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In Reply to: RE: Funny though, I knew there was something wrong with your argument posted by Ralph on August 03, 2007 at 14:26:56
Are you sure you were driving only one grid of the diff amp? If you were driving both (maybe looking at a down stream stage) then I would expect to see little or no signal at the connected cathodes.
Anyway, since you called me out on it, I performed some measurements of my own. The only PP amps I have around are guitar amps. The one I looked at uses a 12AX7 LTP with a comparatively 'short' tail resistance of about 11k along with unequal plate resistors to compensate for the short tail (82k and 100k.) With 2Vpp applied to the grid I saw almost exactly 1Vpp at the cathodes, maybe a tiny bit more. That surprised me; considering the small tail resistance I expected to see less than 1/2 the input voltage there. Perhaps the unequal plate resistors are a little too unequal... (In case anyone is wondering, I did disconnect the NFB that is applied to the tail.)
While I was at it, I looked at the output voltage: roughly 32Vpp at each plate giving a circuit gain of about 16. This is a little lower than I expected. I would expect a 12AX7 at this operating point and with this plate load to have a gain around 40 in an ordinary grounded cathode circuit, so I was expecting a circuit gain of around 20 in the LTP.
I also put together a circuit in Spice with 6SN7 whose behavior was dead-nuts on what I expect, but at this point I see no reason to share the details.
I'll try to help make sense of the signal at the cathodes. Consider a simple cathode follower with a CCS under the cathode. How would you measure the output impedance? The usual way is to load it down until the output signal is 1/2 the unloaded output. Whatever load was needed to accomplish that is the output impedance of the cathode follower.
Well, the LTP (when used as a phase splitter) can be thought of as a cathode follower driving a grounded grid amplifier. If you think about it you'll realize that the input impedance of the grounded grid amp is equal to the output impedance of the cathode follower. Given what was said above, we should expect the signal at the cathodes to be 1/2 the signal at the input grid (ignoring the fact that a CF actually has a gain slightly less than 1.)
Ralph, I would really like to put an end to this discussion. I remain completely confident in the facts as I have been presenting them. You've learned something; that should make you happy.
-- Dave
Follow Ups:
Thanks to everyone for your posts here! Sometimes the debate on such issues can become a little contentious on these boards (to say the least) and your civil discussion of the relative merits of these circuits has made it MUCH easier for someone like myself to learn something.
For instance, Dave mentioned above that the use of unequal plate resistors in a LTP is an effort to compensate for a lower than optimum value on the tail resistor. I was wondering about that because I noticed in the Eico circuit I posted that one plate resistor was 33k and the other was 28.75k.
I've also read that, optimally, the plate resistor should be about 2x the plate resistance. In the case of the 6SN7, that would be 13.4k, so a 15k (or 18k or 22k) would be better than a 33k as long as the supply voltage could be adjusted to get the proper plate voltage. Correct?
The power transformer I'm using has several secondary taps so I should be able to adjust the supply voltage somewhat. I may try to compare the paraphase with the differential (LTP) by building each channel differently. I don't have a scope so the comparison will be by ear only.
So, one question is: What size cathode tail resistor would I need in order to use equal value plate resistors? I'm thinking that, because of the versatility of the PT, I may be able to come up with enough of a negative supply voltage to use a 'longer tail'. Any suggestions?
The other question: On the Eico schematic, the grid of the AF Amp section of the 6SN7 is 120v which is 10v higher than the (110v) grid of the Inverter section. Why is this?
On the LTP channel I'd like to take mikeyb's advice and adjust the voltages downward so I can use a single 6SN7 per channel with no input stage preceding it. Then, to preserve the operating points from the Eico, I'd have 0v on the grid, 180v on the plate and 5v on the cathode (by using whatever combination of negative supply voltage and cathode resistor that would preserve the 7ma of total current).
If I could accomplish this, how would I keep the grid of the inverter portion 10v higher than the other grid? Or would this be necessary/important?
Or would an entirely different set of 6SN7 operating points be better? Suggestions?
I've got a few more holes to drill on the chassis and then, if it quits raining long enough, I'll put some paint on the top of the chassis and start putting it all together.
Thanks . . . Charlie
You can set up a negative supply from most transformers with a seperate rectifier, creating a seperate leg of the supply. I would consider it as you'll have to have a substantial B- for the bias of the output tubes.
You are not stuck with the plate resistors set at 2X the plate resistance- you can go a lot higher. I would recommend though that you will want some current if you want bandwidth, so 47K might be a nice value to work with.
The cathode resistor can be a bit tricky, as with a scope, a sine generator and a means of changing the cathode resistance, a range of output gains can be seen. I have found that it is best to set the cathode resistance high enough that gain is optimized, and then just a little higher to allow for variance in tubes. This gives you the best differential (phase splitting) effect.
Effectively this often places the plate voltage about 1/2 of the supply voltage or about 10% higher, so the rest of the voltage will be dropped across the plate resistor. So the cathode resistor will be chosen to accomplish that.
Looking at the Eico schematic, I think the 10V difference must be a typo. The way the circuit is designed I would expect both grids to be at the same potential.If you're interested in a LTP with a negative supply, look at what Poinz has done with his Musical Machine: http://www.audiotropic.net/Projects/machine1.html
The schematic is above. That basic circuit is a good starting point, though with different tubes the component values will be different....
The size of the tail resistor is easy to calculate if you know how much negative voltage you have and how much current you want to run through the pair of tubes. Just use Ohm's Law: R = V/I
Don't know how much current you want to run? Draw a loadline.
Don't know how to draw a loadline? Read some books. I suppose I could write a tutorial for you, but it's really beyond the scope of an asylum post.
I got my first introduction to tube electronics from the intro pages of an old RCA Receiving Tube Manual. I think Pete Millet has a copy for download on his site as well as a whole selection of other texts.
EDIT: Probably the most important thing for you to do right now is to just get something up and running. Don't worry about it being the best possible design; it won't be. Even if it is, you will (should) want to change it.
As far as I can tell, there are two ways to build something that will work. (1) Copy an existing design verbatim, or (2) if you modify an existing design, make sure you understand how the design works and understand how your modifications will change things. If you don't understand the design first, then you have some work to do before you start making changes.
Consider a LTP with both input grids driven with equal but opposite phase signals. In that case both tubes would be operating much as they would in an ordinary grounded cathode arraignment (with the same operating point, load, etc.) In particular, the voltage gain from grid to plate would be the same.
If that gain happened to be 20 and each grid sees a 2VAC signal WRT ground, then each plate would swing 40V WRT ground. Measured from grid-to-grid the input signal would be 4V and the output would be 80V plate-to-plate. It so happens that the residual signal voltage at the cathodes would be zero in the ideal case.
Now consider driving only one grid with 2V. Since the other grid sees zero AC, the grid-to-grid signal is now only 2Vpp; half of the 4Vpp in the first example. Just by viewing the diff amp as a black box, without any knowledge of how it works internally, we would expect the output to be 1/2 what it was in the first example. If the first example had stage gain equal to the tube gain, this example must have 1/2 as much.
By comparison, the paraphase provides full 'tube' gain with one (its only) input driven.
-- Dave
The circuit I ran was more like the ideal circuit that you posted, with a constant current source and running B-.
One grid was driven only.
The circuit that ran your tests on had no CCS. It seems that the thing to learn is how important the CCS can be; in fact we have found that a 2-stage circuit is required in order to optimize the performance of the diff amp. Since you posted a circuit with a CCS, it is fair to use an example with an optimized CCS.
What we found from examples with and without the CCS, but otherwise employing a B- equal in magnitude to the B+ is that gain is improved and distortion is reduced. In the examples I have given (both with B- and B+ of equal magnitude), the overall gain of the circuit with one grid driven is certainly nearly the same as the same gain stage executed single-ended.
Operating the diff amp with less than an optimal CCS and also lesser magnitudes of B- definitely degrade gain performance. Its my guess (since we've not messed with such circuits) that it is probably possible to approach the 1/2 gain you describe. There is no need for the driver circuit in this particular case to operate at that level of degradation, as to set up the 6B4s requires a substantial 'B-' voltage for the bias network. This can be used for the B- of the diff amp as well; we make such a circuit for a guitar amplifier that we produce. It has substantial gain.
So it would seem that 'what we learn' from this exercise is that sub-optimally set up circuitry will also perform with sub-optimal characteristics. I did in fact know that before this, but did not know how substantially the gain of a diff amp could be decreased on this account.
I have to admit that your findings RE:gain are mystifying. If they are real and true they apparently violate the laws of physics!
Anyway, since you don't want to believe my words I was tempted to refer to higher authorities. In other words, start quoting respected texts etc, reproducing the relevant parts here. But I really don't like doing that.
What I did instead was put together a circuit in Spice. Really, two circuits. One is an ideal LTP with CCS, perfectly matched tubes and plate loads etc. The other circuit is a simple grounded cathode stage with the same tube, operating point, load, etc.
Note than in Spice CCS's are perfect; they have no dropout voltage or compliance issues. They need no negative supply.
Let's look at the outputs when 2 Vpp is applied to the grids.
It's pretty easy to see that the grounded cathode output (green) is about 31 Vpp while the LTP output (blue) is only half that at little over 15 Vpp. All of this agree with what I expect from a tube with a mu of 20.
Let's look at the input signal and the signal at the shared cathodes.
The 2 Vpp input signal clearly results in a 1 Vpp signal at the connected cathodes.
Now, it's true that Spice mimics reality only within certain limits, but the fact that the results are exactly what I know to expect leaves me feeling that we are not flirting with those limits. Honestly Ralph, if you want to insist that you are seeing something completely different on your bench, I'll let you do that. But you should know that I don't believe it.
-- Dave
Dave, I want you to understand that I am not here just to make you wrong! I always prefer to get to the bottom of it (even if I have to eat crow), and I think I may have found it- I did my measurements using a 10:1 probe. *That* puts the measurement within the range of your comments.
Crow is always better while still warm :)
FWIW, I maintain that the differential driver is the more elegant way to go, as you don't have to run feedback (which cuts gain) to obtain low distortion. This is nice when you are working with a tube like the 6B4; run in class A push pull the load lines are a thing of beauty.
Hey, getting to the bottom is my goal too. It's too bad that these things start to feel personal after a while. I try my best to avoid that, but I admit that I am not completely immune.
Anyway, I'm glad that it's settled.
And, I agree (and always have) that the LTP does seem more elegant.
-- Dave
So are you saying that if I use the paraphase circuit shown above with the PP 6B4Gs I will have to use feedback?
Check out Pete Millet's 6B4G circuit at the link. How would you characterize this circuit?
The cathode resistors on the 6SN7s are much smaller than those in Eico HF-87 I posted. It does not use a negative voltage supply and it does not use feedback.
Pete's circuit has half of V4 operated as a voltage amp, which is directed coupled to the other half, which is operated as a split-load invertor. This alone is enough to drive tubes like a 6v6 or el84 but not a triode like a 6b4g. The split-load invertor is cap coupled to V5 which is operated as a push-pull amp. Its plate are cap coupled to the 6b4g grids and drives the output tubes.
Notice one other big (and important) difference. V4's first section is running at 3.6ma, its second section at 4.5ma and V5 is running both sections at 6ma. The schematic you have posted is running about 1.3ma. This is important because it takes current to make a voltage change in the face of capacitance. The grids of the 6b4g have a good deal more capacitance the the grid of a pentode like a 6v6, el84, el34, 6l6, etc. So if you want to use a vintage schematic look for one that uses 6b4g/2a3 as output tubes and not a pentode/beamer.
If it was me I might make the wattage higher on some of the resistors in Pete's schematic. Of course I'd also suggest you take a look at the link I provided above under the "Try looking here" post I made.
Russ
I'm a fan of keeping things simple. The paraphase has somewhat more distortion than the differential amplifier and if you look at the original RCA schematic, the feedback is drawn in, coming into the cathode of V5.
I'm not a fan of feedback either- it compounds odd ordered harmonics in the range that the human ear uses as loudness cues, IOW humans are very sensitive to the resulting odd ordered enhancement, and describe it as a 'sheen' or 'hardness.
So I would run the amp zero feedback if I could. This will be a snap if you keep the output section in class A.
A single stage of gain, direct-coupled to a differential driver that employs a cathode resistor tied to B-, will give you more than sufficient gain to easily clip the power tubes with any preamp. Three audio-quality film caps would be needed, 2 for coupling caps between the driver and power tubes and the 3rd to bypass the 2nd grid of the driver stage. Overall pretty simple, and low distortion for a zero feedback amplifier.
Pete Millet's amp is not bad but seems overly complex to me, but as I said I'm a minimalist.
Now that we can all agree on (I hope)! Might be nice if we could put the paraphase in spice too and see how things compare if tubes are "unmatched". The floating paraphase might end up doing better.
I know some guys don't like this forum as they feel we fight/argue too much. Personally I love it cuz I think we try hard to help each other reach a real understanding.
Might as well give me some of the crow too. After all I was the guy knocking the paraphase first and didn't see the point Dave Cigna was making at first.
Russ
Russ, if absolute balance is a priority, then it's hard to beat the LTP with a CCS in the tail. So long as the load resistors are matched so is the balance. Kinda like a concertina in that respect. Mismatched tubes should not affect the balance in any way, except perhaps to the extent that they present different capacitance to ground at their plates, but that's probably just academic in any real case. (W.T. Cocking wrote a five part series on phase splitters published in Wireless World back in 1948. He invested three or four pages analyzing the imbalance in signal that results from an imbalance of capacitance in the LTP. I had to ask myself why he went to the trouble. But, he considered the circuit unsuitable for audio because of the 1/2 gain thing. He was promoting it mainly as an vertical deflection amplifier for oscilloscopes where presumably precision at high frequencies is important.)
It's tempting to think that with mismatched tubes the impedances at the two outputs would be mismatched. I'm not sure that even that is true, considering the concertina and the fact that both of its outputs are balanced despite the fact that one comes off the cathode and the other off the plate.
I might get around to Spicing the paraphase. I admit that despite the seeming inelegance it can be made to sound good. Years ago, when I was experimenting with PP amps (EL84 and 6V6) I tried all of the common phase splitters. They all sounded good, especially compared to the SS Harmon-Kardon receiver that I had been using for years and felt pretty good about up until then. But, the paraphase seemed to give more of whatever was good about tube amps. The music came out of my little mini-monitor bookshelf speakers bigger with more dimension. I later found that SE amps, and then SE directly heated triodes, could give me even more of that. The rest is history as they say, and I haven't gone back to investigating PP amps very much. Lately I've been feeling the urge to retrace my steps, so I might take another look.
SwitcherCad/LTSpice is very easy to use. If you're really interested in modeling then you should download it and give it a whirl. Steve Bench has put together a few tips (not quite tutorials) on using it for practical modeling of tube circuits at the intact audio webs site.
-- Dave
Yes, I do need to drag myself out of the dark ages and learn it. Mr. Slage encouraged me too, but I chickened out. Too many irons in the fire thang. Maybe I will have a chance in the coming days. I'll need the right mix of ass kicking and encouragement. In particular, I don't really listen/learn until I understand how things work/interact on a root level. Looking at the spice programming I see I don't have a clue on that part.
We will talk about the balance if tubes were badly mis-matched when I make it over yonder, which will be at best a few days.
Russ
Russ,the beauty of LT spice is it is really easy to use. If you have a working design you learn by playing and asking just like modding a ST-70. you are welcome to anything i have and i'm sure others will share too. I have found it to be a very valuable teaching tool. I used to wonder what would happen if I tried something, and now i can sim it and see function prior to building it.
Just do it... Steve bench has a nice little jump start to get you going. The below link will get you started, and then read the rest of his posts in the section as you climb the curve. You will hit some quirks, but all you need to do is ask and someone will certainly help out.
dave
dave
Guys - So I'm still wondering if it's possible to accomplish this:
Run PP 6B4G in Class A (no feedback), cathode bias, using only four 2C22s (equivalent to one 6SN7 per channel) ? Keep in mind that I will be using a preamp that provides gain, not just a passive.
Paraphase or LTP with negative supply is fine with me. I don't want to use an interstage transformer since I'm trying to keep the cost down. I see how Poinz' LTP is built using a negative supply. And mikeyb explained how I could use the Eico HF-87 schematic and adjust the voltages downward to keep the same operating points by using a negative supply. And I did check out the PP1 schematic that Russ pointed me to which uses a CCS. My understanding is that a negative supply and a CCS are two methods of accomplishing the same goal. Correct?
Russ' comments help me understand what I need. . . I think. Are my interpretations correct?
"Pete's circuit has half of V4 operated as a voltage amp, which is directed coupled (actually, isn't it cap coupled? see C8) to the other half, which is operated as a split-load invertor. This alone is enough to drive tubes like a 6v6 or el84 but not a triode like a 6b4g. The split-load invertor is cap coupled to V5 which is operated as a push-pull amp."
So Pete's circuit uses V5 to provide additional gain. Will my use of a preamp be able to make up for the proposed elimination of this stage?
Also: "Notice one other big (and important) difference. V4's first section is running at 3.6ma (actually 3.0 I think - see R13), its second section at 4.5ma and V5 is running both sections at 6ma. The schematic you have posted is running about 1.3ma. This is important because it takes current to make a voltage change in the face of capacitance. The grids of the 6b4g have a good deal more capacitance the the grid of a pentode"
So I need more current than the 1.3ma provided by the paraphase. How much is needed for PP 6B4Gs in Class A? Is the current sent to the outputs the sum of the currents from those previous stages?
Can this be provided by the four 2C22/two 6SN7? In the Eico HF-87 schematic I posted the single 6SN7 per channel is running about 7ma. Is 7ma enough?
I've learned a lot from this discussion and I can build from a schematic but, frankly, I get confused by some of the math and charts. Ohms law I understand and I've read about drawing load lines in my tube manual and some online but it still confuses me. I'll keep trying but math is definitely not my forte.
So, is it possible? Can anyone suggest some decent operating points?
Thanks . . . Charlie
Charlie, here's a sample loadline analysis using a 6SN7's to drive 6B4's. The assumptions that I've made are that the 6B4's are biased with about -45V on their grids (relative to the cathodes) and that a supply of about 300V is available for the 6SN7's
The loadline in red is 47k ohms. One end is at the supply voltage of 300V and the other end is at 300/47k = 6.4mA. I chose an operating point of 3.5mA which puts the plate of the tube at about 135V. That's indicated in blue. The difference between supply and plate voltage (300-135=165V) is dropped across the 47k plate resistor. It all checks out with Ohm's Law: V=I*R=0.0035*47000=164.5. Close enough...
Green indicates the part of the loadline that will be traversed when driving the 6B4's to full power. Since the 6B4's are biased at -45V the plates of the 6SN7's will need to swing +/- 45V peak. The quiescent plate voltage is 135V, it will need to swing up 45 to 180V and down 45V to 90V. What is also shown in green, but is hard to see, are the corresponding 6SN7 grid voltages (really, grid to cathode voltages.) Just reading them off the graph, the quiescent grid voltage is about -4.5 and it needs to swing between about -1.9 and - 7.5 to drive the 6B4's fully.
Ok, that means you'll need about 3 volts peak input to the 6SN7, or twice that at about 6 volts peak is they are configured as a LTP. If you have a preamp that can deliver that, then you're golden. If you don't need full power, then you might get along just fine with less input signal.
-- Dave
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