Yet another load-line question

[eljefe3126]

Greetings. I'm having a little difficulty with this and wondering if I'm doing something wrong. Hopefully, someone here can set me straight.

I'm trying to wring every last watt out of a pair of KT-88s. The project is a musical instrument amp, so distortion levels aren't critical (although I can't have crossover distortion, so class B is out). I'm considering going to Class AB2, but even if I stick with Class AB1, the question still stands.

The attached pic shows what I'm trying to do. The red line is the "Class A" loadline, with Vo at 800 Volts and Ipo biased to 43-44 mA. The other half of the push-pull pair cuts off and the active tube goes into "Class B" operation at about 468 Volts and 87 mA. Finally, the system peaks at about 35 Volts and 200 mA (Class AB1) or 0 Volts and 210 mA (Class AB2).

The green curve is constant plate dissipation at 42 Watts.



Now, from the blue "Class B" line, I get an impedance of 3800 Ohms. If my true impedance is 4x that, that seems absurdly high at 15,200 Ohms. To put that in perspective, the Marhall Major produces about 200 Watts from 4 KT-88s in ultralinear, with an output transformer of 2,250 Ohms, equivalent to 4,500 Oms for a single pair. The Hiwatt DR201 is in the same class, 200 Watts from 4 KT-88s in Pentode mode, from an output transformer at 2,200 Ohms, equivalent to 4,400 Ohms from a single pair. Hiwatt also made the DR405, which extracted something north of 300 Watts from 6 KT-88s, with the output transformer at 1,700 Ohms (equal to 5,100 Ohms for a single pair).

5,100 Ohms is a long way from 15,200 Ohms. Before I start winding a transformer, I want to be sure that I'm not committing some gross error.

There is also output power to deal with. I'm using P = (HT-Vmin) * Ipeak / 2, which in Class AB2 gives P = (800-0) * 0.210 /2 = 84 Watts. I should be somewhere north of 100 Watts for the pair, which is what the reference circuits in the GE manual show. Is that for one side only? If so, 168 Watts is really good news, as the Class B reference circuit generates 150 Watts into a 6,000 Ohm load "for speech only", and apparently something less into 7,500 Ohms "for other purposes".

In Class AB1, power should be P = (800-35)*0.200 = 76.5 Watts per tube, which would yield 153 Watts without the Class B crossover distortion, which would still be a good day.

Using P = 2 * (HT-50)^2 / Rload, I get 2 * (800-50)^2 / 3800 = 296 Watts, which seems wrong. I presume the 50 is an assumed Class AB1 peak at Va = 50 Volts, Ia = 0. If I double 3800 to 7600, I get 148 Watts, which seems reasonable.

I'm just looking for a sanity check here, thanks.

 

[maxhifi]

Usually load lines are drawn such that you can use the tube all the way to 0V on the grid. Your load lines are insufficiently steep, wasting most of the current delivery capability of the tube, with it cutting off at -15V. This is why your load impedance is so high. I suggest to try and reverse engineering the data sheet suggested operating points as an exercise, once you see what those load lines look like you will get an idea where you need to go.

As for class AB2, it is simpler and will be more reliable to just double the output tube count and stick with AB1. Four tubes is still cheaper than two, if you operate them conservatively and nearly never need to change them.

The correct load impedance for getting 100W out of a pair of 6550 or KT-88 is around 5000 ohms, with 600V on the plates, and 300V on the screens. If you can regulate the screen voltage, even better. It's good to do the analysis, but ultimately the tube data sheet suggested operating conditions are suggested for a reason, they really do the best job of flattering the tube's capabilities.

 

[dcgillespie]

I think your first mistake is assuming that the Class B load line must never cross the max dissipation curve. Under quiescent conditions, this is certainly true -- but the quiescent operating point is only one point on the load line. Under dynamic conditions, it is quite normal for the Class B load line to cut through the max dissipation curve, based on the use of sinusoidal wave forms. The plate dissipation rating is hardly exceeded when this happens, as under dynamic condition, the element of time must also be factored into the equation: instantaneous plate dissipation is only exceeding the "constant" max dissipation curve for a fraction of the time over the course of one complete cycle -- typically, less than 50% of the time. Much of the time, the plate is not conducting any current at all. Therefore, understand that under dynamic conditions, it is the average plate dissipation that must not exceed the max plate dissipation rating for the tube. As maxhifi correctly stated, your load lines are not steep enough. Understanding the average plate dissipation relative to the max plate dissipation rating of the tube lets you then see why your load lines can in fact be safely much steeper than they are.

Dave

 

[eljefe3126]

Well, the reverse engineering could be going better. I'm going from the 150-Watt Class B circuit in the June, 1961 GEC Valve supplement for the KT-88. The circuit description starts on page 16. The screen and grid are connected to each other (the grid gets a current-limiting resistor), forming a high-impedance triode. However, that's not the part that has me confused.

Va(o) = 850 Volts (yes, I know this exceeds the Absolute Maximum Rating of 800 Volts, but it's in the manual, so it must be ok)
Ia(o) = 2 X 10 mA (so, 10 mA for each side)
RL(a-a) = 6,000 Ohms (just like you said, or at least close enough).

So, when I draw my "Class A" loadline, that should be an equivalent impedance of RL/2 = 6000/2 = 3000 Ohms. Since this is Class B and biased all the way down to 10 mA, it shouldn't be on this line for very long.
My "Class B" loadline should be an equivalent impedance of RL/4 = 6000/4 = 1500 Ohms. Let's just draw a 1500 Ohm loadline through 850 volts and 0 Amps, and see where it goes.

850 / 1500 = 567 mAmps. Ok, great, that avoids "wasting most of the current delivery capability of the tube". Except...

The manual lists RL(max sig) = 2 X 140 mA. Hmm. If I draw a loadline from 850 Volts, 0 A using a 6,000 Ohm slope, it goes to just about exactly 140 mA. And on the very next page is a graph that shows Ia going from 10 mA to 140 mA as output power goes from 0 Watts to 150 Watts.

And this is how I stumbled in, all confused. Do I need to divide by 2? By 4?

The rest of the manual doesn't help much. On page 2 under "Typical Operation", it lists "Push-Pull. Class AB1. Fixed Bias. Tetrode Connection.":
Va(o) = 552 Volts
Vg2 = 300 Volts
Ia(o) = 2 X 60 mAmps
Ia(max sig) = 2 X 145 mAmps
RL(a-a) = 4500 Ohms
P(out) = 100 W

Yeah, I can't get from here to there, either. According to the graph on page 6, tetrode connection with Vg(2) = 300 Volts and Ia = 145 mAmps means Vg(1) is anywhere from -15 to -23 Volts. And for the life of me, I can't draw a 4500 Ohm, 2250 Ohm, or 1125 Ohm loadline from 552 volts, 60 mAmps to anyplace that makes sense to me.

So, while I'm trying to comprehend all that, let me shift gears, back to my original concept. Suppose I did hook up a 15,000 Ohm transformer to my tubes and drove each one to cut off at 200 mAmps, say, -12 Volts on the grid. Would anything bad happen? I would be happy with 153 Watts output (if that's what I woudl really get), and if I need current, I can get that out of the output transformer with enough turns on the primary. I'm aware that I'd have to insulate the output transformer windings for some 1130+ volts due to back EMF, and that was a problem for the Marshall Major with one particular brand of output transformer that wasn't rated for that, but aside from that, is there a problem with that idea? I can always write whoever ends up owning GEC an apology for "not using all the current capacity of their tubes". Will the tubes mind?

 

[maxhifi]

Well, the reverse engineering could be going better. I'm going from the 150-Watt Class B circuit in the June, 1961 GEC Valve supplement for the KT-88. The circuit description starts on page 16. The screen and grid are connected to each other (the grid gets a current-limiting resistor), forming a high-impedance triode. However, that's not the part that has me confused.

Va(o) = 850 Volts (yes, I know this exceeds the Absolute Maximum Rating of 800 Volts, but it's in the manual, so it must be ok)
Ia(o) = 2 X 10 mA (so, 10 mA for each side)
RL(a-a) = 6,000 Ohms (just like you said, or at least close enough).

So, when I draw my "Class A" loadline, that should be an equivalent impedance of RL/2 = 6000/2 = 3000 Ohms. Since this is Class B and biased all the way down to 10 mA, it shouldn't be on this line for very long.
My "Class B" loadline should be an equivalent impedance of RL/4 = 6000/4 = 1500 Ohms. Let's just draw a 1500 Ohm loadline through 850 volts and 0 Amps, and see where it goes.

850 / 1500 = 567 mAmps. Ok, great, that avoids "wasting most of the current delivery capability of the tube". Except...

The manual lists RL(max sig) = 2 X 140 mA. Hmm. If I draw a loadline from 850 Volts, 0 A using a 6,000 Ohm slope, it goes to just about exactly 140 mA. And on the very next page is a graph that shows Ia going from 10 mA to 140 mA as output power goes from 0 Watts to 150 Watts.

And this is how I stumbled in, all confused. Do I need to divide by 2? By 4?

The rest of the manual doesn't help much. On page 2 under "Typical Operation", it lists "Push-Pull. Class AB1. Fixed Bias. Tetrode Connection.":
Va(o) = 552 Volts
Vg2 = 300 Volts
Ia(o) = 2 X 60 mAmps
Ia(max sig) = 2 X 145 mAmps
RL(a-a) = 4500 Ohms
P(out) = 100 W

Yeah, I can't get from here to there, either. According to the graph on page 6, tetrode connection with Vg(2) = 300 Volts and Ia = 145 mAmps means Vg(1) is anywhere from -15 to -23 Volts. And for the life of me, I can't draw a 4500 Ohm, 2250 Ohm, or 1125 Ohm loadline from 552 volts, 60 mAmps to anyplace that makes sense to me.

So, while I'm trying to comprehend all that, let me shift gears, back to my original concept. Suppose I did hook up a 15,000 Ohm transformer to my tubes and drove each one to cut off at 200 mAmps, say, -12 Volts on the grid. Would anything bad happen? I would be happy with 153 Watts output (if that's what I woudl really get), and if I need current, I can get that out of the output transformer with enough turns on the primary. I'm aware that I'd have to insulate the output transformer windings for some 1130+ volts due to back EMF, and that was a problem for the Marshall Major with one particular brand of output transformer that wasn't rated for that, but aside from that, is there a problem with that idea? I can always write whoever ends up owning GEC an apology for "not using all the current capacity of their tubes". Will the tubes mind?

Where did you get curves, for this sort of operation? The curve you showed is not valid for this circumstance. Also, I would not trust non NOS GEC tubes to stand up under those conditions.

 

[kward]

Assuming you really want 300V screens and you are settled on 800V quiescent voltage on the pate, you've got to get your class B line up to the point on the graph of approximately 60V, 350 mA. Meaning much much steeper than currently drawn. Just eyeballing it, that would put each primary side under class B operation at (800V - 60V) / 350 mA = 2114Ω, thus full plate to plate winding impedance 4 * 2114Ω = 8.4KΩ.

That will deliver plate power of about:
(800V - 60V)/1.414 * (350 mA/1.414) = 129 Watts (combined across both tubes in the PP pair)

And deliver power at the speaker terminals of about:
129 watts * 0.85 = 109 Watts.

I dunno if the KT88s can handle that much poop. If you can back it off to between 90 and 100 watts plate power, I think you would have a longer term reliable amp. There have been several threads on this forum running the KT88's (or maybe they were 6550s) that way that you could look at to see the loading conditions they used. Those threads were in regard to the Bogen 200 sound distribution amps, but I don't remember the thread names exactly.

 

[eljefe3126]

Where did you get curves, for this sort of operation? The curve you showed is not valid for this circumstance. Also, I would not trust non NOS GEC tubes to stand up under those conditions.

I understand that tying the screen to the grid to form a "high impedance triode" is very unusual, and I don't know that curves of plate current vs. voltage exist anywhere for it. It was just a 150-Watt Class B example given in the manual, complete with circuit description, parts list, schematic, a table of operating conditions, and a graph of plate current, plate dissipation, grid-to-grid rms input voltage, and common grid and screen input current all vs. power output. The manual says that this is for ICAS operation, and that "a typical use would be in a speech modulator for a transmitter".

I don't presume to fully understand this circuit, which is why I started to reverse-engineer it. I presume that since RL(a-a) is 6000 Ohms, that it operates on a 6000 Ohm, 3000 Ohm, or 1500 Ohm load line (probably the latter, as it is push-pull Class B). Everything else in my last post was straight out of the table of operating conditions on page 18 of the GEC KT-88 supplement. I'll be happy to reproduce any of that here, or you can look it up yourself at https://frank.pocnet.net/sheets/084/k/KT88_GEC.pdf.

The second set of operating conditions that I tried to reverse engineer is much more conventional, a simple Class AB1 push-pull pair, tetrode-connected and putting out 100 Watts. That starts at the bottom of page 2 of that same reference, and continues on to page 3. RL(a-a) there is listed as 4,500 Ohms and Vg2 is 300 Volts, so I should be able to plot in on the same set of curves as in my first post. Yet nothing I can plot goes anywhere near the 145 mA listed for Ia(max sig).

 

[eljefe3126]

I think your first mistake is assuming that the Class B load line must never cross the max dissipation curve. Under quiescent conditions, this is certainly true -- but the quiescent operating point is only one point on the load line. Under dynamic conditions, it is quite normal for the Class B load line to cut through the max dissipation curve, based on the use of sinusoidal wave forms. The plate dissipation rating is hardly exceeded when this happens, as under dynamic condition, the element of time must also be factored into the equation: instantaneous plate dissipation is only exceeding the "constant" max dissipation curve for a fraction of the time over the course of one complete cycle -- typically, less than 50% of the time. Much of the time, the plate is not conducting any current at all. Therefore, understand that under dynamic conditions, it is the average plate dissipation that must not exceed the max plate dissipation rating for the tube. As maxhifi correctly stated, your load lines are not steep enough. Understanding the average plate dissipation relative to the max plate dissipation rating of the tube lets you then see why your load lines can in fact be safely much steeper than they are.

Dave

I think it prudent to expect that someone will send a highly clipped instrument signal into the output stage, essentially a square wave, and that invariably they will adjust the signal level so that the peak of the waveform sits right on the part of the load line that has the highest plate dissipation. I think I could assume a 50% duty cycle for that waveform. If that's the case, could I have the load line just kiss the dissipation curve for 2X the listed max, assuming all other cases would be slightly less brutal? Of course the quiescent point would have to be outside 1X the listed plate dissipation curve.

If so, that definitely allows me to drop the load impedance to something closer to typical practice.

 

[eljefe3126]

Assuming you really want 300V screens and you are settled on 800V quiescent voltage on the pate, you've got to get your class B line up to the point on the graph of approximately 60V, 350 mA. Meaning much much steeper than currently drawn. Just eyeballing it, that would put each primary side under class B operation at (800V - 60V) / 350 mA = 2114Ω, thus full plate to plate winding impedance 4 * 2114Ω = 8.4KΩ.

That will deliver plate power of about:
(800V - 60V)/1.414 * (350 mA/1.414) = 129 Watts (combined across both tubes in the PP pair)

And deliver power at the speaker terminals of about:
129 watts * 0.85 = 109 Watts.

I dunno if the KT88s can handle that much poop. If you can back it off to between 90 and 100 watts plate power, I think you would have a longer term reliable amp. There have been several threads on this forum running the KT88's (or maybe they were 6550s) that way that you could look at to see the loading conditions they used. Those threads were in regard to the Bogen 200 sound distribution amps, but I don't remember the thread names exactly.

I'm not married to 300V screens, I just don't have curves for higher screen voltage. For example, the Hiwatt DR201 runs at 450V screens, and I suspect the Reeves 225 goes even higher. I have data for a Marshall Major which was connected in ultralinear with B+ of 666V (measured, probably 650 nominal) and over 640V quiescent on the screens (bad, bad, bad!) Even sagging under load, the screens still took over 600V, which is why the Major has a reputation as a tube-eater. Gorgeous sound, though.

More importantly, then my original power calculation of 84 Watts Class AB2 and 76.5 Watts Class AB1 was correct? That certainly explains why 15,200 Ohms is wrong, and all those other amps use much less impedance on the output transformer. Back to the drawing board...

I think some of those other threads are what originally drew me here, but I may not have seen all of them. Thanks for the tip.

 

[dcgillespie]

Referencing your post #4, again remember the time element of this exercise: Since this is nearly a perfect Class B exercise, each tube is only conducting 1/2 of the time, making that .567A only .284A over time. Then, because the the figures we are dealing with at this point are peak figures, the current draw over time will will be an average of that number, based on a sinusoidal waveform -- which all such exercises are classically based on. An average of .284A peak current over time will be approximately 142 mA -- or very close to the 140 mA shown in the curve for 150 watts power output.

Dave

 

[maxhifi]

In

Greetings. I'm having a little difficulty with this and wondering if I'm doing something wrong. Hopefully, someone here can set me straight.

I'm trying to wring every last watt out of a pair of KT-88s. The project is a musical instrument amp, so distortion levels aren't critical (although I can't have crossover distortion, so class B is out). I'm considering going to Class AB2, but even if I stick with Class AB1, the question still stands.

The attached pic shows what I'm trying to do. The red line is the "Class A" loadline, with Vo at 800 Volts and Ipo biased to 43-44 mA. The other half of the push-pull pair cuts off and the active tube goes into "Class B" operation at about 468 Volts and 87 mA. Finally, the system peaks at about 35 Volts and 200 mA (Class AB1) or 0 Volts and 210 mA (Class AB2).

The green curve is constant plate dissipation at 42 Watts.

View attachment 1353360

Now, from the blue "Class B" line, I get an impedance of 3800 Ohms. If my true impedance is 4x that, that seems absurdly high at 15,200 Ohms. To put that in perspective, the Marhall Major produces about 200 Watts from 4 KT-88s in ultralinear, with an output transformer of 2,250 Ohms, equivalent to 4,500 Oms for a single pair. The Hiwatt DR201 is in the same class, 200 Watts from 4 KT-88s in Pentode mode, from an output transformer at 2,200 Ohms, equivalent to 4,400 Ohms from a single pair. Hiwatt also made the DR405, which extracted something north of 300 Watts from 6 KT-88s, with the output transformer at 1,700 Ohms (equal to 5,100 Ohms for a single pair).

5,100 Ohms is a long way from 15,200 Ohms. Before I start winding a transformer, I want to be sure that I'm not committing some gross error.

There is also output power to deal with. I'm using P = (HT-Vmin) * Ipeak / 2, which in Class AB2 gives P = (800-0) * 0.210 /2 = 84 Watts. I should be somewhere north of 100 Watts for the pair, which is what the reference circuits in the GE manual show. Is that for one side only? If so, 168 Watts is really good news, as the Class B reference circuit generates 150 Watts into a 6,000 Ohm load "for speech only", and apparently something less into 7,500 Ohms "for other purposes".

In Class AB1, power should be P = (800-35)*0.200 = 76.5 Watts per tube, which would yield 153 Watts without the Class B crossover distortion, which would still be a good day.

Using P = 2 * (HT-50)^2 / Rload, I get 2 * (800-50)^2 / 3800 = 296 Watts, which seems wrong. I presume the 50 is an assumed Class AB1 peak at Va = 50 Volts, Ia = 0. If I double 3800 to 7600, I get 148 Watts, which seems reasonable.

I'm just looking for a sanity check here, thanks.

For your load line, the class AB1 power is as follows:

1. The slope of that blue load line gives an impedance of 800 / 0.210 = 3809 ohms
2. This gives a plate to plate load impedance of 3809 x 4 = 15238 ohms
3. Power output is the maximum RMS plate to plate voltage squared, divided by the plate to plate load resistance = ((0.7071*(800-50))^2 / 15238 = 18.43W

Your mistake is confusing peak voltage, peak to peak voltage, and RMS voltage. You can see how such a mismatched load impedance is leaving a lot of power on the table here.

 

[kward]

In Class AB1, power should be P = (800-35)*0.200 = 76.5 Watts per tube

More importantly, then my original power calculation of 84 Watts Class AB2 and 76.5 Watts Class AB1 was correct?

The 76.5 watts estimate seems correct to me, and the result you stated is correct but you need to divide by 2 to get that number. (i.e. divide each voltage and current swing estimates by 1.414 to convert from peak to RMS.)

Power output is the maximum RMS plate to plate voltage squared, divided by the plate to plate load resistance = ((0.7071*(800-50))^2 / 15238 = 26.10W

But in class B, each tube is only conducting every half cycle, so each sees about 3.8K load, so power calculated in the V Squared over R method is nearly identical as the I*E method.

 

[kward]

I found the thread I was referring to earlier. See this post:
http://www.audiokarma.org/forums/in...hty-bogen-mo-200a.556575/page-15#post-7518740

Note in the attached schematic on that post that the primary impedance is 4750 ohms (plate to plate), center tap voltage is 690V, screen voltage is 340V. In that setup, 100 watts was achieved. As compared to data from your opening post on this thread, the Bogen has less voltage swing but more current swing while still achieving 100 watts output.

Some data anyway for you to compare and consider....

 

[eljefe3126]

First, I'd like to thank everyone for the help. I spent some time playing around with load lines that just kissed the 84 W dissipation curve, and a lot of things came into focus. Mainly, I need to trust that operation is going to follow the load line, and that the tube characteristics control what it CAN do, but the output transformer characteristics control what it DOES do.

Let's start with that strange 150 W Class B circuit from the handbook. We know that at idle, it's at 850 V, 10 mA. We know the transformer load is 6k Ohm. So, in order to produce 150 W rms, the signal peak has to be located at 179 V and 447 mA. I don't know what the characteristics of the tube are when operated as a "high impedance triode". But I think it's fair to assume that whatever they are, a plate current of 447 mA can flow at a plate voltage of 179 V. Maximum instantaneous plate dissipation occurs at 425 V and 284 mA, and is 120 Watts! No wonder this circuit is recommended for "normal speech applications only"!

For "other purposes", a plate load of 7.5k Ohm is suggested. Let's look at how that works. I'll assume that we can still swing down to 179 V, which given that it's acting like some kind of triode probably isn't too far-fetched. Max current drops to 358 mA and power output drops to exactly 120 Watts. Max instantaneous dissipation is still high at 96 Watts, but maybe not that much higher than the 84 Watts I was playing with (still handles worst case of a square wave parked right at the worst point on the load line).

Let's try another one from the manual, Class AB1, Tetrode-connected, fixed-bias, 100W, screen voltage is 300. Va(o) is 552 V, Ia(o) is 60 mA. RL(a-a) is 4.5k Ohms. Signal peak is 78 volts, 422 mA. We're close...the best I can do on the charateristics graph is 80 V and 420 mA, so call it 88 V and 412 mA using the chart. That gives me 82 Watts instead of 100. Either way, the maximum instantaneous plate dissipation is 68 watts which should be quite manageable.

Next, let's try the Marshall Major, with 2 push-pull pairs of KT-88s in parallel, connected in ultra-linear fashion. These are measured values, not nominal or calculated. Va(o) is 666 V, sags to 626 V under load. Vg2 is 660 V (!!),. sags to 605 V under load. Pscreen is 20 Watts under load, which is probably why this amp is a notorious tube-eater. Pout is 221 Watts for 2 pair, or 110 per set. Primary impedance is calculated from voltage measurements to be 2311 Ohms, or 4622 Ohms equivalent for one push-pull pair.

So, I calculate a signal peak at 162 V and 436 mA. Maximum instantaneous plate dissipation is 96 Watts. Looking at the ultra-linear chart in the manual, this seems reasonable, although the current level is off the top of the graph. If I project the Vg1=0 line up, I don't think this is to the left of it.

It's not clear to me why this amp is designed this way, since the manual has a 100 W version of the fixed-bias ultralinear circuit at 560 V with a 4.5k Ohm load. Why go over 600 V on the screens if you don't need it?

Finally, the various Hiwatt and Reeves (high quality clones) KT-88 amps. These either have 2 (DR201, Reeves 225) or 3 (DR405, Reeves 400) push-pull pairs in tetrode connection, fixed bias. For the Hiwatts, Va(o) is 650 V, and Vg2(o) is 410 V. I am told the Reeves run higher voltages in order to get more output. Replacement output tranformers for the Hiwatts are listed at 2250 Ohms (2 pair) and 1450 Ohms (3 pair), which work out to 4500 Ohms and 4350 Ohms respectively. Power outputs are "at least 200 Watts" for the DR201, 256 Watts measured for the Reaves 225, "over 300 but not 400" for the DR405, and anyone's guess for the Reeves 400.

At Va(o) = 650 and Pout = 100, and RL(a-a) of 4500, I get a signal peak at 176 V, 422 mA, and a maximum plate dissipation of 94 Watts. That seems possible even at Vg2 = 300 V from the graphs.

Same thing, RL(a-a) of 4350, signal peak is 184 V, 428 mA, maximum dissipation is 97 Watts. Still seems doable with Vg2 = 300 V.

Let's try to reverse engineer the Reeves 225. First guess is Va(0) = 650, Pout = 128, RL(a-a) = 4500. Peak is at 113 V, 477 mA, dissipation 94 W. I don't think that can be done at any Vg2 level. If you project the cutoff line up as a straight line, you get to 120 V before you get to 477 mA. Admittedly, this is a bit of a guess since I'm off the top of the chart. Perhaps this is why it is claimed that "you can't get those power levels with those voltages".

Next guess, same thing, Va(o) = 700 V. Peak is at 163 V, 477 mA...this seems much more doable, given enough Vg2. Peak dissipation rises to 109 Watts, starting to get to scary territory for music.

Let's try the 3-pair circuit at Va(o) = 650 V with the 4350 Ohm transformer, aiming for 128 Watts output. Peak is at 122.3 V, 485 mA. That might or might not be possible, depending on what the tube characteristics look like with Vg2 at 500 or 600 volts. Dissipation is 97 Watts. There's not much point in seeing if we can hit 133.3 Watts from 650 V, we will clearly be left of the cutoff line.

Split the difference between 650 and 700 with the 4350 Ohm transformer, Va(o) = 675 V. Peak is at 136 V, 495 mA. Pout is 133.3 Wats, maximum dissipation is 105 Watts. If 400 Watts is possible from 6 KT-88s, that's probably how it is done.


So, this has all been very helpful, and I have a much clearer idea of where I'm going. One thing on my to-do list is to see if I can find the RL(a-a) that Reeves uses, since nobody else seems to push KT-88s so hard. They advertise that they use transformers wound to the original [Hiwatt] Partridge specs, but that could mean identical quality specs with a slightly differenct turns ratio. I'd also like to see what Va(o) they use.

I'd also like to know how one guestimates characteristic curves that are past the edge of the graph. I don't know what design reference the engineers at Hiwatt would have used back in the 1960s beyond the GEC tube manual, and that has no curves for the sort of screen voltages they used.

And finally, I'd like to explore Class AB2 operation. If anyone can point me to some references, I'd be grateful. I don't understand how to design anything that takes me to the left of the Vg1 = 0 line, but I know people do it. For now, I'll take a look at the Fender PS 300 and PS 400, which I am told were Class AB2 amps.

Thanks again.

 

[kward]

I'd also like to know how one guestimates characteristic curves that are past the edge of the graph.

One of two ways probably:
1. Run the tube on the test bench using a bench supply and see what current it draws at various points. Thus plotting your own characteristic curves for the operating conditions you want.

2. Guess. Take a look for example at the "transfer characteristics" chart on the 1972 GE 6550A data sheet:



You can estimate operating conditions above 300V Ec2 by guessing at where the 350V Ec2 curve would be, for example, by drawing it in at roughly the same distance above the 300V curve as the 300V curve is above the 250V curve. You can use this chart with your estimated Ec2=350V curve to draw the plate curves for Ec2 = 350V.

 

[cozido512]

Welcome to the 21st century...
Interactive valve datasheets.

 

[eljefe3126]

Welcome to the 21st century...
Interactive valve datasheets.

Sweet!

 

[kward]

Welcome to the 21st century...

I'm not sure how I would make that work with my slip stick:

 

[dcgillespie]

Remember that in any kind of tapped screen design, as the plate voltage swings lower, so does the screen grid voltage by the amount of AC voltage present at the screen tap, determined of course by the percent of the winding it is placed at. In many KT88 designs, this was placed at 40% in high quality applications. I don't know where the taps are placed when used in production amplifier service (i.e. guitar amps), but in view of keeping some check on screen dissipation, and that a high level of power output is safely obtainable even in UL mode with these tubes, it's not unreasonable to assume that they might be placed at 40% in this type of service as well - or at least at 33% anyway. KT88 tubes respond well to a wider range of screen tap positions than other tubes, where the precise position is of much more importance. Speaking of Fender, I always marvel at all of the talking heads that refer to their tapped screen grid designs as "Ultralinear" designs. UL operation occurs at a very precise screen tap placement in most tubes, and varies with each tube type at that. For the 6L6 family of tubes, this occurs with the screens tapped rather precisely at 43%. In the Fender designs using a tapped screen OPT transformer, the taps are placed at just 12.5% -- high enough to keep screen dissipation in check when operated from a 500 volt supply, and low enough to prevent any significant power loss. It also allowed Fender to conveniently do away with using a choke on those models as well. If the Push-Pull-Parallel KT88 designs from Marshall do in fact use a 40% screen tap (or close to it), then the tubes in those units are truly operating in Ultralinear mode. But if much less, then the design is merely a tapped screen grid design.

Dave

 

[eljefe3126]

If the Push-Pull-Parallel KT88 designs from Marshall do in fact use a 40% screen tap (or close to it), then the tubes in those units are truly operating in Ultralinear mode. But if much less, then the design is merely a tapped screen grid design.
Dave

The OT is described as "ultralinear" by a replacement parts vendor. That and $5 will get you a Starbucks.