Understanding "grid drive" to drive, 2, 4, 6 power tubes

[Helmholtz]

Trying to create a visualization:

1) Draw a first-order pentode Eb·Ib curve using a simple straight line to substitute for the diode-cutoff line; that's the left-hand leg of the "triangle", ie: " / "
2) Draw the load-line from the idle point (conjunction of dVp=0 and Ipq) up toward the knee; that's the right-hand leg of the "triangle", ie: " \ "
3) Draw a "base" line from the idle point horizontally (parallel to X-axis) over to the vertical (Y-axis); that's the base of the "triangle", ie: " _ "
4) Now, drop a line vertically DOWN from the intersection of steps #1 and #2, where diode-cutoff line and load-line cross, and read the dIp and dVp values at that point and calculate P, the area of the triangle.
5) Slide the load-line DOWN the diode-cutoff line slightly (while holding idle point constant) and read the dIp and dVp values and recalculate P...it should be getting slightly greater.
6) Repeat step #5 and you'll see that although dIp is decreasing, that dVp is actually increasing (slightly).
7) At some point dIp and dVp will simultaneously reach "common" optimum values, yielding maximum output power under the given conditions of XX-load-line and (assumed) constant idle point (because tube and B+ haven't been changed).
8) It's all about maximizing the AREA of the Eb·Ib triangle under the load-line and its slope...which occurs then the load-line slope is 45-degrees and dVp and dIp have equal lengths.

Just did that for 4 different loadlines using real tube charts and got opposite power results: Max. power is achieved for longest loadline at the knee. From there loadlines having lesser slope (meaning higher load impedance) give less area/power.

[Old Tele man]

...and I bet all of those load-lines encroached well into and above the 30W plate dissipation curve, didn't they. Put that constraint on your curves and see what happens, although, I WILL concede that not everyone operates below the 6L6GC's 30W rating.


Disregard post per Old Tele man!

[Helmholtz]

...and I bet all of those load-lines encroached well into and above the 30W plate dissipation curve, didn't they. Put that constraint on your curves and see what happens, although, I WILL concede that not everyone operates below the 6L6GC's 30W rating.

Exceeding the plate dissipation limit in (class AB or B) PP operation is acceptable as each tube only conducts half the time.

But to make sure, I repeated the graphical procedure with 6550 charts and class A operation (fixed operating point). Same result: Power drops if the loadline intersects the Vg = 0 curve below the knee.

The "longest useable loadline" method may not work satisfactorily with pentodes having very rounded characteristics (not showing a sharp knee). But there are still the datasheet recommendations for optimal load.

[Helmholtz]

My contributions have to be seen within the context of nsubulysses' adventure to squeeze out the most of 6 x 6CA7s.
(Not what I would do, especially as 1 dB (26%) more power is barely audible.)

[Old Tele man]

NOTICE to HELMHOLTZ and J M FAHEY -- After some memory (and notes) searching, I've come to the conclusion that I must concede to both of you that my earlier posts (#143 and later) are indeed WRONG...and that your explanations are CORRECT.

So-o-o-o-o, I will attempt to remove the offending posts above, but will leave this one. Thanks for hearing me out, nevertheless.

[nickb]

Here's a plot of a 6L6 at Vsuppply= 450V, Vg2=400v, Vg1=0 . Of particular interest is red line showing the max peak power out.

Plate current (GREEN)
Max (peak) power out (RED)
1.58k Load line (BLACK)
30W tube dissipation line (BLUE)

Of note:
1) In this particular case max power occurs at the top of the knee. Whether this is necessary that this is the greatest distance from the Iq point, I'm not certain.
2) The max power point, i.e. the highest point of the RED line occurs where it's slope is zero.
3) The max power point is just that, is has nothing to do with the load line per se.

Just for simplicity I chose Iq=0. I don't think that affects any conclusions.

[Helmholtz]

Thanks, interesting.

3) The max power point is just that, is has nothing to do with the load line per se.

This I don't understand, as it is the loadline that determines the input for power calculation.

As mentioned result is peak power corresponding to twice real (RMS) power.


Using the 6L6GC GE datasheet chart and a 1.58k loadline at same conditions I meet the Vg = 0 curve markedly below the knee. Your saturation voltage seems much lower than datasheet.

[Old Tele man]

Yes, that plot shows the 6L6 "knee" too low to my eyes too, I vaguely remember it being something closer to 80Vp.



ADDENDUM -- I found this RCA 6L6GC datasheet Eb·Ib plot which seems to confirm nickb's "knee" at 50Vp:

[J M Fahey]

Your results match my experiments.

I start with 430V raw which drop under load, we also drop a few volts across primary DCR, I am satisfied with getting net working 400V at bplates, real world full drive.

And some 50V saturation voltage, so voltage swing is 350V peak.

Supposing we want maximum output and make a pure Class B amplifier (0 A at idle) and considering 260mA peak current for a good normal tube (what the datasheet shows by the way), so current swing is 0.26A peak.

So peak power into load is 350*0.26= 91Wpk so 45W RMS . No smoke and mirrors but Real World.

And load impedance would be 350/0.26=1346 ohms, single plate, or as commonly expressed: 5385 plate to plate.

This matches my experiments to a T, so I chose 5500 ohms (and actually wind my OTs for that value) , first because along time tube emission will be lower, so a slightly higher impedance is gentler to a tired tube, but also because being in Argentina I can´t be picky about tubes, all shops carry "the tube du jour" , meaning whatever was in the last tube container which got through Customs, so much so that sometimes there are no "original" tubes available but "branded" ones , meaning "Fender" or "Marshall" brand, with the odd Sovtek.
Other brands such as Svetlana, JJ, Tesla, etc. come and go according to wind and the phase of the moon, you can´t *count* on having them on a particular day.

And to be realistic: 40W RMS is almost same as 50W RMS and pushing 1 or 2 good Guitar speakers are more than enough to play in any Club situation.

As of 4 bottle tubes, think Twin or Plexi type, I have measured many putting out 40W RMS with a set of 4 original factory tubes, 20 or 25 years old , at which point owners complain that "the drummer is giving me trouble, while years ago *I* drove him crazy" and bring it to me to check what´s going on.

Of course, good quality non abused tubes only sllloooowwwwllllyyyyyy lose emission, nothing is noticed day-to-day .

Stan, a San Francisco Tech who moved to St Petersburg, Russia, and does some servicing on the side just not to get bored, often mentions getting amps for servicing with 30 or 40 years old original tubes even under weekly Pro use (think "Uncle Vanya and his merry Accordion") .
Surprised, he found those tubes are used under Datasheet conditions, say 25W RMS from a pair 6L6, 360V on plates and 250V on screens.
Maybe those who wrote the original datasheets knew "something" about tubes?

Now the Krazy Kapitalistic Amerikansky certainly abuse tubes, same way as they abuse their proletariat

[Old Tele man]

1) The RCA datasheet Class-AB1 'example' is 55W from Zoo = 5600, 450Vp, 400Vs, Vg(DC bias) = -37VDC and Vg(AC signal) = 35Vpk. Sound familiar?



2) Today's tubes are not the same as old original NOS tubes...most are merely 'close' to the original specs, quality, and reliability.

[nickb]

Thanks, interesting.



This I don't understand, as it is the loadline that determines the input for power calculation.

There is a max power than can be achieved for a given tube as specified voltages and idle current and ignoring tube limitations. Look at the max power curve. There is no load required to be specified to produce that line. OTOH it does imply a certain load in order to hit that point. The distinction is subtle I know but I think it helps to look as it this way.

As mentioned result is peak power corresponding to twice real (RMS) power.

Quite so. I used peak to keep the lines separate to make it easier to read.

Using the 6L6GC GE datasheet chart and a 1.58k loadline at same conditions I meet the Vg = 0 curve markedly below the knee. Your saturation voltage seems much lower than datasheet.

This is a red herring and is hardly a surprise. The data was extracted from one 6L6CG sample. As might be expected they never correspond to the data sheet values. The difference is not significant to the general picture.

[Helmholtz]

There is a max power than can be achieved for a given tube as specified voltages and idle current and ignoring tube limitations. Look at the max power curve. There is no load required to be specified to produce that line. OTOH it does imply a certain load in order to hit that point.

You are right, the power curve and thus max. peak power are completely defined by the plate curve and the operating point.
What I meant is that each pair of dVp/dIp values used to generate the power curve also defines a load line.

The difference is not significant to the general picture.

Not regarding the principle you stated above and your method.

But, of course, peak power will depend on the specific shape of the plate curve. Seems you picked a "better" than average 6L6 for your measurement.
I would expect a tube having a saturation voltage of around 80V to produce less peak power than your example. Which probably also means a different optimal loadline.

This is a red herring

Had to look this up. My comment was not meant to distract from the useful principle you showed. Just wondering about the unusually low saturation voltage and sharper than average knee.

[Old Tele man]

re: "There is no load required to be specified to produce that line."

That's not true, a load is always stated in the data sheet at EIA industry-standard (at the time) operating conditions of Vp=Vs=250V. However, that load is user selectable:



At 'other' operating conditions, a different load is appropriate, but its selection is left to the tube user, which is why data sheets include "example" circuits in different configurations.

[nickb]

re: "There is no load required to be specified to produce that line."

That's not true, ...

Sorry, no. The statement is a fact. It follows from how the max power line is defined. End of story. You'll just have to take my word that I didn't use the load resistance in deriving it.

...a load is always stated in the data sheet at EIA industry-standard (at the time) operating conditions of Vp=Vs=250V. However, that load is user selectable:

[ATTACH=CONFIG]52071[/ATTACH]

At 'other' operating conditions, a different load is appropriate, but its selection is left to the tube user, which is why data sheets include "example" circuits in different configurations.

Each point on the max power line does correspond to a load resistance is required to achieve that power. You are indeed are free to choose any load you wish. Just note the Vp where the load line intersects the Vg=0 curve and read the power off the max power line. Easy peasy!

This isn't a bad time to point out that the power is just an estimate based on some simplifying assumptions.

[Old Tele man]

re: "Each point on the max power line does correspond to a load resistance is required to achieve that power."

Exactly, since voltage-times-current defines power, but likewise simultaneously defines the load resistance (R = V/A) at that instant.