On a Bode Plot (idealised frequency vs gain plot), a pole is where the response changes its slope negatively by 10dB per decade.
The grid capacitors thus introduce an LF pole in the loop gain.
What is important is that BallPencil's design has two coincident poles - LF roll-off corner frequencies if you prefer. So with only a tiny bit of phase shift somewhere else, the system is unstable. Otherwise it just has a huge LF peak.
True.
I assumed the pk-pk bit was a typo and he meant pk. That fits the discussion.
Koonw's pretty smart - he will be well aware of pk and pk-pk. It's all to easy to mistype in this forum. I do it all the time.
His 2 in the demonimator is harder to regard as a typo - its actually an error.
Yes, I mean p-k.
Now do you mean you want to do high voltage OTL with EC + forwards correction? Actually it sounds good not solely due low distortion but also due forwards correction (positive feedback). I do have some experience on OTL with positive feed for some form of EC in front stage, and using low mu tube like 6sn7, 5687.. If you have read my two threads, tell me your options. Maybe in another thread?
Yes. The client has requested an all-tube OTL design for 400 ohm speakers and a matching (to the degree possible) transformer design for 8 ohm speakers. He has 2 sets of Philips speaker systems identical except that one is 400 ohm and the other is 8 ohm. (Im hoping his 400 ohm tweeters are ok - none of mine are). I am requested to produce designs that match the best contempory solid state amps on test, to the extent reasonable possible (do my level best in other words).
Both amps will use feedforward error correction - different forms.
A misconception here - feedforward is not positive feedback. It is about deriving an output error signal (without the correction) by resistive mising, then amplifying it by a fixed amount, and subtracting it right at the output. Actually my design is slightly more devious than that.
I haven't read your other threads.
If you want to dicuss options, best to start another thread I think. Start the thread if you want and send me a private message so I don't miss it. I don't usually join threads unless they have about 4 answer or fewer, but if you ask me to join I will. Once I've joined a thread I usually continue with it. I am more likely to join a thread if I sense a newbie needing valid advice.
Plase bear in mind I cannot reveal all about any of my designs in diyAudio because they are commercial in confidence and adverse comments, justified or not, may affect the client's sales. If the client is happy, my two designs will be published in one of the electronics magazines, together with my and hopefully the client's views on whether it has been worthwhile doing an OTL.
Last edited:
I've now had a quick scan of one of your threads.
The technique of applying postive feedback aroud early stages is another way of reducing distortion. Quite different to what I'm doing, but just as valid.
To my knowlege, it was first written up Wireless World in the 1940's by someone in the editorial team.
Hmmm... Lower anode peak current for a given power output? Since we are talking class AB or class B here, the power at the peak of the waveform is entirely supplied by the one tube that is conducting, with the other tube essentially non-conducting. Surely the power is dictated entirely by the peak current that the one tube that is conducting is passing? So why would biasing to a lower quiescent current affect this in any material way? What matters, for a given power, is the anode current in the conducting tube at the peak.
Likewise, to achieve a given maximum power, the determining factor is to achieve the required peak anode current that achieves that power. That means achieving the sufficiently small |Vgk| (but hopefully Vgk is still negative) in order to reach that required peak anode current. It doesn't matter where the grid bias voltage starts from, if you want to reach, say, 2.5A peak anode current in order to achieve 25W rms into 8 ohms, you need Vgk to head sufficiently upwards in the positive direction (while still, hopefully, remaining negative) to achieve 2.5A anode current. The answer for what Vgk will achieve this is independent of how negative the grid bias voltage may have been in the quiescent state. Mutatis mutandis for whatever maximum power you are hoping to achieve.
Chris
Hi Keit,
Before i go about changing the idle current value, let's say, for the sake of argument, i agree this amp is Class B. Also, let's say the tubes actually put out 2A at Vgk=0. So, currently this amp is 16W Class B with idle current of 200mA. I know that if i increase the idle current to 1000mA, it becomes a 16W class A amp. So, is there a class AB somewhere between that? If yes, what would be the idle current value? i.e, by only changing the idle current, can this amp be a Class AB? Or is it stuck in Class B until it eventually becomes class A?
Refer to the definitions of Class AB and Class B. An amp is Class B if each tube is cutoff for approximately half a cycle. It doesn't matter how that comes about - either with a high standing bias voltage or by use of grid capacitors to raise bias voltage during signal.
Also recall that whether you should call and amplifier Class AB or Class B depends on what measues you need to take to get it to work properly.
If you were to increase the standing current to 1A, then (forgetting for momnet that teh tubes couldn't take it), you would have near zero bias voltage. So you would have extreme problems with grid current. The grid caps would charge up and bias the tubes back, so you'd end up with Class B (each tube cut-off for approx half a cycle just the same). It will not be a Class A amp just by dropping the bias, it will remain Class B.
With an idle current of 200 mA, for all but small signals your circuit does cut each tube off for approximately half a cycle, and the tubes draw grid current causing problems with the grid capacitors.
If you increase the standing current to say 400 mA (which would require approx half the bias voltage) the situation isn't really different. Because of the reduced bias, the tubes will be driven to grid current at lower drive levels, but as both tubes are conducting for a greater time, you actually need less drive at moderate sigbnal levels. At high levels, one tube is still cut off for approx half a cycle, and the amp is still Class B.
Essentailly, with 6C33C tubes feeding an 8 ohm load, you have a Class B amp regardless of wht the standing current might be.
To convert the amp to Class AB1, you would need to do one of three things:-
1. Use multiple 6C33C in parallel, say three, so the same signal grid drive gives three times the total anode current excursion. This will allow you to completely avoid grid current. It is the approach taken by the Chinese firm Bezdz.
2. Find an enormous triode that can dissipate 200W and with a gm of 130 mA/V
3. Increase the loudspeaker impeadance to 150 ohms.
All three methods enable AB1 operation.
Last edited:
At Vgk=0, the current is 2A.. so, with current of 1A, it's near zero bias? Surely there has to be enough bias voltage margin between 1A (near zero) and 2A (actual zero) although it's on "uncharted teritory". By my approximation as shown below, Vgk is still around -20 to -30 volts.
An externally hosted image should be here but it was not working when we last tested it.
Found this 6C33C Ep-Ip characteristic from Sanei Radio (Japan), which used to produce a kit for 6C33C. Please note that the Eg=0V curve was added by someone else, since the original Sanei data only went to Eg=-2V. In any case, use the data at your own risk...😉
An externally hosted image should be here but it was not working when we last tested it.
The point I was making is this: If grid current is a problem with an idle current of 200 mA, then if you bias for higher current, the bias voltage is going to be nearer zero. And if it's nearer zero then you have bigger problems with grid current and bias shift due to grid capacitor charge up.
BallPencil, it doesn't matter how much energy you invest in your own unique definitions and looking for loopholes. Your amp is a Class B amp, draws grid current, and suffers from grid shift.
You need to stop looking for loopholes, look at it with more self honesty, and adopt the measures you have been advised to take. Get the bias voltage up to near cutoff and it will still be Class B, but at least it will be a relatively good Class B. Get rid of those capacitors.
You didn't answer it. There is a Vgk gap of 20 to 30 volts at idle current of 1,000mA before Vgk goes positive causing grid current. I don't think this classifies as near zero bias. This gap is actually the grid drive required for the amp to reach 16W (at 2A peak current).
Not looking for loopholes. Not my own unique definitions, SY backed me up by saying it's Class AB (sorry for dragging you into this discussion again SY). I am ready to erase years of understanding of Amplifier Classes if i am convinced by your explanation. My number of years understanding obviously pales in comparison to your years but.. this is is irrelevant.
I would like to keep grid current away from the discussion, so let's say the power supply for the op-amp is adjusted so that the op-amp clips first just before Vgk goes positive, i.e whatever i feed the the amp, there will be no grid current. Hopefully this keeps grid current issue to rest.
Again, assuming at Vgk=0, Ic equals 2A at Vak=150V.
Then from idle current 200mA at Vgk=-47v (16W class B, per your definition), i then raise the idle current to 1,000mA at Vgk approx -25v. This should make it 16W class A, using the same logic that 200mA bias current means 0.63W class A.
So, is there class AB in there somewhere between 200mA (class B) and 1000mA (Class A)? If yes, what is the idle current?
Not looking for loopholes. Not my own unique definitions, SY backed me up by saying it's Class AB (sorry for dragging you into this discussion again SY). I am ready to erase years of understanding of Amplifier Classes if i am convinced by your explanation. My number of years understanding obviously pales in comparison to your years but.. this is is irrelevant.
I would like to keep grid current away from the discussion, so let's say the power supply for the op-amp is adjusted so that the op-amp clips first just before Vgk goes positive, i.e whatever i feed the the amp, there will be no grid current. Hopefully this keeps grid current issue to rest.
Again, assuming at Vgk=0, Ic equals 2A at Vak=150V.
Then from idle current 200mA at Vgk=-47v (16W class B, per your definition), i then raise the idle current to 1,000mA at Vgk approx -25v. This should make it 16W class A, using the same logic that 200mA bias current means 0.63W class A.
So, is there class AB in there somewhere between 200mA (class B) and 1000mA (Class A)? If yes, what is the idle current?
No disrespect meant for Sy, but finding someone to agree with you doesn't make you right. You need to look at the logic.
Years of misconceptions perhaps. According to definitons in standard textbooks, your amp is a Class B amp. I have explained why.
That's like being told your pet parrot's dead, and you say "I would like to keep death out of the discussion..."
The occurence of grid current is the crux of what's wrong with your amp. And easily prevented by simple changes, you don't seem to accept.
As I pointed out before, this will unnecessarily considerably reduce power output - in an amplifier that in any case has barely adequate power output.
As I explained before, it won't be Class A, because the low bias voltage will considerably increase grid current. The grid current will charge the grid capacitors, restoring the bias voltage and restoring operation to Class B.
With a -47V bias supply (200 mA idle current), you have under signal about 80% of the working bias comming from teh bias supply and about 20% comming from teh grid current.
With a -27V bias supply (1000 mA idle current), you will have the contribution of the bias supply droping to about 60%, and grid current now contributing 40%, with the total working bias much the same as before.
This ability of grid current to automatically regulate working bias has long been used in radio transmitters.
I already told you. No. Becuase at 1000 mA you woudn't get Class A. It's the effective working bias voltgae and conduction angle that counts, not the idle current.
If you want AB1, you have to do one of three things:-
1) Using three or mor 6C33C in parallel each side;
-or-
2) Find a MUCH bigger tube;
-or-
3) Use a 150 ohm speaker system. This will also dramatically increase power output.
May I point out you are being rather silly. You have made this thread some sort of debate, looking for loopholes. You should instead use this as a learning exercise, look at it objectively, and learn how to do OTL right. If you want to persist with 2 x 6C33C and 8 ohm speakers, say to contain costs, fine. But you then need to recognise that it can then only ever be Class B, and you need to design it on a Class B basis. That means -70V bias and no grid caps.
Last edited:
Exactly.
Okay, i will not carry on debating this and will sadly have to end this by saying: let's agree to disagree.
I tried adjusting the bias in a Spice simulation of a 6C33C OTL this morning, setting the quiescent current to about 1A. (Of course, in the spirit of the previous posts above, we agree to draw a veil over the fact that the tube will not survive long in practice, if one did this!) This indeed required a Vgk bias of about -22V. Going for the same 25W into 8 ohms as before, the required anode current in the "more conducting" tube now had to peak at somewhat over 3A (as opposed to the 2.5A that was needed in class AB or B), since now, at this point in the sinewave cycle, the other tube was still conducting (as per the definition of class A). In fact, the "less conducting" tube was still passing about 0.5A or so when its anode current was at its minimum. Hence, to get the 2.5A through the 8 ohm load, it now takes about 3A through the more-conducting tube in order to offset the 0.5A or so that is still passing through the less-conducting tube.
Of course, because of the need for 3A rather than 2.5A of anode current in the more-conducting tube, this means the Vgk voltage has to be a bit more (i.e. more towards the positive direction) than it did for 2.5A. With the 6C33C model that is used in my Spice simulation, this amounted to needing something like Vgk = -0.5V for 3A, as opposed to about -1.5V for 2.5A. So the grid voltage does have to be a bit more towards the positive territory when operating in class A like that, but only by a small amount. But still, at least with the 6C33C model that I am using, Vgk is negative (just!), even for 3A anode current. (Actually, FWIW, this conclusion, that Vgk is still negative, is pretty much in line with that extended graph that jazbo8 included in his post #193, I think. Suggesting still no grid current.) Now, someone else's 6C33C model might disagree a bit about when Vgk first becomes positive, but the general picture, that the grid drive voltage only needs to be a little bit more towards positive in the class A case as opposed to the class AB or class B case, is presumably still broadly the same.
Of course, the question of whether one calls an amplifier class A or class AB is probably almost as contentious as whether one calls an amplifier class AB vs class B. In the Spice model I described above, the minimum anode current as a tube heads in the direction of non-conduction was about 0.5A. This presumably, in anyone's definition, is unambiguously class A. But suppose we increase the negative grid bias voltage, thus dropping the quiescent current, so that the minimum anode current becomes smaller. At what point does one say that the amplifier is out of class A and into class AB? Suppose the minimum current is 100mA; is that still class A? What about 10mA? Or 10 microamps? Obviously at some point it would be absurd to still claim class A on the grounds that the tube never goes into absolutely totally zero conduction. But as with the class AB vs class B divide, it can be a rather grey area, and hence an ideal topic for heated and inconclusive debate!
I suppose the important thing is not so much what classifying label one wants to attach, but how the amplifier actually performs when carrying out its intended job.
Chris
Last edited:
Very early in this thread BallPencill put up graphs that clearly show conduction over about half a cycle. That's Class B - no room for debate.
Very early in this thread Koonw showed with his sims that grid current flows, and hand calculations show that it flows. Koonw also modelled the changes that markedly reduce grid current.
Its not at all grey if you read the standard definitions of Class A, AB, B. Very early in this thread I quoted directly the definitions given in several well known textbooks - all standard work in their day.
A standing current small conpared to the peak curent does not make a Class B amp a Class AB amp. In a Class B amp you cannot have grid coupling caps, as otherwise the tubes will raise their own bias voltage.
Definitely! This one performs as a Class B amp. As in Class B amps generally, this one has grid current, there will be assymetric distortion due to tube missmatch, etc.
Yes, but that is based on a particular Spice model for the 6C33C. Other models for the 6C33C disagree with that. The only real arbiter for this question is to do the real-life measurements.
Well, to quote the definition you gave earlier in the thread, "A Class A amplifier is an amplifier in which the grid bias and alternating grid voltages are such that the plate current of the output valve or valves flows at all times." So what qualifies as "plate current flowing at all times"? A minimum pltae current of 0.5A presumably qualifies. But what about 50mA? Or 1mA? Or 1 nanoamp? Obviously somewhere between 0.5A and 1 nanoamp, one has to draw a line. But that means that a somewhat arbitrary, and hence potentially contentious, line is being drawn.
I don't know why you say that. There is no necessity to drive the output stage so hard that it develops grid current. I maybe missed a point that you were saying before, but it sounded to me as if you were saying that you want some other factor, other than onset of grid current, to be the defining one that sets the "maximum power" of the amplifier. (E.g. clipping in the op-amp driver stage, in the present example discussed in this thread.) A class B amplifier that is not driven so hard that it develops grid current is not going to suffer from the problems that grid current brings. There is no obligation to crank up the volume so high that grid conduction occurs.
Chris
- Status
- Not open for further replies.