LOAD MATCHING 3
TRIODE PUSH PULL AMPLIFIERS.
The content in this page includes :-
A
very 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. Graph of EH6550 triode curves with load lines for 5k
a-a.
Minimum load for PP triode AB amps,
Class AB1 operation
explained.
How to plot a load line to give the outcome for class AB
triodes,
Anode heat dissipation,
Comment on B+ regulation, Biasing the
class AB PP stage,
Distortion, Output resistance, Negative feedback.
Using
a higher RL value,
* Fig 3. 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 4. Schematic for 35 watt class AB1 PP triode amp with
KT90.
Speaker SPLs with 25 watts.
Output transformer
ratios.
Comment on using KT90 or multiple
tubes.
-------------------------------------------------------------------------------------------
A brief history of triode use.
Way back in
about 90+ years ago when triodes had barely been invented, someone must have
tried to analyse what
the distortion was like in a triode and thought that
the distortion is something we could do without.
Negative feedback had yet to
be utilised or understood.
But soon the idea of push pull operation caught
on and apart from sound amps in early radios push pull triode amps
became
very common in the more expensive radios and record player amplifiers.
By
1928, the world greeted the release of the 300B which remains one of 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
which we want to design for will produce 5% of mainly second harmonic
distortion when the tube is optimally set up
as a class A single
tube.
Luckily though the presence of 3H and other harmonics is relatively
low.
PP class A basics.
The
first step towards PP operation is to have two identical triodes each working in
class A with the same load and each giving the same power just as the single
tube is doing, but using a balanced output transformer to combine the power from
each tube to make twice the power of one tube, but with much less THD without
any reliance on applied loops of NFB.
The PP output transformer has a single
primary winding with a CT which is connected to the B+ supply, so dc flows in
both directions along the one winding which is wound in one direction, so two
opposing magnetic fields are set up in the core
and which cancel each other's
presence. The output anodes of the two tubes produce voltages of opposite
phase.
Their grids are driven by oppositely phased input signals.
There
are second harmonic currents generated in each tube but these are in phase in
each tube even though
the fundemental frequency signal currents are
oppositely phased.
The same phase of 2H currents is therefore applied across
the output transformer primary end to end at the anodes
and so no 2H current
flows in the OPT, so no 2H appears as a voltage across the OPT, so there is no
2H in the secondary,
Fig
1.
I have
plotted the *current* wave forms at each cathode.
There are no signal
voltages at the cathodes because they are grounded to 0V.
The tube currents
have oppositely phased fundamental frequency currents distorted as I have
shown.
The distortion is due to the presence of a pure sine wave currents
plus distortion wave currents, mainly twice the
fundamental frequency and as
shown.
If we were to place a small resistance between the CT and
the B+ supply shown as a battery,
then while the circuit works in class A, we
would see a 2H voltage signal across the R but little fundamental F because
each tube is developing an oppositely phased and equal amplitude signal and
the CT at the half way point between
the two anode signals.
What I
have said about 2H also applies to all even numbered harmonics.
This
"cancellation process" thus removes the 2H and leaves only odd numbered
harmonics, so instead of seeing
18 watts of class A with 5% of mainly 2H
from this output stage we see only 1% of mainly 3H with other odd numbered
harmonics, 5H, 7H, 9H etc well below the 1% level.
Of course nothing in
this world is perfect, and unless the output tubes are closely matched there
will always be a slight net
imbalance of 2H current production in each of
the output tubes so in the real world there will be some 2H current across the
OPT so hence some in the speaker output signal. But often its well below the
3H levels.
If there is 1% thd at 18 watts, then at 1 watt there will be
only 0.23 %. In practice I find that the class A triode output stage
is probably the most blameless and wonderful sounding output stage ever
devised by man. The kind of performance
here was first achieved way back in
1928 with the 300B, and in fact the above schematic could use the 300B with good
effect.
The Williamson triode
amplifier.
Mr D.T.N.Williamson in his famous 1947 design used a
similar output stage as this with KT66, and also applied 20dB of NFB so that at
full power the amp made 0.1% THD and only about 0.02% at 1 watt.
There is no
reason to expect any more from any tube power amp with regard to
THD.
Williamson used a common cathode resistance to bias both tubes and
with some adjustments. The principles
are the same as I have outlined.
Class AB efficiency,
preferences.
Not everyone was happy to stay with pure class A
operation.
The day after everyone agreed in 1928 that pure class A class PP
triodes were quite a modern marvel
along came accountant and marketing
dudes to try to reduce the size and weight of a triode amp and the costs
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 where
people can be conned into thinking they won't hear any difference.
Pure
class A1 triode amps have poor maximum efficiency. The A1 denotes pure class A
with no grid current operation.
There is class A2 where a single tube is
forced to swing more anode voltage below the Ra line Eg = 0V region
by
driving the grid positive but nobody thinks the extra few watts is worth the
cost and complexity
of having another extra driver tube set up as a low
resistance drive source in cathode follower.
The class A1 triode can only
convert a maximum of 30% of the anode input power into audio power.
The class
A2 triode might make 40% efficiency, and similar to a beam tetrode.
But many
SE triodes only make 25% efficiency.
In a PP circuit where the operation is
pure class A for the load which is twice that for the SE case, efficiency is the
same.
The only way to increase efficiency is to go to class AB1, or AB2.
Hardly anyone tries to use triode class AB2 because of the
two extra driver
triodes required, the low amount of extra power above AB1, and the added THD of
AB2.
Its very easy to calculate the pure
class A RL for a pair of triodes.
In Load Matching (2) for SE triodes, I showed that the
centre value RL for one class A1 triode = ( Ea / Ia ) - Ra.
This load
formula give a load slightly higher than that used to obtain maximum power but
its good practice
to avoid high THD.
For two tubes in PP pure class A1, the 2H is
cancelled and the RL for two tubes is 2 x ( [ Ea / Ia - 2Ra ] ).
So RL for PP class A triodes = ( 2Ea / Ia ) - 4Ra.
Let us consider class AB1
only.
One could easily make a pure class A PP triode amp into a class
AB amp merely by biasing the output tubes so there is less idle current. But
this will not increase the maximum output power because the negative going anode
swings are still limited by the value of EA and the Ra at Eg =
0V.
Therefore makers of class AB amps choose a working point for each
output tube where Ea is slightly higher than for pure class A, and Ia is
slightly lower than Ia for pure class A so that the Pda at idle is slightly less
than it is for a tube in an SE class A amp.
This will allow a wider Ea swing
in the negative direction as load current increases.
A typical pure class
A triode SE amp or PP amp might have the idle at Ea = 420V and Ia up to 83mA for
a Pda maximum
= 35 watts, from which we may get about 9 watts of SE pure
class A with an efficiency of about 25%.
Virtually the only application for
class AB triode is for hi-fi. Class AB1 with a higher Ea Q than for the SE
tube
means that the pure class A efficiency can be higher at up to 30%, so
if Pda for two tubes = 66 watts we could expect
a maximum of about 20 watts
of pure class A. But there will be much more AB power for transients when the
Class AB1 load is less than needed to produce only pure class A. So a lower
load value than that needed for pure class A
will be well tolerated although
the ampunt of pure class A for the lower load will be less than the 20 watt
maximum.
For this situation the Pda could be say 33 watts maximum, and a
typical operation point might be Ea = 500V, and Ia = 66mA, and where Ra for the
triode = 1k.
The load for pure class A can be worked out on the basis
that Ia will be double the IaQ value on -ve anode swings with a class A load and
the Ea minimum is where Ra curve for Eg = 0V intersects the horizontal line of 2
x IaQ.
In this case the class A Ea minimum 130V, so the Ea -ve swing = 500 -
130 = 370V. Since the Ia change is 66mA
the pure class A load for the single
tube must be 370 / 0.066 = 5.6k.
A load line can be drawn from Ea = 860V on
the Ea axis to Ia = 155mA on the Ia axis and you will see that this line
passes through the Q point at Ea = 500v and Ia = 66mA.
The anode voltage
swings for +/-62V of grid V change is -370V and + 315V giving 10.5 watts at 4%
2H.
However the 2H is cancelled for the pair of output tubes, and so about 20
watts of pure class A with low THD is achieved
with pair of tubes with a load
of 11.2k a-a.
If we were to apply the formula for pure class A1 for PP,
RL = ( 2 x 500 / 0.066 ) - 4 x 1k = 11,151 ohms
or about 11.2k . This load is
well above what we may wish to to use for class AB1 so that a higher max power
is available but with a still adequate class A ceiling. The 11.2k load would
give us 20 watts of pure class A, but nice watts indeed because efficiency is
high at 36%, and distortion is very low at less than 0.7% perhaps, and the OPT
ratio could be 11.2k : 6 ohms which would transform the Ra from anode to anode
of 2k down to only 1.07 ohms.
This loading for pure class A would not need
any global NFB.
But suppose someone said 20 watts wasn't anywhere near
enough for the purpose.
And its possible to use a 5k load anode to anode to
produce over 30 watts of class AB power, and to see how, The load lines and
curves for such a set up need to be examined.
Fig 2.
Here
you can see the curves for ONE of the tubes of the PP pair. In some old books
they endeavour to describe the anode curves for two tubes as they are combined
to make a graph that is very hard to draw and understand.
See page 575,
Radiotron Designer's Handbook, 4th Ed, 1955. I don't know anyone who fully
understands the sketch on that
page or who has drawn a page up like that
one.
We need only the tube curves
for one tube and its operation which will be easily understood and give us all
the knowledge about the PP output stage's working.
We have
established above that the the anode to anode load for class AB1 will always be
lower than the load
for pure class A1 which is reliant on the quiescent Ia.
The approximate class AB1 minimum PP
load for triodes = 5 x Ra, Where Ra is measured for the working
point.
The Ra can be found by drawing a tangent to one of the Ra lines
nearest the Q point and through a point on that curve
where Ia is the chosen
Q point value, and calculating the ohmic value of the slope of that tangent
line.
I have drawn the tangen line of TQS through Q so that there is a
292V change for 300mA of current change
between T and S, so Ra is indeed
close to 1k.
So for a pair of 6550 Ra at Ia = 66mA, Ra
= 1k, so RLa-a minimum = 5 x1k = 5k.
We could try 5k to see if there
are any problems.
In Fig 2 above, the idle
point for a 6550 in triode has been established at Ea = 500V and Ia =
66mA,
with Eg1 = -62V, and Pda = 33 watts.
The class A load of 2.5k for one tube has been
drawn through Q and is the line EFCQG.
Looking at the 2.5k
loadline shows that it it seems a very odd load since there appears to be much
more possible Ea minimum.
Class AB1 operation explained.
The load
which is nominated as being "anode to anode" is difficult for most people to
understand. But where you have two PP
output tubes working in class A for the
first few watts at least, each tube is loaded by 1/2 the anode to anode
load,
which is 2.5k in this case. We need to understand the current flow in
each tube for a complete sine wave cycle
applied to the grid.
As the power
is increased above a few watts, a threshold point is reached where the grid
voltage in one tube moves
negatively enough to cause that tube to cease
conducting any current, and it becomes cut off from the output transformer
until the grid voltage has descended to its negative crest and moves back up
to the cut off grid voltage level.
In the above example, the grid can swing
+62V and a relatively linear current vs voltage outcome is achieved,
but when
Eg = -110V, only 48V into its possible swing the grid has lost control of any
current but the grid still swings another 14V before feaching the -ve
crest.
During this part of the wave cycle, the cut off tube contributes no
current x voltage change to the load connected
through the output
transformer. It is as if somebody removes the tube from the amp for this part of
the wave cycle.
Nothing prevents the negative swing applied to the cut off
triode from moving as negatively as the driver stage permits.
While the cut
off tube is cut off, the only connection the secondary speaker load has to the
PP pair is through 1/2 the primary winding to one tube that is still conducting
because its grid voltage is moving more positively. So the output
transformer
turns ratio is halved from what it is when both tubes contribute
to the power during class A.
When only one tube is conducting, the load
condition for it = 0.25 x RLa-a, and this is equal to the load seen by the
tube
in a class B amp where there is no idle current.
The class B load
line on the above Fig 2 is the line ABCD, and its ohm value is 1.25k.
Each
tube in a PP output stage takes turns at every 1/2 wave cycle to be cut off and
become loaded for the most of the wave cycle with a load which is 1/2 the class
A load it has while in class A which occurs before the AB threshold.
Now
what I have tried to describe with straight lines is an imperfect way of
description of the change of loading that occurs
in a class AB amp. The
change in loading occurs at point C in the EFCQG line. In fact, the rate of cut
off in a triode amplifier is not linear and is somewhat gradual or extended, and
the actual load seen by each triode could better be described by
the initial
part of AB which is straight, then curving down, passing above C and around
towards the bottom left, through Q,
passing above G and finishing so the Ea
axis is a tangent to the load. So in fact the loadline is really a curve.
The change of load in each tube causes quite high distortion currents due to
cut off and load changes but these
are cancelled by action of the output
transformer.
To plot the load line for
one tube and calculate the outcome for 5k a-a :-
(1), Establish the
desired working point for each tube at Q.
Ea = 500V and Ia = 66mA, Eg = -62V,
Pda = 33 watts.
(2), Calculate class B load = RLa-a / 4.
Class B load = 5k
/ 4 = 1.25k.
(3), Calculate Ia for Ea applied across class B load.
Ia =
500 / 1.25k = 400mA.
(4), Drop a vertical from Q to Ea axis to plot point D.
Plot point A which is on the Ia axis and for the Ia current
calculated in
(3).
(5), Draw a straight line from A to D. This is the 1.25k class B
loadline.
(6) Calculate current change for the class A load line =
RLa-a /2 .
Ia = Ea / RL = 500 / 2,500 = 200mA.
(7), Add Ia at Q to Ia from
(6).
66mA + 200mA = 266mA.
(8), Plot point E on the Ia axis for the Ia
found in (7).
(9), Draw the straight line from E through Q and on to G.
Check that Ea at G / Ia from (7) = Rla-a / 2.
665V / 0.266 = 2,500,
OK
(10), Mark the the intersection of the two drawn lines as point
C.
(11), Drop a vertical from point C to the Ea axis and read off the
Ea voltage.
The Ea voltage is 335V.
(12), Calculate class A power
for one tube, P = ( Ea at Q - Ea at C ) squared / ( 2 x RL )
Power = ( 500 -
335 ) squared / ( 2 x 2,500 ) = 5.4 watts.
(13), Calculate class A power for
both tubes at class AB threshold, = 2 x class A power from one tube.
Pure class A Power = 2 x 5.4 = 10.8
watts.
(14), Plot point B where ABCD intersects Ra line for Eg = 0V.
Drop a vertical from B to EA axis and read off the Ea
which is Ea minimum
anode voltage swing.
Ea = 200V.
(15), Calculate maximum peak anode
voltage swing = Ea at Q - Ea minimum.
Peak swing = 500 - 200 = 300V
pk.
(16), Calculate maximum anode to anode swing voltage = 2 x pk swing
at each anode x 0.707 Vrms.
Vrms a-a = 2 x 300x 0.707 =
424Vrms.
(17). Calculate maximum class AB power = max Vrms squared / RL
a-a.
Class AB1 Power max = 424 x 424 / 5,000
= 35.9 watts.
Heat dissipation at
the anode.
The input power to the anode circuit when 36 watts of class
AB power is produced will be more than at the idle condition for both tubes when
the anode input power = 2 x 33 watts = 66 watts.
The power dissipated at the
anodes can be calculated for class AB but I leave that to those of you who are
able to follow the steps in the Radiotron Designer's Handbook. Its safe to
assume that input power may be up double the idle condition
but it will never
be continuous with a test signal designed by Mozart or Beethoven.
The
graph also has a Pda curve showing the maximum limit for Pda = 42 watts. Strictly speaking the class B loadline
should not cross over
to be above the Pda limit line. Where this occurs there is a danger than
the power as heat liberated
at the anode will exceed the pda limit. So it
would be foolish to try to drive RLa-a = 2k, because the tubes would overheat
even at low output
levels.
When testing for full power after completion of an amp at maximum sine wave
class AB power, the dc power going into the CT of the OPT can be measured and
recorded as Idc x Edc.
Power dissipated in
two tubes = B+ power input - audio power output.
In this case
if the Pda for both tubes < 84 watts, all will be well, but even if it
is 20% greater, it won't matter with music
since the average output power
level with music is well below the clipping level using a sine wave which is the
most
arduous of all conditions. Just do not
use low loads.
The 5k a-a load in
the above example is my lower limit for what I would allow to be connected to
tubes.
Regulated power
supply.
Such a class AB amp will potentially draw much more anode dc
from the B+ supply at full power so the B+ must be fairly
well regulated. In
the 1950s a tube rectifier and choke input filter may have been used with a
swinging choke of perhaps 20H and a capacitor of 47uF.
But today any well
rated power transformer with silicon diodes will be better self regulating than
the tube rectifier and choke input filter, and the Si diodes allow a simple CRC
or CLC where C = 235 uF or greater and L = 2 Henrys.
Biasing the class AB PP stage should be done
using fixed bias because with cathode bias there is a big rise in the
Ek when
high level signals are sustained into class AB. But where the amount of class
power is equal to or greater than
half the total class AB power, cathode
bias is fine with music signals. Then the B+ supply needs to be higher to
accomodate
the extra cathode to 0V bias voltage.
Distortion in such class AB triode amps
will reach up to about 4% at clipping.
THD remains low until the AB threshold
when the THD suddenly begins to increase, however such class AB triode
amps
are forgivable since the THD spectra is mainly all 3H with other odd orders well
below.
Output resistance is much
higher for class AB triode than for the class A amplifier because the OPT ratio
for the AB
amp has a lower impedance ratio so the Ra-a is higher at the
secondary. The AB amp becomes de-coupled from one of the tubes while in class AB
so the Rout changes during each full class AB cycle and this is the major cause
of increased 3H
distortion in the THD compared to a pure class A amp.
We
should only be concerned with the Rout during class A because nearly all we
listen to is covered by the
class A power. The slight compression or extra
3H distortion that occurs on transients during AB operation is
negligible
effect in music programme.
The
class A Rout where power < 2 watts = 2 x Ra-a / OPT Z
ratio.
Assume *nominal* speaker = 8 ohms, so use a 6 ohm
secondary winding. OPT ZR = 5,000 / 6 = 833 :1.
Rout = 2 x 1,450 / 833
= 3.5 ohms.
At least 12dB of global
NFB needs to be used to reduce Rout to about 1 ohm.
THD will also fall
to negligible levels.
Using a higher
RLa-a value.
The distortion and output resistance and amount of class
A of the above class AB1 PP amp can be improved by raising the RLa-a to 8k by
using an OPT with a ratio of 8k : 6 ohms for an impedance ratio of 1,333 :
1.
Fig 3.
Maximum power pure class A to 8k : 6 ohms = 13 watts, Maximum power
class AB1 = 30 watts
Distortion, 30 watts = approx 3%,
at 2
watts = <0.5%. Rout 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 : 15 ohms,
pure class A 17 watts < 1.5%,
2 watts < 0.4%.
12k : 9 ohms,
class AB 24 watts <2%,
class A portion 20 watts <
1.5%, 2 watts < 0.6%
8k : 6
ohms,
class AB 29 watts < 3%,
class A portion 16 watts < 2%, 2
watts < 0.7%
5k : 3.75
ohms, class AB 36
watts < 4%, class
A portion 10 watts
<3%, 2
watts < 1%.
4k : 3 ohms,
class AB 39 watts
<5%, class
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 4.
This schematic will sound well with KT90 set up as shown. Although it will
operate with 6550 or KT88, KT90 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.
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, ( and remember there will be two
channels. ) for 95dB, you would need an average of 3 watts per channel. (
this to me is deafening!) 3 watts into 6 ohms is 4.2Vrms of output, and the amp
has a 30 watt ceiling, there is a 10dBV of available headroom for transients. 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 at 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.
Since their Pda rating = 55 watts, we
could set them up at 36 watts of dissipation and get a higher amount of pure
class A,
and greater 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.
Often there can be say 4 secondary layers each with 72 turns
each, and 1 layer with 4 sub sections of 18 turns each.
These can arranged to
make 4 parallel windings of 90 turns to match 6 ohms, or have 5 parallel
windings of 72 turns
which is a match to 3.8 ohms.
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 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.
If more power is needed from triodes, use more output tubes
rather than try to go for class AB2.
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