get the bias.
The measured Ra with Rk fully bypassed will be about 10k, somewhat higher
than the
makers say because at 3.3 mA, Gm is lower than it is when Ia = 10 mA,
and as you should know, for all tubes, Gm = Ra / U and Ra is higher as Ia
is reduced
yet U remains fairly constant, determined by the dimensions of the
electrode structures.
One can draw the load line across a fresh printed copy of the anode curves
for 6SN7
and for the 50k RL, and get some pleasant looking results for gain and
thd. RL = about 5 x Ra.
If the line describing RL is rotated so it is more vertical through the
quiescent idle point
of 3.3 mA and 155v, then the maximum voltage swing is reduced, thd
increases,
and that's as a result of making the RL : Ra a lower ratio.
One could have a 10 ohm load on the 6SN7 and such a small RL:Ra ratio
will give low gain and the highest possible thd, and such a load line
appears almost vertical.
Measuring and observing the above 6SN7 with a CRO then should complete you
mental question about
WTF is Ra?
Gain for all tubes = ( U x RL ) / ( Ra + RL )
So you know why the gain is tiny with 10 ohms.
The Gm character of a tube is not linear, and the current change at an
anode
due to a grid voltage change is proportional to the square root of the
applied voltage cubed.
[[ Sorry, but the day the God of Triodes invented that idea he'd had a row
with the Main God
during a maths class, and he took it out on Residents of the Universe such
as us.]]
However,
While there is this non linear Gm, or transconductance character,
there is also electrostatic NFB acting within the triode, and it has
maximum effect
when no current change occurs, and with a constant current load on the
triode
there is *no* current change at the anode and *only* a voltage change
occurs.
The output from a triode can be purely a voltage one, and then the transfer
function is
at its most linear.
The load line for a CCS is a horizontal line drawn through the operationg
point.
Then the distances along the line between the plate resistance lines drawn
for
various values of Eg1, ie, grid bias voltages remain equal.
And also they remain equal whether Ia is 1 mA or 20 mA, and this shows
that U is constant.
The gain with CCS = U.
If RL is infinite as it is with a CCS, then adding infinity into the above
gain equation gives ( U x infinity ) / (Ra + infinity ) = U.
In olden times CCS loads were rarely ever used for loading signal triodes
due to cost
unless it was special gear for medical or the military.
But CCS loading is a case of a very high RL:Ra ratio.
But for us latter day mortals, 50k isn't a bad load for our triode
situated in a line stage preamp or at the input to a power amp since it
will
produce less than 0.05% of mainly 2H at a volt of output.
The use of CCS might reduce thd maybe to 0.02% at the volt,
depending on the tube sample.
Tubes are not perfect, despite the triode theory which predicts they ought
to be pretty darn good,
which of course they are, used wisely.
And one rarely ever actually loads signal triode with a pure CCS since
there is always usually
a cap coupled bias R or gain pot from the anode.
But I still like to sneak an MJE350 CCS into the anode circuit to
supply the DC to the anode, thus minimising the thd.
With power tubes, you can't have any power if RL is infininite, ie, a CCS,
or if its zero, ie, a short circuit because for power
we need voltage and current change in a load.
If one sets up a 300B at Ea = 300v and Ia = 120 mA,
then about 6 watts is possible is RL = Ra = 800 ohms.
But the thd is high, and nobody recomends it.
Even 200 ohms could be used, but thd is slightly worse, but the voltage
swing is quite restricted
as cut off occurs and PO is quite low.
Its much better to set up with Ea = 400v and Ia = 80 mA,
and then have RL about 5k, and PO is about 6 watts, but thd is a lot lot
lower than
if the tube is set up at 350v and 120mA, and RL = 800 ohms.
The effect is clearly visible when you draw the load lines across a
printout of the tube curves.
With a high value RL, the clipping occurs due to grid current.
There will be one load value where grid current and cut off begin
to occur simultaneously, and that load can be found graphically.
Often that's the maximum useful PO for the idle conditions chosen.
The other major benefit to using a higher value nominal RL is that the OPT
impedance
ratio is higher so that for a match to 6 ohms, 800 : 6 gives 133:1 Z ratio,
so
the Ra of the 300B ( or any other triode ) is transformed to 800 / 133 = 6
ohms at the secondary,
which gives a poor damping factor of 1 with a 6 ohm speaker.
With a 5k : 6 ohm match, ZR = 833 : 1 so Ra is transformed
to 800 / 833 = 0.96 ohms so the DF = about 6.
In practice, the winding resistance of say 0.3 ohms in a good OPT of 5%
winding losses
increases the Ro of 0.96 to about 1.3 ohms, so DF is 4.6, and acceptable to
many people,
even without any loop NFB.
Pentodes have very high Ra, an EL34 typically meausres 15k at idle,
yet the load recommended is from 2k to 4k.
RDH4 says RL should be = 0.9 Ea / Ia so that if
Ea = 300v and Ia was 80 mA, then RL should be 0.9 x 300 / 0.08 = 3.375
ohms,
and in practice this is about right.
There is no benficial relationship between RA and RL
In fact if we stray from the ideal RL = 0.9 x Ea/Ia we get very much
worse thd either way.
Since pentodes have next to no natural NFB within them
we must apply about 16 dB to most power pentodes to get them to
measure as a well as a triode in an optimally loaded condition,
for the first few important watts.
OK, a pentode makes more power, but I'd rather have two EL34 in triode
in parallel SET with very little NFB if any than a single EL34 in pentode
with
the same maximum power and a shirtful of NFB.
However, even though this is my broad choice, I have found that
SE power pentodes and power beam tetrodes respond favourably
to the use of local NFB in the output stage using a wide bandwidth OPT with
a CFB winding.
This is because the CFB acts to reduce odd order distortion products in the
same way
that the UL connection achieves, mainly due to the effective NFB applied
around the screen circuit,
and the spectral nature of the thd is somewhat as good as the triodes
with no local NFB, and if the power ceiling is say 4 times greater than
what is actually
needed, pentode/tetrode output stages can be quite sonically impressive.
Patrick Turner.
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Andy Evans
Guest
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| Posted:
Wed Jan 12, 2005 2:50 pm Post subject:
Re: Basic Question about RL/Ra |
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Dude, you're doubtless a sweet person,
but you seriously need an editor. >
Not for people who enjoy lucid explanations, which Patrick is very good at.
=== Andy Evans ===
Visit our Website:- http://www.artsandmedia.com
Audio, music and health pages and interesting links. |
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Patrick Turner
Guest
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| Posted:
Wed Jan 12, 2005 7:00 pm Post subject:
Re: Basic Question about RL/Ra |
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Chris Hornbeck wrote:
| Quote: | On Wed, 12 Jan 2005 15:49:05 +1100, Patrick Turner
info@turneraudio.com.au> wrote:
something, I guess. Dude, you're doubtless a sweet person,
but you seriously need an editor. "Life is short. Art is long.
Opportunity fleeting. Experiment treacherous. Judgement
difficult." -Archimedes
I'd personally love to read your comments, but(!) a good
ROT on Usenet is to keep it to a single page. If you don't
care about me reading your thoughts, you can, of course,
continue. It's a free world. But I'd like to be able to.
|
Ge at least I made to being a "sweet person", and
I get to be called Dude.
That's no mean achievement on a news group prone to be full of
characters who have given me such a hard time ....
Anyway, quite a lot of illustrious replies were made to Mr Yeager's inital
enquiry, and a short brief reply seemed out of place because it'd leave too
many questions unanswered
for those of us with inquiring minds.
Its normal around here for some of us to try to explain fully how we see
things.
Sometimes the explanations overlap each other a bit....
| Quote: | Please, please don't take this as some kinda harsh criticism.
Anything but.
Good fortune,
|
If only a fortune could be made from knowing something about tubes.........
I should have been around in 1950.
But the competition was much more severe then....
Have a great year,
Patrick Turner.
| Quote: |
Chris Hornbeck
"Happiness isn't something you experience; it's something you remember."
-Oscar Levant |
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Robert McLean
Guest
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| Posted:
Wed Jan 12, 2005 7:47 pm Post subject:
Re: Basic Question |
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Note that the analysis is different for a transformer coupled load, and a
load resistor connected between B+ and the plate. The section 10.3 from
Seeley that John gives is for a transformer coupled load, and give Rl = 2 *
Rp for max power. For a load resistor situation see section 10.1 from the
same book by Seeley, which shows that max power occurs at Rl = 1 * Rp.
The fact that the situations are different may be the source of some of the
confusion, or occassional dispute.
"John Stewart" <jh.stewart@sympatico.ca> wrote in message
news:41E47D64.61FA8553@sympatico.ca...
| Quote: | Ian Iveson wrote:
"John Stewart" <jh.stewart@sympatico.ca> wrote in message
news:41E42C99.A544EC77@sympatico.ca...
For triodes Rl = 2 Rp gives max power transfer.
I've seen this before and you are probably right in some sense, but
how?
Rather than me trying to explain why such a strange thing should happen,
I will leave it to the experts. See the exert from Samuel Seely's
'Electron Tube Circuits' over at ABSE. Basically, even though the triode
is a pretty good voltage source it is not perfect. So to get max power
we need to treat it a little differently.
If you regard the valve as a fixed voltage source with series output
resistance Rp, then maximum power output is when Rload = Rp. To get
2Rp you must be working with a different definition or assumption
somewhere.
Just tried with a simulated triode-connected EL84. Set up with -12V
on grid and 330V on anode via a 1kH perfect choke, with large cap
and resistor load to ground. Measured Rp as 1.7k by injecting 1V
signal at anode and measuring ensuing current variation. Then tried
a range of load resistor values. Compared power outputs from 1V grid
input signal by measuring voltage and current into load resistor and
multiplying together.
Holding constant 1V signal to grid, highest power was with load of
1.7k
So what does "max power transfer" mean?
I like to use lower values of load Rl for pentode voltage
amplifier since
the pentode is a current source.
lower than what, John?
Usually much lower than Rp. To get bandwidth, I like 100K or less.
Unlike a triode, the pentode becomes more non-linear as the load Rl
increases. I have something in mind to have a pentode drive into a
virtual short circuit. You often see that with Op-Amps. That way it
becomes a better voltage to current convertor. So we will see when I've
got some time!!
And why does lower follow from the fact that
it is a current source? Is it a more linear current source at low
Rload?
Yes.
I have wondered for ages...can't remember resolving the matter...why
the common loading for 4 * EL84 pp amp (for example) is 4k no matter
whether it is triode, UL or pentode mode. Exceptions are for class A
"hi-fi" which use 5k.
Most of these bottles will tolerate loads from half to twice of nominal
& give passable results. If you think of it a loudspeaker load does that
anyway. I usually aim for a load somewhat less than optimum required to
drive an R load since the loudspeaker will be all over
anyway. Cheers, John
cheers, Ian
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Ian Iveson
Guest
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| Posted:
Wed Jan 12, 2005 8:23 pm Post subject:
Re: Basic Question |
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"John Stewart" <jh.stewart@sympatico.ca> wrote
Thanks John!
Diagram on one page, explanation on the other, no printer, learning
can be a tiresome process. Doesn't quite explain how that
complicated function has a maximum at 2Ra. Eyes are sore from
flipping between pages...I'll have another look later.
I notice that the example triode on page 2 appears to peak at 2.5
times Ra.
One way of looking at this is that Ra changes a lot through a full
amplitude swing, especially with large values of RL. This is because
the characteristic curves are curved.
The key to the apparent discrepancy is "maximum undistorted power"
rather than "maximum power for a given input signal". Higher loads
allow a greater input signal by the latter definition. They also
imply a higher HT value with a resistive load, or a greater possible
Va swing in any case.
I was surprised by my result, to tell you the truth. For a given
operating point and a given small (ish) signal, power output is
maximum when Rload = Ra, assuming no clipping. Hence the notional
voltage source is quite linear with respect to grid input for small
signals. Shouldn't have been surprised now I come to think...
As for the pentode and points between, I suspect that Rout for an
ideal UL connection is close to the optimum Rload regardless of
mode. Coincidentally, I remember H&K's graph showing max power
output and distortion for different proportions of screen feedback.
Broadly similar to the ones in your excerpt. Did they keep to the
same Rload? The whole UL thing could be seen as an
impedance-matching exercise perhaps?
| Quote: | I like to use lower values of load Rl for pentode voltage
amplifier since
the pentode is a current source.
lower than what, John?
Usually much lower than Rp. To get bandwidth, I like 100K or less.
Unlike a triode, the pentode becomes more non-linear as the load
Rl
increases. I have something in mind to have a pentode drive into a
virtual short circuit. You often see that with Op-Amps. That way
it
becomes a better voltage to current convertor. So we will see when
I've
got some time!!
|
Right, that would be interesting. Should be able to see this on
characteristic curves. A vertical load line gives a far from linear
result...but is it more linear than other angles? I have a suspicion
that this depends on which way the lines curve...triode and pentode
are opposites in this respect.
Don't get *any* power though, so perhaps we stray from the point?
Just wondering in passing what kind of load a top pentode in a mu
stage sees?
cheers, Ian |
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Patrick Turner
Guest
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| Posted:
Wed Jan 12, 2005 8:27 pm Post subject:
Re: Basic Question about RL/Ra |
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Andy Evans wrote:
| Quote: | Dude, you're doubtless a sweet person,
but you seriously need an editor.
Not for people who enjoy lucid explanations, which Patrick is very good at.
|
If anyone else had written something similar to what I wrote today
in 1990, before I had studied and built a few working amps and read quite a few
books,
I wouldn't have understood a word.
But I used to pull apart old radios when I was 17 and build illegal
transmitters and at one stage
I had an illegal radio telephone with a mate set up.
All I had was a crummy voltmeter which didn't last long.
And I had no deep understanding of what I was doing but things sort of worked.
Probably the RF signal I put out at a watt or two from a 6V6 was full of
spurious harmonics,
and tuning a receiver to any of them gave tolerable reception and voice
intelligibility. I never went into the effort of deeper understanding, and
at 18 other things began to interest me more than boring electronics,
such as girls, and motorcycles, and going to college to get real knowledge
of something gutsy and useful in the construction industry.
I thought the blokes who were interested in electronics to be mainly nerds.
When solid state started and then the first logic circuits I got quite bored.
At this later age I do find tube craft fascinating.
If anyone really wants a challenge just try to work out
reactive load quantities and try to sort out the math involved
with J and involving the square root of -1.
I fail dismally in that area, yet the speakers we listen to will vey often
be very considerably reactive and complex in the load they present to an amp.
It can all be worked out on paper, but not by me though.
Is it boring? well, depends on whether you like math.
Math to some people is like the beauty of a well played game of chess.
Anyway, calculations for reactive loads can easily
involve equations that have 50 quantities stretching across the page.
Only one wrong quantity or + or - sign, and a day's calcs are
entirely wasted.
Fortunately, the loads we connect our amps to are little troubled by reactive
elements.
A book by one Cathode Ray from the 1950s tries to make it simple
but I find myself thinking whatever he is trying to explain is beyond my grasp
of logic.
I have tried several times to re-read it and it remains difficult for me.
If only Mr Ray were around the corner from me and was willing to coach
me on a cold winter's afternoon....
I might "get it" then, but alas I feel rather alone trying to grasp it on my
own.
Most folks here are hobbyists, so surely they can absorb some of the simpler
things I have
come to understand as I see them, but should they want to learn more,
they shouldn't hesitate to read more, and they are well in front of me if they
are good at math.
What one sees in the majority of websites about tubes is inspiring but
often there isn't the depth of understanding that existed in the
earlier era of electronics when analog and tubes was all there was.
Where does all the expressed web based knowledge come from?
Out of old books now mainly moulding away in university basements.
I was lucky enough to get a few which finished up in second hand book stores
as deceased estate hoards from trained guys who were aged 40 in 1950,
and employed in the electronics industry.
Now, anything with a hard cover, and about old fashioned electronics
and perhaps used as a textbook in 1950 is not to be seen in the local book
stores and methinks
that they are spirited off to markets willing to pay serious money, rather than
the $10 I was able to obtain them for.
Ian raised the idea of the equivalent model circuit of a triode
being a pefect low output impedance voltage generator with
a high impedance input terminal controlling the output
voltage so that for a volt of grid input voltage we get
U x Vin as the gene output voltage, so the output is U x Eg.
Then we have Ra, the dynamic plate resistance in series with the load.
This is the classic simple equivalent model for a triode and any triode in a
circuit
can be shown as this model. Its very useful for determining circuit gain from
basic principles and the very simple Ohm's law.
It also applies to a pentodes. Except U for an EL34 is
Gm x Ra = 0.01 x 12k = 120.
So for 20 v input to the grid we have 2,400v at the generator output and
feeding
through 12k of Ra to any load we may connect.
Well how come there is 2,400 v of signal inside the tube?
It isn't really there, that is the fact.
But it is there if we allow ourselves to think of our models as just
models, and imaginery device to be of use when cobbling a pentode into a given
circuit.
The EL34 pentode acts "as if there was a gene and Ra" inside the tube envelope.
Another way to consider a tube is where we have a signal
current source generator where the current output between 0V and the output is
Gm x Eg.
The Ra is then between the current gene output, where our load connects, and
0V.
Most ppl find this harder to understand, since they think of circuit working
for tubes in terms
of voltage gains, not current generation.
If you have a gene of 0.01 A/V with 12k between its output and 0V, then you
have a model for an EL34. The gain of the model is where no load is connected,
and output is simply Gm x Ra = 0.01 x 12k = 120.
if we had a 3k load connected between the output of our model
and 0V, then the current gene producing 0.01A/V is loaded by 3k and 12k in
parallel
= 2.4k and the gain for the circuit is 0.01 x 2,400 = 24.
The Ra is high in relation to the load, and with a pentode gain varies almost
as the the load value. NFB is used to reduce the effects of such gain changes
with load variations, and 20 dB of NFB is enough with an EL34
to reduce the apparent Ra from 12k to say 1k.
In fact, Ra' after FB is applied = Ra / ( 1 + [U x B] )
where Ra is Ra with no FB, U is amplifiction factor and B is the fraction
of the output fed back in series with the input the input.
Since U = Gm x Ra, then U in the above equation can be expressed as such,
and when we then gaze at such an equation we can see that if Ra was an enormous
value of ohms, then the Ra after FB is applied is 1 / ( Gm x B )
So if if we had a signal pentode with Ra = 1 megohm,
and B = 0.1, and Gm = 0.003 A/V ( like a 6AU6 ),
then Ra = 1 / 0.003 x 0.1 = 3.33k.
So by reducing the sensitivity of a pentode gain stage by a factor of about 10
with series voltage NFB will convert what is naturally a near perfect
current source into what is a voltage source, if the RL we want to drive
is many times the value of the effective Ra after NFB is applied.
Pentodes generally have far greater gain than triodes, so we can afford to
apply the NFB and still have an acceptably low number of tubes in the circuit.
Dynaco ST70 is a prime example, where the gain of the input pentode
is enormously boosted with bootstrapping from the concertina phase inverter, so
one pentode does what a pair of triodes in cascade would do.
Using a tube as a cathode follower is a case of maximal local
series voltage NFB.
B = 1 because all the output voltage is in series with the input; there is not
a fraction of the
output fed back in series, *all* of it is, or 1.0 times the output is.
Thus our apparent Ra' becomes I / ( Gm x 1.0 ) or simply 1/Gm,
and that for a 6AU6 = 1 / 0.003 = 333 ohms, somewhat lower than the
1 Megohm we started with! But we pay for this with a reduction in gain to
just less than 1.0.
I still prefer the triodes for gain blocks, but that's another story.
We have to make sure the NFB is either shunt negative voltage feedback
or series negative voltage FB.
Series negative current FB, like an unbypassed Rk, increases Ra, although
positive series current FB has the Ra reducing effect, but alas that increases
THD
and instability.
But I digress, that's enough, enjoy your tinkering.
Patrick Turner.
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