Of course you have to account for required swing and the CCS "overhead," but yes, that's a MAJOR advantage of CCS plate loads. I'm a bit uncertain as to why you need that extra resistor.
A few months ago, I breadboarded a version of an LTP with a cascode CCS in the cathodes and simple single transistor CCS plate loads. I was surprised to find that with a little bit of adjustment to the cathode CCS, the plate voltages snapped into place very readily and were quite stable. I had expected it to be much twitchier.
A few months ago, I breadboarded a version of an LTP with a cascode CCS in the cathodes and simple single transistor CCS plate loads. I was surprised to find that with a little bit of adjustment to the cathode CCS, the plate voltages snapped into place very readily and were quite stable. I had expected it to be much twitchier.
Sorry I still do not understand. In a cathode follower, to my understanding, the cathode voltage follows the grid because a change in the grid will induce change of plate current (following the gm of the tube) this will of course cause the cathode voltage to follow due to the change in voltage drop over the load in the tail.
But if we force the current to be constant, what is actually determining the voltage at the cathode?
PeteN said:...An amp should be as close as possible like a lens - totally clear and flawless as possible, as opposed to a coloured filter or a fairground mirror that makes you look really skinny!
(Admittedly using inherently imperfect components makes this a bit of a lofty goal)
That "totally clear and flawless as possible" magnifying glass, that you are describing, would be indeed preferable if you are looking into something real, namely some object or a natural phenomenon. But if you are looking into a printed photography you will observe a grainy structure, because of medium imperfections and technology limitation. In the end, you would be just magnifying noise and picture imperfections with your "flawless as possible" lense, so some kind of blurring (smoothing) would be in order, because "foggy" and "blurry" are always more natural and pleasant to look at then "grainy" or "pixelated".
You are overlooking an important fact that recorded audio is already distorted and inherently unnatural. It all begins with microphones, which all have their own dynamic range (limitation), self-noise, frequency response, etc... And don't get me even started with miking techniques and ambient acoustics... Then, audio engineers start meddling with those signals (be it analog or digital)... mixing them and applying equalization, dynamic range compression, noise reduction and other techniques in order to sonicly and artistically enhance music performance and circumvent channel (medium) restrictions.
So next time you think about "perfect amp", remind yourself of what exactly does it amplifies. How about straight amplification of phono signal, without RIAA equalization? I don't think so... Is the process of preemphasis/deemphasis used in vinyl records a "flawless" (lossless) one? Nowhere near! Does it makes music sound bad?
The other important and often neglected aspect of audio arts is a psycho-acoustic model of human sound perception. Our hearing apparatus and brain are nothing like measuring equipment. Not only that our brain will mask certain low level sounds in presence of loud ones, but it will also produce (try to reconstruct) some sounds even when they are not present in the source/signal. Should we account for that when we are designing audio equipment or should we stubbornly claim that our rig is "totally clear and flawless as possible", according to measurements, and that our ears and brains are the ones that need to be corrected/redesigned?
The most perfect audio reproduction rig for a particular record, would probably be the one in a studio which mastered the record, because it is the one which influenced audio engineer's decisions to alter the "sound". The best way to hear a recorded music, would be to listen a master track on the same rig and in the same post-production room in which it was created, so we could hear it the way it is meant to be heard (and it still wouldn't be a 100% truthful representation of what happened in front of a mic). But that ain't gonna happen for like 99.9999% of us. Besides, high-end studio equipment costs orders of magnitude more then most audiophile rigs.
So... is there a short and less expensive way to a good sound reproduction? There certainly is, but it inevitably considers the type of music we like to listen (ideally specific tracks and performances) and our preferences about compromises that ought to be taken.
And I hope you realize that an ideal amplifier coupled to an ideal loudspeaker would never reproduce 100% accurate sound of guitar, piano, violin, drum, etc... because there was no ideal capture of their properties to begin with. For your philosophy to work, we would need to have ideal microphones and an ideal channel (information carrier), and it's never going to happen. The sooner you realize that, the sooner you'll begin to having fun with audio.
Please understand that I'm not preaching pseudo-magic rituals with chicken legs, magnetic coupling, choke loading and prehistoric DHT tubes, without any sane engineering backing... The point is, we still don't quite understand what is happening between our ears, and a right amount and type of distortion to a recorded music can indeed bring us one step closer to the "real thing", instead of driving us away as we might expect.
Sorry for the offtopic.
Regards,
Marcus
SY, it just seems a handy way to ensure that the plate voltage will be roughly what you want (doesn't have to be dead accurate), by making use of a deliberate mismatch between the tail current and the plate currents. You'll be struggling to match those currents precisely anyway, so rather than worry about the conflict between CCSs, why not make them deliberately unequal and then utilize the difference?
My thinking here is based partly on reading Morgan Jones's 'Valve Amplifiers', which goes some way to explaining it, and a helpful post from 316A earlier in this thread.
Magick, I don't think it's off-topic at all. What you say is essentially correct, though I'm not sure what conclusions can be drawn from it.
What is recorded is, of course, imperfect, for the reasons you state and for other reasons too, but what most of us are trying to do is to avoid making it any worse than it already is when we play it back. Your lens analogy rings true - the imperfections like distortion, background noise, clicks, hisses etc. in a recording, which may go unnoticed with less than ideal amp, speakers and listening room, can become clearly audible with better equipment and listening environment. A number of people have commented that they never realized how bad some of their recordings were until they heard them through a decent amp and speakers.
It's the nature of the people here to try to do the best we can in the areas we can control - the playback side of things - within the limitations of our knowledge, our ability and the amount of time and money we can devote to our hobby. We try to improve by learning from one another and by sharing our knowledge and ideas.
There is such a thing as the law of diminishing returns, and so there comes a point where any further improvement is infeasible/futile. But I personally have some way to go yet.
This is exactly what I don't understand. Well I do understand that with constant plate current the plate to cathode voltage is what changes, what I do not understand is why the cathode voltage is in any way forced to follow the grid with all currents constant.
Take the example by SY: a pair of tubes with a mu of 10 where the input tube is fed a +1V signal. If the only thing that happens is that the plate to cathode voltage changes on the input tube, what is then the reason that there should be a 0.5 V change in cathode voltage and 5V in the P to C voltage and not for example 0V change in the cathode and 10V in P to C voltage? In the latter case there is no driving of the second tube in the pair so this would not work...
I would acutally still think that with an ideal ccs in both plate and cathode circuit the tube itself is actually floating between the + and - rails and the only reason it still works is the imperfections of real world CCS's.
Sorry to bother you with this perhaps uninteresting discussion but the implications of my line of thought is that if you make the CCS in the plate circuits "too good" it will actually degrade the LTP as phase splitter and also its ablility to cancel even harmonics used as a diff amp since you are effectively isolating the two tubes from each other.
Sorry to be a little obstinate on this one
/Olof
hemgjord said:
I'll take a shot at this. Let's start by considering a concertina phase splitter. It has a gain of 1 because of the 100% feedback at the cathode. (OK, really the gain is slightly less than 1, but let's pretend it's 1 for the sake of discussion.) When the grid goes up 1V the cathode goes up 1V. Suppose we put a suitably large capacitor across the cathode resistor. Obviously, the voltage gain at the plate would be that of the tube when used grounded cathode while the gain at the cathode would be 0. That is the 0% feedback case
Now suppose we put a resistor in series with the bypass cap. Depending on the size of the resistor we would get somewhere between 0% and 100% feedback. Clearly there must be some value of resistance that will result in 50% feedback. In that case the voltage gain at the plate will be 50% that of the fully bypassed case. Meanwhile, the gain at the cathode will be 50% that of the completely unbypassed case; when the grid goes up 1V the cathode goes up 0.5V.
It turns out that the right value of resistance is exactly the same as the resistance you would see looking up into the cathode. (Claimed without proof. I suppose I could try to prove it, but I'd rather not try right now.
) You could figure what that resistance should be. (It might be fun to play around in spice.) Or, you could just use another tube of the same type with the same plate resistor. Connect the cathodes together and viola!, you have 50% feedback. The joined cathodes will follow the driven grid with a gain of 1/2. As a bonus, the second tube is now being driven by the first. The signal feeding the cathode of the second tube is in phase with the grid signal at the first. Further, the magnitude is the same but because one is feeding a grid and the other a cathode, one tube will be inverting while the other is not. If to happens to be useful then you can use the plate of the second tube as a second output and you have a phase splitter.
-- Dave
(edited for clarity)
Draw the Thevenin equivalent, assume cathode CCS is perfect (open circuit at AC), then either grind through the math to prove it to yourself or use a symmetry argument (for me, the more elegant way).
Here's yet another way of looking at it: the input tube with the 1V of signal has a cathode load that looks like the load impedance of the other tube (a big number) in series with rp, all divided by mu+1. So throwing in some numbers, let's say that the plate CCS has a dynamic impedance of 1M and the tube has an rp of 10k. Then the impedance that the cathode of the driven tube sees is that sum (which is slightly more than 1M) divided by mu+1 (21, for my example tube). That's a cathode load of 50k, more or less. OK, we apply a volt to the grid, the cathode has a finite load, so the input voltage appears across it because the tube doesn't "know" there's another section there, it just knows that it's got a 50k cathode load. So a portion of the signal appears across the cathode load (what proportion, we'll come back to). But since the other section grid is pinned to ground, it sees this cathode voltage as an opposite signal in its grid-to-cathode circuit, i.e., the cathode goes positive, the grid is grounded, so that's the same thing as the cathode being grounded and the grid going negative.
OK, then, we can see that some proportion of the input voltage is impressed on the common cathodes. Why is it 0.5V? That's because the source impedance of the driven section looking back into its cathode (50k in this example) is the same as the load impedance looking into the cathode of the tube with the grounded grid (also 50k). So... 0.5V, by symmetry.
Now, having a VERY tight cathode CCS insures balance even if the tube mus (and hence the cathode impedances) vary, as long as the plate loads are matched.
Here's yet another way of looking at it: the input tube with the 1V of signal has a cathode load that looks like the load impedance of the other tube (a big number) in series with rp, all divided by mu+1. So throwing in some numbers, let's say that the plate CCS has a dynamic impedance of 1M and the tube has an rp of 10k. Then the impedance that the cathode of the driven tube sees is that sum (which is slightly more than 1M) divided by mu+1 (21, for my example tube). That's a cathode load of 50k, more or less. OK, we apply a volt to the grid, the cathode has a finite load, so the input voltage appears across it because the tube doesn't "know" there's another section there, it just knows that it's got a 50k cathode load. So a portion of the signal appears across the cathode load (what proportion, we'll come back to). But since the other section grid is pinned to ground, it sees this cathode voltage as an opposite signal in its grid-to-cathode circuit, i.e., the cathode goes positive, the grid is grounded, so that's the same thing as the cathode being grounded and the grid going negative.
OK, then, we can see that some proportion of the input voltage is impressed on the common cathodes. Why is it 0.5V? That's because the source impedance of the driven section looking back into its cathode (50k in this example) is the same as the load impedance looking into the cathode of the tube with the grounded grid (also 50k). So... 0.5V, by symmetry.
Now, having a VERY tight cathode CCS insures balance even if the tube mus (and hence the cathode impedances) vary, as long as the plate loads are matched.
hence I totally missed out on the symmetry argument. Its late in Europe....
Interesting idea!! Morgan Jones presents a similar circuit in the third edition. It brought some discussion up here at diyaudio, but I can’t find the thread (was something with a CCS loaded concertina phase splitter)
I still ask myself about balance in this circuit. How bad is it to have a double triode with unmatched halves? And DC coupling to a following diffential pair, would it work? I ask because I imagine that the DC conditions of the plates are kind of prone to change, as those are the only ones that can change in voltage (grids are at 0V, cathodes are tied to each other). Still it would work nice when AC coupled…
Taking the information by 316a and the diyaudio CCS article I threw the following driver stage together. The main idea of this bunch of CCS’s is that changing two small resistors (Rset) allows one to set any desired current. I would use a PCB connector (the ones with screws) to allow choosing different values of Rset for different currents.
The circuit works as follows. The MJE340/BC559 and the LM317 CCS’s should be set for 5mA (the important thing is that they should be set for the same current, but 5mA seems nice, maybe a bit more as the LM317 minimum current is 5mA). This current develops a voltage drop over the Rset resistors. Subtract 0.6V of this voltage drop, divided it by the value of R* and one obtains the current through the MJE350 and the cascaded BC559 CCS’s. Using two paralleled R* in the lower CCS allows it to draw twice the current. The third resistor R** is of larger value, and used to draw that little bit extra of current (as recommended by 316a). Value should be R*x1/% extra current. What is a good value? 5% extra?
The advantage of this circuit is that it allows different settings with the change of two small resistors, allowing one to test different currents through the same tube, or different tubes, without having to match the plate resistors. All MJE’s can be mounted on a heatsink, to allow operation with larger currents.
This is just an idea of an implementation possibility, comments are welcome, as well as some discussion on the matter of balance with real world (mismatches) tubes.
Thanks for reading, Erik
Attachments
Your PNP transistors are upside-down.
I don't think it needs to be quite so complicated.
Imperfect matching of the two triodes making up the LTP can lead to two problems: unequal gain (AC imbalance), which the CCS in the tail will take care of, and unequal standing currents (DC imbalance), which is a ***** nuisance if you want direct-coupling to the next stage. DC imbalance means the plates won't be at the same voltage and this will be magnified by the next stage if it is directly coupled.
However, the disadvantages of direct coupling can be greatly reduced, and the benefits still retained, by use of a step network for the coupling. This will greatly reduce the effect of DC mismatch and will give a sensible voltage at the grid(s) of the following stage so you don't need crazy supply voltages. It also provides stability right down to DC, to assist in the application of negative feedback if you need it.
I have seen step networks alluded to as a 'desperate measure'. In this case, they're not - they are a good, practical solution.
I don't think it needs to be quite so complicated.
Imperfect matching of the two triodes making up the LTP can lead to two problems: unequal gain (AC imbalance), which the CCS in the tail will take care of, and unequal standing currents (DC imbalance), which is a ***** nuisance if you want direct-coupling to the next stage. DC imbalance means the plates won't be at the same voltage and this will be magnified by the next stage if it is directly coupled.
However, the disadvantages of direct coupling can be greatly reduced, and the benefits still retained, by use of a step network for the coupling. This will greatly reduce the effect of DC mismatch and will give a sensible voltage at the grid(s) of the following stage so you don't need crazy supply voltages. It also provides stability right down to DC, to assist in the application of negative feedback if you need it.
I have seen step networks alluded to as a 'desperate measure'. In this case, they're not - they are a good, practical solution.
ray_moth said:
Interesting to read about active loads on diff pairs. I have attached a sketch of a input/driver circuit for the YL1071 mid-range amp I spent a long time on and never finished. I tested this circuit and it swung lots of volts at low distortion; at +40dB out into a pad (around 30k) THD was 0.063%.
It was designed to be DC-coupled to the output stage. The second E810F is not shown but obviously would have to be there!
It's supposed to work as a thumbnail but doesn't - sorry
7N7
Hi Ray Moth
Thanks for replying. Yes it is complicated, and trusting Broskie's words: the perfect and the complex are enemy of the built. A pair of non-cascoded DN2540's or 10M45 at the top and a cascoded BC559C at the bottom would be easier!
More important (for me) is your input on the balance of the circuit. As I understand this circuit would perform well, even without having matched halves. Would have to try it out someday...
I dug up your thread on the amplifier with step networks: http://www.diyaudio.com/forums/showthread.php?threadid=83241&highlight= Do you still have the same setup? As I understand it the resistors between the plate of U12 and B- 1 set the DC operation point of the grid of U15, while the capacitor passes the AC signal. But how this influences stability in comparison with a capacitor to grid and a negative bias supply is beyound my capacity of understanding.
Erik
Thanks for replying. Yes it is complicated, and trusting Broskie's words: the perfect and the complex are enemy of the built. A pair of non-cascoded DN2540's or 10M45 at the top and a cascoded BC559C at the bottom would be easier!
More important (for me) is your input on the balance of the circuit. As I understand this circuit would perform well, even without having matched halves. Would have to try it out someday...
I dug up your thread on the amplifier with step networks: http://www.diyaudio.com/forums/showthread.php?threadid=83241&highlight= Do you still have the same setup? As I understand it the resistors between the plate of U12 and B- 1 set the DC operation point of the grid of U15, while the capacitor passes the AC signal. But how this influences stability in comparison with a capacitor to grid and a negative bias supply is beyound my capacity of understanding.
Erik
ray_moth said:
Any Tube preamplifier would perform better with CCS.
This will reduce the load ( current modulations ) of the tube.
And make it work more 'linear'.
In this respect there is no difference between transistor amps
and tube amps. Especially preamplifiers.
It is all about making the tube/transistor work in a more narrow bit of the curves.
And so the difference in transfer charateristics will be less.
It will work 'more in one point', one smaller area, of the diagram.
And at one point the transfer is always the same.
Even if no semiconductor can be made to work only in one x, y point.
Because if is no difference, there can be no AC amplification either.
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