2 x GU50 + 2 x KT88 for a 40W PSE -- Trouble about OT choice -- Help required

[HotBluePlates]

You forum readers are aware of my suggestion of self-split class A push-pull.

This probably belongs in a separate thread. Our Italian friends want parallel-SE in order to make use of a salvaged industrial transformer with tons of current on tap. So no push-pull.


Self-split push-pull operation is kind of a second-class circuit: it was only ever practically used in the old days in cheap student amps or table radios, when the desired price-point of the finished product didn't allow the extravagance of paying for an actual phase inverter (think about that; they couldn't justify the cost of a single extra triode and some resistors). IT requires double or more of the normal peak input voltage to the output tubes, and never gives the full output power that the output tube and transformer would normally be capable of. Modern use of the self-split stage is mostly in the realm of homebrew as a curiosity, or when full power output is not really a goal.

Has any one heard of any application where an ultra linear o/t is used in a self split application.  A tap at 20% provides more power than a 40% tap.

Ultralinear doesn't apply to this amp project, as it requires G2 voltage be essentially the same as plate voltage. Maximum G2 voltage for the GU-50 is limited to a value well below the supply voltage available from the power transformer Dom and Franco want to use. Limiting the plate voltage to that value (~250v) would also reduce power output, and fail to fully utilize the transformer they have. So they're looking in a different direction.


Self-split ultralinear would be problematic at best, as self-split pentodes/beam power tubes rely on a largish screen resistance for the non-driven tube to help derive the "drive signal" for that tube. You pretty much need/want the tubes to be operating as pentodes (not midway between pentode/triode as with ultralinear) so that screen voltage variation of the non-driven tube will modulate its plate current. Last, if the designer was willing to pay for ultralinear taps and sought the higher output power, he'd also spring for a phase inverter.

Consider using a anode resistor to reduce voltages to the plates on the power tubes.  I have seen a few older amps that have anode resistors.

Reducing supply voltage doesn't seem to be necessary with the power transformer to be used.

Additionally, wouldn't this allow you to artificially raise the reflected primary impedance? 

Yes, but probably not to any meaningful degree.


Any anode resistor will simply turn some portion of output power to heat in the chassis. Even 100-200Ω seems impractically large when average current for a pair of tubes will be ~260mA, which implies as much as 13.5w of possible audio output lost as heat. 200Ω would only change the suggested 1kΩ primary for a pair of tubes by 20%. And if selecting the ideal load was to deliver more power, or hit a specific design goal for power, why waste it in anode resistors? There seems to be nothing to gain with their use.

[jazbo8]

Our Italian friends want parallel-SE in order to make use of a salvaged industrial transformer with tons of current on tap. So no push-pull.

First part of the sentense is correct, but the second part isn't - in fact, it's rather odd to use a SE design to achieve high output power - all the fidgetting we are doing with the OPTs could be easily avoided with a PP design - not to mention at a lower cost and weight.... yadda, yadda, yadda.

[HotBluePlates]

I guess that's true: a push-pull design would simplify the output transformer requirements.


36w+36w from parallel SE is not my idea of fun for my back, but it seems to be what they want to do...

[kagliostro]

36w+36w from parallel SE is not my idea of fun for my back, but it seems to be what they want to do...

Yes, at the moment is so, previously was 40W (or better, 36W) in PSE


K

[HotBluePlates]

Just make a Ferrari-themed cart to wheel the amp around on, and all will be well! 

[jazbo8]

Did someone say Ferrari, va-va-voom!





+

[kagliostro]

Just make a Ferrari-themed cart to wheel the amp around on, and all will be well!

Oh, may be my friend didn't want to move around the amp, it is just a practice amp for bedroom playing

 




may be I can use those layout for the 8 x 6ak5 amp we discussed in one other tread

but as you can see B+ has some difficulties on reaching the tube that is more far from PS

 

K

[jjasilli]

If your friend wants an amp with 2X KT88's to play in the bedroom, I assume he lives alone!   

[kagliostro]

Not so far from GU50 Tank Sockets 


K

[kagliostro]

Of the previous amp I haven't a schematic


but I've of this one


K


p.s.: Ahem, the G2 voltage seems a bit too much exaggerated (or so looks to me)

[Dom]

There is no reason your operating condition cannot work, and well. Sooner or later, you will have to think about designing the system (meaning the whole amp, including power supply), and consider what you have to work with.

Your consideration  about "designing the system" will be  taken into account very seriously from now on.   ...I didn't know that your condition give a power about 18 W with only 300 V on plate! I' have not do graphical's calculation for this condition, but if you tell me this I presume that it's true. Then your condition could it be very interesting.

Now, you suggest ( in a other post about the P.T. ) not use the 95v winding at all. I think understand the reason of this choice, but there are also the particular preamp stages which is required more voltage than 300 V. For an example you can visualize some points of SLO preamp stages: the first 4 stages ( 364 V and 353 V  respectively ).
My project is similar to this for voltage dedicated to stages ( i can post something if is required ), so i think that this winding is necessary. After this B+ dedicated to stages, i must take the voltage for the KT88 and GU 50. That's another reason for "why" KT88's tension was so high: first tension for tubes amp, second tension for preamptubes. Now i would make the inverse.

...Sorry for the late or my answers, but now i have a very very hard week !!! ( i work 13 / 14 hours all days    )

[HotBluePlates]

p.s.: Ahem, the G2 voltage seems a bit too much exaggerated (or so looks to me)

Yes, but he's using ~4kΩ screen resistors. He will lose some of the full power output, but with ~700v plate, 350v screen was probably an easily-derived voltage.

...I didn't know that your condition give a power about 18 W with only 300 V on plate! I' have not do graphical's calculation for this condition, but if you tell me this I presume that it's true. Then your condition could it be very interesting.

Look at the first graph I posted with the blue and green lines. The idle point of the green line (2374Ω per tube) is on the vertical 300v line @ 132mA. The bias is -19v, so with a symmetrical input signal, you use the points on the green line 19v on either side (at G₁=0v and -38v). I labeled the endpoints (59v, 231mA) and (541v, 28mA). The difference of the voltages times the difference of the currents divided by 8 gives output power for a single tube.


[(541v-59v)*(231mA-28mA)]/8 = 12.23w, or 24.46w for the pair.


Sorry, I got confused with my own loadlines... I plotted a condition for 500v plate and 250v screen with a 5kΩ load for a single tube (or a 2.5kΩ primary for 2x tubes) which gave ~18w. You could use that load with the highest voltage you can manage with the power transformer you have (keep the screen ~250v) and you'll get a little less than 18w per tube.

Now, you suggest ( in a other post about the P.T. ) not use the 95v winding at all. I think understand the reason of this choice, but there are also the particular preamp stages which is required more voltage than 300 V. For an example you can visualize some points of SLO preamp stages: the first 4 stages ( 364 V and 353 V  respectively ).

You may not need the high voltage.


The highest "preamp" voltage needed would usually be for the phase inverter, or in your case the last preamp stage. You select a voltage for that stage based on how large an input signal is to drive the output tubes to full power.


The drive signal needed from the phase inverter is equal to the bias voltage of the output tubes. If you take my condition for the GU-50 with a ~1kΩ load for the pair, 300v plate 250v screen and -19v bias, you only need a 19v peak output signal from the preamp. Very easy to drive!


100w amps with 6L6's might have a bias voltage of -40v to -50v. They also have push-pull output stages, so you need a peak-to-peak output signal from the phase inverter of double the bias voltage of each tube. So the SLO would need to develop an 80-100v peak output from the phase inverter. Doing that requires a large supply voltage.


Design the system. So the trick is to design your output tube stage with an eye towards the supply voltages (and available output transformers?) and your required power output. Then you work backwards through the amp designing each stage to meet the drive needs of the stage you just designed, until you get to the input jack.


All this does not mean you cannot use a high supply voltage for the preamp... it is just to explain why you may not need a high preamp supply voltage.

[kagliostro]

Yes, but he's using ~4kΩ screen resistors. He will lose some of the full power output, but with ~700v plate, 350v screen was probably an easily-derived voltage.

For sure that is a reasonable motivation


K

[Dom]

Our Italian friends want parallel-SE in order to make use of a salvaged industrial transformer with tons of current on tap. So no push-pull.

Yes. I know that if I wanted to have a Push pull, many problems and "inanity's ideas " for this project would never have had. But i love the sound of Single Ended and I wanted one amp that could be at the same level of push pull ( ...ok, 40 watt  for a S.E.  isn't 100 W on push pull. But, for me, this is a heavy strike on the stomach ! Now, we can try 72 W !) or next to this, but WITH YOURS DISTINCTIVE SOUND'S.

Just make a Ferrari-themed cart to wheel the amp around on, and all will be well!

Oh, may be my friend didn't want to move around the amp, it is just a practice amp for bedroom playing

 

You are very funny guys !

...But "de gustibus", sometimes, can bring some kind of madness to a difficult mind like mine!

[Dom]

...If you take my condition for the GU-50 with a ~1kΩ load for the pair, 300v plate 250v screen and -19v bias, you only need a 19v peak output signal from the preamp. Very easy to drive!


100w amps with 6L6's might have a bias voltage of -40v to -50v. They also have push-pull output stages, so you need a peak-to-peak output signal from the phase inverter of double the bias voltage of each tube. So the SLO would need to develop an 80-100v peak output from the phase inverter. Doing that requires a large supply voltage.

...All this does not mean you cannot use a high supply voltage for the preamp... it is just to explain why you may not need a high preamp supply voltage.

You have right! I never tinked about this because my aim was to replicate 3  type of famous preamps ( with some different voice and mods respect the originals ) and listen to them on one amp in S.E..
Naively, I copyed this legendary circuits as they was, but not taking account of this difference!

Now, i suppose, i must re-designer my preamp stages's.

[Dom]

Hi everyone.

I've tried to do something after studying and analysed your proposal. With maximum respect for your advice, I've deliberately choosed high anodic tension  because  for 4 type of tube in parallel the current was too much, i thought. But I have also tried to conform all that i learned on this tread ( symmetry of the load impedence, different type of impedence for tube, best compromise between power and sound, ect. ect. )

After so many unsuccessful and successful attempts at mediation to this possibile coexisting of tubes, I think that should be possible doing this solution.

My limit whith the language does not permit me to expain so good my intention, than i hope the schematic can do this.

[Dom]

Stages that I think can do reasonable way.

[Dom]

Whith option on 4 ohm in the 2° secondary of T.U. :

[Dom]

A little gift to all:

[Dom]

Example for KT88

DATA:

VA= 400 V; IA = 95 mA; VAp = 780 V; IAp = 200 mA; Vpp = 18. VAmax = 765V; VAmin = 30V; IAmax = 192 mA; IAmin = 5 mA.

Vg2 = 250 V; Z = 3,9K ; Vg1 = - 18 V;  Ig2Q = about 7 mA; Ig2max = about 95 mA;

WA max = 42 W ; WG2 max = 8 W ;


[( VAmax – VAmin ) X ( IAmax – IAmin ) ] : 8

Wout /8 = [( 765 – 30) X ( 0.192 – 0.005 )]: 8 = 17 W

W% = ( VA x IA ) : Wmax = ( 400 x 0,095 ) : 42 = 90,4%


IG2 AV = Average screen current: ( 0,25 X IG2max ) + ( 0,5 X IG2min )

IG2 AV = ( 0,25 X 0,095 ) + ( 0,5 X 0.007 ) = 27,2 mA

WG2 % = IG2AV X VG2 // WG2 % = (0.00272 X 250) : 8 = 84%

IK = IA + IG2 = 95 + 7 = 102 mA

RK = VG1 : IK = 18 : 0,102 = 176,47 Ohm = 180r

[jazbo8]

Please take a closer look at the datasheets, the estimated screen current at low anode voltage is going to be a lot higher, e.g., the GU50 datasheet used for Ig2 has the condition of Va of 800V, when the anode voltage swings low under maximum drive, Ig2 will not stay at 12mA as you have shown. Likewise, for EL34, you somehow ignored the peak in Ig2 when you use max Ig2 of 28mA. Therefore in both cases, the maximum/average Pdg2 would be higher than your estimates suggest.

[Dom]

Damn! 
You have right.
For the GU50 i have found a graphic similar to the 6l6gc's graphic (...whith curves of screen grid at different values), and i can approssimate maximum peak,  average current and the power of G2.

But i can't find a similar datasheet for EL34. How i can stimate this values of G2?

[jazbo8]

Damn! 
You have right.
For the GU50 i have found a graphic similar to the 6l6gc's graphic (...whith curves of screen grid at different values), and i can approssimate maximum peak,  average current and the power of G2.

But i can't find a similar datasheet for EL34. How i can stimate this values of G2?

Good question, I have not seen one either. But I thought you were going to use the KT88s.

[HotBluePlates]

Please take a closer look at the datasheets, the estimated screen current at low anode voltage is going to be a lot higher ...Therefore in both cases, the maximum/average Pdg2 would be higher than your estimates suggest.

Dom, which of the many posted operating conditions have you chosen?

Average screen current probably has little relevance. You need to know idle screen current (very small) to accurately calculate a bias resistor (but that precision will be swamped when you have to round up to a standard value).

You also need to know the worst-case maximum peak screen current. Because you want to know if the screen will melt when you hit full power output. If that peak screen current * idle screen voltage is less than the screen dissipation rating, you need do nothing. If it exceeds the screen dissipation rating, figure how much the screen voltage would have to drop to just touch the dissipation rating; calculate the implied screen resistor value.

If the calculated screen resistor seems large (3-4kΩ), you may want to investigate more and determine an accurate average screen current. It will be less than (I_{G2 max} + I_{G2 idle})/2, because most of the time the screen barely rises above the idle value and is only very briefly at its peak value.

That accurate method involves finding the screen current for 5 or 9 or 11 equal portions of the input sine signal. Meaning figuring screen current when G1 is at Sin 0 degrees, then G1 = Sin 10 degrees * peak input signal, then G1 = Sin 20 degrees * peak input signal, etc.

Yes, it's a pain in the @$$. But it's what you have to do if the simple approach doesn't work and you can't use a loadline that cut G1=0v above the knee of the curves.

[jazbo8]

Dom - I use the SPICE model to generate the Ig2 curves to give you an idea how high it can go once the load line crosses "below the knee", the grid lines are from 0V to -30V, and Eg2=250V in the chart below.

[Dom]

Tank you very  much for the model, Jazbo8!

...This help me to imagine that the peak screen grid current will be very high, but i can't  figure a sort of approximation for this high value. ... Anyway I think I should safe:

The limiting value for screen grid is 8W.  If I suppose that in a minimum voltage value ( maybe 25V or 35V or 45V ) the max current possible for the screen grid would be 8W:35v = 228 mA, i think is hardly likely that my operating condition could to take the screen current so high. If i suppose the worst-case whith 110 mA of max peak current for the grid and 40 V of min. votage, i have 40 X 0,11 = 4,4 W. Is it correct ?

...But I thought you were going to use the KT88s.

Yes, certainly. I want use KT88 with the operating condition that i've posted. When i want try another tube like 6l6 or EL34, i change the bias ( bias rotary switch on the schematic ) and secondary of T.U. to 4 ohm at the same time. ...Maybe i must post a jpg.

[Dom]

Dom, which of the many posted operating conditions have you chosen?

All of them. Now ( ...with your important advices) with some correction that i propouse in the next attachment.

.....You need to know idle screen current (very small) to accurately calculate a bias resistor

Yes. You have right.
In the schematic "Finale Aggiornato.jpg", i have calculated 10 mA for one screen grid by way of the implied screen resistor value: 260V - 250V = 10v. 10V : 0.01mA = 1K.
EL34 needs about 5 mA ( quiescent ), 6L6gc needs about 3 mA, KT88 needs about 7 mA and GU50 needs about 5 mA.

That accurate method involves finding the screen current for 5 or 9 or 11 equal portions of the input sine signal. Meaning figuring screen current when G1 is at Sin 0 degrees, then G1 = Sin 10 degrees * peak input signal, then G1 = Sin 20 degrees * peak input signal, etc.

Yes, it's a pain in the @$$. But it's what you have to do if the simple approach doesn't work and you can't use a loadline that cut G1=0v above the knee of the curves.

...It's  too complex for me, but thanks for the procedure.

I hope that will be correct new grapich for GU50.

[jazbo8]

If I suppose that in a minimum voltage value ( maybe 25V or 35V or 45V ) the max current possible for the screen grid would be 8W:35v = 228 mA, i think is hardly likely that my operating condition could to take the screen current so high. If i suppose the worst-case whith 110 mA of max peak current for the grid and 40 V of min. votage, i have 40 X 0,11 = 4,4 W. Is it correct ?

I don't think it would be that high... So assuming you are using the same OPT as the KT88, I can try to calculate the Pdg2 for you, but just to make sure, have you decided on the OPT that you are going to use or wind yourself? If so, what's the primary impdeance?

[Dom]

Yes i did. The OPT can be made from a great man that works in Italy.

In view of the points described above, 2 twins OPT will be:

Zprim. = 2K ( for 2 tube in parallel with a relative impedence fo 4K);
max power = 40W;
Max current = 250 mA ( 200 mA is required, but i think that it's better );
1° sec. = 8 ohm;
2° sec. = 4 Ohm ( for el34 and 6l6g.  this secondary reflect primary impedence to 4K with a load of 8 ohm)

[Dom]

Thanks for your patience!

[HotBluePlates]

Dom - I use the SPICE model to generate the Ig2 curves ...

What is the scale for the screen current? It won't be the same as the plate current numbers on the y-axis.

What are the multiple screen current lines? I presume these are for differing values of G1 voltage, but only the G1=0v line is important (well, maybe include the G1=idle bias, to figure idle screen current)

Dom, which of the many posted operating conditions have you chosen?

All of them.

Well, you have to analyze one at a time, and the explanation is a lot of work. PICK ONE, and there will be much less confusion.

For example, look at the EL34 curves Jazbo posted. You need to have a loadline plotted, because screen current will change with the changing plate voltage (which itself is due to plate current variation creating a voltage drop across the selected load impedance).

If the load impedance crosses the G1=0v line at Ip=100mA, that is about 20v or so on the plate. The screen voltage is supplied by a steady d.c. power supply point, and doesn't drop from the G2=250v Jazbo says the curves are for. The Screen Current must have a different scale than the plate current scale marked, because the uppermost G2 curve (assumed to be for G1=0v) also appears to cross 100mA at the same time the plate is passing 100mA. 250v * 100mA = 25w, and the screen would have vaporized before passing that current.

You also need to pick a condition because I'm lost with all the proposed operating conditions, and I assume we have to pick one which makes sense with the power transformer you have available (otherwise, why bother with it when you could pick an ideal power transformer or have one made?). There are too many overlapping steps and considerations to work without any boundaries.

[PRR]

> The limiting value for screen grid is 8W.  If I suppose that in a minimum voltage value ( maybe 25V or 35V or 45V )

The _Plate_ may fall to 25V-50V.

The Screen (G2) stays near 250V (or whatever the Screen supply is).

8 Watts at 240 Volts is 33mA average screen current.

Plotting is difficult and confusing. It looks to me (for EL34) like Screen current is under 25mA for 3/4 of the full-power cycle, peaking to 100mA for 1/4 of the cycle. Assume 15mA for 3/4 of the time and 63mA for 1/4 of the time. It is about 26mA average over the full cycle.

So the rough approximation says "safe" but the counting was very rough. Close enough to go ahead and breadboard; close enough to meter the Screen current on the first full-power test. (The men who designed the EL34 were surely thinking about Screen dissipation, and gave you "enough", but not "too much".)

GU50 seems to be less Screen current, but again there probably is not much safety-factor in the Screen dissipation.

Some series resistance is wise. 100 Ohms won't do much. 1,000 Ohms may reduce maximum power output (but it will be hard to overheat the Screen).

[jazbo8]

What is the scale for the screen current? It won't be the same as the plate current numbers on the y-axis.

What are the multiple screen current lines? I presume these are for differing values of G1 voltage, but only the G1=0v line is important (well, maybe include the G1=idle bias, to figure idle screen current)

Ig2 has the same scale as Ip so a lot of screen current can flow when operating below the knee, however, due to the SPICE model's inaccuracy and the fact that modern tubes' diode line are seldom below 280R, so Ep is unlikely to swing as low as the curves suggest. And as correctly pointed out by you and PRR, the SPICE model assumes zero sag on the plate and screen supplies (nor the voltage drop by Rg2), so in the real circuit, the dissipation tends to be somewhat lower, perhaps this exercise can be viewed as the worst case scenario.

Below is how I go about estimating the screen grid dissipation:
1) draw a 280R diode line and note the intercept with the load line, in this case it falls at 40V;
2) pick out the maximum Ig2 from the Eg1=0V curve at Ep=40V, in this case, it's 61.6mA;
3) from RDH4, Pdg2 can be estimated by, (Ig2 idle/2 + Ig2 max/4) x Eg2, so (0.0072/2 + 0.0616/4) x 250 = 4.75W.

[Dom]

First of all: I’m very sorry for confusion I have caused  and for my stupid mistake about calculation of Wg2!

It’s evident that if the average power of screen grid is calculated regarding a fixed tension ( 250V in this case ), the peak current ( max ) is calculated  with THE same tension. For example 23 mA = 250V * 0.023 = 5,75W. This is not the same thing  like plate dissipation. Ok.

Then… with a peak of 63 mA, I suppose that for ¼ of full power cycle the screen grid will be at 15,75 W ( 0.063 * 250V = 15.75W ), right ? Even if it will be for a little instant, can this current fuse the screen grid ?
If that were so, only 32 mA should be the “max” peak current ( screen grid current )reasonable for a tube like EL34. …This seems more strange for me, and I’ve somes dubt! This means that all operating conditions exceed  this threshold can damage/fuse the screen grid ?

HotBluePlates says “If it exceeds the screen dissipation rating, figure how much the screen voltage would have to drop to just touch the dissipation rating; calculate the implied screen resistor value.”, that’s good. But I presume that it’s good for a “little” drop voltage, because if I decrease VG2 from 250 to 200 ( for example ) I change ALL the operating condition ( VG1 curves will be much squashed and the plate current will much rise at the same load impedance).
In other words, I would minor output and more supply current for the desired result.

For the average screen current and its relative power,  this should not be a big problem : ( 0.25 * 0.063) + (0.5 * 0.005) * 250V = 4.55W.
From here I’m very confused by this question: What is the point for calculating the average screen current if the peak current ( max ) can damage screen grid with a few mA ?
Please, let me understand!

@ HotBluePlate

You also need to pick a condition because I'm lost with all the proposed operating conditions, and I assume we have to pick one which makes sense with the power transformer you have available (otherwise, why bother with it when you could pick an ideal power transformer or have one made?). There are too many overlapping steps and considerations to work without any boundaries.

My apologies for “bother” and bring confusion.
My intentions were not “bother”, but show you last possible solution which I came to that,  ( by your help as to all others ) after reading your consideration and suggestion, for evaluating its various aspects and possibilities and If  it was okay.

I came to the choice of 2 separates  OPT after reflect and read all your opinions about the issue.
I have changed the anodic tension for various tubes ( and I redesigned my power transformer ) for a specific reason:  the operating condition that you suggest can work perfectly ( 300V - 130 mA ), but with 130 mA for a single tube. If 4 tubes works at the same time, we have about 130 mA * 4 tubes, as you have written earlier. I have accepted and verified ( on my limited knowledge ) your operating condition, but I have chose this only for saving current and for a little more watts.

For other suggested operating condition that can work with 500V and 5k of load impedance, I think that I cannot do with my power transformer under full load.
If I bring high tension and lower current a little, maybe I can reach a reasonable compromise between power output and current saving.

After, I tried many and many operating conditions ( for the 4 type of tubes that I want use ) with the aim to coexist KT88, GU50, EL34 and 6L6 with the same impedance, voltage and current employed. This because I would work an amplifier with 2 KT88 and 2 GU50 or 2 EL34 and 2 6L6gc in the same time, with the opportunity to hear how sounds only a pair of KT88 or GU50 ( same thing for EL34 or 6L6gc with the 4 ohm secondary  )
.
My hope was  that can be work without fuse the screen grid with operating condition posted, but now I’ve some doubts!

I just wanted  to show  my proposal to know if it was okay, because I am not an expert on that like you.
Moreover, it seemed to me correct and respectful to all of you show these various proposals and the continuation of the project.
I’m sorry for mistakes, but I'm trying.

@ PRR
Thank you very much for explanations. I begin to understand something.

Some series resistance is wise. 100 Ohms won't do much. 1,000 Ohms may reduce maximum power output (but it will be hard to overheat the Screen).

I don’t know exactly what you mean. Do you mean about some resistance on my schematic ?

@ Jazbo8
Many  thanks for graphic simulation!

[jazbo8]

It’s evident that if the average power of screen grid is calculated regarding a fixed tension ( 250V in this case ), the peak current ( max ) is calculated  with THE same tension.

Then… with a peak of 63 mA, I suppose that for ¼ of full power cycle the screen grid will be at 15,75 W ( 0.063 * 250V = 15.75W ), right ? Even if it will be for a little instant, can this current fuse the screen grid ?

No, it's not quite right... At quarter power, the screen grid current is completely different, it is certainly nowhere near 63mA, so the dissipation is not 15.75W! See if you can figure out the actual dissipation, hint, take a look at the signal swing needed for 1/4 power output.

[PRR]

> Even if it will be for a little instant, can this current fuse the screen grid ?

Tube dissipation ratings may be "averaged over a complete audio cycle".

So if the rating is 8 Watts, and the dissipation is 16W for a short time, and 4W most of the time, you are safe.

Because screen current is highly non-linear, and no simple formula, it has to be plotted from the curves. This is very tedious.

It is better to take the data-sheet conditions for a single-ended amp. A long time ago, some junior engineer worked over a hot tube for hours to find "good" conditions, so his company could sell more tubes.

Yes, factory data-sheets for many large tubes do not show single-ended conditions. Why would you want a huge SE amplifier, when a push-pull could make more power with less cost?? Both in tubes and in the output transformer. So there will have to be Smoke Testing.

A DC meter watching the screen current (or voltage across a small 100r resistor to the screen), time screen voltage, tells you the screen power "averaged over a complete audio cycle".
_____________________________________________

> aim to coexist KT88, GU50, EL34 and 6L6 with the same impedance, voltage and current employed.

Then you need voltage and current, and *power dissipation* (and maybe impedance) which all tubes can work well at.

A good 6550 can stand 42 Watts dissipation (KT88 should be similar).

6L6GC is rated 35W, but I've had trouble at 30W. (A genuine 6L6 from 1938 is 19W; genuine old 6L6G is the same.)

EL34 says 25W but we all know they stand 30W.

GU50 is rated 40W.

So we have 30W and 40W tubes. You *can* run the KT88 or GU50 at 30W dissipation. Then all tubes will make nearly the same power output (about 12W per tube).

6L6GC has the lowest plate voltage rating, 500V.

It is usually NOT necessary to use HIGH voltage in a single-ended amplifier. Power output is limited by dissipation, not voltage losses.

Also the "optimum" screen voltage should -drop- for higher plate voltages. (High plate voltage means low plate current, so we do not need a high screen voltage, and do not want a high screen voltage because it is bad for the screen and we will need more control grid voltage to balance the high screen voltage.)

So I suggest these values for ANY of those tubes:

500V 60mA 8K3 load
450V 67mA 6K7 load
400V 75mA 5K3 load
350V 86mA 4K1 load
300V 100mA 3K load

[Dom]

Tube dissipation ratings may be "averaged over a complete audio cycle".

So if the rating is 8 Watts, and the dissipation is 16W for a short time, and 4W most of the time, you are safe.

Because screen current is highly non-linear, and no simple formula, it has to be plotted from the curves. This is very tedious.

Tank you so much! This is precisely what I wanted to know. It was always doubtful this concept for me.

....Why would you want a huge SE amplifier, when a push-pull could make more power with less cost?? Both in tubes and in the output transformer.

Because I love the sound of Single Ended. Anyway, this project was start with intention to make 2 version with the same Preamp. Obviously, second SON was / is thinked with Push Pull.

...It is usually NOT necessary to use HIGH voltage in a single-ended amplifier. Power output is limited by dissipation, not voltage losses.

Also the "optimum" screen voltage should -drop- for higher plate voltages. (High plate voltage means low plate current, so we do not need a high screen voltage, and do not want a high screen voltage because it is bad for the screen and we will need more control grid voltage to balance the high screen voltage.)

So I suggest these values for ANY of those tubes:

500V 60mA 8K3 load
450V 67mA 6K7 load
400V 75mA 5K3 load
350V 86mA 4K1 load
300V 100mA 3K load

Ok. I should value it very seriously and I'll test with graphics and curves.

...But... Perhaps I did not make myself clear in my first speech ( the language is a limit for me, and I think that can bring confusion ):

With my power transformer  I cannot go further about 440 V. So, I presume, the first operating condition suggested it's not possible for me. But I'll try the other!

@ HotBluePlates

...I have a vision for the complete amp:

[Dom]

Kagliostro will test it with a load very adapted ! 

[PRR]

> Because I love the sound of Single Ended.

Yes, I understand.

However "MOST" designers, back in the time when vacuum-tubes were common, used SE only for "little" amplifiers, to minimize the number of parts to buy and assemble.

A one-6L6 amplifier is about the same cost as a two 6V6 amplifier. While you need a second power tube plus a phase-splitter, the output transformer is much less money (no unbalanced DC), the power output may be a little higher, and the distortion number is "better".

So very few tubes have suggested operating conditions for SE over about 10 Watts.

One exception, though an odd one. The HK-257 has a typical condition for 30 Watts output from one tube in Class A.

http://www.mif.pg.gda.pl/homepages/frank/sheets/114/h/HK257B.pdf  middle-right of page 3

Note that the same tube can be run at 1000V or 500V with very little change of output power, but a large change of load impedance.

Note that two such tubes gives, not twice, but TEN times the output power.

And the sheet says "Audio and Television". Television was new in 1946. The TV video signal has to be impressed on the radio wave. In 1946 this could be a big SE tube modulating the carrier signal-- that avoids having a Modulation Transformer which is difficult with video. However a 30W modulator could only manage a 50W-100W total transmitted power output; enough for testing in 1946 but by 1948 they were using much bigger transmitters and different modulation systems. So "big SE" faded away again.

These big SE amps are fun to think about. A couple of guys here actually built one on the breadboard. One is a professional house/factory Electrician, knows the dangers and how to reduce the risk of electrocution and fire. However I think "Real Life" got in the way of finishing the work.

http://el34world.com/Forum/index.php?topic=10262.msg93751

[kagliostro]

Hi PRR 

just a curiosity, which is the socket type name ?

it is similar to UX7 (that fits 1625)



but seems much more larger



Thanks

Franco

EDIT: Is an U7G socket ?

[Dom]

PRR, I will take account of your professional experience and your warning about this idea for a big single ended. Maybe, one day, I'll tell you "Oh yeah, you were right.", but now I want try this "obsession".

...If we have to speak at it from the perspective of "cost" and "the number of parts to buy and assemble", you are completely right. In fact the whole project looks like madness, but I like this and I make it possible whatever it takes!  Ok, I'm just a  crazy man.

In view of the potential dangers  that it involves, I will try, if at all possible, to make sure everything is going well.

Your suggestion about the HK-257 sounds very good for another project (...maybe with a microwave transformer's ??? ) in Singele ended. I'll think about it. Seriously. It's very interesting!

[kagliostro]

Visto che la la HK-257 va anche con anodica di 500v e rende più o meno allo stesso livello (25W invece di 30W), forse è meglio evitare 1000v di anodica, inoltre con 500v ti basta un TU da 2.6k invece con 1000v ti servirebbe un TU da 12k

Since the HK-257 also goes by anodic 500v and makes more or less at the same level (25W instead of 30W), perhaps it is best to avoid  1000v B+, 500v also requires just a 2.6k OT, instead 1000v requires a 12k OT

K

[jazbo8]

Just to throw more crazy ideas into the mix, if you want high power SE with modest plate voltage, look at the 13E1. Rule of thumb - 1 x 13E1 ~ 3 x KT88/6550 or 4 x EL34! But make sure your PT can supply the high current.

[HotBluePlates]

... The HK-257 has a typical condition for 30 Watts output from one tube in Class A.
...
Note that two such tubes gives, not twice, but TEN times the output power.

You implied this, but casual readers may overlook it... When you note that 2 tubes gives 10x output power, the data sheet gives "Load Resistance, plate to plate" and also describes the condition as "2 Tubes, Overbiased."

The first note tells you this is push-pull, not single-ended anymore.

The second note is very-old-school terminology for "Class AB" rather than Class A. "Overbiased" means "more than class A bias" to keep the plates from overheating by allowing the tubes to cool off by shutting off part of the signal cycle. The increase in supply voltage from 500v or 1000v to 1.5kV is your other clue about this being a class AB condition.

500v also requires just a 2.6k OT, instead 1000v requires a 12k OT

Once you realize that load impedance is selected on the basis of supply voltage available and output power needed (not because the tube directly dictated a needed load) then it becomes clear why the load impedance increases with increased voltage.

Play with the equations Power = Voltage * Current and Voltage = Current * Resistance to prove this for yourself. You can mix the equations to derive Power = Voltage²/Resistance to make it even more obvious. (CAVEAT: "Voltage" in this case really means RMS signal at the plate, not the supply voltage; however, the RMS signal voltage is always less than the supply voltage.)

For example:
500v Supply
25w = Voltage²/2600Ω
65000 = Voltage²
Voltage = ~255v RMS
RMS Voltage = 51% of supply

1000v Supply
30w = Voltage²/12 000Ω
360 000 = Voltage²
Voltage = ~600v RMS
RMS Voltage = 60% of supply

[kagliostro]

Many thanks HotBluePlates

The second note is very-old-school terminology for "Class AB" rather than Class A. "Overbiased" means "more than class A bias"

I'm glad I learned a new thing 

The other stuff you added is really explained in a very easy to understand way

Play with the equations Power = Voltage * Current and Voltage = Current * Resistance to prove this for yourself. You can mix the equations to derive Power = Voltage2/Resistance to make it even more obvious. (CAVEAT: "Voltage" in this case really means RMS signal at the plate, not the supply voltage; however, the RMS signal voltage is always less than the supply voltage.)

For example:
500v Supply
25w = Voltage2/2600Ω
65000 = Voltage2
Voltage = ~255v RMS
RMS Voltage = 51% of supply

1000v Supply
30w = Voltage2/12 000Ω
360 000 = Voltage2
Voltage = ~600v RMS
RMS Voltage = 60% of supply

Franco

[PRR]

> Voltage = ~255v RMS
> RMS Voltage = 51% of supply

> Voltage = ~600v RMS
> RMS Voltage = 60% of supply

More math but perhaps more insight:

Look at Peak voltage.

Ideally Peak voltage is the supply voltage. But the tube gets in the way.

255V RMS is 361V Peak, or 72% of 500V supply. Not great, not horrible.

600V RMS is 848V Peak, or 85% of 1000V supply. Not great, but better.

On the flip side, look at lost voltage. The 500V case loses 139V (at at-least 139mA Peak). The 1000V case loses 151v (at at-least 71mA Peak).

Which suggests they left some power on the table. The diode-line is about 45V at 100mA. It should be able to pull nearly 90% of B+ as peak output. The 72%-85% figures are probably for "low distortion" (however they don't say how low is low). In G-amp service you could probably get more Watts (and chest-impact) before it changed from "flavor" to "fuzz".

Yes, factory data for a 2-tube condition is always push-pull. Only fools (like us) would throw in more tubes without taking advantage of "better" ways to use them.

Yes, the "overbias" case perhaps should not be called "Class A". But H&K were radio-heads, mostly working class C with occasional hard-B (idle near cut-off). Maybe anything softer looked like A to them. The fact it is heavy AB is clear from the DC current change, >1:3 idle to full.

[kagliostro]

Thanks PRR

as always a very good explanation 

Franco