| aletheian |
I am trying to design an output stage with a minimal tube count. I have settled on the concertina splitter since it uses only one tube to so it's dirty work.
My problem is: I am having trouble visualizing I can inject NFB? Preceeding the splitter is a tone network that is fed from a cathode follower, and then a volume control, so I don't want to inject it into that stage since it will alter the foreward gain in the loop as the tone and volume controls are altered.
So I assume that I have to inject it into the concertina itself. I have drawn up a few variations. The .1uF capacitor I suppose is the culprirt in question. |
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| aletheian |
| Here is the third option |
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| aletheian |
| Here is the final thought |
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| Miles Prower |
Be much easier to just use a LTP. The LTP has both inverting and non-inverting inputs, so there's a natural summing node for NFB. Since types like the 12A*7 series, the 6BQ7, etc. are all dual tridoes, the tube count doesn't change.
FWIW: avoid the 12AU7A. That one has horrible linearity and more distortion than you want. |
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| aletheian |
Unfortunately, I am using the other half of the 12au7 for the preceeding stage cathode follower, that's why I went with concertina instead of LTP. I thought about subbing in a 12ax7 or 12at7 for the 12au7, but I'll deal with that later. It is the NFB that has me stumped.
Thanx |
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| SY |
Negative feedback works by trading off gain for distortion/bandwidth/impedance. The problem is that your circuit has no gain to start with...
It's difficult-to-impossible to feedback to a split-load; generally, the split-load is driven by a gain stage and the feedback is brought back to that stage. |
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| Miles Prower |
| If you have to go with the cathodyne, then your best bet would probably be something like what I attached. With the cathodyne, you really don't have a second input for the NFB summing node. So try something like the inverting connection that you'd use with an op-amp. In this case: Avcl= Rf/Rs (or close to it). |
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| SY |
| BTW, when you revise the circuit, make sure that the lower output is taken from the top of the load resistor, not from the load plus cathode resistor. |
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| aletheian |
| quote: | Originally posted by SY
BTW, when you revise the circuit, make sure that the lower output is taken from the top of the load resistor, not from the load plus cathode resistor. |
I have seen them noth ways, with the signal coming off the junction of the loadresistor/cathode resistor, and with the signal coming off the junctuin of the cathode trsistor/cathode.
Is there a benefit in doing it the other way? Possibly an impedance issue that I am missing? |
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| aletheian |
| quote: | Originally posted by Miles Prower
If you have to go with the cathodyne, then your best bet would probably be something like what I attached. With the cathodyne, you really don't have a second input for the NFB summing node. So try something like the inverting connection that you'd use with an op-amp. In this case: Avcl= Rf/Rs (or close to it). |
Sweet, thanx for that one.
I am seeing the issue now with the lack of gain in the cathodyne. Maybe I'll rethink the output stage... possibly using a self-split conficuration for the output tubes and using that half of a dual triode as a gain stage /driver and return the NFB there. |
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| SY |
Balance. Think of the current flow. There's no path other than straight down from the B+ to ground (the grid draws no current). So any ac current is multiplied by the resistance to ac ground (which is the B+ rail and ground). The cathode bias resistor adds an error to the bottom drive voltage.
So since you need symmetry to achieve balance, you can either take the lower signal from the top pf the load resistor or you can take it from the cathode and bypass the bias resistor with a big ol' electrolytic cap.
BTW, a conventional common-cathode voltage amplifier direct coupled to the split-load inverter is boring and common. And it's boring and common for a reason- it works really, really well. |
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| Miles Prower |
| quote: |
I am trying to design an output stage with a minimal tube count. I have settled on the concertina splitter since it uses only one tube to so it's dirty work.
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Giving this some further thought, it looks like this is going to be conected to a fairly high-level preamp. In this case, why not just do away with the cathodyne altogether and use the 6L6's as a differential power amp? That way, it does its own phase splitting. Adding a CCS in the tail will improve balance greatly, and you won't be needing any feedback on this.
Just a thought. |
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| aletheian |
| quote: | Originally posted by SY
Balance. Think of the current flow. There's no path other than straight down from the B+ to ground (the grid draws no current). So any ac current is multiplied by the resistance to ac ground (which is the B+ rail and ground). The cathode bias resistor adds an error to the bottom drive voltage. |
Well, I have mostly seen it in old guitar amps, and early PA systems... probably just bad design then.
BTW, a conventional common-cathode voltage amplifier direct coupled to the split-load inverter is boring and common. And it's boring and common for a reason- it works really, really well.
[/QUOTE]
HAHA! Yeah, true enough. |
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| PRR |
The concertina has one hi-Z input and two odd-Z outputs. The obvious way to take feedback, as Miles says, is like you wire an op-amp as an inverting amp.
The 6L6 will need about 25V peak at the grid, give about 12RMS or 18V peak at an 8Ω load. The concertina has gain just under unity. So the overall amp, before feedback, has 27Vpk in, 18Vpk out, gain is 0.66. Passive feedback factor is 1:1 for non-inverting mode, 2:1 for inverting mode. As SY says: you got no gain and can't really get any feedback at all.
The transformer can be wound to (not necessarily bought) any ratio and output impedance. If the output is 600Ω, 27Vpk in gives maybe 140Vpk output. If you could rig a concertina for non-inverting unity gain, you could get 14dB NFB; since you must rig it inverting, only 8dB NFB. Of course 600Ω is an uncommon speaker impedance. We could wind both an 8Ω for the speaker and a 600Ω winding for the feedback; the two windings won't couple exactly the same thing, and the copper resistance of the 8Ω winding will not be reduced by feedback. Note that unity-gain implies that the input signal at the pot wiper must be like 150Vpk. If we want just 6dB of reserve gain in hand, the preceding stage must produce 300Vpk. I'd prefer to see 10dB or 20dB reserve gain, 450Vpk-900Vpk. That's insane.
Power amps that take a "comfortable" input level (a volt or so) and drive "wires" (voice coils, lines, line transformers) almost invariably need a Power tube and a Volt-amp tube. The concertina is not a volt-amp. You need another tube in there. Either one between volume pot and concertina, or two between concertina and 6L6s. The first way is simpler and also gives a reasonably good place to inject feedback. It is of course THE most common way to wire a push-pull power amp.
The LTP is also an option. If you want a lot of gain or NFB, and have a low final output impedance, note that a LTP gives half (or less) the gain of the same two tubes as volt-amp/concertina. 12AU7 amp/concer gives gain of 16 to each 6L6 grid; 12AU7 LTP gives gain of 8 to each 6L6 grid. The LTP does have an infinite-Z feedback input while the volt-amp's feedback node is under 1K. For hi-fi line-amps with limited output current, the load of that feedback input limits performance. For 8Ω outputs which could easily carry a 100Ω feedback network, the low Z of the volt-amp cathode is no big deal.
The LTP is forced to use similar devices, which must give high gain without high plate resistance. You typically get modest gain and impedance. The concertina has similar output impedance (the bottom output is not so low-Z as it looks) but high input resistance, which means the volt-amp can be worked at higher gain than it could driving a 6L6 grid, and can even be a different device of much higher gain. 12AX7 driving 12AU7 was a popular rig. 7199 is a pentode+triode and can give gain over 200 to each 6L6 grid.
LTP has a natural symmetry and beauty. I used it a lot in my youth, works good. Fender uses a bodged LTP in all their big amps, and they do fine. Yet Ampeg did fine with voltamp+concertina, and I'm favoring that plan now. The "push-pull" of the LTP isn't perfect, especially at high levels. The concertina has so much internal feedback that it can be insanely linear up to higher levels than any pair of plates.
BTW: what is this pot-cap hanging on your feedback loops? Is this a gitar amp? It may help to know what the amp is supposed to be: a hi-fi amp and a gitar amp suggest very different choices.
Oh: in Mile's suggested plan, the "obvious" place for the pot-cap stuff is across Rs, not grid to ground. But coming off a pot wiper, it may not have much effect. Taken to ground, it won't affect gain much, it will reduce NFB at high frequencies (which may have use in screamin-gitar work).
Does this go with your super-gain cascode? I was thinking about that. The only reason to have that much gain is for huge (impractical) feedback, or..... to build a 2-stage guitar amp. Gain of 1,000 into a 6L6 gives input sensitivity of 25mV, a usable gitar sensitivity. Traditionally we run 2 or 3 12AX7 or 12AT7 stages to do the same thing. Reducing the number of stages will change the flavor, though maybe not a whole lot. Fewer stages gives fewer opportunities to insert EQ and gain-structure controls, which are essential to some guitar work. Feel free to explore, but have a look at where you are going.
Oh... you do realize that triode-strapped push-pull 6L6 will work very fine without any NFB at all? Maybe too clean for guitar. Not dead-clean for hi-fi, but less NFB-induced garbage than many plans. |
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| aletheian |
OOOh, yes, it is for a geetar amp. The NFB also allows for the attenuation of NFB at high frequencies...eg a presence control, that's the hanging capacitor stuff. it is being fed from a very high gain preamp...4 cascaded 12ax7 sections into a cathode follower and then a tone stack. The preamp design is not original, but tried and true with some tweaks so that it will make more sense gainwise. The preamp is pretty much set in stone... the values are carefully balanced to clip in just the right way...the cathode follower too! So I can't run any feedback there.
I'm helping a bud out with a few things by distilling a circuit down to it's simplest possible form. I personally like to go more over the top with my own stuff, but he has never even built an amp before, so naturally I thought that the concertina would be the simplest wat from point A to point B in a Class AB1 output stage. I thought if I deleted the typical grounded cathode driver that normally preceeds the concertina, and ran the signal right from the master volume, that I could avoid having to add another dual triode just to use half of it. But then I coildn't figure out where to run the dar-ned global NFB!
I have run NFB into the top grid of a LTP inverter before, with great results, so I thought I'd just do that here, but I started tnkering and had to call on the experts.
It's not going with my crazy cascode of DOOM, that is just a learning experience for me to expand my knowledge of how best to blowing audio amps up ;-) |
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| aletheian |
| quote: | Originally posted by PRR
The only reason to have that much gain is for huge (impractical) feedback, or..... to build a 2-stage guitar amp. Gain of 1,000 into a 6L6 gives input sensitivity of 25mV, a usable gitar sensitivity. |
This guy is SLICK! That is the plan exactly. Which is why I don't give a darn about bandwidth and noise (well...a little bit about noise) Plus I'll be slugging the input wth 2 volts from the pickups, so that gives some room too. |
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| tubelab.com |
OK you need half of a 12AU7. I forgot about this when you asked about 7 pin tubes. 1/2 of 12AU7 = 6C4. Need half of a 12AX7 use 6AV6. I know that musicians like tubes that can be bought at SAM ASH, so you are stuck with 12A*7. I am sure that you can find use for the other half. Vibrato, NAH too out of style. Reverb driver, maybe. Effects loop driver?
I am sure that you have heard of the 12DW7, invented for Ampeg, it is 1/2 of 12AX7 and 1/2 of 12AU7 in the same envelope. The Russians (and maybe the Chinese) are making new ones. |
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| aletheian |
| quote: | Originally posted by tubelab.com
I am sure that you have heard of the 12DW7, invented for Ampeg, it is 1/2 of 12AX7 and 1/2 of 12AU7 in the same envelope. The Russians (and maybe the Chinese) are making new ones. |
Cool. I didn't know that they were back in production. Have you heard them yet? I'd be intereted to see how they turned out.
Now if only someone would remake a nice mercury rectifier, like an 83... |
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| Tubes4e4 |
| quote: | Originally posted by tubelab.com
I am sure that you have heard of the 12DW7, invented for Ampeg, it is 1/2 of 12AX7 and 1/2 of 12AU7 in the same envelope. The Russians (and maybe the Chinese) are making new ones. [/B] |
JJ/Tesla too.
Tom |
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| Tubes4e4 |
About 12DW7:
| quote: | Originally posted by Tubes4e4
JJ/Tesla too.
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It is called ECC832, I think.
Tom |
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| Johan Potgieter |
Aletheian,
You have had so much good advice that you hardly need more, but since I seem to be writing nothing tonight....
I would echo PRR (post #15) and just use the circuit without NFB for your purpose. Otherwise a no go; you will need an input stage for anything proper. I have seen many attempts over a long time at making an over-simple circuit too prosperous - often not worth the effort because of minimal real advantage, if at all. As was said, straight output triodes are .... mmm ... not too bad on their virgin own. |
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| gingertube |
I recently did a restoration on an AMPEG tube GIT Amp - The V9 preamp circuit.
I used new production 12DW7 from EI - the guy was VERY happy with the sound, said it was at least as good as when the amp was new.
Cheers,
Ian |
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| gingertube |
While on the subject of Cathodyne Phase Splitters can anyone tell me what the effective source resistance (impedance) is in terms of tube parameters and RL = RK resistance value.
The 1960 article by Albert Priesman seems to suggest that the way to think of this is that the "effective" source resistance at both anode and cathode is rp + (1+mu)(RL+RK) so long as RL=RK and the effective source voltage is considered to be mu x Vgk.
Since we know that the gain of this thing is actually close to 1 which implies an effective source voltage approx. equal to vgk this requires some mental gymnastics to visualise.
Considering it as an approximately unity gain circuit and adjusting the above to suit would suggest that the effective source impedance is approx rp/mu + RL + RK OR 1/gm + RL + RK
Can anyone confirm the "maths interpretation" of what Priesman wrote and/or my interpretation of the same.
I have a Cathodyne driving 2 pairs of EL34 in Triode Mode. To improve the drive (at high frequencies in particular) I need a lower effective source impedance.
That would suggest either ECC99 (what I'm using now) or 6H30 at higher current setting and correspondingly reduced RL and RK is the way to go. This would seem to be right intuitively.
Does this seem right to you experts or have I missunderstood the Preisman article or screwed up the maths.
Should I perhaps just forget the B maths and analysis - stick on the bench, make the changes and see what happens?
Cheers,
Ian |
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| tubetvr |
Hi,
I have both measured and simulated the Cathodyne Phase Splitter and have confirmed that the Preisman calculations are correct, it is not that difficult to do.
However given that the gain to each output is normally given as u/(u+1) the output impedance will rather be (rp+Zk)/(u+1) i.e. much lower than you would expect, (it is described in the article, look at figure 3 and the surrounding text).
Do what I did, do some simulations in Spice or build one or 2 phase splitters and measure and you will find the results I did.
The reason I did this was because I earlier measured the open loop response of my OTL and didn't discover the poles I expected, the poles due to the phase splitter was much higher which of course was due to that the output impedance of the phase splitter was much lower than what I assumed earlier.
My measurementy method was to use different values of load impedances, (but equal for cathode and anode outputs) and comparing the voltage, from the voltage difference I calculated the output impedance.
The confusion around the cathodyne is probabaly due to that some want to measure the output impedance of cathode and anode independantly and then they are indeed very different, however when the load impedances are seen as independant it is wrong to talk about this circuit as a phase splitter, isn't it?
To your questions:
| quote: | | I have a Cathodyne driving 2 pairs of EL34 in Triode Mode. To improve the drive (at high frequencies in particular) I need a lower effective source impedance. |
| quote: | From the above - that would suggest
1) rp NOT that critical, it will be small compared to the other term
2) lower mu is an advantage
3) BUT mostly lower RL and RK values => higher tube current
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You conclusions are correct but only if you not consider that that particular definition of output impedance also implies that the gain is u*Z/(rp+(1+u)Z+Z).
Given that the gain is most often written as u/(u+1) and the output impdance as I wrote above as (rp+Zk)/(u+1) you can draw the following conclusions:
rp is as critical as the anode or cathode resistors
High u is beneficial, (as long as the load impdances are equal but lower u keeps better balance if the loads are unbalanced, e.g class A2)
I use 12BH7 or 6SN7 as phase splitter as I think that fulfill the above requirements to a reasonable compromise.
Regards Hans |
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| gingertube |
Hans,
THANK YOU !!!
So I want higher mu, lower RK and to a lesser extent lower rp (since for any sensible tube selection rp will be 1/4 off RK or less)
The higher mu will be handy since the whole amp is a bit short of gain and the other half of the tube is the input common cathode amp.
So I'll try
1) 6N1P (mu=35, rp = 8K)
2) 12AT7 (mu=70, rp =8K)
and drop the anode and cathode load resistors to the extent possible consistent with conservative power dissipation limits. Actually will try the 12AT7 first since that doesn't need any heater rewiring.
Have ONLY last night installed B2 Spice on my home machine so may try simulating as well.
Cheers,
Ian |
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| tubetvr |
Gingertube wrote:| quote: | | drop the anode and cathode load resistors to the extent possible consistent with conservative power dissipation limits. |
For me the choice have always been dependant on the output level needed, higher anode resistors of course means higher available voltage swing, but as you say the lower the better for drive capability and frequency response.
Regards Hans |
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| SY |
| There's an argument to be made for low mu if there's a chance that the stage being driven will leave class A. |
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| tubetvr |
| quote: | | There's an argument to be made for low mu if there's a chance that the stage being driven will leave class A. |
Yes that is correct and I also described that in an earlier post:
| quote: | | High u is beneficial, (as long as the load impdances are equal but lower u keeps better balance if the loads are unbalanced, e.g class A2) |
Morgan Jones have made a analysis of the Cathodyne in his book, (3rd edition) where he draw the same conclusion, i.e. that low mu is beneficial in the case there is risk of the output tubes drawing gridcurrent.
Regards Hans |
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| Johan Potgieter |
After that good analysis by Hans perhaps just a practical point, i.e. it must be kept in mind that rp can change with the load resistors in the sense that it is dependant on Ia. I also like using the ECC81 here, but unfortunately do not have parameter/Ia graphs for that. Perhaps an illustration using ECC88 parameters might be interesting (mu about 29 and Gm about 3 mA/V here):
For a tube Va-k of 150V (where my graphs are valid) and h.t. of 400V, it is found that:
For RL=RK=33K, Ia=4.4 mA and rp=5.5K
For RL=RK=47K, Ia=2.6 mA and rp=8.5K
For RL=RK=100K (rather high), Ia=1.3 mA and rp=12K
This means that the (rp+Rk) term does not change all that much.
But Hans: You conclude that rp is as critical as RK (or RL), but according to the above it forms about 12% - 18% only of RK (RL). Do I misunderstand you here?
Also, Gingertube, the ECC81 has a mu of about 57 and rp of about 18K at the currents normally used, which make things a bit more critical - although, as said, I have often used it (I normally do not drive output triodes directly from a cathodyne, which eases matters somewhat for me).
And what about the E182CC? It can be used at quite a high current (mu about 22 and rp about 3K at 9 mA). One can of course go on suggesting tubes ad infinitum..... |
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| gingertube |
The amp in question has a 12AU7 Common Cathode / Direct coupled Cathodyne feeding 2 pairs of triode strapped EL34s into a Plitron PAT4006 (VDV2100) Toroidal O/P Transformer (1900raa : 5 secondary). Cathodyne RA and RK are 22K and stage current is 4.4mA.
NOTE ALL Measurements except max power refered to 6V RMS across 3.5 Ohm load @ 1kHz (approx 10W out) - 0.8V RMS input. These levels set using my multimeter.
Baseline measurements with 12AU7:
Power into 3.5 Ohm dummy load @ 1KHz = 39W
Zout @10W, 1kHz = 4.4 Ohms
Gain = 5.5
High Frequency -3dB point is 225kHz
Assuming approx 100pF each side for the parallel EL34 inputs that tells me straight away that the effective source impedance from the cathodyne is NOT MORE than 7KOhms.
Change to ECC99 - Sound was a lot cleaner (The Common Cathode stage???)
Power Out - unchanged
Zout - some suggestion it was slightly higher at 1kHz (measured a value of 4.9 - shifted pole from cathodyne effective source Z ???)
Gain = 7.2
High Frequency -3dB point is 250kHz
While running the frequency response checks using oscilloscope I could see the effect of the feedback from the Miller capacitance of the EL34s. As frequency went up, so did feedback due to Miller Cap. This reduced EL34 rp which reflected to the secondary as reduced Zout and we saw a boost in voltage across the 3.5Ohm dummy load.
Zout measures (Vout open circuit =88V p-p):
Freq. Vout Zout
20 28 7.5
50 34 5.6
100 35 5.3
200 35 5.3
500 35.5 5.2
1000 37 4.8
2000 40 4.2
5000 46 3.2
10000 48 2.9
20000 49 2.9
50000 49 2.9
100000 42 3.8
200000 28 7.5
These were a consistent set of readings from the oscilloscope so Zout results are correct BUT for some reason it doesn't stack up with the 6V RMS into 3.5 Ohms @ 1KHz set using my multimeter. 6V RMS is NOT 37 V pk-pk????
I havent bothered BUT I think that you could calculate backward from this data to arrive at an effective Source impedance from the cathodyne.
Also: There was some indication (on the oscilloscope) of slewrate limiting starting at about 50kHz
The gain difference 7.2 vs 5.5 can be explained by difference in mu (22 vs 17).
I think the next step for me and this amp is ECC81 front end with 6dB of global feedback.
Hope there is stuff of interest to you in the data.
Cheers,
Ian |
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| tubetvr |
Hi,
Johan, I agree with your post, as rp is dependant on the operating point you have to be a bit careful when selecting anode/cathode resistors.
| quote: | | But Hans: You conclude that rp is as critical as RK (or RL), but according to the above it forms about 12% - 18% only of RK (RL). Do I misunderstand you here? |
No it is not a misunderstanding but rather me being incomplete or unclear with my statement, it depends on what you mean by either rp or RK being most critical, what I meant is that a reduction of rp in ohms is as beneficial as the same reduction of RK, of course you could say that it is better to compare for the same reduction in % and in that case it would be more beneficial to reduce RK rather than rp, RK being larger.
Gingertube, I think you have some interesting results, a 12AU7 at 4.4mA should have a rp of about 13kohm and a mu of about 16 which should give a Zout of ~(13 + 22)/(16+1) or ~2.1kohm which woul give 2 poles close to 760kHz assuming 100pF in input capacitance of the EL34's. Changing to ECC99 but keeping the 22k's would then not change the poles so much but maybe a ECC99 allow for lower RK's as well?
BTW, 6VRMS would be 2*SQR2*6 or ~16.9V Pk-Pk
Regards Hans |
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| gingertube |
Hans,
The one major thin I learned from the experiments last night is that the effective source impedance of the cathodyne splitter is NOT a problem in this amp. It is quite adequately driving the input capacitance of the EL34s. This was worthwhile learning as I'd prviously assumed it would be a problem and I was all set to replace the entire front end.
The major problem with the amp is its Zout - thats why I'm now thinking about some global feedback.
I had thought to manipulate the tube type and RL,RK values to get the Zout of the splitter and the output tubes input capacitance to be my dominant pole and aim to set it at about 100kHz (about 1/2 the output tranny -3dB point). This now seems to be less practical than I'd hoped (since the cathodyne output Z is much lower than I expected) so may have to shift the dominany pole to the Common Cathode Stage. The one good thing about the Plitron output tranny is that its bandwidth is so high that ensuring stability is not a difficult thing.
I'm surprised at how good the amps sound into my nominally 6 Ohm speakers (DF=1.3). There is no evidence of bloated or wooly bass, it sounds a little bright but not objectionablly so.
Cheers,
Ian |
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| SY |
| At the expense of a bit more drive, you might consider feedback from the OPT secondary to the output tubes' cathodes in lieu of global feedback. |
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| gingertube |
SY,
Good idea - unfortunately the Plitron PAT4006 (VDV2100) has ONLY a single 5 Ohm Output winding with no centre tap etc.
I do have a pair of VDV2100-CFB/H sitting on the shelf as well, which have both a centre tapped 5 Ohm secondary AND 2 separate 20 Ohm cathode feedback windings BUT I am saving those for the next project where I want to mess about with Ultralinear + Cathode feedback (Menno's "Super Triode" connection).
Learning some good stuff from this lot though.
Thanks,
Ian |
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| gingertube |
I worked through Morgan Jones treatment last night. Her is the summary of his work.
For equal loads on Anode and cathode
Zout = RL.ra/RL(u+2)+ra
The ra term on the bottom line is insignificant compared to RL(u+2) term so drop it. Then the RL terms top and bottom lines cancel leaving
Zout approx = ra/u+2
At typical values of u (>=20) U+2 approx = u, so simplify again
Zout approx = ra/u = 1/gm
If driving output stage directly The equal loads on Anode and Cathode will NOT be guaranteed if:
1) Output stage strays out of Class A (When a tube cuts off it has no gain so Miller capictance will change, particularly with Triode Mode Output, less so with Ultralinear and less so again in Pentode Mode)
2) Output Stage strays into grid current.
If the Anode load drops significantly then:
Zout cathode = RL+ra/(u+2) x ra/RL - ra/RL term insignigicant so
Zout cathode approx = RL+ra/u+2 - at uual values of u
Zout cathode approx = RL/u + ra/u = RL/u + 1/gm
That is it increases by RL/u
If the cathode load drops sifgnificantly then:
Zout anode = RLxRL(u+1)+RL.ra / RL(u+2)+ra
RL squared (u+1) is much larger than RL.ra and RL(U+2) is much larger than ra so
Zout anode approx = RLxRL(u+1)/RL(u+2)
and at reasonable values of u
Zout anode approx = RL
Summary:
As the loads on Anode and cathode become unbalanced then
Zout anode increases from 1/gm toward RL
Zout cathode increase from 1/gm by maximum factor of RL/u
Thats why its suggested that low u is better when driving other than Class A Output Stage.
Cheers,
Ian |
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