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Tube DIY Asylum Do It Yourself (DIY) paradise for tube and SET project builders. |
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Tube DIY Asylum Do It Yourself (DIY) paradise for tube and SET project builders. |
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In Reply to: Raa/2 and Raa/4, and other questions... posted by Damir on September 15, 2004 at 00:20:22:
… among other useful comments redacted from the fiasco.All,
Be forewarned:
1) It’s not pretty and it’s not easy to read, but it’s also not “gone.â€
2) There was a LOT of rubbish to dig through and it is highly unlikely that any such efforts will be made in the future should a wholesale delete become necessary. That said, some participants may wish to consider the value of saving their work in the event that another round of chaos ensues.FWIW, most Bored Members and/or moderators have full-time jobs, families and pursuits outside of the Asylum. Sometimes they are not able to drop in periodically to do quick clean-ups. Often, outside priorities preclude checking in even daily. Even with regular check-ins, it is rare that moderators have time to read all previously unread posts. When things reach a certain level of chaos, the easiest, least time-consuming, and often most-effective action (in terms of having the participants realize that it’s important for them, too, to self-manage or play a actively constructive role in the development of their “communityâ€) is a mass delete.
All the deleted threads were sifted through. Those containing valuable information were redacted. The format is the usual:
Subject:
Posted by:
In reply to:
____________________________
THREAD ONE
____________________________
And the Answer Is..... (drum roll please)
Posted by Mark Kelly (A) on September 11, 2004 at 22:00:09What were we arguing about?
I set up a balanced quasi CCS into a push pull transformer so that I could directly measure input impedance to each side.
Made some measurements which accorded very nicely with theory.
The 9k A-A transformer measures 4k5 each side when driven in balanced mode.
Drive one side only and it measures a little under 2k (2k2 in parallel with
the 10k output impedance of the undriven side of the quasi CCS).Sat back and thought to myself "Hang on a sec, what were the arguments about anyway?"
Re: And the Answer Is..... (drum roll please)
Posted by Meister (A) on September 12, 2004 at 03:08:00
In Reply to: And the Answer Is..... (drum roll please) posted by Mark Kelly on September 11, 2004 at 22:00:09:I can't understand your measurement. If you are using a common current source for the cathodes you can't drive only one side with the grids. Or do you seperate one plate from one side of the outputtranny?
Not V/Ts involved
Posted by Mark Kelly (A) on September 12, 2004 at 03:42:06
In Reply to: Re: And the Answer Is..... (drum roll please) posted by Meister on September 12, 2004 at 03:08:00:I'm just intereted in the behaviour of the tranformer.
All I am using is a balanced signal source (Old HP 200CD generator) with enough resistance in series with each output (10k per leg) for it to approximate a current source.
Posted by mqracing (M) on September 11, 2004 at 23:19:32
In Reply to: And the Answer Is..... (drum roll please) posted by Mark Kelly on September 11, 2004 at 22:00:09:Hi Mark:
Here's where the confusion stems from I believe.
You measure voltages and currents across one half of a push pull trans... and then compute the the "resistance"... and you get 4.5K... so you figure you add 4.5K plus 4.5K and you get 9K... close enough to 10K to figure it's right.
problem is... this is a PP transformer... not two resistors in series with one another.
the PP transformer has two windings in series with each other. Assuming equal number of turns and etc (all the normal stuff about PP tranneys)...
with two primary windings in series....
-- the voltages add (will be two times the volts on one side)
-- the impedance quadruples
-- the current is cut in halfas contrasted to the voltages, impedances and current in one half of the primary winding.
One of the problems with calling this a 4.5K 1/2 pri reflected impedance
is.... that if you series two coils of the same impedance then the impedance quadruples... so your 4.5K suddenly becomes an 18K CT primary impedance... and your voltages only double (twice the turns)... but instead of the 9K which you thought you had... you really have 18K if each half the primary has a reflected impedance of 4.5K as you suggest... then you will only deliver half of the power you expected. The math doesn't add up. So it violates the conservation of energy law.Measuring the resistance via the voltages and currents on one half of the transformer and thinking you can simply add the resistances together like you would an 8 ohm resistor in series with an 8 ohm resistor equals 16 ohms of resistance
but... and 8 ohm winding put in series with another 8 ohm winding equals 32 ohms not 16 ohms.
Your observation and conclusion would be correct if we were loading and driving "resistors"... but does not work properly with PP transformes.
again, why? Because this violates the most basic premise of transformer operation.... the impedance ratio of a transformer is (e1/e2)^. And the impedance ratio of each half of the primary must be ((e1/2)/e2)^.
In order for a class A PP transformer's function to be properly modeled each half of the primary must be (AND ACTUALLY IS) 1/4 of the whole primary's reflected impedance. And then you do get the proper voltages and impedances... and no violation of the law of energy conservation. And to do this... you cannot take the apparent "resistance" of one half winding (as you measured it) and add it together like you would two ordinary
resistors...Again... if you simplify the complex behaviour of the transformer and the way it 'adds' impedances expoentially by modeling the transformer as two resistors (one for each pri half) in series with each other... it all works *IF* we were delivering power into resistors... but fails when we take into account the actual way a PP transformer "adds" or "sums" voltages, currents and impedances.
msl
It's a 9k transformer.
Posted by Mark Kelly (A) on September 11, 2004 at 23:32:24
In Reply to: Re: And the Answer Is..... (drum roll please) posted by MQracing on September 11, 2004 at 23:19:32:Nuff said.
The whole is greater than the sum of it's parts.
In Reply to: It's a 9k transformer. posted by Mark Kelly on September 11, 2004 at 23:32:24:Hi Mark:
Actually... this is the best most PRACTICAL answer... itis a 9K CT transformer. If it is a 9K CT transformer... and your using the whole thing in a PP amp (say class A to keep it perhaps a bit simpler) then you have your answer and you don't need to know or figure out how the two reflected impedances of 1/4 of Raa are combined through transformer action to get the
9K ct.You can go to the bank on the fact that you have 9kct primary.
Measuring the resistance of one half... ignores the PP nature of the winding and just leads to conceptual errors of how the CT transformer actually "combines" or "sums" the voltages, currents and impedances in the two halves similtaneously.
treating the whole primary impedance (9kct)... as two resistors in series with each other works as far as ALL the voltages and currents (across the whole ct primary) are concerned. It just doesn't capture the electrical operation of the transformer in "transforming" the reflected impedances and voltages and currents of the two SIMILTANEOUSLY driven halves considered as two "lesser" parts which form the whole. Said another way... looking at each half and arguing that it is half of the whole cannot then capture the complex (which I would not have thought so) summing action of the transformer... that you have two half windings each operating similtaneously with 1/4 of the overall impedance from end to end.
From a practical point of view.... if you have a 9kct trans... then you have one of two things... as a WHOLE winding it is a 9kct transformer. And if you use say only one half of the winding... say for a SE class A amp... then it will have an impedance of simply 1/4 of Raa. These two parts being lesser than the whole.
In a sense, the argument that Dave Slagle has made that the operation of the PP winding is not divisable into two parts is true... sort of like that ole cliche that "THE WHOLE IS GREATER THAN THE SUM OF IT'S PARTS"
the only other advice I can make (and which I believe has lead to much of the confusion) is to make sure that you add the right quantities in the right way....
i.e., as just one example... that you don't add the value of a simple resistor straight ahead arithmatically with the value of an inductive reactance (a different kind of impedance)...
i.e., transformed impedances (for lack of a better word) don't add together like simple resistors when describing the functioning (summing) action of transformers.
And on this note... unless there is some new information or new content forthcoming concerning this subject matter I am bowing out and getting back to work.
cheerios,
I didn't meaure the resistance across anything.
Posted by Mark Kelly (A) on September 12, 2004 at 15:31:16
In Reply to: The whole is greater than the sum of it's parts. posted by MQracing on September 12, 2004 at 09:27:58:Except the resistor used to load the secondary.
I measured the voltage present at each primary termination and the current flowing into each of these points.
Since the transformer was acting as a transformer these quantities by definition include the "electrical operation of the transformer"
Re: I didn't meaure the resistance across anything.
Posted by mqracing (M) on September 12, 2004 at 16:32:01
In Reply to: I didn't meaure the resistance across anything. posted by Mark Kelly on September 12, 2004 at
15:31:16:Mark:
if the reflected pri impedance of each half of the transformer is a 1/4 of the whole.
and you have the two windings connected in series.
then what will be your "output"... i.e., after the two halves are combined in the ct primary.
and what transformer action must take place for this to occur?
I've already given you all the info you need to answer this question in the two posts I put up already today to you. Perhaps you will also need to review some of my earlier posts for additonal info and formulas... but's it all there at this point.
but... don't confuse the "output" with the "input data" nor confuse the voltages and imedances which are subject to transformer action to those which are the RESULT of transformer action.
but... you guys are going to have to figure it out... all the info is there to do it. And I just get tens and tens more posts to respond to each time I try to clarify the issues which simply eats up my time.
You or anyone else is free to go on believing what you want... and maybe a genie transforms the 1/4 raa pri half reflected impedance into an output looking something like 1/2 raa... or maybe the transformer "action" that I have described in quite good detail is capable of the same thing... you guys figure it out and tell me if it's the genie or the transformer doing the work...
for sure... at this juncture, if you or anyone else cares enough to work through a complete example... and take the formulas I have already posted... then you to can figure out how a PP trans would work.
msl
Re: Isn't this like a half-wave antenna?
Posted by Matts (A) on September 12, 2004 at 09:40:46
In Reply to: The whole is greater than the sum of it's
parts. posted by MQracing on September 12, 2004 at 09:27:58:Have not been following this discourse to closely, but isn't this the same theory of current/emf interaction that occurs when you set up a half-wave transmitting antenna at 180degrees? The current flows through both halves and gives much greater strength than opposing quarter-waves, for example, where the EMF can cancel the other side out instead of augmenting it.
Thanks for your effort
Posted by Damir on September 12, 2004 at 03:48:19
In Reply to: It's a 9k transformer. posted by Mark Kelly on September 11, 2004
at 23:32:24:If someone is still confused with those contradictory informations, please see RDH, pages 199-203. Let`s say that our secondary load is 8 Ohms. Then our impedance ratio across whole primary is 9000/8=1125 or turns ratio is 33,54. Turns ratios from both halves of primary to secondary is N2/N1 = N3/N1 = 16,77 where N1 is a number of turns of secondary and N2,N3 number of turns of halves of our primary. From W1=W2+W3 (power in secondary is equal to powers in both halves of our primary) we have formula (4) in RDH:
1/R1 = 1/(R2(N1/N2)^2) + 1/(R3(N1/N3)^2)
R1 is secondary load - 8 Ohms, and R2=R3=reflected loads across halves of our primary. Note that N1/N2 = N1/N3 = 1/16,77 = 0,05963. We have 1/8 = 2/(R2,3 * 0,05963^2)
And our R2=R3 = 4500 Ohms! Or, with both halves of primary loaded (class A) load (that every tube "see") of each half of the CT primary is exactly half of the reflected load across the whole primary.
And, from RDH again, page 201:"If however only one half of the primary is used the transformed load presented is 1/4" (reflected load). In that case (say, class B) we have R2 or R3 = 9000/4=2250 Ohms, this is understandable `cos we transferred the power only across 1/2 of the primary, W1=W2 or W1=W3, see above formula, 1/8 = 1/(R2(N1/N2)^2) = 1/(R3(N1/N3)^2).
Re: Thanks for your effort
Posted by mqracing (M) on September 12, 2004 at 10:01:26
In Reply to: Thanks for your effort posted by Damir on September 12, 2004 at 03:48:19:or see RDH p. 576 where RL prime (one half the primary) is said to be raa/4.
And the example on the preceding page of a class A 2A3 PP amp... constructed with composite loadlines each having a slope of raa/4.
OK-you asked for it, the definitive ansver from RDH, page 571 :-)
Posted by Damir on September 12, 2004 at 10:33:02
In Reply to: Re: Thanks for your effort posted by MQracing on September 12, 2004 at 10:01:26:Page 571: "The load resistance R2 is connected across the secondary, and the reflected resistance across the whole primary is:
RL=4R2(N1^2/N2^2)
N1=turns in half primary winding and N2=turns in secondary winding
And the reflected resistance across half the primary is:
RL" = 1/2 RL = 2R2(N1^2 / N2^2)
And on the page 572: "If valve V2 in Fig. 13.31 were removed from its socket, the load resistance effective on V1 would then be
RL` = 1/4RL = R2(N1^2 / N2^2)
which is half the load resistance on V1 under PP conditions. This is the condition which occurs when one of the valves reaches plate current cut-off."
read pages 574 through 576 in RDH
Posted by mqracing (M) on September 12, 2004 at 10:48:27
In Reply to: OK-you asked for it, the definitive ansver
from RDH, page 571 :-) posted by Damir on September 12, 2004 at 10:33:02:where the half primary is described as having an imepedance of a 1/4 of the
whole.then try to harmonize these SEEMINGLY disparate information that you have...
I have provided a means to unify these seemingly at odd statements...
the one is looking at the "whole" while the other describes how the whole is achieved.
see my post "the whole is greater than the sum of it's parts"...
the one set of measurements result (the ones you mention) exactly and only because the pp ct primary combines voltages, impedances and currents in the way that it does....
everything I've said aims us toward a UNIFIED theory... where we do not have
to throw out the most basic laws of how tranneys work... nor be at odds with other electrical theory.and you must not confuse the "product" or end result of a transformer's proper functioning with what it sees as it's raw material (or how it produces this "end result").
it's hard to say all that in the right way....
read my post titled "the whole is greater than the sum of it's parts" and realize that transformers are not simple resistors...
the other mistake people make is to take a simplified model of some behaviour and believe that it describes the actual functioning at issue.
Msl
Re: read pages 574 through 576 in RDH
Posted by Damir on September 12, 2004 at 11:27:18
In Reply to: read pages 574 through 576 in RDH posted by MQracing on September 12, 2004 at 10:48:27:Page 576:"The composite loadline is a straight line through this point, with a slope corresponding to RL`=1/4RL... We may therefore imagine a composite valve, taking the place of both V1 and V2, working into half the primary winding with the other half open-circuited. This composite valve have a plate resistance...which value is aprox. half that of one valve..."
I can`t see anything opposite from page 571-572.
Re: read pages 574 through 576 in RDH
Posted by mqracing (M) on September 12, 2004 at 11:38:27
In Reply to: Re: read pages 574 through 576 in RDH posted by Damir on September 12, 2004 at 11:27:18:Hi Damir:
RDH quoted
:::have a plate resistance...which value is aprox. half that of one valve...":::
they are talking about (in this instance) the internal plate resistance of the tube itself. Not about the reflected primary impedance of the transformer.
bear in mind the 2A3 valve has in internal plate resistance somewhere in the range of perhaps 800 ohms or so...
now we are confusing the plate resistance of the tube with the reflected primary impedance.
Msl
Re: read pages 574 through 576 in RDH
Posted by Damir on September 12, 2004 at 11:47:00
In Reply to: Re: read pages 574 through 576 in RDH posted by MQracing on September 12, 2004 at 11:38:27:No confusion - they simply describe their model/method, in this case the composite valve is just "invented" ONE valve that substitutes TWO valves in PP. Boring already?
and then read....
Posted by mqracing (M) on September 12, 2004 at 11:06:39
In Reply to: read pages 574 through 576 in RDH posted by MQracing on September 12, 2004 at 10:48:27:Norman Crowhurst's article where he is explicit that one half of the PP primary is one quarter of the whole PP ct primary's reflected impedance...
or read the reference that Bas Horneman put up to a transformer design guide.... it sez the same thing...
but you quoting some portions of RDH and me quoting others (and other refs).... this has already been done to death...
there is nothing new here....
msl
____________________________
THREAD TWO
____________________________
What on earth...
Posted by groverg (A) on September 13, 2004 at 10:29:49...is all the fuss about? I'm looking at the Crowhurst article Douglass is referring to, ga400.pdf downloaded from audioXpress.com, and here's what it says:
"Now, that actual load is connected on the secondary of transformer. Suppose, for example, that the transformer is a 40-to-1 step-down--that means 20 to 1 from each half of the primary--and that the secondary is loaded with a 2 ohm resistor. The impedance ratio from one half of the secondary is 20 squared, or 400 to 1, so the impedance presented to one half of the primary by the secondary will be 400 x 2, or 800 ohms. To refer the secondary to the whole winding, it will be multiplied by an impedance ratio of 40 sqaured, or 1600 to 1, and will appear as 3200 ohms. The figure quoted as plate-to-plate load for wach tube would be 3200 ohms, but the load for each tube would be equivalent to 800 ohms."
And again, in the next paragraph:
"But the load for each tube will only be 1/4 of 6800 ohms, or 1700 ohms, when drawn against the composite load line."
And yet again:
"All this may seem a little confusing at first, but the important point to realize is that the practical circuit makes the two loads appear to be in series, because the windings windings of the transformer which combine the ouput are in series. But the magnetizing effect in the transformer is differential, so you subtract the smaller current from the larger. "This means that the current change represented on the curves is double that which occurs in each half-winding, and the voltage difference on the curves is what occurs across one half the winding (either half). Hence the effective impedance considered on the composite load line is 1/4 of the apparent impedance from plate to plate."
So where does anyone get 1/2 Rpp from this? The load lines are simply graphic equivalents of the math, aren't they? According to Crowhurst's Class A PP load lines for the 45, each tube is operating into a 1700 ohm load. But because of the circuit's behavior, the *composite* load line for the *composite* device represented by the tube tubes is 6800 ohms.
I'm not arguing, and I don't care who's right. For my own sake I'm curious as to where 1/2 Rpp comes from, and what difference it makes in this example. Are the curves lying? :-)
Re: Noise
Posted by mikeyb (A) on September 13, 2004 at 19:10:45
In Reply to: Noise posted by Henry Pasternack on September 13, 2004 at 16:07:04:People with knowledge are somtimes perceived as arrogant ...
Just an observation. If paralled devices result in lowered SNR, how come paralleling transistors is a commonly used technique in the semiconductor world? Many years ago, National Semiconducors produced a 'super low noise' op amp by paralleling many transistors in the input stage.
If someone wants details I'll see if I can dig up more info.
Parallel Noise
Posted by Owen (A) on September 14, 2004 at 01:53:51
In Reply to: Re: Noise posted by mikeyb on September 13, 2004 at 19:10:45:Ok.
There is induced noise, and self noise. The former is as a result of the signal being transmitted, and the second is due to the thermal characteristics of the semiconductor.
In a differential input stage, the former to a greater or lesser extent (down to componant matching, which incidentally is significantly better in single die chips than discrete) cancel. Hence less additional signal related noise (distortion, etc). However, the thermal noise SUMS, as it is entirely random in nature. So whilst the signal is in better shape, the noise floor is raised, lowering the SNR.
The tech specs that National were basing thier claims on, will have been based on the signal related noise (including distortion), and will have been deliberately choosen to show a competitive advantage.
Henry Pasternack is indeed correct.
My 2 cents
Owen
Differential stages are different from parallel stages.
Posted by Henry Pasternack on September 14, 2004 at 07:13:39
In Reply to: Parallel Noise posted by owen on September 14, 2004 at 01:53:51:I agree that the noise performance of a differential pair is worse than that
of an equivalent single-ended device.-Henry
Re: Parallel Noise
Posted by Joelt (A) on September 14, 2004 at 06:31:26
In Reply to: Parallel Noise posted by owen on September 14, 2004 at 01:53:51:>> Henry Pasternack is indeed correct.>>
>>
Hi Owen, Henry is actually on the *other* side of the argument from you. He believes paralleling devices lowers noise.Joel
ps. Good post. I agree with you.
Re: Again...
Posted by groverg (A) on September 13, 2004 at 13:14:20
In Reply to: Again... posted by Damir on September 13, 2004 at 11:18:34:Okay, here's the same thing in Landee, et al, Electronic Designer's Handbook:
"If V1 and V2 have equal instantaneous dynamic plate resistances, each tube will operate into and AC load resistance equal to Rpp/2. If either V1 or V2 is at cutoff. the other tube will operate into a resistance equal to Rpp/4. Whenever the dynamic plate resistances of the two tubes are not the same, the tube having the smaller dynamic plate resistance will operate into an AC load resistance having a value between Rpp/2 and Rpp/4 and the other tube will operate into an AC load resistance having a value between Rpp/2 and infinity."
But then they go on to say,
"Consequently, in a push-pull amplifier, neither tube operates into a fixed load resistance since the instantaneous values of tube dynamic plate resistance are a function of the instantaneous values Eb and Ib which change with the instantaneous values of the signal. The actual analysis, however, can be accomplished in a straightforward manner by constructing the composite characteristics for the two tubes."
Then later,
"Consequently, the optimum AC plate-to-plate load resistance is approximately four times the inverse of the slope of the composite characteristics."
So if I were trying to plot a load line for PP 2A3s working into a 5K p-p transformer, and I attempted to plot 2.5K for each tube, I'd get screwed up, right? I have to plot the composite load, which for each individual tube is 1/4 Rpp. In other words, you can't really break it down as 1/2 Rpp for graphical purposes, would that be right?
Re: Again...
Posted by Tre' (P) on September 13, 2004 at 16:54:17
In Reply to: Re: Again... posted by groverg on September 13, 2004 at 13:14:20:From what I understand, and I think it's what NC is trying to say, you are right. You have to use 1/4 a-a to construct the graph. When, if both sides are being driven, the load per tube will be a-a/2. If only one side of the transformer is being driven (using one side for single ended, or Class B push-pull) the load per tube is a-a/4.
I come to this conclusion after reading the posts here, RDH, NC,DOW..etc.. I doesn't necessarily mean I'm right.
Tre'
. Have Fun and Enjoy the Music "Still working the problem"
Re: Again...
Posted by groverg (A) on September 13, 2004 at 19:27:56
In Reply to: Re: Again... posted by Tre' on September 13, 2004 at 16:54:17:That seems to be my conclusion as well. Here's what I get from Crowhurst et al so far:
The actual transformer itself, because of the way it's wound, represents a static or mechanical reflected impedance, for each side, of 1/4 of the full primary impedance. Placed in series, these windings create four times that impedance from end to end. (Crowhurt's N squared example) Putting the transformer in action, with two balanced tubes, creates a situation where each tube is seeing an effective (virtual?) load of 1/2 Rpp. Unbalance the tubes, and this model ceases to exist. As the imbalance increases, the 1/4 Rpp model comes more into play.
The load lines show 1/4 Rpp because they represent what's happening on *one half* of the winding--"the current change represented on the curves is double that which occurs in each half-winding and the voltage difference on the curves is what occurs across one half of the winding (either half)."
One thing that also wasn't clear at first is that Crowhurst seems to be killing two or more birds with one stone--he's also trying to show that with PP you can get MORE THAN 2X Class A power by reducing the load and lowering the idle current, so he's also running the tubes outside of classic SE Class A.
What do you think? Have I got it mucked up? :-)
Re: Again...
Posted by mqracing (M) on September 13, 2004 at 20:09:00
In Reply to: Re: Again... posted by groverg on September 13, 2004 at
19:27:56:Hi grover: you wrote:
::::The actual transformer itself, because of the way it's wound, represents a static or mechanical reflected impedance, for each side, of 1/4 of the full primary impedance. Placed in series, these windings create four times that impedance from end to end.::::
yep. that's what I've been saying.
::::(Crowhurt's N squared example) Putting the transformer in ACTION, with two balanced tubes, creates a situation where each tube is seeing an effective (virtual?) load of 1/2 Rpp. Unbalance the tubes, and this model ceases to exist. As the imbalance increases, the 1/4 Rpp model comes more into play.::::This is also what I was trying to point out. It is the transformer action which transforms the quater Raa into, as an end product a virtual anode to anode imedance whose two halves then look like raa/2. Of course our end product (raa) must be (perfectly) balanced around the center tap if it is truly a differential primary winding.
It is through transformer "action" or it is the magic of transformers that achieves or converts or transforms the raa/4 reflected impedance of each half primary into an end product that approaches the ideals of a fully differential winding balaned around the center tap.
It's when and how the transformer "combines" or "sums" or "integrates" the two halves having only 1/4 of raa... because the two coils are in series... series connect the two coils and you get twice the turns and you quadruple the impedance of the single coil. THEN... but only because before the tranformer action of quadrupling the impedance... you achieve the "full" anode impedance from end to end.
Norm says the same thing of course. But these guys didn't want to allow the PP transformer to do it's job. Nor could some of them understand how a transformer "transforms" impedances and adds voltages and etc. In this context you can think of the transformer action as being an "impedance multiplier".
the key is that the raw material... i.e., each half of the primary has an impedance of only one quarter of the whole PP primary.
And it logically cannot be a half the impedance of the PP primary as some have proposed... because then when you double the turns (put the two coils in series) and thereby quardruple the impedance you would get two times the anode to anode impedance as would be logical.
Each half of the primary has 1/4 of the pp primary's end to end impedance... and this must mathmetically be the case if the two windings are in series to form the PP primary. Each tube sees the half winding (1/4 of raa) as it's load.
If someone wants to argue that the reflected impedance is 1/2 of raa... then each half primary must have 70 percent of all the turns... and your primary must have a total of 140 percent of it's total turns.... ooops. Because, reflected impedances are arrived at in relation to N^2 (N squared) half the turns equals a quarter the impedance.
But with raa/4 on each priamry half... then the transformer can go to work... with TWO SIMILTANEOUS 1/4 raa impedances transformed or converted by the series differential action of the primary widings THEN you PRODUCE a result Or end) that becomes our 10K ct primary.
that's how I see it Grove.
MSL
Re: Again...
Posted by mqracing (M) on September 13, 2004 at 21:24:29
In Reply to: Re: Again... posted by groverg on September 13, 2004 at
19:27:56:Grover wrote:
::::One thing that also wasn't clear at first is that Crowhurst seems to be killing two or more birds with one stone--::::
if you have a chance take a look at the 2A3 PP composite loadline drawn on p. 575 of RDH... it does not have the "distractions" that Norman's composite loadline does. then on page 576 they show a individual loadline for one of the two PP tubes from fig. 13.33 on page 575.
does this make it any easier or clearer than Norm's graphic? Just wondering.
MSL
Re: Again...
Posted by Tre' (P) on September 13, 2004 at 20:45:34
In Reply to: Re: Again... posted by groverg on September 13, 2004 at
19:27:56:NO, you have it right. Now that we have established that in a Class A push pull circuit each tube "sees' 1/2 of a-a, My 5k transformer is not taking advantage of the fact that in PP I could get more power and stay "clean" into a load lower than 2500ohms for each tube. Maybe the Dyna A-470 at 4300ohms would do better. How much less current?
Hmmmm..
Tre'
. Have Fun and Enjoy the Music "Still working the problem"
Re: Again...
Posted by mqracing (M) on September 13, 2004 at 21:11:24
In Reply to: Re: Again... posted by Tre' on September 13, 2004 at 20:45:34:Hi Tre:
you wrote;
:::we have established that in a Class A push pull circuit each tube "sees' 1/2 of a-a,:::
if I may point this out... and I think Grove sees how it works as well (hope I am not putting words in his mouth).
the virtual or end or apparent impedance of each tube AFTER the PP action occurs in the primary appears to be 1/2 of a-a.
but the load that each individual tube works into is 1/4 of a-a.
and then transformer "action" takes place and sums the two halves (each of which are 1/4 a-a) and produces as a result what you have above.
so for your composite loadline you use 1/4 of raa for each tube. the result after transformer action is that each half appears to be 1/2 of raa.
Here's some practicl advice... just a 700 ohm differnce in a-a impedance is really not all that great... and the "effective" impedance (when you take into account inductive reactance in para with reflected impedance) may not be so great.
the dyna is a fine transformer... but so might your 5K unit be a very good unit. And the 700 ohms difference in impedance... by itself... I wouldn't think would be earth shattering.
have fun,
Mike
Re: Again...
Posted by groverg (A) on September 13, 2004 at 20:57:21
In Reply to: Re: Again... posted by Tre' on September 13, 2004 at 20:45:34:Think so? Cool! Thanks for the confirmation.
Hey, if you plot the thing and figure out a good "Crowhurstian" balance between Class A and a more efficient circuit, I'd love to tag along to learn about it. 4300 sounds about right. Have you ever played around with TubeCad's PP emulator?
Re: Again...
Posted by Tre' (P) on September 13, 2004 at 21:09:34
In Reply to: Re: Again... posted by groverg on September 13, 2004 at
20:57:21:No I don't have that program. I do have some Dyna transformers. I think, on a rainy day, I'll just hook them up and listen. Then drop the current and raise the voltage a little and listen some more. That's how I end up doing every thing. My math only get me in trouble. I get some things but only in terms of concept.
Tre'
when the two coils are put in series
Posted by mqracing (M) on September 14, 2004 at 05:25:47
In Reply to: Re: that was such an easy prediction.... posted by groverg on
September 14, 2004 at 04:51:28:the following obtains;
-- the voltages add (will be two times the volts on one side)
-- the impedance quadruples (will be four times the impedance of one side)
-- the current is cut in half (will be half the current of one side)and that is how transformer action "combines" or "sums" the voltages, currents, and impedances from each half winding to the whole. And how you move from 1/4 raa to a fully differential primary winding having a virtual impedance of raa/2 for each half winding.
You can get to this same point using simply voltages/turns and then using the "summing" table above...
for example take the recorded voltages from Dave Cigna's earlier example... and compute the reflected impedances and etc.
primary volts 38vrms
secondary 1.74 vrms
load on secondary 15 ohms
n1/n2 equals e1/e2 are equivalent statements
impedance ratio to the whole primary is 476.9:1
impedance ratio to each half is ((e1/2)\e2)^2 is 119:1
multiply in each case by load on secondary and you get for the whole primary 7153.5 ohms of reflected impedance. And each half has a reflected impedance of 1788 ohms.
If you then compute power... each half winding appears to have twice the current that is needed... but transformer action saves us... when you put the two coils in series the current gets cut in half and the impedance quardruples. And we satisfy the law of conservation of energy as well.
msl
the law of conservation of energy...
Posted by mqracing (M) on September 14, 2004 at 05:42:34
In Reply to: when the two coils are put in series posted by MQracing on
September 14, 2004 at 05:25:47:would only be violated if the reflected impedance were raa/2.
Because then when you series the two windings and the impedance quadruples...
if the argument is that a 10Kct has a reflected imedance on each side of 5,000 ohms this obviously cannot be true since it would require that each side have 70 percent of the turns of the whole winding.
because the voltage (and turns) ratios must be the square root of the impedance ratio.
solve for square root of .5 (the 5,000 to 10,000 ratio) is .7071 telling us that each side of the transformer must have 70 percent of the turns if the reflected impedance of one half is itself half of the reflected impedance of the whole. Which obviously cannot be the case.
now solve for the square root of .25 (the 2500 to 10,000 ratio) and you get half. Meaning that your voltages/turns must be one half of the total on both sides... and that your reflected impedance is raa/4.
again, much of this stems from the confusion that you cannot add up impedances subject to transformer action like you do resistors.
The transformer, in this context, is an impedance multiplier and quadruples the impedances when the two coils are put in series. which gives us the right answer.
If you quadruple the proposed half pri reflected impedance and treat it like it was 5,000 ohms... then when you put the two coils in series then you would get 20,000 ohms... which is not your 10K ct pri impedance.
so... the only way to make all of this work and allow the transformer to function properly (pp action) is to realize and take into account what happens when you combine the two SIMILTANEOUS halves of the pri into a series "whole".
And that is why I said in an earlier post... that the "whole is then greater than the sum of it's individual parts".
Msl
AGAIN^2 - The proper math - the law of energy conservation
Posted by Damir on September 14, 2004 at 07:59:20
In Reply to: further notes on the law of energy conservation posted by
MQracing on September 14, 2004 at 06:39:13:CASE 1) We have transformer with CT primary, both half loaded and 19V across each half or 38V across the whole primary. On the secondary, we have 1,74V and 15 Ohms resistor. Now, our voltage ratio across the whole primary is 38/1,74=21,84 or impedance ratio across the whole primary is 21,84^2=476,95. Reflected secondary resistance across the whole primary is Rp=476,95 *15 = 7154 Ohms. The Law of the energy conservation - power at the secondary = power at the whole primary, or Ws = Wp , Ws=1,74^2/15=0,202W
Us^2/Rs = Up^2/Rp
1,74^2/15 = 38^2/Rp or Rp = (38^2 * 15)/1,74^2 = 7154 Ohms
Now, some brighter fellows allready know - power at the primary is devided in two identical halves of the primary, or in every half power is Wp/2=0,202/2=0,101W and further (Rp1=res. of the half primary winding)
19^2/Rp1=0,101 and Rp1=19^2/0,101=3577 Ohms
But, we want the proper way and again, The Law of energy conservation tells
us that power at the secondary is equal to powers in the both halves of the
primary, orWs = Wp1 + Wp2
Us^2/Rs = Up1^2/Rp1 + Up2^2/Rp2 where Up1=Up2=19V and Rp1=Rp2=Rp1,2
1,74^2/15 = 19^2/Rp1 + 19^2/Rp2
1,74^2/15 = 2*19^2/Rp1,2 and Rp1,2 = 722/0,20184 = 3577 Ohms
Or in another words, with both halves of the primary loaded, reflected secondary resistance across the whole primary is 7154 Ohms, and across each halves of the primary is Rp/2= 3577 Ohms.
CASE 2) We have only one half of the primary loaded and we transfered the energy from half primary to the secondary. Again, our Law:
Ws = Wp1` or Us^2/Rs = Up1^2/Rp1`
1,74^2/15 = 19^2/Rp1` and Rp1` = 19^2/0,20184 = 1788,5 Ohms
Or, in another words, with only half of the primary loaded, our reflected secondary resistance across (only) half of the primary and other half unused is Rp/4 or 1788,5 Ohms.
preservation of the law of energy conservation
Posted by mqracing (M) on September 14, 2004 at 08:32:05
In Reply to: AGAIN^2 - The proper math - the law of energy conservation
posted by Damir on September 14, 2004 at 07:59:20:Damir writes:
:::Or in another words, with both halves of the primary loaded, reflected secondary resistance across the whole primary is 7154 Ohms, and across each halves of the primary is Rp/2= 3577 Ohms.:::
what you show is the product of the reflected impedances and voltages of each half primary after they been combined through transformer action. Your mixing up the PRODUCT of the PP transformer action (having the two halves operate in series differential) with the "input data" (the reflected voltages and impedances) BEFORE they are combined through the series differential integration of the PP primary.and your answer is correct AFTER we combine (through the series connection of the two windings) the two individual PP primary halves through transformer action.
so, what you have shown... is that the power is divided evenly btwn the two halves AFTER the transformer has taken the REFLECTED voltages and impedances as presented by the loaded secondary and has added them in the following manner;
1) that the voltages add (2x)(as compared to the reflected voltage on one half of the primary winding)
2) and that the impedances quadruple (as compared to the reflected impedance of one half of the primary winding).
and of course the power must be divided evenly btwn the two halves AFTER their reflected impedances and voltages have been "processed" by the series connected differential primary.
No argument there... but it confuses the PRODUCT of the transformer doing it's job... with what the transformer sees as it's raw material (i.e., the reflected voltages and impedances) from the loaded secondary).
Don't confuse the reflected impedances and voltages with the virtual impedances and voltages that are the product of the transformer doing it's job.
And I have shown demonstrably compliance with the law of energy conservation and that this same law requires that the reflected impedance of the each primary half can ONLY be raa/4.
raa/2 as a reflected impedance for each half of the PP primary would only give you half the power... and thereby violate the law of energy conservation as I have shown (quantatively) in one of my most recent posts.
msl
one last word this evening....
Posted by MQracing (M) on September 13, 2004 at 21:48:52
In Reply to: Re: that was such an easy prediction.... posted by groverg on
September 13, 2004 at 21:12:44:Good points Grover about... the individual loadlines and "depends where your at" on the loadline...
that's why I tried to keep all extraneous stuff out of the thread... so that we could focus on the transformer action and see how a transformer take you from two similtaneoous 1/4 raa all the way to a fully differential PP primary.... which has combined the windings and electrical activities of the two series coils.
And to see how impedances are transformed from the smaller to the larger
values... you can just use voltages that you read off of a subject
transformer.there are plenty of distractions and complications in real world circuits that make more complex the ability to elucidate even the basic functions of the transformer.
Msl
Re: Again...
Posted by Henry Pasternack on September 13, 2004 at 13:53:44
In Reply to: Re: Again... posted by groverg on September 13, 2004 at 13:14:20:The standard way of doing a graphical analysis of push-pull operation is as follows. Note, it works equally well for Class A, Class B, linear, non-linear, and everything in between.
1) Start with the plate curves for a single tube.
2) Pick you operating point. For simplicity, ignore the resistance of the output transformer so that the static plate voltage equals B+.
3) Plot a second set of plate curves, flipped top-to-bottom and left-to-right. Line them up so the horizontal (Vp) axes coincide at the chosen operating point plate voltage. Notice that the reversal of the Vp scales means that as the plate voltage goes up on one tube, it goes down an equal amount on the other tube. This reflects a constraint imposed by 1:1 coupling between the two halves of the primary.
4) Determine the grid voltage that corresponds to the chosen operating point. For the purposes of instruction, it would be helpful to round this voltage to a whole number corresponding to Vg on one of the plate cureves.
5) Identify the pair of plate curves, one each top and bottom, with Vg equal to the voltage selected in the previous step. Construct a new composite plate curve by summing the values of Ip at each horizontal position along the curves. The composite plate curve represents the net output transformer primary current as a function of voltage with the grid voltage held constant.
6) Group the remaining plate curves into top-and-bottom pairs. As you make positive grid voltage steps on the top set of curves, the corresponding curves make negative grid voltage steps on the bottom. For instance, if at the operating point Vg = -40V, then the remaining pairs might have grid voltages of [-30V, -50V], [-20V, -60V], etc.
7) Finish drawing the remaining composite plate curves. The resulting family of curves is a representation of the current-vs.-voltage characteristics of the two tubes operating together into a single transformer primary.
8) Draw the composite load line. Its slope should correspond to a load resistance of Rpp / 4. This is not a function of the tube characterstics or the operating class. The factor of four comes from the fact that we have transformed the original circuit containing two tubes operating in series with the entire transformer primary into an equivalent parallel circuit with a single composite tube operating into effectively half the number of turns (hence one-quarter the impedance).
9) To draw the individual load lines, do the following. For each composite plate curve, draw a vertical line at the point where it intersects the composite load line. Extend this line to the intersections with the corresponding individual plate curves. Plot the intersection points for every plate curve and then connect the dots horizontally, top and bottom.I'm sorry I don't have a picture to post, but I'm sure you can find one if needed, or draw one yourself. The point is that each and every step in the process, and each feature on the graph is there for a reason that should be clear and easy to understand. Once you thoroughly understand the composite curves and how and why they are constructed, all the mystery goes away and it becomes easy to see who knows what he's talking about and who is talking through his helmet.
-Henry
Re: 2nd paragraph...
Posted by groverg (A) on September 13, 2004 at 13:36:28
In Reply to: 2nd paragraph... posted by Sector-7G on September 13, 2004 at
11:07:19:Oh, okay. I get you. Since it's curved, though, at what point did you calculate the current/voltage slope ratio?
This is also interesting, even futher down:
"Therefore, the load lines as drawn represent the working impedance as measured across one half of the transformer primary winding."
And I see further down that Crowhurst repeatedly reminds the reader that each tube's load will *appear* to be Rpp 1/4 on the composite graph. So Rpp 1/4 is a function of the graphical representation, not the reality?
Re: 2nd paragraph...
Posted by Sector-7G (A) on September 13, 2004 at 16:40:38
In Reply to: Re: 2nd paragraph... posted by groverg on September 13, 2004 at 13:36:28:HEy Grover....calculate the slope yourself( of the dotted line), all along the dotted line...go between each set of data points. When you have a set of answers, we can proceed, until then, it is a waste of time. regards, Douglas
and since this probably got responded to in the wrong order...
Posted by Sector-7G (A) on September 13, 2004 at 17:19:01
In Reply to: Re: 2nd paragraph... posted by Sector-7G on September 13, 2004 at 16:40:38:Hey Grover, still, run through the dotted line NC mentions. slope in Ohms is change in plate V/change in plate current, volts per ampere, from Ohm's law, V=i*r. but you probably got it already...
regards, Douglas
Re: and since this probably got responded to in the wrong order...
Posted by groverg (A) on September 13, 2004 at 18:07:29
In Reply to: and since this probably got responded to in the wrong order... posted by Sector-7G on September 13, 2004 at 17:19:01:Yes, I'm there, but since the line is curved, the load seems to change--is this right? So how do you calculate the overall load? Now Doug you know I'm no math whiz so go easy on me! You want me to see that the *working* load line, as opposed to the composite line, approaches 1/2 Rpp, is that right? Just gimme a few hints...:-)
Re: and since this probably got responded to in the wrong order...
Posted by dave slagle (M) on September 13, 2004 at 18:59:29
In Reply to: Re: and since this probably got responded to in the wrong order... posted by groverg on September 13, 2004 at 18:07:29:grover,
i swear this is the next can of worms and it WILL be productive... if can get past this trivial one and all agree on the ideal case, maybe we can move on
i really want to get to the consideration of the "oft ignored" and "non-linear" reflected load of the other tube you suggest.
this is often incorrectly called the autoformer effect, yet it still models the same as the transformer effect which must be considered as another parallel load. the worst part of this is, that the reflected load by the other tube is nonlinear.
hopefully someday we can get there, since i have never seen it discussed publicly.
Dave
Re: and since this probably got responded to in the wrong order...
Posted by groverg (A) on September 13, 2004 at 22:03:13
In Reply to: Re: and since this probably got responded to in the wrong
order... posted by dave slagle on September 13, 2004 at 18:59:29:Well, I'm game. Personally I feel I'm agreeing pretty well with myself :-).
You might lose me on transformer theory, but I'm interested in the
"oft-ignored" and "non-linear" part. Care to expand on that in a new thread?
Here's the answer.
Posted by Henry Pasternack on September 14, 2004 at 11:13:44
In Reply to: I have the answer. posted by Henry Pasternack on September 14,
2004 at 06:02:21:Remember the composite plate curves and loadline are the graphical representation of an imaginary transformation of the push-pull output stage into an equivalent single-ended circuit. A load line is not a resistor. It is a constraint on plate voltage vs. plate current imposed by a resistor -- a subtle difference, but an important one. The load line tells us what the plate voltage must be given a particular B+ voltage and plate current.
In the composite model, the load resistance transforms out to be Rpp / 4. As I've explained before, this has NOTHING to do with "transformer action" because you get the same result even if you replace the transformer primary with a center-tapped resistor. Because the resistor is fixed and linear, so is the composite load line that represents it.
The individual load lines indicate how voltage and current vary on the plate of a single tube. Bear in mind the individual tube doesn't exist in the composite model -- the point, after all, of the composite model is that the two tubes have been melded together into a single imaginary unit. So, to get the individual load lines we have to deconstruct the composite model.
The individual load lines indicate how hard each tube is working at any point in the signal cycle. Straight lines indicate the tubes work equally hard at all times. Curved lines mean the output is varying. The sum of the individual curves is the composite curve, which is straight. This means that as one tube slacks off, the other one has to pick up the effort to keep the total constant. This is reflected in the slopes of the individual load lines. As one load line shallows out, the other steepens. The work available from the tubes varies because the transconductance isn't constant with plate current.
This is one interpretation of the curvature. The other interpretation is that the individual load line is simply a function of the impedance looking into one end of the primary. From this vantage point, you can see the reflected secondary load, but also the nonlinear voltage generator and plate resistance of the opposite tube. It's that nonlinearity that accounts for the curvature. With perfectly linear tubes, the individual load lines are straight as well. From the point of view of tube "A", though, the curvature of its individual load line is due to the nonlinearity of tube "B" -- not a reflection of its own nonlinearity.
-Henry
Re: Here's the answer.
Posted by groverg (A) on September 14, 2004 at 13:34:36
In Reply to: Here's the answer. posted by Henry Pasternack on September 14, 2004 at 11:13:44:Thank you, Henry. That's extremely clear and very helpful. Gives me a lot to
think about and research.
Re: I have the answer.
Posted by mqracing (M) on September 14, 2004 at 06:47:00
In Reply to: I have the answer. posted by Henry Pasternack on September 14, 2004 at 06:02:21:Henry writes:
:::Michael's answer is wrong.:::
which answer is wrong Henry. I welcome you to take the substantive text that I have written and show me where my errrors are. So far you keep calling me names and saying I am wrong...
but, so far, you have not once shown me any logical inconsistencies with my formulas and\or their application as I have actually presented them.
Instead... we get grand pronouncements and a lot of abusive, condescending remarks from the great one.
And, oh, by the way... the composite loadline does not necessatate the the two tubes be modeled in parallel as you suggested in your post on constructing composite loadlines. Actually, that it is a composite loadline with the end to end impedance being four times greater in magnitude than the half pri impedande demonstrates that the two tubes are in series.... since the composite loadline impedances are in series with each other.
though I think that was just a deliberate argumentive trick... designed to throw people off the track.
Msl
Supercalifragilisticexpialidocious...
Posted by Henry Pasternack on September 14, 2004 at 10:06:10
In Reply to: Re: I have the answer. posted by MQracing on September 14, 2004
at 06:47:00:Michael, to respond effectively to your arguments, I have to restate them in comprehensible terms so I can make a point-by-point rebuttal. I have tried and failed to do so. Basically I have found that your remarks fall into two categories. The first is valid restatements of the basic formulas for transformer voltage and impedance ratios. The second is incomprehensible gobbledegook typified by phrases like "transformer action", "complex summing action", and "virtual anode-to-anode impedance." It's clear to me (and anybody else that knows what real engineering analysis looks like) that you have convinced yourself that "If you say it loud enough, you'll always sound precocious"...
Nobody disputes that Z1 / Z2 = (N1 / N2) ^ 2. The interesting part of this problem is where you go from there. I am quite certain you are sincere in your belief that you understand this stuff and are doing a useful job of explaining it. All the more power to you, Micheal.
Time permitting, I will try to explain things in my own way and then people can judge for themselves whether or not my remarks have merit. I don't have the time to reconstruct your baloney just so I can tear it down again.
There's nothing in it for me.-Henry
____________________________
THREAD THREE
____________________________
Why 1 + 1 = 1.4
Posted by Mark Kelly (A) on September 14, 2004 at 04:34:20At the risk of a flameout, here's a simple demonstration of non-correlative addition. This is germane to the ongoing discussion regarding noise.
If you are reading this you have a computer. You therefore probably also have a spreadsheet like Excel. This means that you can do numerical things that would have been impossibly tedious a few years ago, like demonstrating non-correlative addition, with just a few keystrokes.
This is a little tedious but it's not too bad.
The rand() function is excel's random number generator.
In cell A1 type: =rand() then copy this to B1
In cell C1 type: =2*a1-1 then copy this to D1
In cell E1 type =c1+d1
In cell F1 type =c1-d1
Now copy that row and paste it as far down as you like - I got reasonable accuracy with 1000 rows.
What we have done is created a pair of non-correlated strings with value varying from +1 to -1 (columns C and D) and their sum and difference terms (rows E and F)
Now in cell c1001 type =stdev(c1:c1001). If you used more than 1000 rows adjust this accordingly.
Copy this across d1001, e1001 and f1001.
The standard deviation of a gaussian distribution equals its root mean square if the distribution is symmetrical about zero (which ours are), so we now have the rms values of our non-correlated strings.
You will note that the sum and difference strings have rms values which are 1.41 (sqrt 2) times the original strings AND they are near enough equal to each other. If you do not understand why this is so, ask someone who does but for heaven's sake don't go claiming that it isn't so.
BTW the values for the SDs should be around 0.58 for the original strings and 0.8 for the sum and difference strings.
BTW if you want to run it again just type anything into a cell not in use and press enter. Excel automatically establishes new random numbers.
are you trying to get me to pull my hair out?
Posted by Joelt (A) on September 14, 2004 at 13:02:07
In Reply to: Let me try again, or how to add voltage sources in parallel
posted by vry on September 14, 2004 at 12:37:33:>> Now, if you take the average of 2 uncorrelated noise signals with rms 1, you
>> get... wait for it.... 0.707V>>
>>
No. You are WRONG. Plain and simple. If you add two uncorrelated noise voltage sources of 1v, you get:E=sqrt(1^2+1^2) E=1.414V
A 3dB increase over the value of one source alone. How many times do we have
to go over this? Did you read Mark's post?
a hint (again)
Posted by Dave Cigna (A) on September 14, 2004 at 15:11:01
In Reply to: are you trying to get me to pull my hair out? posted by Joelt on September 14, 2004 at 13:02:07:"If you add two uncorrelated noise voltage sources of 1v, you get:
E=sqrt(1^2+1^2) E=1.414V
A 3dB increase over the value of one source alone."
Right, but in every case we're talking about improved signal to noise ratio. What I, and Mark, and vry have been trying to tell you - what has been in all of those 'moving target examples - is that while the noise increases by a factor of 1.4, the signal increases by a factor of 2. That means the SNR is better, not worse as you have tried to claim.
It doesn't matter whether you look at cases of summing voltages or summing currents. The basic result is always the same. It's certainly posible that an error or two crept into one or more of the examples presented to you (who can be sure?), but for the most part they've been pretty accurate. What I think you should do now is take some time and think hard about what we've been telling you. Assume that we have something worth listening to and try to make sense out of it. We've all done a LOT of work. Now it's time for you to figure it out, because I know you can.
Remember: it's not really about tubes or transistors. It's about adding two (or more) signals that have some part in common (the music?) between them and some part that is completely random with nothing in common between the two. The two parts add differently. Both parts get bigger when you add them, but the music part get's more bigger than the uncorrelated (random) noise.
Thus an improvement in SNR.n Dave Cigna
and a thoughtful reply
Posted by Joelt (A) on September 14, 2004 at 15:45:13
In Reply to: a hint (again) posted by Dave Cigna on September 14, 2004 at
15:11:01:>> ...is that while the noise increases by a factor of 1.4, the signal increases
>> by a factor of 2...>>
>>
Hi Dave. Please explain how the signal at the input of the two paralleled tubes has 'increased by a factor of two'.(Of course, it hasn't - rhetorical question)
And it seems to me that if you're saying the noise voltage at the OUTPUT of the amplifier has increased by 1.414 times, then the noise voltage at the INPUT of the amplifier must also have risen by 1.414 times (this follows logically).
So... if the input 'music part' is the same as when you use a single tube, and the 'noise part' has increased by 1.414 times, then we have to say that the SNR is 3dB worse. Don't we??
>> Remember: it's not really about tubes or transistors. It's about adding two
>> (or more) signals that have some part in common (the music?) between them and
>> some part that is completely random with nothing in common between the two.
>> The two parts add differently. Both parts get bigger when you add them, but
>> the music part get's more bigger than the uncorrelated (random) noise.>>
>>
Again you're talking about the signal voltage adding. What TWO signal voltages are adding? Both parallel tubes have their grids tied together - the literally see the same single input signal voltage. There is no 'adding' of input voltages. You are talking about the output perhaps...? If you use 1/2 the load of a single tube (which is how this is always done with parallel tubes) then the signal voltage at the output of the amplifier will be exactly the same magnitude that it was with a single tube. You acknowledge that the noise in the output is up by 1.414 times, and if the signal is ampified by the same amount as a single tube - then again, it logically follows that the SNR is worse by 3dB.Joel
Re: and a thoughtful reply
Posted by Dave Cigna (A) on September 14, 2004 at 17:56:36
In Reply to: and a thoughtful reply posted by Joelt on September 14, 2004 at
15:45:13:Joel, are you really trying to make sense of this? Are you really trying to understand what everyone has been trying to tell you? If you are then you'll work hard to answer these questions yourself rather than make us work hard to answer them for you.
Mark has done a great job describing the subtleties of what goes on between grid and plate with a resistive load, so I won't try to repeat that. I'm going to stick to a really simple model of a tube as a simple transconductance device. In other words, it creates current. It doesn't matter where the current is coming from. Don't bother yourself with the grid and the plate resistance, etc, etc.
So there's two tubes... err... I mean there's two current sources. They might not be ideal current sources, but that doesn't matter. Each one is producing the same music signal in the form of current. Both are also producing about the same amount of noise, but we understand that the two noise currents are unrelated.
Ok, so the currents sum directly; it easy! The total music current is 2x that of one tube alone, but the noise current is only 1.4x. So the signal to noise ratio has improved by a factor of 2/1.4 = 1.4
If you want to convert that current to voltage by passing it through a resistive load, that's your business.
n Dave
voltage "sources" in parallel AVERAGE (rms x0.7), they add in SERIES (rms
x1.4)
Posted by vry (A) on September 14, 2004 at 14:05:45
In Reply to: are you trying to get me to pull my hair out? posted by Joelt on September 14, 2004 at 13:02:07:I'm glad I'm pretty much bald...
What are you disputing here?
That 2 batteries (with non-zero impedance) if paralleled will give a voltage that is the average of their individual voltages?
Or that averaging two uncorrelated gaussian variables *reduces* the variance?
Re: voltage "sources" in parallel AVERAGE (rms x0.7), they add in SERIES
(rms x1.4)
Posted by fatbottle (D) on September 14, 2004 at 15:00:11
In Reply to: voltage "sources" in parallel AVERAGE (rms x0.7), they add in SERIES (rms x1.4) posted by vry on September 14, 2004 at 14:05:45:Hello ,
'I'm glad I'm pretty much bald...'
...I can remember the soft glow of 212 filaments reflecting off the said bonce ;) What are you trying to do here ? I don't see anything to worry about . Nothing morally wrong has happened and no-one's going to die . Noise ? Pah ! This is just another self-induced '2x 4x' transformer type thread :)
...FB
Indeed - no need to worry
Posted by Jim Dowdy (M) on September 14, 2004 at 16:59:58
In Reply to: Re: voltage "sources" in parallel AVERAGE (rms x0.7), they add in SERIES (rms x1.4) posted by fatbottle on September 14, 2004 at 15:00:11:Thanks to PhatB for injecting a bit of hedonistic common sense here!
ENDFax mentis incendium gloria cultum, et cetera, et cetera...
Memo bis punitor delicatum! It's all there, black and white,
clear as crystal! Blah, blah, and so on and so forth ...
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- Damir, your math lies within ... - The Bored 16:58:35 09/16/04 (0)
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