Grounding
https://el34world.com/charts/grounds.htm
https://www.valvewizard.co.uk/Grounding.pdf
Triode Amplifier Design
https://www.valvewizard.co.uk/Common_Gain_Stage.pdf
Transformers
Faraday Shield https://canadatransformers.com/faraday-shield/
A Faraday shield in a transformer is a grounded conductive layer, often copper foil, placed between the primary and secondary windings. It acts as an electrostatic shield, reducing capacitive coupling and diverting noise and transients to ground, thus improving power quality and protecting sensitive electronic circuits.
https://www.hammfg.com/files/products/1594rfi/emi-rfi-shielding-info.pdf?v=1697662069
Resistance-Coupled Amplifiers
http://www.tubebooks.org/tubedata/hb-3/Receiving_Tubes_Part_1/Resistance-Coupled_Amplifiers.PDF
Biasing
Knowledge from HotBluePlates El34 https://el34world.com/Forum/index.php?topic=29807.msg328357#msg328357
Let's not focus on "biasing" because your original analogy was about "limits." That is, the ocean wave propagated fine until the "floor moved up," running into the bottom of the wave, and caused the wave to fall apart ("distort"?).
I swiped a graphic from Merlin's explanation of a basic gain stage to show "limits."
Merlin put a vertical line at 350v on the plate, because data sheets give a plate voltage limit of approx. this amount.
Merlin also drew a red curve corresponding to 1w of plate dissipation, again because data sheets have roughly this limiting value.
I added the Blue cross-hatched area denoting where grid current is certain to happen if input signal ventures to this extreme along a load line. This is certain-distortion on one side of the signal cycle, where the signal will be abruptly clipped.
I also added a light gray box down at the low plate current region, where the grid lines get very-bent. If the input signal drives a single-ended tube all the way along the loadline to zero plate current, the tube will clip. But the light-gray area represents increasing distortion before actual "clipping."
Now if we design a gain stage, we can bias anywhere inside that "Safe Operating Area" and our loadline can be literally any angle.
Afterwards, the 2 ends of our loadline will run into one of the four Limits that bound our operating area:
Grid Current
Low (or Zero) Plate Current
Plate Dissipation Limit
Plate Voltage Limit
One end of the loadline is then our "Floor Limit" and the other end is the "Ceiling Limit" (though they're generally positioned "upper left" and "bottom right").
Along the way, the angle of our loadline (which is the amount of plate load Resistance) will determine how much Output Voltage change we get for an Input Voltage change (our "Gain"). And that is going to depend a lot on the relative slope of the grid curves (which is really the Internal Plate Resistance; we're balancing external load resistance against internal resistance to garner "Gain").
Where we Bias just amounts to where we start along the loadline, and whether we've optimized the stage to accept a large input signal.
If tubes were perfect devices, there would be Zero Distortion until the signal hit the ends of the loadline & clipped. But they're not perfect, and there will be some distortion before "clipping." York's Amplifiers does a good job in Chapter 3 of explaining how to perform graphical analysis, and how to recognize the "distortion before clipping" (which is unequal amplification on each side of the signal cycle, resulting in an output waveform that is not the exact same shape as the input waveform).
It will greatly benefit you to have a death-grip on Ohm's Law, the Equation for Power, and how to convert between Peak and RMS values of a sine wave.
When designing from a blank sheet of paper, we start with the speaker & a desired power output, and work back towards the input jack.
Assume we're making a 40w Super Reverb. The voltage across the speaker terminals is
Volts = √(Power x Impedance) = √(40w x 2Ω) = 8.944v RMS
The output transformer is 4kΩ to 2Ω. The Voltage (or Turns) Ratio is the square-root of the Impedance Ratio:
4kΩ / 2Ω = 2000 : 1 ----> √2000 = 44.72 : 1 volts ratio
Let's use the transformer's Volts Ratio to figure the RMS volts on the primary side for our 40w output:
8.944v RMS x 44.72 = 399.98 volts RMS
The figure above is the voltage swing across the entire primary, or from "plate to plate." But we commonly calculate the required supply voltage & power output with respect to one side of the push-pull output stage. So let's divide our RMS Volts in 2 parts, to get the part contributed by one side:
399.98v RMS / 2 = 199.99v RMS
But we were talking earlier about "limits" and that the signal is only 1/√2 (0.7071 or 70.71%) times as big as its limit. So let's convert from RMS sine volts to a peak value:
199.99v RMS x √2 = 282.83v Peak
We now know that to make our 40w of output, the 6L6 on one side will reach a peak plate voltage swing of nearly 283 volts. But the supply voltage needs to be "Tube Output Volts" + "Tube Volts." That is, there needs to be some voltage left across the tube for it to continue working & amplifying. There might even need to be some extra "safety margin."
The upper graph of Page 6 of the 6L6GC Data Sheet shows the limiting 0v gridline for a number of screen voltages. The "Knee" of the curve is where it goes from being mostly-horizontal to mostly-vertical. Our loadline should stay just above the knee on the horizontal part, and this implies a lower-limit of plate voltage swing.
The knee of the curve for Ec2 = 400v lands pretty close to 100v on the Plate Voltage Axis. If there were an Ec2 = 450v curve, we might guess we would need closer to 120v at the 6L6GC plate.
Now we add our required minimum plate voltage of 120v to our peak signal output:
120v + 282.83v Peak = 402.83v required
Now let's look at the Super Reverb schematic: 460v plate & screen. About 50v of extra margin allowed for supply voltage sag when the amp pushes maximum output power.
I know people mostly bias by plate dissipation today, and that generally works. But what did Fender proabbly do to determine the required bias for the Super Reverb?
Grid Input Volts result in a Plate Current Output of the 6L6GC. This plate current is pulled through the output transformer's primary impedance to create the "plate volt output" we calculated earlier. How much plate current? We calculate by looking at one side of the push-pull power section:
Transformer Impedance at Max Power is 1/4 the plate-to-plate impedance: 4kΩ / 4 = 1kΩ
Peak Plate Current = Peak Plate Volts / Transformer Impedance = 282.83v Peak / 1kΩ = 282.83 mA Peak
How much plate current a tube pulls for a fixed amount of grid-voltage change is the tube's "transconductance" (or "Gm"). Gm is not a constant value, but changes with plate current. But we usually see 6L6 Gm figures cited between 5-6 "micromhos."
Mho is the inverse of "Ohm" so it is "Conductance = Amperes / Volts" (where Ohm's Law is Resistance = Volts / Amperes)
Europeans say "5 micromhos" as "5 milliamps per volt" or "5mA/volt"
This means 1v of change at the 6L6 grid yields a 5mA change of plate current
How much Grid Voltage Input do we need to get ~283mA of Plate Current Change? Let's guess a middle-Gm of 5.5mA/volt or 5500 micromhos:
283mA Peak / 5.5mA/volt = 51.45 Volts Peak
Look again at the Super Reverb schematic:
If the peak of the Signal Input exceeds the bias voltage, we push the grid positive of the cathode & draw grid current.
That grid-current case is a "limit" we mentioned earlier.
We see Fender chose a bias voltage of -52v, just a bit over the needed signal input calculated above.
The above should show why Fender landed on 450-460v of supply voltage for their Super Reverb, and why they selected a particular bias voltage.
Transconductance = 1/impedance
Example 6D6
Gm = Trans = 1600 uMHOS; 1/1600*1e6 = 625ohms
Plate resistance (spec @ 250v 8.2ma) = 800k
Mu = Gain = Amp factor = 800k/625 = 1280
For 1ma Plate R = 100k; Mu = 100k/625
Example 6SJ5
Gm = Trans = 1650 uMHOS; 1/1650*1e6 = 606ohms
Plate resistance (spec @250v, 3ma) = 1Mohm
Mu = Gain = Amp factor = 1M/606 = 1650
True, as defined, and at high current (8.2mA, 250V). Gm is 1600uMho or 625 Ohms, rp is 800,000k. The quotient is 1280. https://frank.pocnet.net/sheets/127/6/6D6.pdf
But (as you know) we almost never see this quoted on datasheets because amplification-factor of a pentode is never a useful number. Especially in audio. The external resistor to feed DC to the plate is always much less than the pentode rp.
Also the datasheet numbers are show-off values, and especially this RF/IF tube which can use like 8mA. In Timbo's plan with a 250k plate resistor under a maybe 250V-350V supply the tube current is more like 1mA. Gm is likely to be near 500-600uMho or 1800 Ohms, rp well over 1Meg, so stage gain may be (250k||500k)/1.8k or 91.
Audio (HiFi) Amplifiers
https://www.worldradiohistory.com/BOOKSHELF-ARH/Circuits-For-Audio-Amplifiers-Mullard-1963%20.pdf
HiFi/Radio Museum
https://www.stonevintageradio.com/description.php?II=964&UID=
