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More help with low watt output stage please

Hi all,
I recently posted for help designing a tube output stage that would exploit the unused power from my preamp PS. Several contributors helped identify the current limits of my supply, and shared advice on what output tubes might meet my design criteria. I have been reading up on amp basics, limited yet to SE triode mode configuration.

I took two recommended tubes and plotted some load lines. The 6l6 will be underutilized in my build, and the ECC99 maxed out. I am attaching a schematic of my PS which should allow me to continue to supply my 4x6sn7 Aikido preamp with the 40mA it requires, leaving an optimistic 100mA for the output stage, given my transformer is rated at 150mA, and is capacitor loaded. Furthermore, it has a 5A 6.3V secondary, which should be enough for 2 6l6 or ECC99 tubes.

I have modeled the PS in PSUDII, and my voltages are somewhat higher than estimated, probably due in part to my house supply frequently surpassing 121V. Using the software, I have determined that as it is shown gives me 390V with a 40mA load and 4uV ripple, and at 150mA, voltage dips to 340V and 15uV ripple. This seems pretty good to me.

Would you be kind enough to look at my load lines and offer advice on which will likely be a better fit given the PS limitations. I should add that I am not intent on squeezing as much wattage as possible from this amp, and I plan to listen only at modest levels close to the speakers, which are relatively efficient. With this in mind, preserving the character of the preamp (low distortion) is preferable. Another thing I should mention is that the Aikido gives a voltage gain of 9. I plan to use it with a CD player that quotes its output at 2Vrms. Given these, I calculate that I might have max voltage to the output of +-25Vp-p.d My choice of grid swing reflects this, and with the ECC99, I realize that I will not be using the full output of the preamp. 25 V seems really high to me, am I figuring that correctly or is that too literal?

I realize there are many details I have neglected, but thanks for any help with this.
 

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In your situation I would try 6L6 in triode mode with a high-ish plate load impedance (6-8k) and cathode feedback from the secondary winding. When it works (it doesn´t always work well, depending on the OPT), CFB is an elegant way to improve damping factor, distortion and bandwith without adding a single extra component.
 
My thinking is that the 220 and 240Vdc values are based on my B+ of 350-390Vdc with different plate resistor values.


This would be true for an actual plate resistor, but output stages are normally loaded by a transformer, which has much less resistance than its reflection of the speaker load impedance. Maybe a coupla hundred Ohms, so usually ignorable for output stage loadlines.


To draw your output stage loadline, first subtract from the estimated B+ an estimated voltage loss across the output transformer's primary winding resistance. For instance, you measure or read from spec sheet a primary DC resistance of 200 Ohms. You then guesstimate a plate current - let's say 50mA. Multiply to get a DC voltage of 10 Volts, and subtract it from B+.


Second, guesstimate a cathode DC voltage, the output valve's bias. Let's say 30 VDC for a 6L6. This is also unavailable to the output valve, so we subtract it too. Gives us B+ minus 40 Volts.


Our first cut at idling point is now 50mA and, say, 330 Volts. We mark this point on the curves, draw several possible loading slopes, observe and iterate.


All good fortune,
Chris
 
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Thanks Chris for reminding me that this is not a driver stage. I found an inexpensive output transformer on ebay that serves as a suitable load for this project. It is rated for 5watts, with an input z of 5k. (It claims to be good for 20-20kHz.) It has a primary dc resistance of 315 Ohms.

The 50 mA goal maybe at the limit of what my PS can supply, but lets go with it. I decided to bias at -20v to be closer to the Vmax/2 mark. Using your maths 380 Vdc - [(315 x .05) + 20]=344vdc.

Attached is a new load line. It looks pretty asymetric at this low current range. How will that affect the sound?
 

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This is not your loadline, which must pass through the idle point (50mA and 344 VDC). Mark your idle point first, then draw a line through this point of slope 5K Ohms.


A trivial point but you don't really get to choose bias voltage; it falls out from your choice of idling voltage and current. Fortunately, that stuff doesn't really matter enough to worry about - none of these specific numbers are carved into three, sorry, two, stone tablets. Tolerances, etc.


The important bit is to start with the idling point and to remember that all signal must pass through this point (for resistive loadlines). All possible loadlines must pass through this point, and vary (for resistive loadlines) only in slope.


All good fortune,
Chris


ps: Is the term "slope" of the loadline understood?
 
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This is very helpful. Attached is an updated load line following your instructions. I understand slope from HS algebra to mean the ratio between x and y axes, in our case V and I, which are modulated with R, in this iteration, a 5k OT. Attached is a load line that subtracts the voltage drops discussed above, in this case about 36 Vdc, arriving at 340. We guestimate a 50mA idle current, which is too much for my PS powering 2 6l6s, but let's dream a little. What I don't understand is that this then shows that with decreasing current, voltage goes way up above the PS rating. Or do we then shift the load line down, keeping the slope, along one of the bias lines, until it fits into the constraints of 390 Vmax and 50mAmax?
 

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What I don't understand is that this then shows that with decreasing current, voltage goes way up above the PS rating

This is the "magic" of inductive loads such as transformers. I´m sure someone can explain this in more technical terms but in short, transformers and plate chokes allows the plate voltage to swing up to 2*B+
 
When current changes in an inductor there is an induced voltage generated .
That's why the voltage swing can be double B+ .
Of course even the common sense will tell you that the voltage swing must go up and down from the idle point which is +B on the plate of the tube minus a few volts losses in the resistance of the primary .
 
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I like magic (well, not magic smoke), and I wish I had more common sense.

So, if I am correct, the constants in this proposed circuit are
1. The impedence/load of the OT of 5k
2. The idle voltage of ~340vdc
3. The max available current of PS, minus preamp tubes, of 100 mA, or 50 mA per 6l6

Do I now calculate the cathode resistor to pull the loadline down to accommodate an input voltage swing that stays within the 50mA ceiling? Like in the new graph?
 

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Your loadlines seem a bit off, they look more like 10k loadlines to me? Low distortion but also very low power.

If you have 100mA available, set each tube to 50mA and let the last capacitor in the power supply take care of thesignal current variations.
A 560R cathode resistor (for each tube) should be about right.
 

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Hi Fuling, So not only do we get "magic" induced voltage when current goes low, but also magic current when input signal voltage goes positive?

And how did you get the load line slope in your graphic? It seems the slope reflects a lower load resistance than I am dealing with. (The OT I am looking at has a 5k impedance.)
 
also magic current when input signal voltage goes positive?

Exactly:D
With a 5k load and 300Va-k (assuming 40V lost in cathode resistor and OPT copper losses) the plate current will swing from 50mA at idle up to just over 80mA and down to 20mA att full power. The power supply will "see" a fairly constant 50mA current draw per channel since the output stage is decoupled by a fairly large capacitor after the choke(s).
(I feel that this could also be explained in more scientific terms and in better English but I hope my point gets through:) )

Unless I´ve made some mistake, my drawing shows a 5K loadline with a 300V 50mA OP. Patrick Turner describes the concept of loadlines very well here:
loadmatch-1-SE-triodes
 
And you are going to use a driver that provides +/- 30V peak (60V peak to peak).
That will maximize the output, but keep the 6L6 from drawing grid current (more power but more problems).

Many circuits and topologies do not do well, or do not perform gracefully when the output tube draws grid current.

Drawing grid current can be a problem for at least these and other circuit parameters and topologies, but there could be more to add to this list:

Medium output impedance, or high output impedance driver stage
RC coupling from driver to output stage
Self Bias on the output tube (and I always use self bias whenever possible, so I prevent the output stage from drawing grid current).
Some interstage transformers
 
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Fuler commented that some current can be accounted for by the capacitance of the PS. Does this mean that the 150mA rating on my Xfmr reflects 150mA RMS, and poses neither a hard limit, nor a likely noticeable one if crossed infrequently? I guess what I am really asking is: In the real world, can I assume that two 6l6 tubes will seldom create a saturation condition with the operating point we have established? Maybe I can answer my own question by getting several different cathode resistor values and find out.

I just saw 6a3sUMMER's post. To answer, plan to use an Aikido preamp that provides a gain of 8.8. I was wondering about this, but hadn't considered what current is involved. Earlier I worked out that a 2Vrms max CD player input signal could create a +/- 25Vp-p voltage at the grid of the output tube. Maybe one solution might be to limit the supply further to the preamp from say 300V to 200. I will search and read more about the implications of this, especially as you forecast problems.

Thanks for all the help, I think I am catching on. Can someone recommend or link to a schematic or post that explains the cathode feedback design mentioned earlier?

Thanks again.
 
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A CD player like mine that produces 2.1Vrms full scale on the DAC . . .
Produces 3.0V peak.

3.0V x 8.8 preamp gain = 26.4V peak, or 52.8V peak to peak.
20 Log (52.8/60) = -1.11dB
Your amplifier will be 1.11 dB less power at 52.8V peak to peak, than it would be at 60V peak to peak.

Suppose an amplifier produces 8 Watts with 60V peak to peak drive, at only 52.8V peak to peak drive, it will produce about 6.2 Watts.
You might hardly notice the difference. But the distortion will be a little lower, and the amplifier (if properly designed) should not clip under those slightly lower drive Volts condition.

However, if most of your CDs have maximum peak recording levels of -6dB or -10dB full scale, you will want more gain from the 'preamp' used as a driver, or more gain from a driver. (unless your speakers are very efficient).
So many variables.
 
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A single ended Triode Wired 6L6 will not produce 8 Watts.

But the ratio of power at the two drive levels of 52.8 and 60V, will be:
77.4% of the power when driven fully (60V peak to peak).

I have used the 6L6 Triode wired in a single ended amplifier.
And I used a similar 807 Triode wired single ended.
I like those amplifiers.
But of course they are power limited.