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.
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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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