Understanding Amp Load Lines

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NB: 6/1/14
When I started this Post I didn’t expect how long but interesting and rewarding it would be – to study this whilst applying the Load Lines to a real world scenario – the Maggie amp.
Trying to solve its early volume level distortion problem was the perfect opportunity for this topic as I gained a much better understanding of loads of stuff – grid input limits, bias points and grid to anode swing limits, changing load resistors to alter stage gain, how changing the cathode resistor in a power stage radically alters the whole amp voltages, and more.
Home experimentation with thought and reason is highly recommended – you will understand and remember far better how and why things work.
If you come across something you can’t explain or doesn’t seem to make sense it is because you don’t know enough yet, or can’t see something staring you in the face. There WILL be a reason for it – its science.
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I have wanted to get to grips with this topic for a while now, as my understanding of load lines is rudimentary, with some of the many concepts lost on me.
I will use the Maggie amp as an example so that I can relate it to something I can actually measure and compare to theory, and hopefully have an empirical representation of this amp’s characteristics to see and understand the flaws that I know exist in this amp due the “experimental” approach I took when building this in stages, rather than just following a schematic as a “join the dots” build.
This was the first amp I built in stages to gain a better understanding of a valve amps operation. There was always a low volume start of distortion problem with this amp that I never understood at the time apart from it possibly being related somehow to a bias issue. Hopefully this Post will clarify this. (It did!)
The original 5 Posts start here:
http://stevepedwards.com/ElectronicsStuff/?p=3617
The project idea and parts values came originally from the Guitar Amp Handbook by Dave Hunter.

This “Handbook Amp” became the Maggie when I found this circuit was an almost identical schematic because it was derived from the book project also, as far as I recollect:

First I will take as many fresh measurements from my amp build (some component values are slightly different from the schematic) as I know or think will be relevant for drawing and understanding their function both in the circuit, on the Load Line graph, and as a function discussion recap in general.
Note there are silicon rectifiers in my build instead of the GZ34 tube rectifier.
The important quiescent DC voltages are in RED.
A+ R16, 220k; C7, 47uF = 355V (discharge to 5V in 5 secs)

B+ R17, 10K; C8, 16uF = 236V

C+ R18, 68k; C9, 8uF = 12AX7 triode 1 HT = 163V

12AX7 triode1+2 1oad resistor R6, R7 = 100K
12AX7 triode1 pin1 1oad resistor voltage = 108V

12AX7 triode2 pin6 1oad resistor voltage = 111V

12AX7 triode1+2 cathode bias R8 = 1.5K
12AX7 triode1 pin3 cathode bias R8 voltage = 0.8V

12AX7 triode2 pin8 cathode bias R8 voltage = 0.8V

The first thing to understand is a valve’s characteristics themselves. Different valves were designed historically for different purposes – audio, radio and power applications – for example. This means that the same parts of different valves – anode, cathode, control grid etc. – will have different characteristics and limitations depending on usage design.
Different family’s developed for different applications from many manufacturers worldwide so many numbering schemes and names exist for valves of equivalent use specification – the ECC83, 12AX7, 7025 etc:
http://en.wikipedia.org/wiki/12AX7
Each manufacturer type has a model specific datasheet, usually available as a PDF from the web:
12AX7 PDF here:

The first things that need demystifying are the abbreviations which are self explanatory electrical unit values, but relating to a particular part of the actual tube – but what are those parts? The ones I know are in red, or a best guess.
R. F. DOUBLE TRIODE

Base: NOVAL (9 pin)

Uf = 6,3/12,6 V (heater filament voltage ac/dc)

If = ca.300/150 mA (heater filament current ac/dc)

Typical

characteristic:

Ua = 250 V (anode voltage, Vak)

Ug = -2 V (grid to cathode voltage, Vgk)

Ia = 1,2 mA (anode to cathode current, Iak)

S = 1,6 mA/V (transconductance (Siemens) inverse of resistance)

Ri = 62,5 k???(anode (plate) resistance)

??= 100 (max gain – per triode)

Limiting values:

Ua = 300 V (max anode voltage)

Wa = 1 W (max anode dissipation power)

Ik = 8 m A (max cathode current?)

Ug = -50 V (max reverse grid voltage?)

Rg = 2,2 M??(max grid pull up resistance?)

Uk/f = 180 V (max cathode to heater filament voltage?)

Rk/f = 150 k??(max cathode to heater filament resistance?)

Capacitances:

system I. system II.

cg/k = 1,6 1,6 pF (grid to cathode?)

ca = 0,33 0,33 pF (anode to where?)

cg/a = 1,7 1,7 pF
(grid to anode?)
Maybe more will come to light as I proceed…
So, to get the logic process going, I’ll refer to a typical ECC83 example from Blencowe:
http://www.valvewizard.co.uk/gainstage.html
http://www.valvewizard.co.uk/Common_Gain_Stage.pdf
then compare with my amp values.
The first thing to understand graphically is the relationship between anode voltage variation and the current that flows to the cathode through a valve (or transistor as this is a fictitious entity below) as explained in Blencowe’s PDF:

There is a very linear part of the curve between the 110V-200V range which can be used for proportional amplification.
With the addition of a grid, fine control of this anode to cathode current can be realised – if the grid becomes more positive it draws more electrons from the cathode which then become drawn to the anode, so increase total current flow, but if the grid becomes negative relative to the cathode then electrons are repelled and kept trapped at the cathode so can cut the current off completely.
At this point it seems logical to assume that the best voltage value for the grid to have would be a value that would allow only half the maximum current rating of the valve to flow in so that when an AC signal is applied to the grid it would cause equal amounts between none and full current values for each half of the cycle. This would be a centrally biased grid point for a particular fixed anode voltage.
This then leads to the first graph in understanding Load Lines which is a family of curves (grid curves) plotted for different combinations of anode voltages for 9 fixed grid to cathode bias voltages:

You can see that for any fixed grid bias voltage line there is a fairly linear part of it that would be a uniform proportional change that would cause the anode voltage and its corresponding anode to cathode current to vary quite proportionately on a small enough scale – which is any point on an ac signal – but not over a larger scale that includes the curvy section and a straight section as this would deform a signal asymmetrically.
A tangent to any grid curve taken on a small enough scale it is a straight line, so for each line where that tangent point is the same value but on a different grid curve, it is approximately constant as seen below – the first constant that can be defined for a valve – the anode resistance Ra. This is the inverse of this tangent slope gradient at any point on a grid curve e.g. for these red lines depending on accuracy:
50V/0.0007A = 71428 ohms or 71.4k plate resistance or (350-225 = 125)/0.0017 = 73.5k ra

This observation leads to the second constant definition – the amplification factor, Mu or gain – or the ratio between the amount of control the grid voltage has compared to the anode voltage on affecting the anode to cathode current.
Below, it is obvious that any horizontal line represents a constant current value, but as the grid curves are quite uniformly spaced apart in places, this means that only a small change of grid voltage from say –0.5V to -1.5V (1V change) for a fixed anode voltage of 80V, would mean a drop in current from 1mA to 0.1mA (pink dots).
For this 1mA current value to be restored would mean raising the anode voltage to from 80V to 180V (100V change). The ratio of these voltage changes of anode to grid is the gain = 100V/1V = 100.
A small change in grid voltage makes a very larger change in anode voltage.
This value is a fixed maximum for this valve model because of the spacing between the grid curves. If they were further apart, a higher anode voltage would be required to re-instate current so the valve would have a higher gain, if closer together, then less gain would result.

The third constant is derived below from the vertical separation of the grid curves, representing the control the grid voltage has over current for a fixed anode voltage.

For a 1V change in grid voltage from -1V to -2V at a fixed anode of 230V, a drop of 3-1mA = 2mA occurs. This is a 2mA/1V value Standards Institute (SI) unit that is the inverse of V/I =R from Ohms Law, called Transconductance, and measured in Siemens.
http://en.wikipedia.org/wiki/Siemens_(unit)
You could think of it similarly as how much a resistor resists the flow of electricity but instead, how much a valve conducts electricity at a fixed voltage. This is an important distinction as the design would dictate how well a power tube for example, emits free electrons at a particular cathode temperature and maximum design limit HT voltage – its efficiency at passing current maybe.
The way I visualise Transconductance being the inverse of resistance is thinking that when the grid voltage increases, more current flows through the valve, causing a corresponding increase in cathode voltage because of the voltage dropped across the cathode resistor R8 (the voltage has to increase here to “force” more current through R8 (Rk).
This same “current forcing” applies at R6 (Ra) also. As these two resistors now drop a higher proportion of the available fixed 163V HT across them, there is less voltage available across the valve, yet current in the circuit has increased overall. This means the voltage across the valve itself has dropped yet current has increased (more electrons are being emitted from the cathode). This is transconductance – the inverse of resistance. This is also why this stage is a phase inverter, because as grid voltage goes up, the pin1 anode voltage goes down.
All three constants are related by ? = gm × ra.
Note the small r for anode (plate) resistance – not to be confused with Ra for an anode load resistor value.
Jumping ahead a bit now, as we know the amplified signal is dropped across the load resistor, Ra or for the Maggie case, R6 and R7:

Doing the maths for triode1 – the quiescent voltage across triode1 is 108V-0.8V = 107.2V.
VRa = 163-107.2 = 55.8V. The current through Ra = 55.8V/100K = 0.00056A = 0.56mA.
This should be the same as through the cathode resistor R8:
Cathode current = 0.8V/1.5K = 0.5mA
A further check may be to see if the valve current of 0.56mA through the valve anode at 107.2V with the spec sheet anode resistance of 62500 ohms equals the spec. transconductance of 2mA per Volt.
107.2V / 62500 ohm = 0.0017A – near enough!
The two clipping distortion extremes of voltage swing are when the grid voltage cuts off current completely so no current flows, so no voltage is dropped across Ra so the total HT of 163V, 0mA current is across the valve. This will be the first point marked on the Load Line graph.
The second point is derived by thinking that the valve is at full conduction so theoretically a short circuit. Then there is 0V at the anode and all (nearly all – the cathode resistor drops its 0.8V) the HT is across R6 (Ra). This would be a current of 163V/100000 ohms = 1.6mA.
The Load Line for the 100k resistor can now be plotted by joining these two points:

From this gradient it is now possible to see what changes in grid voltage cause what value changes in anode voltage and so the gain of the triode.
A change of grid voltage from 0V to -2V would cause an anode swing from about 50V up to 140V or a 90V change. This is a gain = 90V/2V = 45 gain.
It also shows the stage as an inverter, because the grid voltage moves positive from right to left as the anode voltage goes down.
As I know the cathode is biased at 0.8V I can roughly guess this grid curve point above – green circle.
I can also see what size grid signal will cause the current to stop – about -2.5V grid curve at about 150V HT. This would dictate the maximum positive anode swing before clipping as the current cuts off first – not at the 163V HT.
Realise that the grid lines are closer together near the bottom right than toward top left, so there is less anode voltage swing for a given grid bias lower down, so the gain will be less for a biasing in this lower region and lower C+. The original Weber Maggie design used a different mains transformer that drove a valve rectifier, so the HT and so C+ may have been different.
Triode1 gain = 45 as per graph above and scope below. 10 X probe at 1V/div with 1Vpp FLS test tone of 1 kHz.

An important and useful thing about the load line graph is that it can tell what type – whether symmetrical or asymmetrical – and limits of distortion up to clipping may occur to both halves of an input signal over a range of grid voltages. This is useful for deliberately clipping by design for an overdriven pre amp stage. I’ll look at triode 2 for that later.

Above is a triode1 grid input signal maximum size of about 2Vpp max before clipping should occur so is well outside of a Humbucker 200mV signal.
This is about right on the scope with 2V/div X10 anode voltage, with soft deformation starting at the max Scarlett phones output of 2Vpp giving about 64Vpp at the anode:

If I could inject a larger signal I should see both sides start to clip as one reaches current cut off at -2.5V grid curve and the other reaches the 0V limit – Blencowe’s grid-current limiting – more below.
This stage can cleanly amplify guitar signals well over the 200mV+ Humbucker signals to pass a clean amplified signal of about 45 times larger (4.5V or so) to triode 2.
This order of magnitude difference is important to realise as the guitar signal voltage swings at this scale are very small at 1/10thV grid input, shown below, compared to triode2:

Triode 2 Load Line and Distortion

As the anode resistor here is the same as triode 1 with about the same voltage (111V) at the anode then the triode response should be almost identical. So I thought… but I got really hard to explain measurements which it took me all day to see the grid current cut off explanation I gave above.
The difference in this stage is the up to 45 times larger signal from triode 1, which is enough to distort at the grid at this stage as will be seen.
Starting with a fresh Load Line for this triode2 voltage value, I drew the Load Line for a 100k anode resistor at 163V HT cut off voltage again, the 1.6mA Y axis value and/or the 0.8V cathode bias voltage and that corresponding current value (0.8/1500) at the grid curve:
C+ = 163V

12AX7 triode 1 pin6 anode load R7, 100k = 111V

12AX7 triode2 pin8 cathode bias R13, 1.5k voltage = 0.8V

The cathode current is 0.8V/1500 ohms = 0.000 53A. This is equal to anode current (163V – 111V) / 100000 ohms = 0.000 52A = 0.5mA

So to repeat the above logic as for triode1:
The two clipping distortion extremes of voltage swing are when the grid voltage cuts off current completely so no current flows, so no voltage is dropped across Ra so the total HT of 163V, 0mA current is across the valve. This will be the first point marked on the Load Line graph.

The second point is derived by thinking that the valve is at full conduction so theoretically a short circuit. Then there is 0V at the anode and all (nearly all – the cathode resistor drops its 0.8V) the HT is across R6 (Ra). This would be a current of 163V/100000 ohms = 1.6mA.

The Load Line for the 100k resistor can now be plotted by joining these two points:


This may make you think there would be an anode voltage swing before clipping of the bottom half of about 163V – 93V = 70Vpeak or 140Vpp either side of the orange bias line anode voltage. The problem here is that the grid curve at -2V reaches cathode current cut off before this can occur so I assume this is why the signal is clipped at a much lower voltage of only about 10Vp each side of the orange bias line that I recorded.
For scope measurements I first I had to remove the NFB link from the 16 ohm tap, as I thought the small gain of 20V/1.5V = 13.3 I got initially was due to this – it wasn’t.
This start of distortion occurs at 20Vpp anode (lower trace) voltage for a 1.5Vpp grid input:

Full volume distortion below gives a scope voltage of 4Vpp grid voltage for a 50Vpp distorted anode signal, though the grid curve lines reach 0mA around here so things get a bit messy to correlate and the valve behaviour here is sure to be unpredictable I guess.
The grid curve at -2V has already hit 0mA not far from the bias line anode voltage at around 93V so this would explain only seeing a 1.5Vpp grid voltage when distortion occurs as the cathode current is reaching cut off only about 10V either side of the 93V bias line. NOTE I am not really sure this explains graphically what scope values I actually took. I could be wrong, but the overall principles work.
At full volume below there is clipping of one side only of the anode signal.


Above – full distortion at about 50V anode with the 4V at the grid

Notice that as triode2 anode also inverts the grid signal (top channel) then the anode signal is cut off at the anode positive going cycle above.
The 1.5V grid curve swing here is about 0.75V each side of the 93V anode line Q point from grid curve 0.8V to 1.5V one side and 0.8V-0.75V = 0.05V the other side. Drawing this area on the graph explains the undistorted low gain results of 20V/1.5V = 13.3.
This is the only way I can think how these results are explained – may be wrong:

The bias point is way too low to the left for the signal levels at triode2?
There is clean amplification over a very small range of about 20Vpp either side of the orange 93V anode bias voltage line, but at full volume, the positive side (graph negative) is clipped and the full signal size of 50V appears offset on the scope, along the lower left side of the X axis where cathode current and grid voltage still work?

So how can this stage be improved from this poor behaviour?
Idea 1: Increase stage HT voltage?
28K R18 Change

The logic behind this was to raise the Load Line away from the curvy part of the grid curves by reducing the voltage drop across R18 at C+.
After drawing a Load Line for a theoretical but easily achievable small improvement of raising C+ to around 200V, to allow for a cleaner 2Vpp grid signal centred at -1V grid curve:

I decided to try it by putting another 5W, 56K resistor in parallel with R18. This was about right, as C+ is now 197V and the cathodes of both triodes are now at 1V. This should allow a 2Vpp grid swing from 0V grid curve to the -2V grid curve before cathode current is cut off too much?
Note the 2mA value on the Y axis comes from the new C+ of 197V/100k = 2mA. Also, the new current through R18 = 236V-197V = 39V/28000 = 1.4mA.
All new HT voltages are below:

Things are still a bit messy in this region it seems with these overall lower C+ voltages (Blencowe’s examples are at 300V), but it has improved the maximum distorted voltage to 80Vpp from 50V, but the start of distortion still starts at 20Vpp at about volume position 4. It has raised the power cathode bias a tiny amount from 9V to 9.5V also.
I will change the 100R R15 I have here back to the 500R value it should be now I understand what I’m doing a little more than I did when I built this amp back in June 2013. I can’t remember what flaky logic I used for reducing it in the first place now…Still, heads nicely to the next section which I need to fix first before coming back here.
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OK, experimentation is good, as you may find out things you didn’t expect.
Before I started looking at the Load Line for the EL34 I thought I would change the bias resistor to closer to the schematic value of 500R, though this is for a 6V6 really.
This caused a large unforeseen increase in the whole amps HT values below!
Power Tube EL34 with 550 ohm R15 Cathode Resistor

A+ 376V

B+ 308V

C+ 260V

Anode pin3 = 370V

Cathode resistor R15 550R pin8 = 21.6V

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Rk is called the cathode resistor or simply bias resistor. This form of biasing has several advantages:

It is self adjusting. If the average current through the valve increases, so

does the bias voltage, which will oppose the increase in anode current, and

vice versa. This means that the bias will adjust itself naturally to changes in

HT valve aging and variations in manufacturing tolerance. There is also

less chance of the valve going into runaway and over dissipating in the event

of some failure, which can be useful when dealing with power valves.

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This had the unexpected bonus of raising the C+ voltage that I was looking for earlier by another 60V, so it now looks like I can compensate back again by removing the extra 56K I put in parallel with R18 to make 28K there. This should be back to a more middle ground for both stages hopefully.
So a possible problem before was my mistaken low bias value for the power cathode value of 100R?
It is enlightening how much a seemingly small component value change in a stage can alter the whole amplifier.
As Blencowe states about lack of knowledge in these areas:
“It very rare that a single component is solely responsible a particular aspect of the tone, so achieving a particular sound will normally require an understanding of the preamp (and the rest of the system for that

matter) as a whole. This ‘global’ appreciation of the system marks the difference between the competent designer and the amateur circuit bender.”

OK, so I’m still a novice circuit bender…
First I’ll look at the grid curves for the EL34. The cathode current is 21V/550 = 0.04A or 40mA. This looks a crappy area for this 25W valve also if this is how you plot a load line for a power valve with no load resistor?

A bit different graph from a pre amp! Current values up to 0.28A and bias voltages from 0-25V. I can see now that my original 5V cathode voltage was using 0.2A just sat there quietly.
First I’m going to remove the extra 56k R18 and re-measure all amp values:
A+ 376V

B+ 310V

C+ 216V

Anode pin3 = 370V

Cathode resistor R15 550R pin8 = 21.6V

Unfortunately a small change to the resistor R15 550R pin8 in the wrong direction = 21.8V

This returned the pre amp cathodes back to 1V from 1.2V also.
I can re-draw the triode load line again to tidy it for the record as I will keep these values for now as I don’t have other 10W R15 cathode resistors at about 300R which I would like to try to reduce the power tube cathode to about 16V. So back to the triodes new Load Line.

Cathode current = 216V / 100000 ohms = 2mA still and the slope gain now becomes about 185-65V = 120V/2Vg = 60. No good – I want half this gain, not an increase!
A slight problem here also is that this gain value is from an anode voltage swing that is slightly asymmetrical for a symmetrical 2Vpp grid input. A slight decrease in bias point to say 0.8V could equalise this asymmetry. For the same 2mA anode current this would require a 0.8Vx 1500 = 1200 ohm resistor.
Also, note the reasoning behind grid curves not going above 0V on these Load Line plots – from Blencowe:
There is an obvious question that some readers have probably been asking since the start: why are there no more grid curves shown to the left of the Vgk = 0V curve? Curves for positive Vgk values do indeed exist, but they are not usually included on the graph because it is not practical to operate the valve in this region.

The reason is that when the grid voltage approaches the cathode voltage, electrons being drawn from the space charge get attracted to the grid rather than to the anode (since the grid is very much closer), rather as if we had placed a diode between grid and cathode. Conventional current flows in the opposite direction –into the grid and down through the cathode– so we have forward grid current. This current causes a voltage drop across whatever resistance happens to be in series with the grid, making it harder to drive the grid beyond Vgk = 0V. To put it crudely, less of our input voltage actually makes it to the grid. The more we attempt to make the grid positive the more current flows into it to prevent us from doing so, as if some invisible volume control is being suddenly turned down. In more technical terms, the input impedance of the valve suddenly falls from many megohms to a few kilohms or less, and this effect is known as grid-current limiting.

So, has the gain of the pre amp stages changed in reality on the scope?
For a 100mV input at triode 1 I get a 5V anode = 5V/0.1V = 50 gain. Increase from 45 before. I need less gain!
For a 2V input at triode1 I get 80V out = 80/2 = 40 gain.
For a 1V input at triode2, I now see GRID noise starting at 1V. I get 20V out. 20 gain.
I see anode distortion starting at 30V.
At a 3V grid voltage I see noise that I don’t know the reason for:

I now think I understand much better a designer’s problem here from looking at Load Lines…Yay! Am I learning??
After reading Blencowe’s words above about not going beyond 0V curve at the grid, it implies no more than a signal peak to peak voltage from 0V grid curve to -2V grid curve, centrally biased at -1V grid curve should be applied at a 12AX7 grid for an anode HT voltage of 216V? Yeah, I get that…
This would be a maximum input of 2Vpp signal. It would seem I don’t want a triode1 stage that goes to 5Vpp anode that can then be fed to the grid of triode2 at full volume, as this would be too large for the grid at triode2 as explained above surely? I measured the grid voltage at full volume and it is 4Vpp – twice too high!
This values implies this stage would need to be biased at -2V to handle this 4Vpp swing across the grid curves without going beyond the 0V curve but I wouldn’t have the anode swing available anyway there at -4V:

Less gain is required at triode1 so that only the maximum grid input at triode 2 can be sent at full volume – 30 should be ample giving 3Vpp at the grid – the gain being altered by changing load resistor values and/or HT voltage for triode1 only. Bear in mind I am getting 5V anode out for only a 100mV input here! This isn’t even a Humbucker 200mV guitar level which would be worse.
Both triodes are at C+ so changing HT for the triode1 stage only is more difficult to achieve than changing its gain.
The distortion of a stage comes from the asymmetry caused by the bias point, load and cathode resistor values – not by overloading the grid input.
I must remember also that this amp was designed with NFB which would also reduce gain of triode2 (but not reduce the actual grid input signal).
As it was designed for a 6V6 I want to see what difference changing power valves makes to overall HT voltage…
No significant change – power cathode is now 20.3V from 21
A+ 376V no change

B+ 310V – down 13

C+ 216V – down 1

Cathode resistor R15 550R pin8 = 20.3V down 1.3

But – what does that bias voltage mean for a 6V6? Well, as they are interchangeable in most cases you would think not much. I’ve drawn a load line for my anode voltage 376V with a cathode current of 20V/550 ohms = 0.04mA or 40mA – not perfectly symmetrical swing wise:

So before going any further, I want to reduce the gain at triode1 to about 30 to give a wider turn of the volume dial before distortion starts. From Blencowe (p21) the lower value the anode resistor the lower the gain and vice versa, as well as the ratio of Rload to Rk as this decides the voltage across Rk and so its bias point.
This is apparent in his example of the load lines of a 220k, 100k and 47k loads with -1.5V bias point for each case – something like:

It is obvious that the slopes over the same grid curve voltage gives different anode voltage swings – larger swings with a 220k than a 47K for the same grid input – 1Vpp in the above case.
For 47k say 40V/2V = 20 gain.
For 100k say 105V/2 = 52 gain
For 220k say 125V/2V = 62 gain
If I change my load resistor of triode1 to a 47K from 100k, I can I reduce the gain, but for an overall increase in quiescent current.
For my C+ this would be a max current of about 216V/47000 ohms = 4mA giving a new Load Line:

This is a bit asymmetrical but over the 100mV input range this would be insignificant for a few volts output. I may have to alter the cathode resistor to get a 1V bias point again. Actually, no I wouldn’t because the increase in current to 4mA from 1.6mA is also insignificant with a 1.5k resistor.
This would still allow a 2Vpp grid without clipping – just asymmetrical. The gain should be about 90V/2V = 45 still though. Doh!
This is not much of a reduction, but it can be seen why when you look at the 100k also line for this 216V HT:

The difference is quite small – say 110/2 = 55 gain.
I have an 82K I can use which in parallel with the 100k gives a 45k load, so I’ll try that.
This reduced to a 38 gain with a reduced C+ to 206V from 216V, and A+ to 366V from 376V.
This now gave a max triode2 grid input of 3Vpp below – with no noise! Looking good!

Anode distortion now starts at 48Vpp – up from 20Vpp – at about volume position 4 – much better.
Full volume anode distortion signal is at 80Vpp still:

Does it still sound OK with guitar? Yes, but I want to make it more bassy like the PP18. This is another area for research – tonal change – for another Post.
The distortion starts a little early still at volume 4, but it was more distorted before.
The 6V6 anode symmetry is clean up to 50V with the OT primary causing 90 degree phase shifting shown:

The 6V6 anode distortion starts at 200Vpp below:

I need to iron that out. I have seen pics of similar in Briggs but can’t remember the name or cause – bubbles on the sine wave.
I want to look at the R5 grid 120K stopper value also, as it may be too high to stop radio noise. I get a HF sine wave close to distortion volume shown at the T2 grid and anodes.
That’s it for this Post finally, and I’m happy with this triode1 gain improvement for now.
I will try to improve the remaining problem areas in another Post.
This will cover exploring frequency band limiting with capacitors at the triode cathode, and checking whether that HF on the OT is a triode2 grid stopper value or not and why.