LOAD MATCHING 3, edited 2012.
TRIODE PUSH PULL AMPLIFIERS.
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.
DESIGN METHOD, PP CLASS A1,
EH6550 IN TRIODE MODE.
Fig 2. EH6550 triode curves
with load line.
Design Steps (1) to (4).
METHOD, PP CLASS AB1, EH6550 IN TRIODE MODE.
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
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
someone became aware of the unwanted harmonic distortion generated
by all the early amplifiers which were single triode class A1
Like everything that is invented, uses and "apps" were
and very someone applied the "what if we try this?" principle and
the world had push pull amps which allowed class AB1, AB2, and B
because as always, MORE power and higher circuit efficiency was
in a world where vacuum tubes, electricity, and all other parts
radios and amplifiers were terribly expensive. The average man had
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
or understood. But soon the idea of push pull operation caught on
common in the more expensive radios and record player amplifiers.
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
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
But the very tiny number of hi-fi enthusiasts recognized that
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
with low THD, IMD, and NFB was optional, not a necessity when one
pentode or beam tetrode mode amp with 6V6, 6L6, 807, KT66.
movie theatres had SE 300B amps to power their very large horn
and some theatres had PP amps with 300B, many of which remained
until the 1970s. Bandwidth was poor though, because movie sound
and was not hi-fi.
So by about 1930, there were numerous power triodes which gave
and excellent sound and these remain among the most linear
in the universe.
However, as my page on SE triode amplifiers shows, even good SE
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
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
produce 10Watts of power and THD has reached 5%. If more than 10
is wanted, distortion rises exponentially to a maximum of maybe
But below the "clipping level" of 10 Watts, amplifier THD
increases in proportion
to output voltage level.
Graph 1 shows curve A for an SE 10Watt amp with 1 x 300B and curve
shows the use of two 300B to make 20Watts. Both A and B are
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
87dB/W/M and if you sit 3 metres away from 2 speakers in an
room you may get the levels in the graph.
I have not included a curve for the SET amp with 12dB global NFB
THD would then become about as low as the PP amp without NFB,
The curves are for class A1, and if the PP tubes were in low bias
their distortion could double, and THD then becomes only
than the SE amp.
SE amps produce mainly 2H and the resulting IMD is less
the IMD produced by a lesser amount of 3H which is the main HD
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.
fail to understand that our ears can tolerate 0.5% THD before we
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
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
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
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
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
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
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
the same phase in each tube even though the fundamental frequency
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
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
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
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
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.
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
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
at each anode just before clipping and if the load remains a
of 4,000 ohms, 4k0.
However, like most things in the Universe, tubes don't act
A 6550 or a 300B or KT90 or KT120 tends to produce more Ia change
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
Ia varies above and below Iadc at idle causes second harmonic
to be generated.
These currents are plotted in graphs in Fig 1 and labelled as
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,
10 ohm x 5Watt resistors between k1 and k2 cathodes to 0V and hook
an oscilloscope, and then you will see the distorted cathode
between each cathode and 0V. If you analyzed the current wave you
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
the same phase even though the phases of the wanted two
have opposite phase. The 2H currents which flow at the cathode in
are applied to each end of the OPT primary, but because they have
phase and amplitude they cannot produce a voltage across the load.
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,
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
signal is common to both ends of the winding. The result here is
"common mode" current flows to 0V via the capacitor between CT and
and then AROUND the circuit as all signals must flow, and into
and hence you see the waves as I have pictured them. The result of
common mode 2H current being unable to flow in the OPT load is
usual 5% of 2H seen in a class A SET amp seems to miraculously
Is magic perfect? Well, no, never.
The THD of the PP class A triode amp is remarkably linear without
on applied external loops of NFB, because the 2H distortion
currents are the
majority of distortion currents, and when 2H currents are
appearing in the speaker secondary winding and speaker, THD is
from a typical 5% to maybe 1% just below clipping. But there are
harmonics generated besides the 2H, and in any tube you find
all at various levels but even at just below clipping when all HD
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
amplitudes. At 1dB below clipping, the 1% THD would perhaps
3H, with 5H perhaps 10dB lower and so on. In practice, ie, in the
any two triodes are not exactly matched, ie, equal in all their
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
which also produce 2H and 3H etc, and usually the amount is below
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
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
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
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
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
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
any better THD performance any tube power amp because the IMD
from the slight non-linearities of a PP triode amp is benign and
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.
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
everyone agreed in about 1922 that pure class A class PP triode
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
All pure class A1 triode amps have poor maximum efficiency. The
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
has a maximum of about 30%, so a KT66 or 6L6 or 807 in triode
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,
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
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
cathode follower with low output resistance able to cope with the
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 +/-
of 55mA. So pure class A2 PP could give 18 Watts with efficiency =
almost as good as beam tetrode mode which gives up to 44%.
95% of all manufacturers of quality amplifiers since 1930 have not
extra few Watts gained by triode class A2 is not worth the cost
of having another extra driver triode set up as a low resistance
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
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"
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
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
so common in class A amplifiers where it works so well.
AB2 is rarely ever used for the same reason class A2 is not used -
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
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
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.
DESIGN METHOD, PP CLASS A1,
EH6550 IN TRIODE MODE :-
(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 =
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
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
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,
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.
DESIGN METHOD, PP CLASS AB1,
EH6550 IN TRIODE MODE :-
Establish the desired working point for each tube at Q point. Ea =
and Ia = 50mA, Eg = -64V, Pda = 25 Watts.
(2) Draw tangent line to Ra curves at Ia = 50mA, S - T through
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
(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
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
Let us choose RL a-a = 5,000 ohms.
(6) Draw the load lines for single tube on one side of the PP
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
+500V on the Ea axis.
Plot point B where this load line intersects Ra curve for Eg1 =
(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 =
Calculate Power in RL a-a =
V squared / R = [ 2 x (Vpk Ea change at 1 tube) x 0.707 ] squared / RL
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
(8) The class A loadline of 1/2 RL a-a for each tube of the PP may
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
at point C, and where Ia = 2 x idle Iadc. Point C is at Ia =
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
along the Ea axis the Ia increase is 340mA, giving Ra average =
= 735 ohms.
Therefore class AB1 RLa-a should never ever be less than 2,940
(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
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
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.
Class A1 PP RL a-a = ( 2 x 500 / 0.05 ) - ( 4 x 1,100 ) = 15,600
(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
= 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
= 1/2 the chosen RLa-a. As power output increases, the load for
changes from 1/2 RLa-a to 1/4 RLa-a because the tube current
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
During this part of of this tubes cut off region the other tube
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
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
Regardless of the bias current level in class AB1 amps, at some
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
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,
(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
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
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
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
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
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
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
make Rout very nearly constant.
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
measured at the OPT secondary.
class B load line will be well under the 42 watt limit for Pda
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 <
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
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
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
to say 3 ohms, then the tubes could be overheated if loud levels
schematic will sound well with KT90 set up as shown. Although it
with 25 Watts available.
with 6550 or KT88, KT90 or KT120 would be the best choice of
The power supply required for two such channels can be seen in
my page on the
If the speaker has sensitivity = 90 dB/W/M measured in an anechoic
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
are 3 way and have SEAS 200mm woofer, 140mm midrange, and 25mm
and have a Z about the same as the Mozart. I find 25 Watts is
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
There will be 30 Watts available into 6 ohms with a lot of class A
With the Vienna Acoustic Mozart, the impedance is more like an
4.5 ohms in the 100Hz to 1kHz frequency band, and if building an
handle such speakers the ideal impedance match is 8k : 4 ohms.
The only beam tetrodes capable of providing better load tolerance
loads is the KT90 or KT120.
KT90 have Pda rating = 55 Watts, KT120 even higher, so we could
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
be able to be changed to suit various speakers. There is much more
doing this in my page on output transformers.
With a PP amp with 60 Watt ability using a quad of 6550, I would
tubes to 5 ohms and have the tubes working under the same anode
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
nominal load level from 3 ohms and above much better than just
using a pair
One customer with an 8585 amp has KT90 instead of the 6550, and he
100 watts into 4 ohms with CFB windings but if he went to triode
he would still
get a very nice 60 watts.
Education and DIY.
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