How a Class A Output Section Really Works

[HotBluePlates]

I started this topic rather than hijack another thread, with the intent of clearing up misconceptions about output tubes and the output transformer. If you want to eventually design rather than copy, you'll need to understand how this works. Hopefully others can absorb the info faster the way it's presented here, rather than wait a decade for the lightbulb to turn on (which is what I did).

Your OT choice depends more on what output tubes you will be using more than how much power you want the amp to put out.

Your amp understanding will go much farther, faster, when you come to realize this statement is backwards.

The OT primary impedance and the power supply voltage/current determine the power output. The chosen output tubes simply have to be able to swing the voltage/current that the OT and power supply require to achieve the designed power output.

The error in perspective arises because amp makers typically use the fewest number of output tubes of the size/type appropriate for a given power output. Then when you go to clone the amp (say a Fender AB763 with 2x 6L6's), it's natural to notice that all models with that same output section have 4kΩ OT primaries. The obvious conclusion is that 4kΩ is the "impedance used with 2x 6L6's" and that the 6L6 dictated that impedance.

If you cloned a Princeton Reverb instead, you would have had an 8kΩ OT primary with 2x 6V6's. The obvious conclusion is 6V6 needs an 8kΩ primary. But where does that logic leave you when you find out the 2x 6V6 Deluxe Reverb uses a 6.6kΩ OT primary impedance? Or that with the same 2x 6V6 and essentially the same supply voltage, the Deluxe Reverb outputs more power than the Princeton Reverb?

More to the point, the "old way" of thinking of the OT-output tube pairing is a certain mental model of how these things work together. But the model fails when you consider the Princeton Reverb & Deluxe Reverb above, or when the question is raised about why Marshall amps with 2 output tubes might have a 4kΩ OT or a 6.6kΩ OT, yet you could use KT66/6L6/EL34 in either setup.

So, what happens when you want to use 6V6's to be the output section of a Super Reverb, with its 4kΩ OT primary for 2x 6L6's? Assuming you can add 2x more sockets, you could use 4x 6V6 in that amp with no other changes. The simple (but incorrect) logic would say, "2x 6V6's need 8kΩ, but 4x 6V6 is like 2 parallel pairs of 6V6's, so the OT impedance is half."

The correct interpretation is:

• The ~40w Super Reverb circuit is about double-power of the Princeton Reverb, but uses the same supply voltage.
• To get 2x power with same-voltage, the equation for power tells you that current must double (Power=Voltage*Current).
• Ohm's Law tells you to get double current when voltage stays the same, you need half-impedance (Current = Voltage/Resistance).
• So you need half the OT impedance (4kΩ in the Super Reverb rather than 8kΩ in the Princeton Reverb).
• But 2x 6V6 isn't capable of flowing double the current they pass in the stock Princeton Reverb, while remaining within plate dissipation limits.
• So doubling tubes to 4x 6V6 will control the needed amount of current in the 4kΩ OT.

I'll post additional info shortly to help build output stage understanding from the ground up.

[HotBluePlates]

To start off, consider the simple a.c. circuit below.



On the left side is a squiggly line in a circle; this is an a.c. generator. On the right is a resistor. To fully analyze this circuit, you would want to know the resistance of the resistor, and the voltage output of the a.c. generator, and the circuit current implied by these two things. From there, you could easily determine the power dissipated by the resistor. If you know any 2 of these items (resistance, voltage, current or power), you can calculate everything else.

At this point, if you don't have a firm grasp of Ohm's Law and the Equation for Power, it would be a good time to download and read NEETS Modules 1 & 2. This is a pretty user-friendly presentation of basic electronics. Also, you may want to download the Radiotron Designer's Handbook, 4th Edition (RDH4), which has all the facts I'll present though perhaps in a form that's hard to decipher until you already know the material. RDH4 is hosted in the Library of Information on this site, along with RDH3 which is a simpler presentation of some of the same material.

So let's add some numbers to the diagram. See Example 1 below.

The a.c. generator outputs 447v RMS, and the resistor is 4kΩ. We find the current in the circuit with 447v/4kΩ = 111.75mA.
The power dissipated in the resistor is 447v * 0.11175A = 49.95w. We could just round that up to 50w.

You could skip a step and calculate power directly from the voltage and resistance. The form is Power = Voltage²/Resistance, so 447v²/4000Ω = 49.95w, same as before. But sometimes it's important to know the current in the circuit.

Now let's see what happens if the same 447v RMS is applied to an 8kΩ load. See Example 2 below.

The a.c. generator outputs 447v RMS, and the resistor is 8kΩ. We find the current in the circuit with 447v/8kΩ = 55.875mA.
The power dissipated in the resistor is 447v * 0.055875A = 24.98w. We could just round that up to 25w. So double-resistance gave half-power dissipation.

Skipping the current calculation, we have 447v²/8000Ω = 24.98w, same as before.

[HotBluePlates]

Now let's turn things sideways a little. Look at the Example below.

A voltage for the a.c. generator is not specified, but the current in the circuit is shown as 93.5mA. We also have a known load resistance of 4kΩ.

The RMS voltage across the load is then Voltage = Current * Resistance = 0.0935A * 4000Ω = 374v RMS.
The power dissipated by the resistor is 374v * 0.0935A = 34.97w (round that up to 35w).

You could skip a step by calculating power dissipation as Power = Current² * Resistance = 0.0935A² * 4000Ω = 34.97w.

There's an important reason I switched to a known circuit current and load resistance. The a.c. generator represents the power supply and output tubes, while the resistor represents the output transformer and speaker.

• The power supply sets up a d.c. voltage applied to the output tube(s) through the output transformer. The supply has a maximum average current is can provide before the voltage output of the power transformer starts collapsing (this might be specified in transformer specs as "300-0-300v, 120mA d.c., or simply 120mA).
• The output tubes accept a voltage input at their grids, and pull some amount of current through the output transformer primary in response. While we may speak of a "voltage output" at a tube's plate, this is always indirectly created. The tube can't create a voltage at its plate, but can change the current flowing through the tube. This changing current can create a changing voltage drop across a connected resistance/impedance according to Ohm's Law, which then results in a changing plate voltage.
• The output transformer has no impedance of its own, but it does have a turns ratio between its primary and secondary. The speaker has an intrinsic impedance, so when connected to the OT secondary the speaker impedance is reflected by the OT's impedance ratio (the square of its turns ratio) to reflect a set impedance to the primary. So when you attach the specified speaker load to an OT secondary, you will get the specified primary impedance reflected.
• We can use a resistor as a model for the OT and speaker because we are simplifying to assume we're operating "mid-band" where transformer losses and parasitic effects aren't significant, and also that the transformer is operated within its ratings. You will find in actual practice that this correlates well with reality, especially when the transformer is rated for full power at a bandwidth wider than you'll be using. The speaker's impedance is not constant, but we pretend it is for now so that the focus is on output stage operation.
• So when we calculate power dissipated in the resistor, this matches the power we would calculate when looking at an output stage and the OT primary impedance. Actual output transformers take a power at their primary and transform it into an equal power at their secondary, with an efficiency in the range of 90-97% depending on the excellence of the transformer. With this in mind, we can save the mental gymnastics of what's happening at the speaker, and how the transformer transforms voltage & current from primary to secondary, so that we can simplify calculations to voltage, current and OT primary impedance with the knowledge that the power calculated there as "dissipation" as actually passed to the speaker to pump air in our ears.

[HotBluePlates]

Let's move on to an example push-pull output stage.

Look at the 1st diagram below. This is exactly the same as our previous equivalent circuit, except for the new ground symbol. This is the center-rap of the output transformer, which is connected to the B+ supply. The power supply is the same as ground, as far as a.c. is concerned, because large-value filter caps (which pass a.c.) connect from the B+ to ground.

Some folks will be troubled by that ground symbol; they shouldn't, as it is not a magic hole down which signal disappears. It's just a reference point. It is shown because the a.c. voltage across the entire primary may be split into 2 equal halves on either side of the symbol (the OT center-tap).

Now look at the 2nd diagram. Everything is the same as we've previously seen, except we have some new numbers and a ground symbol at the mid-point of the load. Let's calculate as before.

346.4v RMS / 8000Ω = 43.3mA
346.4v RMS * 43.3mA = 15w, or
346.4v RMS²/8000Ω = 15w

So there will be 346.4v RMS across the whole OT primary as long as 43.3mA RMS flows through it. What about each half of the primary?

Well it's simple: 43.3mA is the current through the total circuit, including all of the load (the OT primary). If you look at only 1/2 the primary, there will be half the total resistance of the full primary (4000Ω in this case), so the voltage across each half-primary is 4000Ω * 0.0433A = 173.2v RMS. The total voltage is divided evenly between each half primary (see the 3rd diagram).

"RMS" is a convenient way to describe a.c. in terms of the equivalent amount of d.c. which would cause the same heat dissipation in a resistor. So before we can apply it in a useful way outside of power calculations, we need to know how much peak voltage this implies. We assume sine waves for a.c. to make the math simpler.

346.4v RMS (whole primary)  =>  173.2v RMS (half primary)  =>  173.2v * √2 = ~245v peak.
Recall √2 = ~1.414, and is the value used to convert RMS to the peak of a sine wave. To go from peak voltage to RMS, you would divide by √2 (or multiply by 1/√2, which is the same). So Peak * 1/√2 = Peak * 0.7071 = RMS for a sine wave.

Let's find peak current, too. 43.3mA RMS * √2 = 61.2mA peak

[HotBluePlates]

Now let's take what we've figured out and design an output stage.

In the last example, I cheated a little. I picked a Hammond 15w OT with an 8kΩ primary. The RMS voltage applied by the output stage was found by working backwards from the known 15w of output and the OT primary impedance. The issue is that the OT does a great job of transferring power from the primary to secondary with almost no loss, unless you apply more power than the core can handle. Then the OT losses start rising, and the output stage no longer acts the way you might have planned.

So given our Hammond 1750E, let's figure out what kind of output stage (and power supply capability) will be needed to enable it to deliver its rated 15w to the speaker. Power = Voltage²/Resistance, so the formula can be rearranged to Voltage² = Power * Resistance, or Voltage = √(Power * Resistance).

Voltage = √(15w * 8000Ω) = 346.4v RMS

And we found this means we need 173.2v RMS per half-primary, or 245v peak. At the same time, this half of the output stage needs to deliver 43.3mA RMS or 61.2mA peak. An important point is while there may be half-voltage across the half-primary, RMS or peak current needed remains unchanged. That's because in the original series a.c. circuit, there is only 1 path for current to flow, so all current goes through all parts of the primary. We indirectly checked this fact earlier finding the half-primary voltage while using the total output section current.

We should quickly look at how RMS power relates to peak power for sine waves. We know our output stage makes 15w RMS of output power to the speaker when we have 346.4v RMS and 43.3mA RMS; we can easily find this is the same as 489.88v Peak and 61.2mA Peak. 489.88v Peak * 0.0612mA Peak = 29.98w Peak (really 30w Peak, due to rounding several times). So Peak Power = 2 * RMS Power.

We know our tube will need to swing 245v peak and 61mA peak to make our 15w of output power. A helpful relation between the different values needed arises, if we use peak values across 1/2 the primary. Since peak power is twice RMS power as just shown, we can calculate what happens on only one side of the OT and determine RMS power output. Proof: the power indicated from our half-winding values will be 245v Peak * 0.0612mA Peak = 15w, which is our RMS power output.

This fact is doubly helpful, because we need peak values to figure out how the tube should operate. We will assume our output tube is a single 6V6, and to make data sheet curves easy to use, let's assume there will be 250v on the screen.

A tube can't pull its plate all the way to 0v, so there is some minimum voltage required above the value we calculated for peak half-primary voltage. To find this, look at the plate characteristics, specifically at the uppermost E_c1=0v grid line. The point where this line goes from being mostly-horizontal and bends downward on the way to 0v and 0mA is called the "knee" of the curve. In general, the tube should not be operated with the loadline running below the knee (cutting the vertical portion of that grid line), So look for the plate voltage at which the knee occurs on that 0v gridline. Look below at the attached 6V6 curves, and you see it happens around 40v or so when the screen is at 250v, though the knee will happen at a slightly higher plate voltage as screen voltage is increased.

If we assume the 6V6 needs about 50v on the plate minimum, then we need a supply voltage of at least 245v peak a.c. + 50v minimum plate voltage = 295v. Let's round that to 300v for the remainder of our design.

[HotBluePlates]

Can we operate this output stage in Class A? Recall that the theoretical maximum efficiency of Class A is 50%, and we have 2x 12w tubes, so it seems the maximum output power to stay in Class A would be 12w. The answer appears to be "No."

But let's look into it further. Our output section is to be 15w using an 8kΩ OT, with a supply voltage (B+) of about 300v and 61mA peak current. What idle current is needed?

The rule of thumb to use is that for Class A operation, the tubes must idle at a current equal to or greater than the needed peak current. Recall that a.c. swings both positive and negative around a reference point which is 0vac or 0mAac. And given the OT does pass any output power to the speaker in response to d.c. in the primary, the tubes' idle current is our zero-reference for our a.c. output.

So when we calculated our peak current and peak voltage, those were in one direction away from the reference. They were not peak-to-peak, which which account for both positive and negative excursions away from the reference. Further, since in Class A all tubes are conducting all the time,
our tubes can only ever instantaneously touch 0mA during their negative excursion, or else they have crossed over to Class AB.

Look at the 1st diagram below. The "ground point" at the middle of the OT has indicated 61mA idle, and is the current which will be our "zero reference for a.c." The upper half-primary is shown at the moment it reaches the positive peak current of 122mA (61mA peak above the reference), and has a solid green arrow showing current directed away from the mid-point of the OT towards the tube. The lower half-primary is shown at the moment it reaches the negative peak current of 0mA (61mA peak below the reference), and has a dashed green arrow showing current directed away from the tube towards the mid-point of the OT.

Now the choice of how to depict the direction of the arrows in relation to the tube current was purely arbitrary, so don't get too attached to those. However, what is important to notice is that while each half of the push-pull output stage is making the opposite current-change from the individual tube's perspective, from the load's perspective there is but one current flowing through the whole OT primary in the same direction. This matches the  earlier statement that the calculated current for the designed power output for the whole primary is equal to the current in each half-primary.

Look at the 2nd diagram below. This shows the opposite half-cycle of the a.c., and is identical to the 1st diagram, except that the tubes have changed direction of their current swing away from idle, and as a result the current through the OT primary is moving in the opposite direction.

Now you should see where "push-pull" comes from.

Now that we understand why the tube must idle at or above the value of peak current to operate in Class A, will our output stage work in Class A? We need at least 300v of B+, and an idle of 61mA or greater. 300v * 0.061A = 18.3w. Since we're using 6V6's with a 12w plate dissipation rating, we cannot idle the tubes this hot, and cannot get 15w from a pair of 6V6's in Class A.

In the next installment, I'll show how we'll need to adjust the output stage so that we can operate in Class A with 6V6's.

[HotBluePlates]

For the time being, please consolidate replies in this thread. I'm trying to maintain continuity in this thread, and it will take some time before I get everything out there for Class A & Class AB.

[HotBluePlates]

So let's continue by setting up our 6V6's to run in Class A with the 8kΩ Hammond 1750E.

As it turned out, we couldn't make 2x 6V6 produce 15w to fully drive the OT and stay in Class A. However, we should be able to get 2x Class A 6V6's to output 10w. Earlier, we found we could find the RMS voltage across the primary with Voltage = √(Power * Resistance).

Voltage = √(10w * 8000Ω) = 282.8v RMS; continuing with the other steps previously shown:

Current = 10w / 282.8v RMS = 35.36mA RMS
Half-primary RMS voltage = 282.8v RMS / 2 = 141.4v RMS
Half-primary Peak Voltage = 141.4v RMS * √2 = ~200v Peak
Peak Current = 35.36mA RMS * √2 = 50mA Peak

We previously found 50v as a safe minimum plate voltage for 6V6's with 250v on the screen, so:
Required B+ = Half-primary Peak Voltage + Minimum Plate Voltage = 200v Peak + 50v = 250vdc

And we saw why, for Class A, idle current is numerically equal to the required peak current, so
Idle dissipation = B+ * Idle Current = B+ * Peak Current = 250vdc * 50mA = 12.5w

That seems like it's too high for a 12w dissipation tube. It is (or is pushing the envelope), but I'm gonna cheat a little because we also know 6V6's later carried a 14w plate dissipation rating.

Let's assume the plate and screen voltages are exactly the same at 250vdc. This allows easy use of the published plate curves. But what bias do we need to get 50mA idle at 250v plate & screen?

The easiest way, if the plate & screen are the same voltage, is to use the triode curves for the tube. See the lower graph on page 4 of the 6V6GT data sheet, and the 1st graph below. Triode operation of a pentode/beam power tube assumes the plate is tied to the screen, such that both are the same voltage. This exactly applies to our situation at the moment, as we've assume plate & screen are both 250vdc. In fact, this technique can be used at other times when plate & screen aren't exactly the same, as the driving factors of plate current are screen voltage and control grid voltage.

• This is an object lesson in the value of using a choke between plate and screen nodes. The choke has little voltage drop for the d.c. screen voltage, but provides great filtering for ripple.

So erect a vertical line at the B+ voltage we calculated, and a horizontal line at the calculate idle current. The intersection is the bias voltage needed to idle the tube. In our case, the lines cross almost exactly between -10v and -15v, so we can call that -12.5v bias.

We need to account for idle screen current to get accurate cathode biasing. Look at the 2nd diagram below. The dashed lines indicate screen current when the control grid (G1) is a specified voltage. Conveniently for us, there is a line for E_c1 = -12.5v (the red arrow). If we see where this line crosses our plate voltage, then read the screen current of the scale on the right side of the graph, we see it's almost 5mA (and we'll call it 5mA for simplicity).

The required cathode bias resistor will be the required bias voltage divided by the sum of plate and screen current. -12.5v / (50mA + 5mA) = 227Ω. There are 240Ω resistors available but 250Ω is fine, is common and allows non-critical tolerance. You'll want to round up to the nearest standard value. The resistor is assumed to dissipate (50mA + 5mA) * 12.5v = ~0.69w, so a 2w or bigger resistor is plenty.

You can use a cathode bypass cap, but if the tubes are reasonably matched the cap is unneeded because in Class A the increase of current in one side is offset by the decrease in current on the other side.

The actual total B+ is raised by the amount of bias voltage. The tube still only feels the voltage from plate-to-cathode and screen-to-cathode, and in any event a 12v discrepancy is not enough to upset operation of the output stage. We'd like to have a B+ of 262.5v, but 250v is close enough (235v is probably close enough, too).

So our total output stage specs are:
Push-Pull Class A 6V6's
OT Primary Impedance: 8kΩ plate-to-plate
B+ = 250 to 262vdc
Screen voltage = Plate voltage (as near as practical)
Idle Plate Current: 50mA
Idle Screen Current: ~5mA
Bias voltage: -12.5v
Cathode Resistor: 240-250Ω

[HotBluePlates]

One other tidbit for those whom this previous example was all child's play...

Look at the data sheet entry for a single 6V6 with 250v plate and screen. Notice it lands on the same bias voltage and near-enough same plate current. Take a look at the value for transconductance (G_m).

The American unit for G_m is the "mho" which is "Ohm" spelled backwards. And when you rearrange Ohm's Law to solve for resistance, you get Resistance = Voltage / Current.

Well, the Mho is the inverse of the Ohm, and so Conductance (Mho) = Current / Voltage.

If you also remember the "micro" before a unit is one-millionth, while "milli" is one-thousandth, then you'll notice something else... On the data sheet, under these conditions, the 6V6's G_m is 4100 micro-mhos, which is the same as saying 4.1 milli-mhos. And since Mho = Current / Voltage, this is the same as saying "4.1mA / 1v" or 4.1mA/v.

This latter expression is how the Brits (and maybe the rest of Europe) specify G_m. So what does that mean to us?

We have a bias of -12.5v. It is assumed that fully-driving the output tube rasies the G₁ voltage momentarily to 0v during the positive peak of the input signal. This is born out in that the data sheet says the peak G₁ voltage is also 12.5v.

If our peak signal input is 12.5v, and we have a G_m of 4.1mA/v, then
12.5v * 4.1mA = 51.25mA peak

And that near-enough exactly matches the peak tube current we need for our full 10w of output power across 8kΩ with a 250-262v supply.

[HotBluePlates]

Before continuing to other examples, we should look at RDH4 page 561 to understand how real pentodes differ from an "ideal pentode." You can follow along by downloading RDH4 from the link I provided earlier, though I have posted the relevant text & graphs below.

First off, why use an "ideal pentode" and what is different about it than a real tube? In a nutshell, an ideal pentode allows allows the design process I showed earlier to be easily derived with a knowledge of Ohm's Law and the Equation for Power, as well as simple facts about how to convert RMS sine voltages/currents to peak voltages/currents. The ideal pentode allows a quick, mathematical process to generate all the information about a proposed output stage to figure out if it will meet your output power needs, and whether you can reasonably use off-the-shelf output and power transformers to bring it to life. The only things to know about an output stage design using ideal pentodes is that real pentodes will result in less output power, more distortion, or both.

All this is readily presented in RDH4, though if you already didn't know the facts shown, a translation would help. Here, I'll present that translation.

"An ideal pentode is one having infinite plate resistance, a 90° angular knee and equally spaced characteristic curves." (text and "Ideal Pentode Graph")

Looking at the graph, "infinite plate resistance" means that the gridlines on the graph of characteristic curves are exactly horizontal. This means that as long as the tube's control grid (G₁) has the voltage indicated by that line, the plate current is one value regardless of how high/low the plate voltage is.

"90° angular knee" means the grid line stays at its constant current as plate voltage drops until it reaches the knee, where plate current immediately drops to 0mA. Rather than a rounded downward turn, this is a sharp corner, and its effect is to increase the area of the rectangle formed by the plate current and plate voltage changes. You'll see shortly how that also means increased power over real pentodes.

"Equally spaced characteristic curves" means that if you start at any grid line (a value of G₁, or "control grid" voltage; G₁ is also written as C₁ in some texts) and look to the lines above and below it, that each represents exactly the same amount of current change. For example, say you look at the -15v grid line, and see it indicates 30mA of plate current. You then look to the -13v grid line and see it indicates 34mA of plate current, while the -17v gridline indicates 26mA of plate current (equal voltage change in either direction yields exactly the same current change). Similarly, our example curves would show 60mA for the 0v grid line, and 0mA for the -30v grid line.

• This matters because it implies zero distortion, and a perfectly linear tube. Every part of an input signal results in exactly the same corresponding change in plate current, which creates exactly the same corresponding voltage drop across the load impedance.

[HotBluePlates]

Continuing on page 561:
"The zero bias characteristic is OAC, the -E_c1 characteristic is ODE, while the -2E_c1 characteristic is OB."

Look at the 1st graph linked below. OAC is traced in blue (though it extends all the way down to 0mA; it is just covered by the green line). ODE is traced in green. OB is traced in red. So what do these tell you about the tube?

The blue OAC line shows the tube with the 90° knee, and the knee's position shows that plate current stays a constant value until plate voltage hits 0v, at which point the plate current drops immediately to 0mA. In other words, the ideal pentode can "pull its plate down to 0v."

The green ODE line also has the same 90° knee, also position right at 0v on the plate. What this implies is that plate voltage is irrelevant with respect to the tube's plate current unless the plate voltage is 0v, and that the only controlling factor for plate current is control grid voltage.The red line OB is at 0mA. But it is also "-2Ec1" which means that if -Ec1 is our idle grid voltage, then -2Ec1 is double that voltage. For example, if idle grid voltage is -15v, then -2Ec1 is -30v. This line is also implying that the ideal pentode can "turn off" easily and linearly.

"The operating point is Q, the optimum loadline is AQB where AQ = QB and the distortion is zero." (2nd graph below)

Q is selected as the operating point for maximum output power in Class A because it is exactly halfway along the current axis between 0mA and that shown for 0v on the control grid ("full-on"). Though a plate dissipation curve is not shown, we could assume that point Q sits right on maximum plate dissipation.

AQB is the optimum loadline because the area of the rectangle bounded by points O, A, C & B is as large as possible. Current swings from idle to double-idle and to zero-current, and the voltage drop created across the load resistance causes plate voltage to swing from its idle value to 0v and double-idle plate voltage.

The distortion is zero because the excurisions along the loadline are equal in both directions. AQ is shown as the blue line, and QB is the red line. Plate current increases because of the grid voltage increase; the current increase causes a voltage drop across the load's resistance in accordance with Ohm's Law, and that voltage drop subtracts to leave less voltage on the tube plate. Our load is linear, like a resistor so with equal plate current changes in either direction, we equal plate voltage changes in either direction.

• If AQ was a different length than QB (which would happen because the current changes are unequal), then the amplification or output would be different for the positive half of the signal from the negative half of the signal.

[HotBluePlates]

"Optimum Load" was described as the load which maximizes output power by making the rectangle formed by the loadline as large as possible. But what does that mean?

Look at the graph below. The black loadline is the original ideal load, but now I've added some numbers to the graph.

The ideal load line idles at 30mA at 200v on the plate. It swings to double-idle current (60mA) and to zero current. This line represent a load resistance of 6.67kΩ. Know for now this represents a power output of 3w (you'll see the math later).

The blue line shows a reduced load of 4kΩ. The line is steeper, so for the same current change of 30mA in each direction, we only get a plate voltage change of 30mA * 4kΩ = 120v in each direction instead of the original 200v. The faint blue lines show the rectangle formed is as-tall as the ideal load, but is not as wide. The power output is reduced to 1.8w.

The red line shows a increased load of 8kΩ. The line is shallower, which means it reduces current (as you would expect more resistance to do). For our 200v change in either direction, we only get 200v / 8kΩ = 25mA of current change in either direction. Said another way, we can only increase the plate current by 25mA before the line hits a plate voltage of 0v and current drops to zero. So a equal, distortionless excursion in the other direction implies a corresponding 25mA change. The faint red lines show the rectangle formed is as-wide as the ideal load, but not as tall. The power output is reduced from 3w for the ideal load to 2.5w (but obviously not as bad as the 4kΩ load).

Let's say you want to verify the new loads really are 4kΩ and 8kΩ. How would you do that?

Ohm's Law can be written 3 ways; one of them is Resistance = Voltage / Current. Since 2 points define a straight line, we should pick any 2 points on a given loadline where we have numbers available.

For the 4kΩ line, we know the tube idles at 200v & 30mA, and has endpoints at 80v & 60mA and at 320v & 0mA. Take the change in value between the 2 points (subtract the smaller from the larger) and use Ohm's Law to find the load resistance. Say we use the idle point and 80v & 60mA. (200v-80v)/(60mA-30mA) = 120v/30mA = 120v/0.03A = 4000Ω. You could easily use the 2 endpoints; the result will be the same.

For the 8kΩ line, we have endpoints at 0v & 55mA and at 400v & 5mA. (400v-0v)/(55mA-5mA) = 400v/50mA = 400v/0.05A = 8000Ω.

[HotBluePlates]

Now let's continue with the equations presented on page 561 (lower portion of the text attached earlier).

1. Optimum R_L = E_b / I_b

We just showed how load resistance can be calculated by change of voltage divided by change of current. The tube idles at E_b (plate voltage) and I_b (plate current). Moving  along the loadline we looked at above towards the left, voltage drops to zero (a change equal to E_b) while current rises to 2I_b (a change equal to I_b).

This is also proved graphically in the diagram below.

2. P_o = 1/2 E_bI_b

As mentioned above, E_b & I_b are the change in voltage and current from the idle point to an endpoint. If we think of the control grid voltage input signal creating these changes, then E_b & I_b are the peak values resulting from that input signal. But we usually specify RMS power not peak power. We can arrive at RMS power by dividing peak power by 2 (which is how the forumla above is written).

Or we can convert each peak value to RMS.
RMS volts = E_b / √2, and RMS current = I_b / √2.
1/√2 * 1/√2 = 1 / (√2 * √2) = 1/2
Which is the same result that we got earlier in this thread when comparing peak power to RMS power.

3. D.C power input = E_bI_b

This is simply idle dissipation, plate voltage times plate current. However, since both voltage and current swing between zero and 2E_b and 2I_b, then E_b & I_b are also the average values for the a.c. voltage and current swings (this is only ever true when the tube has zero distortion and an optimum load).

4. Plate circuit efficiency = 1/2*E_bI_b / E_bI_b = 50%

Simply, this is "power output / power input". The upper term is 1/2 the lower term, so efficiency is 50%.

5. E_c1 = E_c2 / 2μ_t
where E_c2 = screen voltage
and μ_t = triode amplification factor = μ_g1g2

This formula indicates the bias voltage of the tube with an optimum load. The triode amplification factor (μ_g1g2) is listed on pentode and beam power tube data sheets. The amplification factor (μ or mu) is a measure of how much more effective a change of grid voltage is at causing a plate current change, than is a change of plate voltage.

Since pentode plate resistance is very high (often treated as if it were infinite), a pentode's screen voltage is much more effective at controlling plate current than is its plate voltage. And the control grid voltage is more effective than screen grid voltage.  So triode amplification factor (μ_g1g2) is a measure of how much more effective the control grid is over the screen grid at controlling plate current.

Okay, so what?

Let's say our tube's μ_g1g2 = 10, and the screen voltage is 250v. If we applied -25v to G1, it has the effect of an applied screen voltage of -25 * 10 = -250v. But screen voltage is 250v. The tube feels a screen voltage (250v) which would increase plate current exactly offset by a control grid voltage (-25v, μg1g2 = 10) which decreases plate current. So this is the cutoff bias of the tube.

The grid lines on our graph of the ideal pentode show an idle bias halfway between cutoff and 0v on the grid. So we need to idle halfway between E_c2 / μ_t and 0v, which is where the 2 in the denominator comes from.

Equations 6 & 7 are really not for our optimum load case. They actually provide shortcut forms for the equation for power when you have a load higher or lower than optimum (like I showed in the last post). They use either E_b or I_b when the value for the non-optimum load matches that of the optimum (meaning, for the lower-than-optimum load, you get 100% of the current variation of the optimum case, so I_b is used in the calculation). This simply prevents you from having to use all steps to calculate output power the way I did earlier.