Parallel Connection of TriodesCopyright © June 2005. Author: Dmitry Nizhegorodov (dmitrynizh@hotmail.com). My other projects and articles
2. Singleended output stage with 3pin 2A3 model 3. The Five Triodelets 4. 5 little, equal triodelets 5. Kg1 values in a range 6. 5 little triodelets, Mu is not constant 7. Some interim lessons 8. Mu compensation with Vg 9. 5 equal triodelets, input signals are not the same 10. Filament wire as voltage generator. 11. References
1. What is this about?In this article, we SPICE power triode stages containing several triodes running in parallel. The main areas of interest are: what are the properties of the sum of equal and notsoequal parts? how changes in triode parameters affect signal quality? which parameters are critical and which are not? What tricks, if any, can help to neutralize mismatches? We also present the results in a format directly relevant to "composite triode models", discussed on other pages on this site, see for example [1] and to the issues of hum in directly heated triodes (DHT).It seems as these simulations can have various useful practical implications; at least these spice experiments helped to understand how to properly implement parallel singleended output stages. The experiments also helped to accurately calculate the level of harmonic hum generated by ACfedd filaments of DHTs.
2. Singleended output stage with 3pin 2A3 modelFor these experiments a 3pin Koren triode model with parameters computed for audiomatica's 2a3 curves is used as a "reference":
** 2A3CRV ************************************************************ * Created on Fri Mar 12 00:55:11 PST 2004 using tube.model.finder.PaintKIT * curves URL: audiomatica 2a3crv.gif * .SUBCKT TRIODE_2A3CRV 1 2 3 ; P G K ; + PARAMS: RGI=2000 + CCG=7.5P CGP=16P CCP=5.5P ; ADD .7PF TO ADJACENT PINS; .5 TO OTHERS. + MU=4.58 EX=1.512 KG1=1710.0 KP=40.8 KVB=1188.0 VCT=2.24 ; Vp_MAX=600.0 Ip_MAX=0.2 Vg_step=12.0 * E1 7 0 VALUE={V(1,3)/KP*LOG(1+EXP(KP*(1/MU+(VCT+V(2,3))/SQRT(KVB+V(1,3)*V(1,3)))))} RE1 7 0 1G G1 1 3 VALUE={(PWR(V(7),EX)+PWRS(V(7),EX))/KG1} RCP 1 3 1G ; TO AVOID FLOATING NODES IN MUFOLLOWER C1 2 3 {CCG} ; CATHODEGRID C2 2 1 {CGP} ; GRID=PLATE C3 1 3 {CCP} ; CATHODEPLATE D3 5 3 DX ; FOR GRID CURRENT R1 2 5 {RGI} ; FOR GRID CURRENT .MODEL DX D(IS=1N RS=1 CJO=10PF TT=1N) .ENDSwhen loaded with a 3.5k OPT, this model running from 350V B+, with bypassed 1K autobias resistor (typical conditions), gives the following distortion data:
We'll be using this data plot as a yardstick, to compare other configurations against.
3. The Five TriodeletsFor these experiments, five tubes were wired in parallel. Initially, each receives a an independent Sine signal source; all cathodes and anodes are connected in parallel.For each tube, we use the same 2a3 model with a small twist: instead of wiring 5 devices each representing a 2a3, and thus observing an odd combined result that runs 300mA if idle current, etc., we wire in parallel 5 "triodelets" each being 5 times wimpier than a 2a3.
Behind each triode is the same model; for convenience of experiments, Mu and KG1 are made visible. Each triode has KG1 = 8550 which is 5 times of KG1 of real 2a3. This is pretty much all that is required to scale a 2a3 model down 5 times. This is so because the parameter KG1 relates to triode's conductance: G1 1 3 VALUE={(PWR(V(7),EX)+PWRS(V(7),EX))/KG1} The Koren model is phenomenological, yet KG1 has pretty obvious physical interpretation: each electron contributes to total dynamic conductance delivered by a triode; each square milimeter receives an average the same number of hits as any other, and 1/KG1 corresponds to the size of the electrodes. Thus switching from KG1=1710 to KG1=8550 is like cutting a 2A3 5 times, preseving its Mu, linearity, etc. To represent a 1/5th of 2a3 in an even better way, we'd also need to reduce all interelectrode capacitances 5 times. We may also want to change the leaking resistances just in case. It was not done simply because for these specific experiments doing that was not necessary.
4. 5 little, equal triodeletsFirst data is for all such "triodelets" being equal. If we are right about each being exactly 1/5th of 2a3, we should obtain data exactly equivalent to a single 2a3. Indeed, it is so:
5. Kg1 values in a rangeWhat happens if instead of 5 triodelets each having KG1 = 5 * 1710 = 8550 we use 5 triodelets with varying KG1, still totalling to the same conductance?It turns out that the distortion data is essentially the same. We do not even show the data, it is equivalent to the plots above. If you are surprised, do not be. Since each of 1/5power triodelet represents a portion of 2a3's structure area, then what if the first one represents 5% of the plate, the second one  10%, the third one  20%, the 4th  30% and the 5th  45%? The sum is 100%, and we get the "same" composite as the result.
6. 5 little triodelets, Mu is not constantWhat if the triodelets have different Mu? For instance, what if one among these 5 has mu that is 3.5, and the rest 4 have Mu=5?
Intuitively, this should worsen the spectrum but to what degree? Anyone expecting something like 30% change to worse will be surprsed, then, as 2nd and 3rd go up 5 times:
when each triodelet is biased individually, linearity improves, but not very much. Why change of the amplificaltion of 1 out of 5 has bad effect? One explanation for this may be: "the first triode is "behind" in mu, and struggles against the rest of the hurd". It is not very correct. A better way to explain this is to point out that the first triode has lower resistance than the rest as thus "shorts" them; this is not only because it runs higher idle current; it does, but that alone is not the main problem; the main problem is that the first triode behaves as a dynamic load for the rest. It is so because it changes its resistance less eagerly. Another, but less correct way to say it is to describe it as delivering lesser plate voltage swing. Imagine now that we discovered that a reallife tube has internal structure where 1/5th of it has lower Mu. This experiment suggests that any such triode will be less linear than a triode all parts of which have the same mu. How this happens in real tubes? Consider tubes with rectangular plates, elliptical grids, and cylindrical cathodes as an example. Consider also triodes with planar structure, but with boundary areas affecting the total too much. Consider also 1950+ highgm triodes that have very small interelectrode distances, where several microns of error matter. If we draw combined plate curves, we could see that the lines are more roundedup in the lower right area of the plot  something we associate with nonlinear tubes. To study how 2 or 3 models combine, I wrote the follwing 2 applets: you can enter two identical models in each, and then see what heppens when Mu changes. More on ths subject is in [composite triode models].
7. Some interim lessonsThere are 2 lessons to learn here. One is that tubes that have irregularities that affect mu of elementary fragments are less linear, and vice versa.The second one is for design of parallel SE output stages. To achive good linerarity, it is very important to select tubes for parallel operation having the same Mu. Failure to do that will increase distortion. There are also some unexpected recipes: if one tube is wimpier (higher plate resistance) but has the same Mu and biasing it is OK to parallel them! What it means in practical terms, when 2 nonideallyequivalent tubes must coexist? If mu differs just a little, have each tube in the parallel SE stage biased individually (it is alsoways a good idea). When retubing, select biasing in such way that each tube amplifies the same way. If one of them is out of reasonable range of bias resistor values (say, off than 50% change) then attenuate the input,so that each triode develops the same amplitude on its plate. This can be checked if each tube is loaded separately. Another interesting trick is to wire a small resistor in series with each tube and check voltage signal shape on that (which means, current). If the tubes are not in sync, the voltage signals will be visibly distorted [a spice pic may be needed here]
8. Mu compensation with VgHere is something practical to consider when tuning parallel stages  give the tube that lags in amplification more grid signal and vice versa. The idea is very simple: each triode, if performing independently, should develop the same amplitude swing on its plate.Voltage attenuation is one very effective way to match otherwise unequal tubes to work in the same carriage. This recipe may sound heretic to those who get use to current matching tubes in PP amps, but it is in fact correct view on matching of tubes that work into the same load, in fedbackless circuits and when core saturation is not part of the equation. Yet it may be a valid idea even when core staturation is part of equation.
9. 5 equal triodelets, input signals are not the sameInput signal attenuation is a practical way to equalize tubes with unequal Mu. How about a reverse experiment: what is we use identical triodes [again], but this time give each one an individually attenuated signal? This may sound as pretty retarded experiment, but hold on, it has some interesting derivations; thus, this configuration offers some insights into ways how "real" filamentary triodes operate and offers a way to explain some properties of DHTs.To get data from a setup with varying inputs comparable with our benchmark, we increase the signal for U1 and U2, and proportionally decrease for U4 and U5. Think of these voltage levels as rheostate taps; U5 gets a zero level signal:
It seems as U5 should be clipping but it is not yet. For convenience, the results are shown next to 5andallequal data:
As you can see, this configuration dramatically increases distortion, and not only when U5 approaches high swing. Why distortion is high? One quick answer may be: it is high because U1 and U2 work harder (2x and 1.5x) and as result, distort more. This appears correct, and it is, but this effect by itself is not the whole picture. With this effect alone, distortion would increase only by a small factor. The correct answer is that all triodelets work harder, much harder. U1 and U2, are most to suffer, but each one develops its own conductance and therefore each triodelet sees a fairly difficult dynamic load. Worse, the load is nonlinear because each subtriode is not linear.
10. Filament wire as voltage generator.First, as a transition, consider he same idea of amplifying signals of different amplitude, but this time injecting the signal in cathodes. Imagine a configuration where each triodelet operates as groundgrid amplifier on common load. We skip simulation for this as the data is pretty similar to the previous setup.Next, lt's apply yet another twist here: borrow a humbalancing cathode circuit of an autobiased, bypassed directly heated output stage. With that, let's add a chain of lowohm resistors and feed the triodes off taps in such chain:
The chain R19R24 represents the filament wire. Each triodelet sees cathode voltage as it develops on a section of filament structure. If you have difficulty with this schematic, consider this as a help aid:
When we let this circruit to run, it develops somehwat small, but substantial signal on the 8ohm load:
Let's attempt to find out how this hum relates to the data presented in the previous section. We change the frequency from 60hz to to 1khz (this gives more accuracy to FFT setup), and also "shift the balancing pot" to one of the corner positions, namely, change R25 to 100 and short R26. We also give V1 the same run we used for the inputs in the previous section.
For convenience, here we switch to plate voltage measurments, and compute the data for "typical" filament voltage values: 1.4, 2, 2.5, 4, 5, 6.3 VRMS:
This predicts that a 2a3 will develop 5 * 0.63 * 0.01= 32mV RMS hum on its plate. A 6B4G or a Soviet 6C4C will produce 12.4 * 1.6 * 0.01 = 198mV. Indeed, these value match to experimentally obtained hum levels, see [1] !! This simulation also gives a ballpark value for harmonic hum levels produced by other DHTs; the values would not be precise because the linearity of other DHTs may be different; yet, assuming they are as linear as 2a3, we estimate the following. For 300B: 9.8 * 1.25 * 0.01 = 122mV RMS. For AVVT AD100: 8 * .01 = 80mV RMS.
11. References[1] compositetriodemodels.htm Composite Triode Models[2] filamentacharmonic.htm Harmonic Hum in DHTs
Author: Dmitry Nizhegorodov (dmitrynizh@hotmail.com). My other projects and articles
