LOAD MATCHING 3, edited 2012.
This page is about...
Brief history of triode use, Class A Push Pull  basics.
Fig 1.  Schematic of basic PP triode output stage with current
waveforms to explain 2H cancellations.
Comments on class AB1 amps, Williamson's amp, 
Class AB efficiency, preferences, Class AB1 basics.
Fig 2. EH6550 triode curves with load line.
Design Steps (1) to (4).
Fig 3. EH6550 triode curves with load lines.
Design Steps (1) to (8).
Relevant notes (A) to (O)
Fig 4. Graph for PP 6550 triode class AB1 power output vs RL values.
Class AB power and portion of class A power listed for 8k : 6 ohms loading.
Fig 5.  Schematic for 35 watt class AB1 PP triode amp with KT90.
Speaker SPLs with 25 watts.
Output transformer ratios.
Comment on using KT90, KT120 or multiple tubes.
A brief history of triode use.
Way back in about 90+ years ago when triodes had barely been invented,
someone became aware of the unwanted harmonic distortion generated
by all the early amplifiers which were single triode class A1 amplifiers.
Like everything that is invented, uses and "apps" were investigated
and very someone applied the "what if we try this?" principle and soon after
the world had push pull amps which allowed class AB1, AB2, and B and C
because as always, MORE power and higher circuit efficiency was desirable
in a world where vacuum tubes, electricity, and all other parts needed for
radios and amplifiers were terribly expensive. The average man had to save
for a long long time to buy a radio in 1925. And what he bought had very bad
performance by today's standards.

Negative feedback had yet to be invented, and when it was, it was little used
or understood. But soon the idea of push pull operation caught on and became
in the more expensive radios and record player amplifiers. Even by
1930 the millions of poor in the world rarely heard anything amplified by PP
amps, they all mainly heard sound from single ended output tubes in mantle
radio sets, and awful it was. PP amps didn't outnumber SE amps until the 1960s

when it became possible to build a cheap solid state amp which cost less to
instal into a radio, radio-gram, or TV set than using a 6V6 or 6BQ5.
But the very tiny number of hi-fi enthusiasts recognized that triodes were
the best, used in SE or PP. Directly heated triodes like 2A3, 300B, 45, 845
had been invented by 1930 and these set the gold standard for sound quality
with low THD, IMD, and NFB was optional, not a necessity when one used a
pentode or beam tetrode mode amp with 6V6, 6L6, 807, KT66.  Countless
movie theatres had SE 300B amps to power their very large horn speaker,
and some theatres had PP amps with 300B, many of which remained operational
until the 1970s. Bandwidth was poor though, because movie sound was limited,
and was not hi-fi.
So by about 1930, there were numerous power triodes which gave good linearity
and excellent sound and these remain among the most linear amplifying devices
in the universe.
However, as my page on SE triode amplifiers shows, even good SE triodes do
produce distortion and often the load value we want to use will produce 5% of
mainly second harmonic distortion even when the triode is optimally set up as a
class A single tube. Luckily though the presence of 3H and other harmonics is
relatively low.

But it must always be remembered that although all distortion in all audio / video
electronics is evil, in audio amps it is extremely unlikely anyone would find their
music is ruined with 5% distortion. Usually the 5% quoted THD figure is at the
maximum output level just before the amp wave form starts to turn into a square
wave with flats on the positive or negative sine wave peaks. An SE 300B might
produce 10Watts of power and THD has reached 5%. If more than 10 Watts
is wanted, distortion rises exponentially to a maximum of maybe 40%.
But below the "clipping level" of 10 Watts, amplifier THD increases in proportion
to output voltage level.
Graph 1.
Graph 1 shows curve A for an SE 10Watt amp with 1 x 300B and curve B
shows the use of two 300B to make 20Watts.  Both A and B are with zero
global NFB.
Curve C shows the effect of a modest 12dB of global NFB.
The curves are for ONE amp and ONE 5 ohms speaker with sensitivity of
87dB/W/M and if you sit 3 metres away from 2 speakers in an average lounge
room you may get the levels in the graph.
I have not included a curve for the SET amp with 12dB global NFB but its
THD would then become about as low as the PP amp without NFB, curve B.
The curves are for class A1, and if the PP tubes were in low bias class AB,
their distortion could double, and THD then becomes only marginally lower
than the SE amp.
SE amps produce mainly 2H and the resulting IMD is less objectionable than
the IMD produced by a lesser amount of 3H which is the main HD product of
the PP amp. The fact is that excellent sound is possible with either SE or PP
circuits, despite these graphs indicating quite high THD levels compared to many
modern solid state amps with a measured 0.001% THD at 10 Watts. Many people
fail to understand that our ears can tolerate 0.5% THD before we notice its
presence or the presence of resulting IMD. 

The simplest Push Pull operation has two identical triodes each working in class A
and with the same load conditions as one SE tube working alone. But the two
tubes each giving the same power use an output transformer to combine the
power from each tube which each work with oppositely phased signals so that
the total power is twice that of one class A triode.
All tubes work to produce audio
power which is measured in Watts and which are calculated as the product
of the load voltage x load current. In one tube of a PP pair, the power could be
produced by a positive going voltage x negative going current, and this product
is the same as the power produced in the other tube with a negative going voltage
and positive going current. The action is like two men using a long bush saw with a
handle at each end. One man pushes the saw while the other pulls the saw, and at
the end of the saw stroke they change direction, so each man is applying the same
power but always in opposite direction to each other. But although each man tries
to act with equal force on the saw in each direction, his muscle structure and
physiology mean that he can pull the saw harder than push it, so the forces of
push and pull are different. But with a man at each end of the saw, the imbalance
of push and pull forces of each man sum together to give the saw an equal rate
of log cutting in each saw direction. The variable forces applied by each man is
like the changing tube current, and
the change of position of the saw blade is like
the change in voltage at the anode. Each man sawing produces saw blade
motion like a sine wave, and the forces applied to the saw are a sine wave, but
one with bigger swing in one direction than the other, and this force difference
is second harmonic distortion. In any SE tube, current swing also varies in peak
amplitude for + peaks of waves and - peaks in waves.

So there are second harmonic currents generated in each triode but these have
the same phase in each tube even though the fundamental frequency signal voltages
are oppositely phased.
I do hope everyone understands the difference between Voltage and Current,
because if you don't know this basic electronic fact, all I am saying will be as
easy to drink as a glass full of mud.

Each tube of the PP pair attempts to produce the opposite phase of signal at their
anode connections to the two ends of the OPT primary winding, and the voltage
change is like two children on a see-saw, and the fulcrum of the see-saw is the
centre tap of the primary winding. The CT is kept at zero signal volts by means of the
power supply capacitors between CT and 0V. A typical 470uF capacitor has reactance
of only 34 ohms at 10Hz, so the CT may be considered "shunted to" or "short circuited"
to 0V. Each 1/2 primary winding is coupled magnetically to each other, and each 1/2
primary "collects" the power produced by each output tube's change of voltage and
current and the two phases of power are summed together in the magnetic field to
transfer power magnetically to the secondary winding which is connected to the
loudspeaker load.

The primary can ONLY accept power into its primary load if there is a voltage
difference applied to each end of the primary winding. If TWO EQUAL amplitude
voltages with the same phase are applied to each end of the primary, there is no
voltage difference across the winding, so no current can flow due to such voltages.
Fig 1.
Fig 1 shows the relative voltages in a typical PP triode output stage.
Now this schematic is here to describe the basics so some details I would use
in a real amp are missing, but first you must consider the simple picture or else
you will be overwhelmed by the final complexity. The circuit shows 6550 in triode
mode with screens connected to anode, usually via a 220 ohm 1/2W resistance.
Each 6550 has Ea = 420Vdc and Ia = 70mA, so Pda = 29.4Watts.
The load for perfect pure class A1 right up to clipping and for each triode tube
RLa = Ea / Ia - (2 x Ra ) = 420/0.07 - ( 2 x 1,000 ) = 6,000 - 2,000
= 4,000 ohms.

Each tube theoretically produces 10 Watts of power if there is 200Vrms
at each anode just before clipping and if the load remains a constant value
of 4,000 ohms, 4k0.

However, like most things in the Universe, tubes don't act perfectly linearly.
A 6550 or a 300B or KT90 or KT120 tends to produce more Ia change per volt
applied if grid becomes positive than if the grid becomes more negative. In other
words the transconductance, gm, is higher when Ia is above the idle Iadc and
when Ia is less than idle Iadc. The change in gm during each sine wave where
Ia varies above and below Iadc at idle causes second harmonic signal currents
to be generated.
These currents are plotted in graphs in Fig 1 and labelled as "cathode currents"
The graphs are a little exaggerated to enable you to visualize. If you doubt such
currents exist, then you should build a sample circuit as above, and install
10 ohm x 5Watt resistors between k1 and k2 cathodes to 0V and hook up
an oscilloscope, and then you will see the distorted cathode current flow
between each cathode and 0V. If you analyzed the current wave you would
discover there is a distortion current flow with twice the frequency of the applied
signal at grids, and the magnitude of such 2H currents will be between 4% and
8% of the undistorted current wave.

In an SE amp with just one triode tube the undistorted signal current plus the
2H current both appear applied to the load.

But when you have 2 tubes in PP, the 2H currents produced in each tube have
the same phase even though the phases of the wanted two undistorted waves
have opposite phase. The 2H currents which flow at the cathode in each tube
are applied to each end of the OPT primary, but because they have the same
phase and amplitude they cannot produce a voltage across the load. So if
distortion currents flow in/out of each end of the OPT winding, where do they
go? They most obviously cannot just disappear, now can they? Well, when you
have two equal amplitude and equal phase signals applied to each end of an
OPT winding, they are said to be applied "in common mode", because the one
signal is common to both ends of the winding. The result here is that the
"common mode" current flows to 0V via the capacitor between CT and 0V,
and then AROUND the circuit as all signals must flow, and into each cathode
and hence you see the waves as I have pictured them. The result of the
common mode 2H current being unable to flow in the OPT load is that the
usual 5% of 2H seen in a class A SET amp seems to miraculously vanish.

Is magic perfect? Well, no, never.

The THD of the PP class A triode amp is remarkably linear without
any reliance
on applied external loops of NFB, because the 2H distortion currents are the
majority of distortion currents, and when 2H currents are prevented from
appearing in the speaker secondary winding and speaker, THD is reduced
from a typical 5% to maybe 1% just below clipping. But there are other
harmonics generated besides the 2H, and in any tube you find 3,4,5,6,7,8,910H etc
all at various levels but even at just below clipping when all HD artifacts are
beginning to increase. Often the 3H is the next highest HD after 2H, followed by
4H, 5H etc. Even numbered HD artifacts are mostly all eliminated by the PP action
in the OPT, but none of the odd numbered H are suppressed, and the 1% of THD
produced by a typical PP triode amp will be 3H, 5H, 7H, 9H etc, in declining
amplitudes. At 1dB below clipping, the 1% THD would perhaps comprise mostly
3H, with 5H perhaps 10dB lower and so on. In practice, ie, in the real world,
any two triodes are not exactly matched, ie, equal in all their electronic properties,
ie, parameters. So 2H, 4H etc are not perfectly "cancelled" in the OPT, and in fact
some 2H and 4H is always to be found to generated in a PP OP stage. Of course
in the real world, the output stage must have an input stage and driver stage
which also produce 2H and 3H etc, and usually the amount is below whatever is
generated by the output stage. But the 2H, 4H etc of input stages is NOT cancelled
by the output stage action, because the input stage 2H is applied in two phases
to the 2 grids of the output tubes, ie, the 2H is applied in "differential mode" like
the two wanted undistorted signals, also applied to the output grids with opposite
phase. See Fig 1, and Voltage wave forms are shown for grids and anodes for
V1 and V2.

The other thing worth remembering about PP output stages is the way one tube's
anode load varies because of the presence of the other tubes load. If you could
phone one 6550 in Fig 1 to chat about how it feels about its load, it would tell you
during low level signals, say less than 20Vrms at its anode, the load hardly varies
much away from being 4k0. The other tube would have the same story. But whatever
one tube does with change of current and voltage, it is transformed to all other things
connected to that OPT via a common magnetic field, one that has only one version
and complete with all wanted signals and the harmonic distortions allowed.

Now when one tube grid goes sufficiently negative its current can become cut right
off, and when that happens, it is as if somebody pulled it out of the amp. And when
one tube of the pair "cuts off", then the class A action ceases and class AB begins
to happen and the turn ratio between primary and secondary is halved, because
only 1/2 the winding is used, so that one remaining tube will tell you its load has
reduced to 2k0. Now in fact, the load each tube sees when power is high is not
a constant value, so it cannot be accurately shown as a straight line on a loadline
graph. It is in fact a curved line, because the load changes during each large wave
cycle and this increases THD. So while talking to one tube, you would hear it say
"Well, I can only do the best job if the other tube shares the work."

The only way to ensure you get 1% THD at a dB below clipping with triodes is to
ensure each tube has positive and negative going peak Ia changes slightly less
than the idle current for each tube. If you were to apply a more negative grid bias
voltage to 6550 in Fig1, and reduce idle Iadc to say 25mA, you could still get 20
Watts of power to the same load, but THD will have risen to 3%, mainly 3H and 5H,
and the music would sound worse.

The Williamson triode amplifier.
Mr D.T.N.Williamson in his famous 1947 design used triode strapped KT66 with a
PP OPT and also applied 20dB of NFB so that at 12 watts of full power the amp
made 0.1% THD and only about 0.02% at 1 watt. There is no reason to expect
any better THD performance any tube power amp because the IMD which results
from the slight non-linearities of a PP triode amp is benign and mainly inaudible.
Williamson used  a common cathode resistance to bias both tubes and with some
adjustments, and to help the balancing of 2H currents in class A. His circuit
could be much improved, but I won't dwell on that now.

Class AB efficiency, preferences.
Not everyone was happy to stay with pure class A operation. The day after
everyone agreed in about 1922 that pure class A class PP triode amps were
indeed a modern marvel along came accountants and marketing dudes to try to
reduce the size, weight, cost of its construction and costs of running it. Less size
and weight in anything is always easier to sell for the same price as the bigger
heavier one if people can be tricked into thinking they won't hear any difference.

All pure class A1 triode amps have poor maximum efficiency. The "A1" denotes
pure class A with no grid current flow during normal operation, and peak Ia
increase limited to no more than 2 x idle Iadc. Hence class A1 triode efficiency
has a maximum of about 30%, so a KT66 or 6L6 or 807 in triode idling with
Pda = 22 Watts may make only about 6 watts of class A1 audio power, and a
pair in PP drawing 44 Watts of idle power can make only 12 Watts, giving class
A1 efficiency = 27%, and winding losses in the OPT of 5% would reduce this a

There is class A2 where a single tube is forced to swing more anode voltage
below the Ra line for Eg = 0V. Instead of only 6 Watts in class A1 for a single
KT66 etc, with load of 4k0, about 9 Watts in class A2 is possible by driving the
grids positive above their normal limit of 0V potential by using a direct coupled
cathode follower with low output resistance able to cope with the low grid
input resistance of less than 2k0 when grid current flows. The load must also
be raised to 6k0 to ensure the peak Ia change is not more than +/- idle current
of 55mA. So pure class A2 PP could give 18 Watts with efficiency = 40%,
almost as good as beam tetrode mode which gives up to 44%.

95% of all manufacturers of quality amplifiers since 1930 have not thought the
extra few Watts gained by triode class A2 is not worth the cost and complexity
of having another extra driver triode set up as a low resistance cathode follower
to drive the grids above 0V. Whenever class A2 drive was ever used it was usually
for non hi-fi applications where tubes worked in beam tetrode mode with a fixed
screen voltage at say +300V, and anode supply could be +600Vdc, and PP output
power max with 807 could be 80 Watts, when THD without NFB was 14%.
Such arrangements were notoriously unreliable.
In such cases the extra pair of
driver tubes such as the two triodes in a 6SN7 "earned their keep" by much
increasing the possible power output.

The simplest way to increase efficiency is to use class AB1. Usually this is done
using the same circuit as the PP class A amp, but the B+ anode supply is raised
by 25%, and the OPT primary load is reduced and the triodes are biased with
less idle current so that Pda is less than for class A conditions. Usually there is a
fixed grid Vdc bias applied instead of the common self regulating cathode biasing
so common in class A amplifiers where it works so well.

Class AB2 is rarely ever used for the same reason class A2 is not used - more
expensive and and unreliable. 
It is very easy to calculate the pure class A RL for
a pair of triodes. In Load Matching (2) for SE triodes, the formula to calculate the
SE class A triode RL is RL = ( Ea / Ia ) - Ra,
Ea and Ia are idle V and I conditions, and Ra is the anode resistance at the Q
point. This load formula give a load slightly higher than that used to obtain
maximum power but it is good practice to avoid high THD. For maximum triode
SE power, RL = (Ea / Ia) - 2Ra.  PP tubes are effectively working in series with a
load that is twice that of one tube. This load is the "anode to anode" load presented
to the anodes by the whole of the OPT primary winding. For PP pure class A1,
the 2H is cancelled and the RL for two tubes is 2 x ( [ Ea / Ia - 2Ra ] ).
This reduces to RL a-a = ( 2Ea / Ia ) - 4Ra, where Ea is the idle Vdc and Ia is
the idle current of each tube and Ra is the anode resistance at the Q point which
is the same for both triodes.


Fig 2
(1)  Nominate idle conditions for each triode. Pda at idle, = 28 Watts.
Ea = +425Vdc, Iadc at idle = Pda / Ea = 28W / 425V = 66mAdc.

(2)  See Fig 2 above. Plot a point Q for Ea = +425V and Ia = 66mA.
This will be at Eg1 = -50Vdc.
Draw tangent line S to T through Q.
Ra at Q = amount of Ia change for a given amount of Ea change
for the
tangent line S-T.
Ra = ( 500V - 367V ) / 150mA = 886 ohms.

(3)  Calculate
RL a-a = ( 2Ea / Ia ) - 4Ra = ( 2 x 425 / 0.066A ) - ( 4 x 886 )
= 12,878 - 3,544
= 9,334 ohms.

NOTE. Each triode has a load of RLa-a / 2 = 4,667 ohms.
This load may be plotted on the anode curves.
Calculate Ea / RL = 425V / 4,667 ohms = 91mA.
Add Iadc at Q, +66mA = 157mA.
Plot point A on Ia axis at 157mA. Calculate V across load
= load x Ia at point A
= 4,667 ohms x 0.157A = 733V.
Plot point D on Ea axis. Draw line A to D.
This line should pass through point Q.
If not, check all calculations. Plot point B and C for SE class A load swings.

(4)   What is maximum power output at clipping?
From load line graph Fig 2, read Ea minimum = 125V.
Vpeak negative Ea swing = Ea - Ea min
= 425V - 125V = 300Vpk. Vrms at each anode
= 0.707 x Vpk swing = 212Vrms.
Va-a = 2 x Va at each anode = 2 x 212V = 424Vrms.
Power Output maximum = Va-a squared / RLa-a = 424 x 424 / 9,334
= 19.3 Watts.

Another useful simpler way to calculate output power is :-
Power Output maximum =
0.5 x Iadc squared x RL a-a, where Iadc
is for one tube.

PO = 0.5 x 0.066 x 0.066 x 9,334 = 20.3 Watts.
Notice that the quick easy method is a little optimistic compared to the load
line derived PO.


Fig 3.
Graph for PP AB1
        triodes, 5k load lines.
(1)  Establish the desired working point for each tube at Q point. Ea = 500V
and Ia = 50mA, Eg = -64V, Pda = 25 Watts.
Draw tangent line to Ra curves at Ia = 50mA, S - T through Q.
Calculate Ra from slope of S-T.
Ra at Q = amount of Ea change / Ia change for line S to T.
for the tangent line S-T, Ra = ( 615V - 450V ) / 150mA = 1,100 ohms

(3)  Draw a line from Ea to the Ia axis so that it just touches the curve for Pda
= 42W of the triode.
This line will appear from Ea = 500V to point H on the Ia axis and its
resistance value = V / I
= 500V / 330mA = 1,515 ohms.

(4)  Calculate minimum class AB1 RLa-a = 3.2 x R value calculated in step (3).
RL min = 3.2 x 1,515 = 4,848 ohms.

(5)  Choose any class AB1 RLa-a above RL minimum in step (4)

Let us choose RL a-a = 5,000 ohms.

(6) Draw the load lines for single tube on one side of the PP circuit.
RL a-a. RL a-a = 5,000 ohms. 1/4 RL a-a = 1,250 ohms.
Calculate the I change of the loadline for
Ea at Q point, I = 500V /1,250 ohms = 400mA.
Plot a straight loadline from point A at 400mA on the Ia axis to point D,
+500V on the Ea axis.
Plot point B where this load line intersects Ra curve for Eg1 = 0Vdc.

(7) Calculate class AB1 max power for chosen RL a-a.
Ea minimum is found vertically below point B on the Ea axis.
Ea minimum = +200V. Ea swing = Ea - Ea minimum = 500V - 200V = 300V pk.
Calculate Power in RL a-a =
V squared / R = [ 2 x (Vpk Ea change at 1 tube)
x 0.707 ] squared / RL a-a.
This can be simplified,

Class AB1 Power = 2 x (Ea - Ea min) squared / RLa-a. 

Class AB1 PO for 2 x EH6550 triodes = 2 x 300 x 300 / 5,000 = 36 Watts.

(8) The class A loadline of 1/2 RL a-a for each tube of the PP may be drawn.
The class A load = 5,000 / 2 = 2,500 ohms.
Calculate Ea / ( 0.5 RLa-a ) = 500V / 2,500 ohms = 200mA.
Add the idle current of the tube, 200mA + 50mA = 250mA.
Plot point E at 250mA on Ia axis.
Draw loadline from E through Q to G on Ea axis.
The intersection of the two loadlines of 1/4 RL a-a and 1/2 RL a-a will be
at point C, and where Ia = 2 x idle Iadc. Point C is at Ia = 100mA.
Class A PP power may only exist where peak load current change
= Idle Ia at point Q.   

Read these notes A to O........

(A) The AB1 RLa-a should not be so low to cause Pda to rise above the rated
Pda max for the tube at clipping with a sine wave signal. To satisfy this condition,
the minimum RL a-a is calculated as :-
RL a-a = Ea squared / Pda max for one tube of the pair.
For example, if Ea = +500V, and Pda max = 42 Watts, RL a-a minimum
= 500 x 500 / 42 = 5,952 ohms.

(B) Class AB1 RL a-a should never be less than 4 x Ra where Ra is the
average Ra at Eg1 = 0V, or the quoted Ra from data sheets.
For this example, Ra at Eg1 = 0V for the EH6550 shows that for a +250V rise
along the Ea axis the Ia increase is 340mA, giving Ra average = 250V/0.34A
= 735 ohms.
Therefore class AB1 RLa-a should never ever be less than 2,940 ohms.

(C) The minimum Ea for class AB1 may be calculated :-
Ea minimum = square root of (4 Ra x Pda max )

where = Ra is average Ra at Eg1 = 0V or data value for Ra. In this example,
Ra av = 735 ohms, Pda max = 42 Watts,
so Ea min = sq.rt ( 4 x 735 x 42 ) = +351Vdc.

(D) With music signals the average Pda increase generated by signals is
much lower
than with a sine wave so RL a-a may be safely
= 0.8 x Ea squared / Pda max. In this example, we would calculate 4,762 ohms.
Any value of RLa-a may be chosen for class AB up the load for maximum pure class A.

(E) RL a-a load for maximum possible class A1 power is calculated :-
RL a-a = ( 2Ea / Ia ) - 4Ra
where Ra is measured from the tangent line at point Q. 
This example,
Class A1 PP RL a-a = ( 2 x 500 / 0.05 ) - ( 4 x 1,100 ) = 15,600 ohms.

(F) Maximum class A1 portion of total class AB power
= 0.5 x Iadc squared x RL a-a
for all AB1 loads up to the load for maximum
possible class A power.

(G) For maximum 100% class A power for this example,
PO = 0.5 x 0.05 x 0.05 x 15,600 = 19.5 Watts.
For RL a-a = 5,000 ohms, class A portion = 0.5 x 0.05 x 0.05 x 5,000
= 6.25 watts.

(H) Harmonic Distortion for 100% maximum possible class A1 power
= 1% approximately, mainly 3H.
Harmonic Distortion for this example at maximum class AB1 power
= 4% approximately, mainly 3H, 5H.

(I) During class AB operation, the first few watts of power are produced by
class A1 operation of both output tubes. The class A1 load for each tube
= 1/2 the chosen RLa-a. As power output increases, the load for each tube
changes from 1/2 RLa-a to 1/4 RLa-a because the tube current becomes entirely
cut off for a portion of each tube's current wave cycle. While one tube is cut off,
it has no connection to the OPT primary so no connection to the load.
During this part of of this tubes cut off region the other tube remains connected
to the load via 1/2 of the OPT primary so the turn ratio halves and the tube
load becomes 1/4 of the RLa-a. Each tube does the same job as the other on
successive wave peaks.

(J) If the tubes were perfectly linear devices, the idle bias could be reduced
to 0.0mA for class B operation. Each tube would then only be loaded with 1/4 RLa-a.
The value of 1/4 RLa-a is sometimes known as the "class B tube load".
Regardless of the bias current level in class AB1 amps, at some intermediate
power level the load on each tube during wave cycles falls from 1/2 RLa-a to
1/4 RLa-a, and with triodes the transition is gradual, and in fact the load line
for each triode is a curved line, but one we do not want to have to draw accurately.

(K) The class A loadline of 1/2 RL a-a or 2,500 ohms for each tube shows there
may be much more possible Ea minimum Ea swing to point F. But this is impossible.
At point C, one tube cuts off leaving the other to power the load on its own with a
changed load value of 1/4 RL a-a, or 1,250 ohms. The use of class AB exploits the
easy capability for a 6550 to produce a much larger current increase than 2 x idle
current which occurs in pure class A. Maximum peak load current is at point B,
= 240mA.

(L) For class AB, the power dissipated at the anodes can be calculated and
graphed for sine waves and for various power levels and load values.
There is no need to calculate such Pda variations or draw graphs if you follow the
above methods. But some of you may be able to understand the concepts and
equations for Pda in the Radiotron Designer's Handbook. Its safe to assume that
input power from the PSU to tubes may be up double the idle condition but it will
never be continuous with a test signal designed by Mozart or Beethoven, or
Mick Jagger, who is Mozart's great-great-great grandson :-)

(M) I base all my load line analysis and calculations around the action going on
in ONE output tube. In some old books the author describes the anode curves for
two tubes as they are combined to make a graph that is very hard to draw and
understand. A classic example is on page 575, Radiotron Designer's Handbook,
4th Ed, 1955. I don't know anyone who fully understands the sketch on that page
or anyone who has drawn a page up like that one.

(N) Regulated power supplies are recommended for class AB amps.
But even in this example where the output tubes may draw twice the idle power
with a sine wave at clipping, in practice the amount of power variation with music
is minimal so the B+ voltage will barely move especially when the total capacitance
of B+ rails is very high compared to amplifiers designed in 1955. So I never need to
regulate B+ supplies in class AB power amps.

(O) Output resistance of the class AB amp varies between lowest while working
in class A and highest when at maximum class AB power. But the NFB used will
make Rout very nearly constant.

Fig 4.
graph for PP 6550
          triode AB1 power vs RL

Fig 4 graph shows class AB and A output power for various RLa-a with
PP 6550 in triode, Ea = 500V, Ia = 66mA, Idle Pda = 33 Watts.
Maximum class AB power, OPT = 8k : 6 ohms, 28 Watts.
Class A PO = 16 Watts.
Distortion, 28 Watts = approx 3%, at 2 watts = < 0.3 %. 
R out at 2 Watts = 1.6 ohms, including 0.24 ohms of winding resistance
measured at the OPT secondary.

The 2k class B load line will be well under the 42 watt limit for Pda so there
is no danger of exceeding the Pda ratings.
Power output into a list of loads and power and thd will be :-
20k : 15r,  A 17 watts < 1.5%,                                              2 watts < 0.4%.
12k : 9r,    A 23 watts < 2%,                                                 2 watts < 0.6%
8k : 6r,      AB 28 watts < 3%,     A portion 16 watts < 2%,   2 watts < 0.7%
5k : 3.7r,   AB 36 watts < 4%,     A portion 10 watts < 3%,   2 watts < 1%.
4k : 3r,      AB 38 watts <5%,      A portion 8 watts < 4%,     2 watts < 1.5%.

The above power allows for output transformer winding resistance losses which
are based on 4% at 8k : 6 ohms. Therefore at 4k, losses are 8%, and at 2k
are 16%, and at 1k = 32%. If the output transformer had two choices for the
turns on the secondary winding to allow for 6 ohms and 2.66 ohms, the losses
at lower load values would be less and a good load match to the tubes is possible.
Loads below as 5.3k : 4 ohms are not recommended, despite the high power
available, because the distortion is much higher at ordinary loud 2 watt or less
listening levels as the above distortion figures indicate. There is a high danger
that if 4 ohms was connected to the amp which has a dip of impedances down
to say 3 ohms, then the tubes could be overheated if loud levels are sustained.

Fig 5.
Schematic for 35w
          pp class ab1 amp.

This schematic will sound well with KT90 set up as shown. Although it will operate
with 6550 or KT88, KT90 or KT120 would be the best choice of tube.
The power supply required for two such channels can be seen in my page on the
5050 integrated amplifier.

Speaker SPLs with 25 Watts available.
If the speaker has sensitivity = 90 dB/W/M measured in an anechoic chamber,
then in a room at 4 metres about 1/3 a watt average is plenty for an average SPL
of  85 dB including both channels. For 95dB, you would need an average of 3 Watts
per channel which may seem deafening. 3 Watts into 6 ohms is 4.2Vrms of output,
and if the amp has a 30 Watt ceiling, the peak SPL = 105dB. I find that a pair of
25 Watt amps are plenty with a pair of Vienna Acoustic Mozart; a friend has no
problems with headroom for jazz and orchestral works. My own Sublime speakers
are 3 way and have SEAS 200mm woofer, 140mm midrange, and 25mm tweeter
and have a Z about the same as the Mozart. I find 25 Watts is enough.

The loading for the pair of PP triodes as above should ideally use an OPT with
8k : 6 ohm ratio where 8 ohms is the nominal speaker impedance and allows for a
dip of impedance to 3 ohms without causing a serious increase in THD.
There will be 30 Watts available into 6 ohms with a lot of class A power.

With the Vienna Acoustic Mozart, the impedance is more like an average of
4.5 ohms in the 100Hz to 1kHz frequency band, and if building an amp to
handle such speakers the ideal impedance match is 8k : 4 ohms.

The only beam tetrodes capable of providing better load tolerance with low
loads is the KT90 or KT120.
KT90 have Pda rating = 55 Watts, KT120 even higher, so we could set them
up at 36 Watts of dissipation and get a higher amount of pure class A, and
better ability with low value RL down to 3k a-a.

It is not always easy to arrange secondaries on the output transformer to
be able to be changed to suit various speakers. There is much more about
doing this in my page on output transformers.

With a PP amp with 60 Watt ability using a quad of 6550, I would match the
tubes to 5 ohms and have the tubes working under the same anode loading
condition as the above pair does with 8k a-a. Therefore the quad of tubes would
have a 4k : 5 ohm OPT. Such an amp with the 4 output tubes will tolerate any
nominal load level from 3 ohms and above much better than just using a pair
of tubes.

One customer with an 8585 amp has KT90 instead of the 6550, and he gets
100 watts into 4 ohms with CFB windings but if he went to triode he would still
get a very nice 60 watts.

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