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In Reply to: RE: You don't wanna use that posted by Russ57 on July 31, 2007 at 16:45:45
Hi Russ - Educate me please. The only differences I notice from one side to the other are the 30 pf cap on pin 5 and the additional 27k resistor. I've seen other schematics that use the two 270k resistors only and I wondered why that 27k would be there.
Is this what makes it a paraphase? Would removing one or both of those parts make it more suitable for use with NOS PP 6B4Gs? Is there a circuit that uses a single 6SN7 per channel that would be better?
Any idea how much input voltage it would take to get full output with this?
Follow Ups:
Signal enters the grid of V5A on pin 4 and the amplified, and inverted, signal exits C16 and goes to drive V6's grid . R29 and R30 form a voltage divider so that we "pick off" a part of that signal and route it to the grid, pin 1, of V5B. V5B amplifies that and also inverts it so that V7's grid, through C17, sees a mirror image of the same amplitude as V6's did.
But notice that only V7's input signal has gone through V5B and balance depends upon the sections of the 6SN7.
Oh, the 30pF cap "rolls off the highs" of the V5A. Typically it is sized to work with the feedback loop. Don't worry about it (now) as it would depend on the parts used as well as the "layout".
Take your time and I understand all this may be Greek to you. I'd encourage you to look up "long tailed pair" and "Hedge cascode". As they say, measure twice and cut once:) I have heard the Seth, which is mentioned above, and it is another way to skin the cat. Sadly, there is no right/best way in audio. At best one can say something is perhaps "the wrong way" and even that is open to debate (often). I know this is frustrating and all I can offer is that many times more is learned by failure than success. Big help huh:)
Russ
It's an Eico HF-87, an amp that I really like. I'm referring to the 6SN7 section only. Is this a better alternative to paraphase?
I won't have 420v but the plates only get 300v which is about what I should have for the B+. I could just leave out the plate resistors if these operating points are desirable. Correct? Or will/should the voltages on these be completely different since there won't be a 12AX7 stage?
I'd like to at least try using one 6SN7 per channel because, if it works out well, I can use some 2C22s that I just got and I've heard they are supposedly even better than 6SN7s. That's why I'm being so pig-headed about this :^) So I'm trying to find the optimum operating points and phase inversion scheme for that scenario. I know Pete Millet is a sharp guy and, if the one 6SN7/two 2C22 per channel setup doesn't cut it I can always build it according to his schematic.
Comments/Suggestions??
Thanks . . . Charlie
Well, yes, IMO. This is basically a 'long tailed pair', or a differential amplifier (cathode coupled) to my way of thinking. Notice that the plate resistors in this circuit have a much lower value (33K versus 100K) than in the other circuit - this will lower the output impedance, which is a good thing when driving a DHT. My problem with the paraphase splitter is that it's balance is highly dependent upon the mu of V5B. V5Bs gain must be precisely 11 in order to achieve balanced outputs (the ratio of it's input divider is R30/(R29+R30). Not to mention that the output of V5 will always be delayed in time relative to the output of V5A.
There was an excellent article about the relative merits of various phase splitter designs (I can't find it now), and it seems to me that the paraphase splitter was rated dead last.
By the way, a dissadvantage of the Eico LTP circuit is that it requires a 'pre driver' - the ECC83. Notice that the grids of the 6SN7 are DC coupled to the plate of the ECC83, and sit at +120 volts.
You say: "Notice that the plate resistors in this circuit have a much lower value (33K versus 100K) than in the other circuit - this will lower the output impedance, which is a good thing when driving a DHT".
If I only have about 300v B+, can I omit the plate resistors entirely or put in a much lower nominal value (like 1k) and still have the lower output impedance? Or does this create some other problem?
Also: "a dissadvantage of the Eico LTP circuit is that it requires a 'pre driver' - the ECC83. Notice that the grids of the 6SN7 are DC coupled to the plate of the ECC83, and sit at +120 volts."
Yes and I'd like to eliminate that tube. Is this possible? Does the 120v have to come from a 'pre-driver' tube or can it be supplied from a cap in the PS and have the input signal come from the normal RCA jacks? Maybe (probably!) a stupid question, but I'm trying to learn :^)
Thanks . . . Charlie
"can I omit the plate resistors entirely or put in a much lower nominal value (like 1k) and still have the lower output impedance? Or does this create some other problem?"
No, you cant't do either one. The plate resistor is part of the 'load' seen by the 6SN7s, and shouldn't be set much below a value of 2*Rp (about 15K Ohms).
"Yes and I'd like to eliminate that tube. Is this possible? Does the 120v have to come from a 'pre-driver' tube or can it be supplied from a cap in the PS and have the input signal come from the normal RCA jacks?"
It is possible to eliminate the predriver, but the grid to plate voltage and grid to cathode voltages in the LTP should remain the same. With a 300 volt B+ supply, one alternative would be to connect the plate resistors to +300, tie the grids to ground via a 220K resistor, and move the resistor in the cathodes of the 6SN7s to -115 volts. Yes, that would require an adiitional supply, but that would also allow you to use fixed bias on the 6B4s.
So I need at least a 15k load on each plate.
Then you said: "It is possible to eliminate the predriver, but the grid to plate voltage and grid to cathode voltages in the LTP should remain the same. With a 300 volt B+ supply, one alternative would be to connect the plate resistors to +300" . . .
The same voltages or the same ratio? If I have a B+ of 300v and 15k (or more) on each plate my voltage at the plate will obviously be less than 300v.
". . . tie the grids to ground via a 220K resistor" OK, that I get.
Then: ". . . and move the resistor in the cathodes of the 6SN7s to -115 volts."
I'm confused here. I thought the goal was to keep the voltages the same, why would I need to change the cathode voltage from +125v to a -115v? If that's what's needed though, what size resistor would I need?
And: " Yes, that would require an adiitional supply, but that would also allow you to use fixed bias on the 6B4s."
So just another R-C section or does this supply need to be negative voltage? Where would it be connected? to the grids? Also, would this require the use of fixed bias or could I still use cathode bias as planned?
Also :^) . . . regarding Dave Cigna's defense of the paraphase below, if the LTP has only half as much gain, how much of a problem is using a LTP likely to create since I'm already eliminating one gain stage by using only one 6SN7/two 2C22s per channel?
Thanks for all your collective wisdom and (especially!) patience guys! . . . Charlie
Ok I understand all that. Thanks for the clear explanation! But what about the gain issue which was raised? If this does produce half the gain of the paraphase version, will this be an issue considering that I've removed a stage?
. . . . Charlie
Try the PP-1C schematic for a "concept". It is, of course, only one of many possible ways to go.
Russ
Actually, one of the things that I like about the LTP (differential amplifier) is that it is fairly easy to understand. Referring to your original schematic - the 18K resistor that is common to both cathodes has 125 volts across it, so the current through it is 7 mA. This is split equally (in theory) between the two 6SN7 triodes, so the current through each plate resistor is 3.5 mA. So the drop across the 33K is about 115 volts, which explains why the plate voltage is 420 - 115 = 305 volts. But the grids are at potential of +120 volts, so the plate is about 180 volts higher than the grid. If we move the grids to a zero volts and the supply to 300 volts and leave the plate resistors and current the same, the plates will be at 185 volts and the grid to plate voltage will be the same - 185 volts.
In the proposed design, we need to get about 125 volts across the 18K resistor common to the cathodes. The grids will be about 5 volts negative with respect to the cathode, or in other words, the cathodes will be 5 volts positive with respect to the grids (0 volts). So we need to connect the bottom (previously grounded) end of this resistor to -120.
“So just another R-C section or does this supply need to be negative voltage?”
It needs to be a negative supply, but only needs to supply about 7 mA of current.
“Also, would this require the use of fixed bias or could I still use cathode bias as planned?”
You would not have to go to fixed bias, but it would be a possibility.
Ok, following your suggestion, the grids have been dropped to 0v and the difference (180v) between the original grid (120v) and plate (300v) has been roughly maintained by dropping the supply voltage to 300v and keeping the 33k plate resistor. Actually 295v supply would be exact and is the B+ in Pete Millet's schematic.
We now want to have the cathode at 5v (+5v relative to the grid and -175v relative to the plate). Couldn't this be accomplished by simply replacing the 18k resistor with a 720 ohm resistor?
Or am I missing something here?
. . . . Charlie
Your math is correct but, unfotunately, there is an additional requirement. Since the coupling between the stages is via the cathode, the resistor in the cathode is a 'load'. The term 'long tail' reflects the requirement that this resistor (the 'tail' of the differential amplifier) be 'long', i.e., have a high value. In fact, the lower the value of the cathode resistor, the poorer the signal balance between halves. This is because the circuit relies on the cathode resistor being a 'constant current' source. The higher the resistance, the closer to a 'constant current source it becomes.
The article below shows an alternative approach that does not require a negative voltage supply. By the way, the 6SL7 will provide plenty of gain for your application. For an even better tube (the 6SL7 is an excellent sounding tube), try the 6188.
if the output of the inverted side of the paraphase is delayed, how can the output of the LTP not be? Can't the LTP be viewed as one tube cathode driving another?
in any event the non inverting signal traverses only one tube but the inverting signal must traverse two. How is that materially different than what happens in the paraphase?
dave
"Can't the LTP be viewed as one tube cathode driving another?"
Absolutely true. But the cathode has a lower driving impedance than the plate. In the paraphase case, the grid of the second tube is being driven from the plate of the preceeding tube. Assuming that the capacitance of the cathode is the same as the grid, both bandwidth and phase shift are improved by having the lower drive impedance of the cathode coupling.
"How is that materially different than what happens in the paraphase?"
Other than the impedances, it isn't. But driving impedance does effect bandwidth and phase shift.
I hear what you're saying, but I'm not sure I buy it. At least, I don't think the impedance issue is serious. The image above is the same one that FlaCharlie put in his first post. Look at R30; it's only 27k. That is essentially the impedance that the inverting stage sees at its grid. That should be plenty low enough for a 6SN7 grid (without doing the math.)
In any case, I think the paraphase inverter has gotten a bad rap over the years simply because it looks inelegant on paper. The fact is, it can sound very good. It just *looks* clunky and awkward on paper, so everyone passes it over for more technically elegant solutions. Technically, it has two advantages over the LTP: (1) it has twice the voltage gain, and (2) no supply voltage is 'wasted' in a tail resistor. It seems that both of those advantages might be important in FlaCharlie's situation.
It is true that the paraphase is trickier to balance than a LTP (and a LTP with a 'short' tail resistor is not as balanced as we might like to believe), but my personal opinion is that perfect symmetry is way down on the list of things that make an amp sound good anyway.
-- Dave
The 'LTP' (an antiquated term at best) can be set up very nicely in 6B4/2A3 amplifiers as the bias voltage on the power tubes has to be rather negative due to the low Mu of the tube. You can use the same supply for the bias as the cathode resistor of the differential amplifier. Its just a matter of working with the divider networks.
Makes for a lot simpler circuit, and you can get the same gain as the rat's nest driver in the RCA schematic.
"The 'LTP' (an antiquated term at best)... Makes for a lot simpler circuit, and you can get the same gain as the rat's nest driver in the RCA schematic."
Not with the same tubes and operating points. The LTP will have 1/2 the voltage gain. If you have trouble with this, read up on how differential amplifiers work.
-- Dave
Maybe you should try building up the two and see for yourself. The RCA driver has a certain amount of gain and you can set up a differential driver to have exactly the same amount of gain- not half. The RCA driver has a propagation delay between the non-inverted and inverted phases; the differential amplifier has none.
FWIW I've been doing differential amplifiers for years- follow the links associated with my moniker and you will see.
Differential drivers have the added benefit of having greater power supply noise immunity FWIW. I really can't think of a reason to use the RCA driver.
That's the same circuit that I put in my post above. Look at the AC signals indicated on the diagram. A 2VAC input signal results in 1VAC at the shared cathode connection. That's how a LTP works when one grid is driven and the other is grounded (either connected directly to ground as in this case or AC grounded through a capacitor.)
If you recognize and accept the AC signals at the points shown on the diagram - both their magnitude and phase as depicted by the sine wave inside the circles - then look carefully at the signal between grid and cathode of each tube. I don't need to remind anyone that that is the signal that actually gets amplified. For V1, Vgk is 1/2 the input signal. The 1V at the cathode effectively is subtracted from the 2V at the grid. It sees only 1/2 the input signal and consequently only 1/2 the output signal appears at its plate. In other words, the circuit has only 1/2 of the voltage gain that the same tube would have if it were in an ordinary grounded cathode stage (same operating point, etc.)
In the drawing above both V1 and V2 have voltage gains of 20; 1VAC between grid and cathode results in 20VAC at the plates. But, the circuit, with only one grid driven, has a voltage gain of only 20/2 = 10.
-- Dave
sorry about that.
If you had the 2nd grid float though, you could direct couple the input. I should point out that the 1/2 voltage idea assumes that only one output is used. Both outputs are used- the gain works out the same. This is particularly useful if everything downstream is differential also, or at least push-pull, which in this case, it is.
"I should point out that the 1/2 voltage idea assumes that only one output is used. Both outputs are used- the gain works out the same."
You mean if you measure between the outputs rather than from one output to ground? Even so, I don't see how it changes anything. All else being equal, the LTP will have 1/2 the gain of the paraphase if you measure both by the same method.
So I did some measurements. The problem is that the *signal* voltage on the cathode of the tube is likely to be about 1/20th that of the grid! I did this measuring actual signal on one of our MA-1s, which use differential amplifiers with a very effective vacuum-tube CCS.
The other half of the diff amp gets its signal not so much from the cathode, but *the effect that that cathode has* on the total current through *both* tubes, which is constant.
So the effect is that for all intents and purposes, a differential amplifier has if anything, very slightly less gain (certainly *way* more than half). With a higher mu tube the gain gets closer to being identical.
I recall measuring this about 10 years ago, so I was a little hazy on it and just accepting my recollection. Hard measurements work better.
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
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
So you're saying that the biggest drawback is that the paraphase requires the two triode sections to be closely balanced and that, if they aren't, some 'smearing' of the sound would occur. Correct?
actually the seth has one half of the signal cap coupled through a grid choke (call it a magnetic load) and the other half cap coupled through an inverting interstage transformer. So one half of the signal relies magnetic coupling and the other half doesnt.
dave
The Iron Concertina? I was always curious to give it a try. Perhaps you have heard it?
Russ
That looks like a too cool circuit! What type of iron is the interstage?T Willman
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