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Old 30th December 2011, 06:31 PM   #1
hummel is offline hummel  Germany
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Hi all,
I posted here a while back regarding building a small amp using 6AS7's (which I have a number of) and realized from the answers I got that I knew next to nothing about the subject. So, I have spent the last few months, off and on, reading and experimenting using a breadboard affair I built so I could (relatively) quickly wire up stuff
The results so far are a two-tube amp that uses a 12AX7 (6H2Л, actually), and a 6AS7 (6H13C) in P-P. A picture of my breadboard, the schematic and what I think is the load-line is attached. I want to say up front that this was (and is) just for learning and doesn't represent something I'll build "for real".
The 12AX7 is a Schmidt phase-splitter. A 2V peak-peak input gives about 80V p-p output. This directly drives the grids of the 6AS7. The output transformer is a toroid 115/115-7/7 that calculates to a 8640 ohm primary with 8 ohm load on the secondary.
One question is regarding the load line. Is that correct for this circuit? The slope is based on the anode load of 2160 ohm (8640/4) and the line's position on the anode to cathode voltage of 170V and the quescient current of 36mA. The bias is effectively -80V. (If I got that right, I think I have a handle on load-lines, maybe.)
I think I want to get the line "higher up" to take more advantage of the tube. That would mean a different transformer impedance, right? And is the bias at -80V appropriate for this tube?
The circuit looks good on the 'scope and sounds pretty good, too. Quite good, actually, but the distortion measure needs a notch filter and I'm working on that.
So, any other input you might provide would great. Like, what should I try next? 80V (p-p) seems rather high from a 12AX7. Should I do a voltage stage between? Or maybe try a cathodyne splitter but how do I handle the next stage? This is fun and there's a lot to learn. I really want to understand this stuff.
Thanks!
- Michael (aka Hummel)
loadline.jpg

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Old 31st December 2011, 01:57 AM   #2
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Lieber Michael,
I don't know about your first posting simply cause I don't drop in here regularly.
I....don't know where to start. Well maybe the Output transformer. I hope you know that these toroids aren't made for that job. They're lackin inductance which is needed for low frequency transformation. Etc.
Lets move to the real problem, the output tube and your calculation. You made a huge mistake here. Why does your loadline reach zero at 250V when you're OPERATING it at exactly that voltage ? An inductor is no resistor....NO resistor ! Why ? And Inductor can do something like this: Click the image to open in full size. with Φ (our friend, the magnetic flux) being a linear function of I. The rest depends on some geometric sh*** but since we don't disassemble our OPT this is constant. And we all know what happens when we derivate constants, don't we ?
On the other side a resistor only does something like that: Click the image to open in full size.

You see the difference ? This has nothing to do with the inductor being a complex load and the loadline being transferred to a load-ellipsoid (Uaaahhh) or something. The load is still completely resistive and the tubes is dynamically working into that load. But our DC conditions changed drastically. You can see this from 2 sides. Because of the Derivation of the flux (and thus the current), you loose every information of your quiescent current and you only see the change of your current. And, as you see this is multiplied by N, the number of turns. Of course since the number of turns concludes our ratio and thus the steepness of our loadline.
Maybe easier for you: When you have Ub=250V you also have Ua=250V (when there is no DC-loss in the OPT). So there is NOTHING interesting without a signal applied here. It even seems that the OPT wouldn't even exist.
More important is the negative sign ! If your anode current changes to a lower value, voltage is "created" in the inductor. (Where does it come from ? You need energy to magnetize the core, to put all/most of the 3d electron spins of the iron in one direction). This voltage now adds up to your supply voltage.
A nice point of view is, that an inductor always wants his current to be constant. Every error will be compensated with a change in voltage across ist. Discontinuous current changes will result in infinite Voltage which is the case when you know how a interruptor-contact-car-ignition or a ingition coil works. And yes, that's the problem why you always should connect a load to a tube-amplifier output.
This means, that your tube can be driven far more than just to 250V.
Sice you want to keep your steepness which should be optimal for the selected to, you normally draw your loadline from 2xIa to 2xUa. So your loadline should end at 500V. Of course, this operating point is crap for a 6080.

Next thing is the load impedance. Your 8640 are way off the optimum.
For max. Output Power, a Triode should have a R-load ~~ 2*R-out. I won't explain why, just buy some books and read it. The theory is explained in Vorhoeve-Niederfrequenz-Verstärkertechnik and Rothe-Kleen-Elektronenröhren als End- und Senderverstärker which are both extremely abstract.
Any value bigger than 2*R-out yields less output power but also less distortion. The rule of thumb is between 2 and 3 times R-out.
The 6080 has a R-out of ~286Ohms. So I suggest about 800Ohms. For a Class A SE and thus 1.6K for Class A PP

Your driver isn't really good, too. For those triodes you should consider 2 stages. The symmetry of your phase splitter configuration can be much better when you place a constant current source in the tail. You can read about that in here or buy a modern book like Morgan Jones - Valve Amplifiers.

You have lot of work ahead when you want to understand everything from a theoretical point of view. Literature will help a lot. This Forum is also great. Don't expect an answer from me tomorrow (or even the day after tomorrow). Chances are high that I'm lying drunk in some fireworks and don't remember anything
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Old 31st December 2011, 06:08 AM   #3
godfrey is offline godfrey  South Africa
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Quote:
Originally Posted by hummel View Post
The output transformer is a toroid 115/115-7/7 that calculates to a 8640 ohm primary with 8 ohm load on the secondary.
Yes, if you're using a single 7 volt secondary (or 2 in parallel).
Quote:
Originally Posted by hummel View Post
One question is regarding the load line. Is that correct for this circuit? The slope is based on the anode load of 2160 ohm (8640/4) and the line's position on the anode to cathode voltage of 170V and the quescient current of 36mA. The bias is effectively -80V. (If I got that right, I think I have a handle on load-lines, maybe.)
The way I figure it is: If you removed one output tube, then the remaining one would see a load of 2160 ohms, as you say. However with both tubes installed, each one only has to deliver half the signal current for a given output voltage, so the effective load seen by each tube is 4320 ohms. I'd expect the load line to be something like what I've drawn in green. Looks OK.

I wouldn't panic too much about the transformer inductance. They're intended to work at full voltage at 50 or 60 Hz, so the bass can't be that bad. If I was building a valve amp, I'd be using toroidal mains transformers for the output, mostly because they're obtainable and affordable.
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Old 31st December 2011, 06:49 AM   #4
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Originally Posted by godfrey View Post
Yes, if you're using a single 7 volt secondary (or 2 in parallel).

The way I figure it is: If you removed one output tube, then the remaining one would see a load of 2160 ohms, as you say. However with both tubes installed, each one only has to deliver half the signal current for a given output voltage, so the effective load seen by each tube is 4320 ohms. I'd expect the load line to be something like what I've drawn in green. Looks OK.
Quite the opposite. When they are both in the circuit, each sees the 2 windings in AC parallel. When one is cut-off, the other sees only one side of the transformer primary.
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Old 31st December 2011, 07:58 AM   #5
godfrey is offline godfrey  South Africa
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...which is why he divided by 4 in the first place.
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Old 31st December 2011, 10:06 AM   #6
hummel is offline hummel  Germany
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Thank you Manta und Guten Rutsch!
I appreciate your pointing out that the current characteristic of the OPT (or any xformer for that matter--learing more about inductors is now on my to-do list) causes voltage swings "around" the voltage at the anode. (Not phrased well, I know.)
But I think that my mistake on the load-line is not that great. The plate potential is relative to the cathode and the cathode is at 80V. So, plate voltage is 170.
And this brings me to an observation/question. Why not use negative supplies for the grid bias (cathode at ground and appropriate voltages used)? You even mention a current source for the input stage. Isn't that current source just another way to get a constant (negative) voltage potential from grid to cathode. I noticed that before I added the so-called "bypass" capacitors to the 6AS7's cathode resistors that there was ~20V p-p "wasted" there. That bypass is really just a way of turning the top of the cathode resistor into a voltage rail, same as filtering a power supply, right? So, the circuit would be the same if the cathode were at ground (i.e., 0V), the plate at 170V and the grid at -80V. Getting a clean -80V at a few mA would be easy. But would that be better? For example, a recent post from Jebem in the thread Fixed Autobias? says:
"I use fixed bias when I want to get the all the available power from the output tubes, and no cathode resistors at all; however one needs to adjust the bias time to time to mantain the correct anode current (or use servos).

Cathode bias (auto bias) is nice as it don't ask for periodic bias adjustments, and it overloads nicely (sound wise); however output power will be smaller, and large cathode resistor bypass capacitors are required to mantain good low frequency response as well.
Because these caps are in the signal path and large values are required (the lower the cathode resistor, the larger value the caps should have to maintain low frequency response), normally electrolytic caps are used, and I really hate to use these in the signal path.
So in the end, I tend to use fixed bias."


And to Godfrey, yes, toroids are cheap (in comparison) and for a treble amp in a by-amp system (the goal) should be OK. In fact the lame circuit I posted is flat (on the scope) at ~5 watts from about 80Hz to over 30.000. But Manta is right and I should understand more about what's going on.

Last edited by hummel; 31st December 2011 at 10:10 AM. Reason: formatting
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Old 31st December 2011, 01:07 PM   #7
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Hi Michael,
Well, that's another issue here.
Of course, automatic bias is not really economic. For the simple reason to create a potential difference you have to waste Supply Voltage at full cathode current.
Fixed bias needs an additional power source, but the current is so low, it shouldn't be a problem. I already did those by voltage quadrupling from the 6.3V AC Heaters.
But this only counts for a static/DC point of view ! I don't know how much you know about control/servo loops. But just imagine a little error in the tube which lets the current rise without any outer interference. At fixed bias, this rise in current goes on unnoticed. It could rise so high till the tube is destroyed. Now think what happens when automatic bias is applied. The rise in current is instantaneous noticed by the cathode bias resistor which now has a bigger voltage drop. (Ohms law) But this voltage is across the Grid and Cathode and has the right polarity to cease that increase in current. The cathode current is "trapped" in a well. Every disturbance is lead back to the control grid and will be blocked.
Now if you don't put a capacitor across that cathode resistors, this happens to EVERY disturbance, including your audio signal. This is called current feedback
Putting a capacitor across that resistor will short out any AC signals and thus eliminate the feedback for AC signals. BUT it is still active for "slow" disturbances like drifting cathode current and that is the reason why this method is used so much. Especially with tubes like 6080, 6S33, ED8000 etc. which are drifting like hell (because those weren't designed for audio. They were used in PSU's and of course they were used in a regulating loop ! It simply didn't matter.) So the caps forms a high-pass filter. Blumlein and Garter even thought of a symmetrical version of this:Click the image to open in full size. You can read about that here: Blumlein’s Garter Circuit Revisited
The use of a constant current source is whole different story. Just imagine what would happen if you use a CCS in simple cathode amplifier. A change in current wouldn't even be possible ! The slightest current change would be answered with every voltage needed to block exactly that change. You wouldn't get any AC power out of it. (Maybe now, you get the clue how a regulated tube PSU works, hehe)
A CCS in the cathode path only makes sence in differential amplifiers like your driver because it keeps up symmetrie.

Your citated text not really helpful. Because the most important thing (control loop) isn't even mentioned. You shouldn't give a damn about peoples preferences for certain components. It is well known that some individuals here can't sleep when thinking about electrolytic caps. Other don't like metal film resistors etc. But that's no academic approach to learn and solve problems. These are experiences which everybody has to work out for himself.
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Old 31st December 2011, 02:56 PM   #8
hummel is offline hummel  Germany
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Mantra, thanks again!
OK, I see how fixed bias could be a problem without some kind of feedback (servo) to keep things in control. BTW, apropo drift, is this why I see the voltage moving around some when I look at, say, the top of the cathode resistor (not 50/100Hz ripple but some mV of very low frequency, seeming random, well, drift I guess)?

I did run across an argument (and circuit) for a splitter with CCS here Practical Phase Inverters.. (This site helped a lot.)

FYI, my background is in IC design, i.e., logic and peripherally, some fabrication, so math, a bit of physics and programming. I have built stuff from schematics (e.g., The Belfry In the Glow) but wanted to learn more.
And I understand what you mean about separating the "wheat from the chaff". But my HiFi system really does sound better with those zigtausend Euro palladium interconnects. (nicht ernst!)

Noch mal, guten Rutsch!
Tschüß
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