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

[Dave H]

Does the picture show cathode currents?

No, sorry. It's plate current. I should have done cathode current.
The plot below is the current through 1 ohm cathode resistors.

[Helmholtz]

Thanks!

[nsubulysses]

After some thought, the best answer I can give you is, "I really don't know."

But you can find out easily enough and at minimal cost. Stick with the transformer you have and just tweak the value of the dummy load. So, if you have 1.6k transformer running into 8 ohms, then for a 1.2k transformer make the load 8 x 1.2/1.6= 6 ohms by whatever means you have to hand. Running at 1Khz means there won't be an issue with the transformer saturating at the higher power. There may be a difference in transformer losses but I'd expect it to be insignificant.

Wow!!

[nsubulysses]

Wow again! with 1.2K primary it works like I hoped it would.

This solves the case. It's not grid drive, not the resistor in place of choke, not PT. It's the output transformer primary impedance that is too high

I dont think I would have thought to manipulate the OT primary impedance by mismatching the load. And I certainly don't know the math to do it. BUT, if you say hook up a 6 ohm load and use the 8 ohm tap and test I can do that. Luckily I found a few of these slide resistors at a garage sale a while back

Measuring over 1 ohm 1% resistor in cathode it reads what I would roughly expect at clipping from viewing other amps, ~130mA. With 1.6K primary it will read ~110mA

it does about 35.3V RMS (bottom right of scope picture) on 8 ohm tap into a 6 ohm load -- 207.7W

with 1.6K primary it does 27V into a correct 4 ohm load on 4 ohm tap for 182W





smocking gun as they say

[nickb]

That was a lucky find at the garage sale. Is there such a thing as resistor envy?

Personally, I would have i set to only just clip when measuring the power, but if what you have is your personal landmark who am I to disagree?

[nsubulysses]

Do you think that 0452 on the resistor is the date code and it's from 1952? Pretty sure it had never been used before when I bought it. It was in a white box the size of the resistor and it had the ohms and watts printed on it. It smelled strange when it heated up when I was using it. maybe it had some 70 year old dust on it.

I got a few of these weird old things . the guy had a box with many dozen in there. Most were weird resistances but I did find one 4 ohm and a few that were 16 ohms or less with the slider adjustment . Were these just old dummy loads of were they supposed to be for something else ?

[Helmholtz]

I also have some of these resistors. They were used in all kind of power electronics. Are intended for free air mounting and can get very hot. So be careful when clipping your probes to the hot lugs. Can make the plastic sleeve melt (guess how I know , but that was 35 years ago).
Don't know about date code.

Regarding power measurement with "mismatched" load: If OT has a specified impedance ratio of 1.6k/8 Ohm, this means that the reflected primary load will always be 200 times the secondary load. So 6 Ohm reflects to 1.2k.
1.6k and 8 Ohm are NOT properties of the OT, these values just specify a ratio, given by the the square of the turns' ratio.

I told you before that 1.2k will increase output power.

[Old Tele man]

Sorry, but OUTPUT power increases with higher Zoo -- assuming there's sufficient B+ available -- it's distortion that decreases with lower Zoo:

The basic P = R*I^2 formula for PP-AB1 using tube parameters: Po ≈ (Zoo/4)*(gm*Vg)^2

• Thermionic: http://thermionic.info/mccaul/McCaul...Value_2012.pdf

Disregard post per Old Tele man!

[nickb]

Sorry, but OUTPUT power increases with higher Zoo -- assuming there's sufficient B+ available -- it's distortion that decreases with lower Zoo:

The basic P = R*I^2 formula for PP-AB1 using tube parameters: Po ≈ (Zoo/4)*(gm*Vg)^2

• Thermionic: http://thermionic.info/mccaul/McCaul...Value_2012.pdf

The equations are small signal (linearized) and just won't work for large signal conditions. If what is said is true we'd all be using as high an impedance load as we could. It's self evident that an infinite load impedance cannot pass any current and so no power will be dissipated.

The reality is that for a given tube at a certain supply and screen voltage there is a load impedance at which the power output is maximized. It's pretty close to the point at which the load line crosses the knee of the Vg=0 line on the plate current vs plate voltage plot.

[Helmholtz]

The reality is that for a given tube at a certain supply and screen voltage there is a load impedance at which the power output is maximized. It's pretty close to the point at which the load line crosses the knee of the Vg=0 line on the plate current vs plate voltage plot.

Exactly!

Higher than optimal load impedance means that the loadline intersects the Vg = 0 curve below the knee, which reduces available plate current swing. Also increases average screen dissipation.

Too low load impedance reduces available plate voltage swing and may result in excessive plate dissipation.

[Helmholtz]

As Zoo is increased, the RL line pivots UPWARD, past the "knee" and runs out of 'available' plate voltage.

Should read: As Zoo is decreased...

Lower Zoo means a steeper loadline and vice versa. The slope of the loadline is inversely proportional to Zoo, as I is plotted over V.

The principle of optimal loadline construction is to make its useable part as long as possible, thereby allowing for large plate voltage AND large plate current swing. And this is achieved when it meets the Vg = O curve in the knee.

[Old Tele man]

I held off for a couple days to let the 'dust settle' before I responded:


1) Yes, "load-line through knee" yields the longest load-line length, but it does NOT guarantee maximum output power; that occurs when maximum plate-load voltage (dVp) swing and maximum plate-load current (dIp) swing occur simultaneously.

2) Maximum plate-load voltage (dVp) swing and maximum plate-load current (dIp) swing occur simultaneously when: "load-line orthogonally intersects the diode-cutoff line at a 90-degree angle."

3) Thus, maximum output power (dVp×dIp) is a function of (a) load-line length AND (b) the slope of the load-line, with power being the dVp·dIp AREA of the triangle enclosed beneath the load-line and its slope.

4) The slope of the load-line is typically (but not always) bounded by (a) the "knee" (maximum dIp) and (b) the quiescent/idle point (zero dVp), thus these two points establish the load-lines' slope.

5) When the load-line lies ABOVE the orthogonal 90-degree point on the diode-cutoff line, the plate current swing increases, BUT the plate voltage swing decreases...reducing output power from maximum.

6) When the load-line lies BELOW the orthogonal 90-degree point on the diode-cutoff line, the plate voltage swing increases, BUT the plate current swing decreases...again, reducing output power from maximum.

7) The location of the diode-cutoff "knee" is a function of the tube and its operating plate- and screen-grid voltages; the quiescent idle point, however, is user defined but is limited by the tubes plate power dissipation rating.

8) All the above apply equally to PP operation of power TRIODES, TETRODES, or PENTODES; the only difference being that TRIODES have no "knee" on their "Vg=0" diode-cutoff line.

9) So, *IF* the plate-load line in use *happens* to intersect the diode-cutoff line "knee" at 90-degrees, then YES, maximum power is occurring, but that seldom occurs in real circuits, which is WHY most PP amps are operated with load-lines somewhere BELOW the "knee"...closer to their orthogonal intersection point.

10) Also, the sloping "J-shape" of the diode-cutoff line means the orthogonal intersection point MUST occur above the lower curving area, where the diode-cutoff slope is most linear.

11) Now, *IF* the load-line is slightly ABOVE the orthogonal intersection point (typical "close-to-thru knee"), then slightly INCREASING the load-line value WILL increase power because it moves the intersection point back DOWN closer to the actual orthogonal intersection point, slightly reducing dIp while simultaneously increasing dVp...getting closer to maximum AREA beneath the sloping load-line.

12) So, for maximum output power to occur, dVp and dIp must be maximized, which ONLY occurs when the load-line orthogonally intersects the diode-cutoff line, maximizing the AREA of the triangle enclosed by the horizontal (Vp) and vertical (Ip) "shadows" cast by the sloping load-line.

13) Thus, it's NOT just about the length of the load-line -- it's about the maximization of the AREA of the Eb·Ib triangle created by the load-line and its slope!


Disregard post per Old Tele man!

[Helmholtz]

I completely agree that max. output power is determined by the (dVp*dIp)*0.5 area below the loadline. For pentodes this area correlates well with the length of the load line.

2) Maximum plate-load voltage (dVp) swing and maximum plate-load current (dIp) swing occur simultaneously when: "load-line orthogonally intersects the diode-cutoff line at a 90-degree angle."

I can't find a justification for such 90° condition. For pentodes and fixed plate supply voltage this would typically result in reduced dVp*dIp area, reducing available output power.

Also this angle would change with a different current scale. Defining an angle only makes sense if voltage and current scales are specified.

[Old Tele man]

I completely agree that max. output power is determined by the dVp*dIp area below the loadline. For pentodes this area correlates well with the length of the load line.



I can't find a justification for such 90° condition. For pentodes and fixed plate supply voltage this would typically result in reduced dVp*dIp area, reducing available output power.

Also this angle would change with a different current scale. Defining an angle only makes sense if voltage and current scales are specified.

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, idle plate dissipation, 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 on the graph.


Disregard post per Old Tele man!

[J M Fahey]

You are agreeing on the basics but focusing onb one or the other main factors.

Sad truth is that real world does not behave exactly like the ideal world, so we must start with ideal conditions, simply because we have simpler, usable models for them, the full model/equations minutely describing real world are unusably complex , so we start with simplest ones to get in the ballpark and then measure and tweak, measure and tweak, until we are reasonably happy/close.

For power we need both voltage swing and current swing.

Both are limited, but for different reasons.

* Voltage can go no higher than +V and can not get closer to ground than tube saturation voltage which in due turn depends on tube current at that point.

* current can not go lower than 0 mA , that´s easy to see, and can not go larger than what tube can put out **under saturation**

So best practical design method to maximize power is the graphical one, where "we see all curves at once" and can make decisions as to which load line gives us best results.

Design is a compromise, and plate current is very dependent (the wrong way) on saturation voltage so in general we can compromise on one or the other but never maximize both.

There are a few load lines (plate impedance values) which yield roughly similar power out, where you increase current a little but with a heavy saturation voltage penalty which eats the power you "would" have gained and viceversa.

That's why "just lower Z to increase power" in general does not apply, you lose significant voltage swing ... and make tubes suffer by trying to extract way too many electrons from that poor cathode ... one reason "modern tubes last little" .

Personally I prefer slightly higher rather than slightly lower Z load, simply because to begin with there is not a big power "loss" (which was never there in the first place) , but mainly because tubes last way longer and as a special bonus: amp is not picky/choosy about tube hyper quality/NOS/etc. , anybody works unter those relaxed conditions.

I can get real 40W RMS, any day of the week, from any brand/type/era 6L6 if fed 430/450V and loaded with 5500 ohms, value I got at after long experiments, while loading them with more popular 4500 ohms (or 4000 or 3800) gives me marginally more power (extra 5W? ... 10W on a freak day? ... not worth it), amp becomes more picky about what you plug in those sockets and tubes lose ooommmppphhhh in less than a year.

You can increase +V to 500V or even 550V ... as long as you keep screens at 400V from a relatively stiff supply, either an extra transformer tap or a regulator; just resistors do not guarantee anything, way high screen current swings.