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MATCHING 3, TRIODE PUSH PULL AMPLIFIERS
Last edited 2017.
This page is about...
Brief history of triode use, Class A Push Pull basics.
Graph 1. THD vs Vo for SE triode without and without NFB,
and PP amp with NFB form comparisons.
Fig 1. Schematic of basic PP triode output stage with
current wave forms 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.
DESIGN METHOD, PP CLASS AB1, EH6550 IN TRIODE MODE.
Fig 3. EH6550 triode curves with 4k6 load line.
Design Steps 1 to 9.
Relevant notes A to O.
Fig 4. 6550 triode load lines for 3k4, 8k0, 16k0.
Design steps 1 to 11.
Graph 2. Po vs RLa-a for 6550 or KT120.
Formula for Pda for two PP tubes in class AB.
Related issues A to D.
Fig 5. Schematic for 35W+ class AB1 PP triode
amp with 6550, KT88, KT90, KT120.
Comment on using KT90, KT120 or multiple tubes.
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A brief history of triode use.
Over 100 years ago when triodes had just been invented, someone
found that 2 triodes in a push-pull
arrangement could produce the same class A power as 2 parallel
single ended tubes in class A, but with
less than 1/4 of the THD. Two triodes in Class AB1 or AB2 could
make 3 times the maximum power
of the same 2 triodes in class A.
After about 1915 there was a huge demand for high amp power for
radio transmissions, military and
industrial applications.
Joe Average, the cousin of the Man In The Street, found that the
Roaring Twenties offered some cheer
after WW1, and he could listen to radio broadcasts. Joe was
always being radios he saw that liked, but he
could rarely could afford them so he may have bought a crystal
set and a pair of headphones which were
sensitive enough to be powered by the resonant energy of RF pick
up in an antenna. Joe's rich uncle Bob
had more dough than Joe, so Bob soon bought an AM radio with 1
or 2 vacuum triodes which allowed
hort wave reception plus all the local stations. And there may
have been a horn type of speaker, and this
allowed 150Hz to 3kHz of audio which could fill a room without
needing headphones. Radios were soon
made for Mr & Mrs Stoinking Riche, and they may have had
several single ended triodes for RF amplifiers,
and then a push pull pair of triodes to power a large sensitive
electro-dynamic speaker. This allowed 200
rich ppl in a huge room to dance, and eye each other off for
late at night dalliance, something poor Joe
could never do. Motor cars for the Uber Rich had radios fitted,
and with PP triodes to disturb the posh
neighbourhood late at night, or listen to stock market reports
at a more sober level.
Push-Pull triode circuits could make twice the class A power of
one triode but with less than 1/4 of the
THD of one triode. By 1930, the world had PP amps working in
class AB1, AB2, B, B2, C, and needing less
electricity than any simple single ended class A amp.
A majority of radio listeners in 1930 had a single triode to
power a small speaker. They could keep track of
the stock market crash and otherwise enjoy some music to
distract them from the misery of the The Depression.
Pentodes and beam tetrodes were devised after 1929, and the
triodes in radios quickly all disappeared, and by
1935, a minimal superhet radio had 1 mixer tube, 1 pentode IF
amp, and a 6V6 output tube, and a small 150mm
speaker. The maximal radios for the rich had an extra 2 signal
pentodes and perhaps a pair of 2A3 for a PP 10W
AB1 triode amp. These were quite luxurious, and there was often
a jack to take input from a crystal cartridge
for 78rpm record play. The result with 78 records in 1935 was
not hi-fi, but if you were young, and armed
with a decent bottle of whiskey, you could have fabulous early
rave parties. The PP amp could easily power
the 12" speakers in a large cabinet and bass performance crept
down to 90Hz.
Negative feedback was not much used before 1930, Harold Black
was credited with inventing it in 1928.
The idea was slow to be adopted because triodes have internal
resistance lower than the load they must drive
so they could give a damping factor between 2 and 4 and the PP
circuit kept THD/IMD less than 1% at low
listening levels for say 2 x 2A3, so NFB was not needed, or
wanted, because an extra gain signal triode or
pentode was needed before you could apply NFB which reduced the
gain. NFB seems like a stupid idea
when I say such things, like two steps forward while going 3
steps backward. Besides, all the signal sources
driving triode audio amps before 1947 had far more noise and
distortions of all kinds than the amps.
NFB was never understood by anyone except university professors
or a tiny minority of their students such
as Mr Peter Walker, Peter Baxandal, David Williamson, and a few
others. By the end of 1930s, a tiny minority
of triode SE and PP audio amps had some NFB. But a larger number
of amps using pentodes or beam tetrodes
needed some FB to work properly at all. For the General Public,
PP amps did not outnumber SE amps until the
1960s when it became possible to build a cheap solid state amp
which cost far less to install into a radio,
radio-gram, or TV set where there usually was a single 6F6, 6V6
or 6BQ5 for the audio amp.
The wealthier folks bought Deluxe TV sets and hi-fi sets with PP
tubes from between 1945 and 1960 when solid
state rapidly began to reduce demand for anything with any tubes
in it. Many tube enthusiasts thought most ppl
had no powers of discernment, they though folks were throwing
out the bath water with the baby when they
embraced solid state and stereo. People generally buy anything
that is cheap as long as they are fooled by lies
told by marketing ppl. The desire to have something overwhelms
their ability to wait until they have the dough
saved up and the whole experience of affluence is to avoid using
any judgement at all about what is bought,
not realizing they are being conned by companies setting out to
make garbage for the hungry masses to buy.
There has always been a minority of those who prefer
triode amps. Some are rabidly extreme about triodes
and will not ever accept push-pull circuits. My large number of
customers enjoyed tubes and knew I was not
wasting their time or money with my efforts with beam tetrodes,
pentodes and triodes and all with NFB.
The amps I made advertised themselves; I would lend them a tube
amp and they said it made music better,
so that's why the bought it. They could ignore the cheap
rubbish.
My pages on SE triode amps show that very good sound is possible
with just one triode. The maximum THD of
1% may occur at say 5W from a single EL34 in triode mode with
say 15dB of GNFB. At average levels of say
0.25W, THD = 0.2%, and most ppl think that sounds just fine, and
better than many other amps.
Graph 1. THD vs Vo.
Graph 1 is for 300B, KT88 or 6550 in triode or similar.
Curve A is for one single 300B in Single Ended Triode operation,
with zero GNFB, Curve B is 2 x 300B
in PP, without GNFB. Curve C shows the PP triode amp with 12dB
GNFB, and you can see just how
much less THD is produced with PP and GNFB. When used wisely,
GNFB reduces all types of distortions,
THD, IMD, and extends frequency response easily from 20Hz to
60kHz.
Single Ended operation always means one device is in a circuit
including PSU, triode, and a speaker.
A small input voltage from a CD player etc feeds one grid input
to a tube and its anode current is changed
linearly and powers the speaker between anode and PSU, via an
OPT. There is only ONE phase of signal
in each amp device. The SE triode amp requires only 2 input
driver triodes which may be the 2 halves of
one 12AU7.
Pull or PP operation has two grids of two tubes each driven at
their grids by Vac with opposite phases to
produce two phases of the
same audio signal at the two anodes. The OPT between the two
anodes has a primary winding with B+
from PSU applied at a centre tap. There is balanced Idc flow in
the OPT primary which extends the LF
capability. The secondary of OPT usually produces a single ended
signal with one phase to drive a speaker.
To get the two phases of drive to drive the two PP output tubes,
there are usually 3 input/driver triodes,
such as 3 halves of a 12AU7 or 6SN7 etc.
Graph 1 shows the power output needed for most ppl for ONE amp
and ONE speaker with sensitivity
of 87dB/W/M. With 2 amps and 2 speakers, and while sitting 3M
away in an average lounge room with
plenty of soft furnishings, you may get comfortable non tiring
levels
with less than 1W per amp. 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, the distortion could double,
but it is still well below the single SE triode.
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 10W.
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 uses two identical triodes each
working in class A and with the same
load conditions as one SE tube working alone. Twice the Po of
the SE tube is available and the distortion
is lowered because the PP tubes produce opposite phased Vac and
Iac which contain common mode 2H
distortion currents. Common mode current cannot flow in a PP
OPT, so the common tube current is
excluded from current across the whole OPT primary.
Push pull 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, the pull and push
forces are different.
With two men at each end of saw the sum of the push and pull
forces of each man become equal, and the
motion of the saw is much more even than if only one man was
using the saw. The force produced by each
man in each direction may be described by a sine wave with high
positive peak and flattened negative peak,
ie a sine wave with second harmonic distortion. Each man has the
same force production character but when
combined on the saw, the motion of the saw shows positive and
negative peaks in each direction, so the
second harmonic is not present. The saw wave motion would show
flattening of both peaks, but a smaller
amount than one man on his own. The peak flattening is largely
third harmonic.
There are second harmonic currents generated in each of 2 PP
triodes, and these have the same phase
even though the Vac at each anode is opposite phase. Each tube
applies its 2H Iac to each end of the PP
transformer, but there is no 2H current flow in the winding so
there is no 2H Vac produced in the load.
In class AB or class B the % of distortion currents in can
exceed 25% because the current waves look
like those in a 1/2 wave rectifier circuit in a power supply.
Yet the two tubes, produce remarkably low
THD in the Vac across the whole primary and in the secondary
winding. The primary winding with a
CT can ONLY accept power into its primary load if there is
difference between Vac applied at each
end of the primary winding. If TWO EQUAL amplitude Vac with same
phase are applied to each end
of the primary, there is no voltage difference across the
winding, so no current can flow.
Fig 1.
Fig 1 shows the relative signal voltages and currents in a
typical PP triode output stage.
The schematic describes basic operation. Many details of a real
amp are missing, but you should
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 220r x 1/2W.
In this case, each 6550 has Ea = 420Vdc and Ia = 70mA,
so Pda = 29.4W, or 70% of maximum rated Pda of 42W.
The class A load for each triode = RLa = ( Ea / Ia ) - ( 2 x Ra
) = ( 420V / 0.07A ) - ( 2 x 1,000 )
= 6,000r - 2,000r = 4k0.
The 1k0 for triode Ra is approximately correct for 6550 in
triode mode.
Each 6550 theoretically produces class A Po10 = 0.5 x Iadc
squared x RLa
= 0.5 x 0.07A x 0.07V x 4,000r = 9.8W.
Like most things in the Universe, tubes don't act perfectly
linearly. The change of Ia with linear
grid Vac change follows a formula of Ia = K x ( [ square root of
Vg1 change ] cubed ). K is a constant
for all equations which you don't need to know now because you
just need to know Ia changes
according to a formula that is not linear, because quantities
relate to square root of something cubed.
The Ra curves for any triode will show the non-linear properties
but despite this, triodes make fabulous
amplifiers and are one of the most linear devices in the
Universe filled with most things behaving so
non linearly we cannot figure out formulas for them.
Consider one triode of the above pair. It sits there with 70mAdc
flow with no signal at grid with
-51Vdc applied. If the grid has 10Vrms of say 1kHz sine wave
applied to grid, then gain should be
A = ยต x RLa / ( Ra + RLa ). For the above 6550, gain A = 6.7 x
4k0 / ( 1k0 + 4k0 ) = 5.2,
so we may expect Va = 52Vrms. But the positive halves of input
sine waves will cause more
Ia change than the negative half waves. When Iac is examined, we
find there is a percent or more of
second harmonic present, and if the tube was in a single ended
circuit by itself we could not get the
theoretical 9.8W we calculated above, and 2H at clipping would
be about 6%, and a lesser amount of
other H.
But where we have TWO 6550 in the PP circuit, and with both
having oppositely phased Vac and Iac,
the 2H in the Iac of each triode has the same phase. Therefore
each tube is generating the same phase of
distortion current which is applied to both ends of the primary
coil with its center tap to a low Z PSU.
The primary of the PP OPT is like a see-saw, and if a child sits
at each end with same weight, they
cannot swing up and remained balanced. If one leans back the
other leans forward, then the one leaning
back descends because he applies more rotating force than his
sister. The see-saw only moves when there
is a difference between forces applied at each end. So the 2H
currents in tube are applied, but cause no
voltage change across the OPT, so no 2H appears in any winding
of the transformer. That is the theory,
and in practice the balance of 2H is never perfect, so there is
some 2H from a pair of PP triodes, but the
THD reduction is often more than tenfold. The same applies where
2 mosfets or bjts were used in a PP
circuit.
Another phenomena occurs. Where one tube produces say +60mA and
then - 50mA with equal
+/- Vg change, the other tube is producing -50mA and +60mA at
the same time. The summed current
behavior of both tubes shows the resulting load current change
in secondary is remarkably linear.
When examined further we conclude the load of each triode
changes slightly during each wave cycle,
because one tube affects the behavior of the other. Of course
the tubes are not perfect and produce odd
number H of opposite phase. Most is 3H, and thus 3H current
flows across the primary load between
anodes. 2 x 6550 triodes in Fig 1 will typically generate 1% of
3H at 1dN below clipping, with 2H < 0.5%.
There may be other lesser amounts of all other H, but perhaps
all would have total < 0.3%. All these H
increase very rapidly when clipping and wave peak flattening
occurs.
These currents are plotted in graphs in Fig 1 and labelled as
"cathode currents" but are equal to anode
currents.
The small Iac wave 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 10r0 x 5W between k1 and k2 cathodes
to 0V and examine waves with an oscilloscope, and then you will
see the distorted cathode current flow
between each cathode and 0V. If you analysed the current wave
you would discover there is a distortion
current flow with twice the frequency of the applied Vg at
grids, and the magnitude of such 2H currents
may be about 7% of the total current wave. The analyzer may show
there are smaller amounts of 3H,
4H, 5H, 6H etc, present. But 2H will be the dominant harmonic
produced.
In class AB amps where RLa-a is less than the load for max class
A Po, each tube cuts off for part of
the positive going Va signal, with Ia change limited to idle
Iadc. But on negative going Va signals,
Ia increases to well above Iadc at idle. Each tube has opposite
Va phase so while one tube cuts off,
the other has much increased peak.
For maximum pure class A, the RLa-a for 2 triodes
= twice the RLa for each tube = 2 x ( [ Ea / Iadc ] - [ 2 x Ra ]
).
Ie, 2 x 4k0 = 8k0. Class A Po for 2 tubes = 0.5 x Iadc squared x
RLa-a
= 0.5 x 0.07 x 0.07 x 8,000 = 19.6W.
For class AB the RLa-a is always less than the RLa-a for max
class A.
Consider RLa-a = 4k0.
Each tube has load of 2k0 while it works in class A, but Ia can
increase to more than idle Iadc,
but decrease of Ia is limited to Iadc. The max initial class A
Po is thus 0.5 x 0.07 x 0.07 x 4,000 = 9.8W.
but we may find the max class AB Po = 29W, and between 9.8W and
29W,
each tube has a load of 1/4 RLa-a = 1k0.
The change of anode load is not abrupt because the cut off
behavior of triodes is somewhat gradual,
so THD caused by change of load and gain is mainly 3H and we may
expect 3% at 29W. But at Po
= 5W with RLa-a 4k0, the THD is little different to 5W with
RLa-a = 8k0.
The 1947 Williamson triode amplifier used 2 x KT66 in pure class
A to obtain 12W pure class A.
THD without any negative feedback = 1%. With 20dB global
negative feedback, GNFB, the THD was
reduced to 0.1% at 12W. If you build a Williamson now, you will
get the same excellent results and sound
is excellent where your speakers have high sensitivity.
Intermodulation distortion, IMD is the far more evil
ugly cousin of THD, and the Williamson produced quite adequately
low IMD. Mr Williamson's schematic
can be much improved, but I will not delve into details
now.
For 2 x KT66 with idle Pda = 21W, Ea 400Vdc, Ia = 53mA, and with
Ra = 1k6, for max class A,
RLa-a = 2 x ( [ 400 / 0.053 ] - [ 2 x 1,600 ] )
= 8.69k. Class A Po = 0.5 x 0.053 squared x 8,690r = 12.2W.
If the RLa-a was reduced to say 4k3, there is an initial 6.1W of
pure class A and maybe 16W of class
AB Po.
As many people became wealthier after 1945, demand for more
power increased with the demand for
all manner of goods which were expected to meet ever higher
demands for quality and functions.
So by 1970 those who may bought a tubed Quad-II amp system at
ESL57 then wanted a Quad 405
solid state amp with a pair of ESL63. They'd also want two
garages in their house, so two fancy cars
could be parked under cover - all unthinkable before about 1960
for most ppl. Manufacturers could not
make sensitive speakers of 96dB/W which had a flawless response
and and so speakers became less
sensitive, smaller, but often better, but you needed more power.
A pair of 6550 triodes in AB could make
up to 3 times the maximum class A Po for a pair of KT66 in
triode. 36W from PP triodes usually sounds
fine. I once repaired an ARC 3030 with two channels each 30W
from a pair of 6550 in class AB.
Once repaired, it sounded extremely well.
The power consumption of a PP monobloc triode amp with 2 x 6550
( or 2 x 300B ) including filament
heating and the input driver amp is typically 90W, so 180W for 2
channels. A total of 60W audio Po max
is available so efficiency is low, but most ppl find they spend
far more on everything else, and they can
afford the tube power. Such ppl who prefer vacuum tubes are well
aware that The Final Solution has
been applied to audio amps since 2000 where most amps are PWM,
pulse width modulated, ie,
"digital amps". So they save $4 a week while earning an average
of $1,000 after paying tax.
Better sound? I doubt it.
Maximum class A Po for triode PP amps cannot exceed about 32% of
the idle Pda. The idle Pda
= Ea x Iadc and should not exceed 0.6 x maximum Pda rating. The
triode PP amp should be able to
make initial pure Class A Po = 1/3 of maximum class AB Po. The
rated Pda for 6550 and 300B is 42W,
so idle Pda, ie, heat liberated by anode at idle should not
exceed 0.6 x 42W = 25W. With two tubes
each with Pda 25W, total Pda = 50W, and max class A Po = 16W.
The AB Po may be 35W max but
with a lower RLa-a, and maybe only 6W of pure class A. But most
would find that sounds well,
providing all other design issues are optimized, and accountants
are not allowed to assist design personnel.
The lifetime for most output tubes is short where it is idled at
near the rated Pda. But with Pda at
0.6 x rated Pda, expect 5,000 hours. 365 days at 4hours a day =
1,460 hours, so expect 3.4 years
for output tubes. During my 18 years in the audio trade, I had
customers who I worked with to maintain
their systems and there were 3 who replaced their output tubes
twice. So they got more than 3.4 years
per set of tubes.
To better understand triodes, and design the load for them, one
needs to understand their Ra curves.
The "triode" may be a beam tetrode or pentode which can have its
screen connected to anode, so it
works in exactly the same manner as a real triode such as 45,
2A3, 300B or 845 which do not have
a screen grid. For more about basics, see my pages starting with
basic-tube-1.html
Fig 2. Ra curves for EH 6550 triode.
This is a typical curve for any available 6550 strapped as a
triode. Feel free to copy it, and use it in
MS Paint or other program to draw loadlines.
You may wonder what each Ra curve means. Each Ra curve
represents a value of resistance which
varies. At any one point on the curve, the resistance = minute
Ea change / minute Ia change.
R = E / I , Ohm's Law. The steeper the slope of the curve, the
lower the resistance. A vertical line
has zero resistance. A horizontal line has infinite resistance.
A straight line at say 45 degrees slope will represent constant
resistance = Ea change / Ia change.
One of the curves above has changing resistance for different Ea
and Ia.
Consider graph co-ordinates of 412V x 50mA. This is on the Ra
curve for Eg1 = -50V.
If a straight line tangent to the curve at this point, the
resistance value of line will be about 1,240r.
If we look at 460V x 100mA at higher on the same Ra curve, and
draw a tangent line the R value
= 800r. If we have a 6550 triode with idle condition of 412V x
50mA, then we will find the dynamic
resistance from anode to 0V = 1,240r. If we increased the Ea by
+/- 10V, we would expect Ia change
= 10V / 1,240r = +/- 8.06mA. The tube may seem like a resistance
sitting there with 412V across it and
50mA, ie, like R = 412 / 0.05A = 8,240r, but something causes it
to have lower resistance than the static
measurement of Ea / Iadc.
The triode anode has the property of transconductance, gm, ie,
the voltage field effect on electron
flow causes change of Ia. In the case of 6550, 10V change at
anode causes Ia change 8.06mA, so
anode has apparent gm = 8.06mA / 10V = 0.806mA/V.
Put simply, the gm of anode = 1 / Ra = 1 /1,240r = 0.000806A/V =
0.806mA/V. If we know what
the anode gm is, then Ra = 1 / anode gm.
This is all different to 6550 when set up in tetrode mode with
screen g2 connected to a fixed Eg2.
Then we would find Ra is maybe 30,000r, and the anode gm = 1 /
30,000 = 0.033mA/V.
The Ra curves for tetrode are nearly flat horizontal lines. when
g2 is kept at say +350Vdc, and
change to anode V cannot change the Ia very much because the
screen does what screens always do;
they screen off the field effect of anode Ea change. Its not
perfect screening, but is substantial.
But when screen is connected to anode, it conveys the anode
field effect to lower regions of the
tube as if the screen and anode were one single sheet of metal.
To draw the curves you see here, samples of 6550 were set up in
a test circuit a test circuit with its
cathode at 0V. Its grid was taken to low resistance negative Vdc
supply able to be adjusted between
0V and -120Vdc. The anode is connected to a low resistance Vac
source able to swing from 0V up
to +850Vpk.
The set of Ra curves could be displayed on a CRO screen to show
Ia change vs Ea change. Rapid change
of Ea is repeated for each Eg1 bias value which is rapidly
stepped by the testing circuit so all curves can
appear one after the other. I found a site where a fellow has
adapted a '570' machine previously used for
curve plotting for transistors,
https://www.youtube.com/watch?v=ayLjvgeyfTc
He explains his testing of 6V6 and 12AX7 to see if the curves of
2 sample tubes are equal. I also found a
site with a picture of a guy at Philips in 1951 with a hugely
complex mass of tubed electronics to show
calibrated tube curves. Such massed curves can be photographed
and printed, or converted to simple black
and white drawings similar to above results for 6550.
The actual value of Ra for any given idle condition can be
calculated from measurements taken of the
change of gain for 2 different RLa with the same Vg applied.
This is dealt with in my page basic-tube-4.html
The old text books from before 1960 explain tube testing and
curves more deeply than I have time
for here. After some time, the dedicated DIYer may understand
how to use the Ra curves he sees,
and he does not have any need to plot his own curves for any
tube because it has all been done better
than he ever could, and using a vast amount of time and
equipment he could never afford.
We just need to know how to use the Ra curves, and understand
how a line on a graph describes
resistance.
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DETERMINE ANODE LOAD FOR PP CLASS A1 FOR TRIODES :-
This example uses 2 x 6550 with screens connected to anode in
triode mode.
Fig 3. Loadline for 4.59k for 6550 triode.
Fig 3 shows one 6550 in a pair operating in pure class A1, and
very similarly to 6550 in Fig 1 above.
(1) Nominate idle conditions for each triode. Pda at idle = 28W.
Ea = +425Vdc,
Iadc at idle = Pda / Ea = 28W / 425V = 66mAdc.
(2) See Fig 2 above. Plot point Q for Ea = +425V and Ia = 66mA.
This will be at Eg1 = -50Vdc.
Draw tangent line S-T to nearest Ra curve, ie, for Eg1 = -50V,
through Q.
Ra at Q = Ia change for Ea change along straight line S-T. Ra =
( 510V - 354V ) / 150mA = 1,040r.
(3) Plot point B for 2 x Iadc on Ra curve for Eg1 = 0V.
B is at 122V x 132mA. B is at Va minimum. +/- Va pk swing = Ea -
Va min = 425V - 122V = 303Vpk.
RLa = Va pk swing / idle Idc = 303V / 0.066A = 4,590r.
(4) Va pk max = Ea + Va pk swing = 425V + 303V = 728Vpk.
Plot point C at 728V x 0.0mA.
Draw straight line from C to B and on through Ea = 0V axis. This
line should pass through Q.
Plot point A at about 160mA x 0V.
ABQC is the class A load line for RLa = 4.590k.
(5) RLa-a for PP = 2 x RLa from loadlines above = 2 x 4.590k =
9.18k.
RLa for a single triode in class A may be simply calculated = (
Ea / Ia ) - ( 2 x Ra ) where Ra is
measured from Ra curve for Eg1 = 0mA and point B on Ra curve for
Ia = 2 x idle Idc.
Point B is at 132mA x 122V, and Ra = Ea change / Ia change =
122V / 132mA = 924r.
Therefore RLa for one 6550 = ( Ea / Iadc ) - ( 2 x Ra ) = ( 425V
/ 0.066A ) - ( 2 x 924r )
= ( 6.439k - 1.848k ) = 4.591k, which agrees with load line
analysis.
For two 6550 in pure class A, RLa-a = 2 x RLa for one 6550 =
9.182k, or can be calculated
RLa-a = 2 [ ( Ea / Iadc ) - ( 2 x Ra ) ] with Ra from curve for
Eg1 = 0V and Iadc is for one triode.
(6) Line S-T shows Ra is higher at the Q point, and this Ra
value is used to determine damping
factor = RLa / Ra at Q = 4.59k / 1.04k = 4.4.1
(7) What is maximum power output at clipping? From (3), Peak Va
swing = 303V,
Va = 214Vrms. For one 6550, Po = Va squared / RLa = 214V x 214V
/ 4,590r = 9.997W.
For 2 PP tubes, class A Po = Va-a squared / RLa-a = 428V x 428V
/ 9,180r = 19.96W.
Theoretical max class A Po at anodes depends on low THD with
GNFB.
The Po at output will be less than at anodes because of winding
wire resistance losses in OPT.
Class A Po at anodes may also be calculated = 0.5 x Iadc squared
x RLa-a, where Iadc is for one tube.
PO = 0.5 x 0.066 x 0.066 x 9180r = 19.99W.
This formula is valid for calculation of initial class A for all
RLa-a values less than for maximum
pure class A, ie, for amps working in class AB. If RLa-a = 4k0,
then initial class A = 0.5 x 0.066 x 0.066 x 4,000 = 8.72W.
Calculation of max Po for PP class AB
will be below.
(8) What is output transformer load ratio? OPT wanted should be
9k2 : 4r0, 8r0, 16r0.
(9) If this load ratio is not easily available, what solution is
possible?
Try use of 2 x KT120 with same Ea = 425V, and each with Pda =
0.6 x 55W rated max Pda,
= 33W.
Iadc per tube = 77mAdc. Ra for Eg1= 0V between 0.0mA and 154mA =
800r approx.
For max class A, RLa-a = 2 [ ( Ea / Iadc ) - ( 2 x Ra ) ] = 2 [
( 425 / 0.077 ) - ( 2 x 800 ) ]
= 7.84k.
Max Pure class A = 23W.
OPT for pure class is 7k8 : 4r0, 8r0, 16r0.
But you could use a 60W rated Hammond 1650 series 6k6 : 4r0,
8r0, 16r0.
This will give initial class class A = 19W, with maybe 25W in
class AB. Problem solved.
---------------------------------------------------------------------------------------------------------------
DETERMINE ANODE LOAD FOR PP CLASS AB1 FOR TRIODES.
This example uses 2 x 6550 with screens connected to anodes for
triode mode.
CLASS AB1 IDLE CONDITIONS.
For pure class A, we may use idle Pda = 0.7 x Pda rating. Pda
means Power Dissipated at Anode.
Pda at anodes of pure class AB amps need not concern us much
because as long as the loading
of the tubes keeps the tubes working in class A, the tubes won't
get any hotter. But in class AB,
tubes can get hotter due normal load conditions. So we MUST
worry about Pda.
A pot on a stove feels hot even before you touch it. Heat
radiates to your hand. The pot is hot,
and it radiates heat energy into your kitchen. It is dissipating
heat. The stove is supplying heat to
the bottom of the pot, and the pot absorbs heat and its
temperature rises and then radiates heat
it has absorbed. There is a limit for how much heat can be
radiated by the pot before it begins to
melt. Fortunately, a stove will not be able to supply enough
heat to melt a pot, although eggs in
the pot may be converted to a horrid mess.
All components of any electronic device such as transistor,
mosfet vacuum tube, resistor, wire,
capacitor will have a maximum amount of heating power within
itself which is safe to tolerate.
The heating power = Average Volts x Average Amps, and measured
in Watts. For a 6550, the
heat produced by anode must not exceed 42W where the Volts and
Current are applied continuously.
So for 6550, having Ea = +500Vdc and Ia = 84mAdc, Pda = Ea x Ia
= 500V x 0.085A = 42W.
At 55W the anode may begin to glow red, and if 100W, orange, and
tube will soon destroy itself.
At 42W, the TEMPERATURE LIMIT has been reached for safe
operation. The average lifetime
of a 6550 working at Pda 42W is much shorter than for where it
works at Pda 21W.
But during operation of most active electronic devices including
6550, Pda can be much higher
than rated Pda for a short time.
The 6550 could have 500V between anode and cathode and have Ia =
500mA, and Pda =
250W, but not for long. It could be for 1/5 of the time, ie,
500V x 500mA for 0.2seconds,
then zero V and I for 0.8 seconds, and so on. There are times
when Mr Idiot connects a
short circuited speaker across output terminals and Pda for
tubes goes way above their rating,
and a fuse might eventually blow just after a tube has failed
and OPT ruined.
A number of tubes for use in transmitters use anodes which have
heat exchanges attached which
allow fans to force air through fins to allow the Pda to be much
higher than if they relied upon
radiation of heat through the glass envelope. Tubes run with
cooler anodes if their glass is kept cool,
so convective air flow up and around output tubes must be
allowed to occur. If a tube is close to
surrounding objects which heat up including other nearby hot
tubes, its temperature is hotter. All tube
amps MUST have good ventilation. There are some triodes used for
radio transmitters which use
anodes with fins for air cooling by blower fans. Some larger
triodes have water cooled anodes.
Fan cooled tubes are not used in hi-fi amps because they create
far too much noise.
Music is not affected by temperature of the tube as long as
temperature remains below what it would
be with continuous heating to have Pda = max rating. All active
devices have temperature limits, and
music is fine just as long as the devices run cool enough.
Even with zero heat produced at anode, a tube will feel warm
because it has a heater element inside
the cathode to heat it up to about 900C to get electron emission
in the vacuum to occur. In 6550,
the heater needs 6.3V x 1.8A = 11.3W to get the cathode hot
enough. The cathode is cooler than heating
element, but cathode radiates heat out through grid wires which
get hot, and cathode is hot due to
cathode heating. The additional allowed anode heat = 42W.
The screen also has screen Pdg2, and if
Eg2 = 500V, expect Ig2 = 6mA to screen Pdg2 = 3W, so the total
heat radiated away from tube
= 11.3W+42W+3W = 56.3W. The heat radiated through glass raises
the glass temperature and this should
never rise above 250C, or above whatever the data says.
Overheated tubes melt the glass, or cause
glass to fracture, allowing air entry, which immediately stops
electron flow - forever.
For class AB, idle Ea may be higher and Iadc lower than for pure
class A mentioned above for pure
class A. The example for pure class A above has Ea = +425Vdc.
This suits nearly all octal base output
tubes 6550, KT88, KT90, KT120, KT150, KT66, 6L6GC, EL34 and
300B, an old design 4 pin tube.
Ea max for triodes such as 45, 2A3, or triode strapped EL84,
EL86, 6V6, 6CM5 will be much less
than +425Vdc.
For 6550, KT88, KT90, KT120, KT150, we may use Ea = +500Vdc to
obtain slightly more maximum
class AB Po but with very adequate amount of initial pure class
A working.
For all these tubes, I recommend Pda at idle for pure class A
does not exceed 0.7 x Pda rating.
For class AB1, Pda should not exceed 0.6 x Pda rating.
Fig 4. PP Loadlines for class AB 6550 triodes, 3k4, 8k0,
and class A 16k0.
Fig 4 shows 6550 triode curves with curve for Pda limit = 42W.
There are TWO sets of loadlines,
for RLa-a = 8k0 and RLa-a = 3k4. Loadlines give us a picture to
define what we are doing without
depending on a vague notion of numbers and calculations. The
load lines and calculations lead us to
a suitable output transformer, and this knowledge is the key to
the best spiritual experience of music,
to give us peace for our troubled lives.
The Ra curve for Eg1 = 0V is the limit line for Ea swings in
class AB1, where no grid current flows
to limit the driver tube output. The Ra curves for positive Eg1
values are not shown here, and were not
often included in old data sheets. Class AB2 needs the use of
direct cathode followers to drive grids
and the effort and expense is seldom worth the extra amount of
maximum AB2 power above the
maximum AB1 power. If more triode power is wanted than is
available with class AB1, then use more output
tubes in class AB1 instead of flogging 2 triodes to death in
class AB2.
Idle conditions 6550 here have idle Pa = 0.6 x Pda rating for
the tube = 0.6 x 42W = 25W.
Iadc at idle = idle Pda / chosen Ea. Ea is to be +500Vdc. Iadc
idle = 25W / 500V = 50mAdc.
(1) Plot point Q at 500V x 50mA.
(2) Plot point X at 2 x Iadc on Ra curve for Eg1 = 0V.
Point X is at 95V x 100mA.
(3) Draw straight line from X through Q to intersect 0.0mA axis
line. Plot point Y.
XQY is the load for maximum pure class for one 6550.
Calculate peak Ea swing = Ea - Ea at point X = 500V - 95V =
405V.
Calculate RLa for one 6550 = Ea pk swing / Iadc = 405V / 50mA =
8k1.
Calculate pure class A RLa-a for both tubes = 2 x RLa = 16k2.
(4) Calculate Nominal class AB RLa-a = 0.5 x RLa-a for pure
class A = 0.5 x 16k2 = 8k1,
say 8k0.
Calculate class A RLa with RLa-a = 8k0 = 0.5 x 8k0 = 4k0.
Calculate class B RLa with RLa-a = 8k0 = 0.25 x 8k0 = 2k0.
(5) Calculate Ia on 0V axis line with class A RLa = 4k0.
Ia = ( Ea / ( 0.5 x RLa-a ) + Iadc
= 500V / 4k0 + 50mA = 125mA + 50mA = 175mA.
Plot A at 175mA.
Draw line from A through Q to 0mA axis at B. Plot point B.
(6) Plot C on 0mA axis line at 500V.
Calculate Ia on 0V axis line with B RLa = 2k0. Ia = Ea / B RLa =
500V / 2k0 = 250mA.
Plot D at 250mA. Plot E where CD intersects Ra curve for EG1 =
0V.
(7) Plot F where CD intersects AQB.
(8) Is F at twice Iadc ? F is at 100mA.
Is Ea between FQ = same as between QB? FQ = 200V, QB = 200V, OK.
If FQ is not equal to QB, a mistake has been made; check all and
repeat process.
(9) Conclusions. FQB is the class A RLa = 4k0 for each 6550.
The peak Va swing = 200V = 141.4Vrms. Pure class A Po
produced by each 6550 = Va squared / class A RLa = 1.41.4 x
141.4 / 4,000r = 5.0W.
Pure class A Po produced by both 6550
= Va-a squared / class A RLa-a = 282.8 x 282.8 / 8,000 = 10.0W.
EF is the Class B RLa which loads each 6550 for class AB Po
above 10W. During class AB,
Ia in one tube reduces to zero mA for part of the wave cycle and
anode is effectively disconnected
from one end of OPT. While this occurs, the other tube remains
connected to its half primary winding.
The RLa-a load is across both halves of OPT, so the connected
tube connects to load with 1:2 turn ratio,
so RLa-a is transformed to 0.25 RLa-a = 2k0. Each tube takes
turns to be cut off while the other powers
the low load.
EF is the B RLa for each tube for all Po above 10W.
(10) Calculate class AB Po. Point E is at 152V x 174mA. Max Ea
swing each 6550 = Ea - Ea at
point E = 500V - 152V = 348Vpk = 246Vrms.
Va-a = 2 x Va swing at one anode = 2 x 246 = 492Vrms. Po = Va-a
squared / RLa-a
= 492 x 492 / 8,000r = 30.3W.
(11) what is minimum RLa-a?
Here is a graph of Po vs RLa-a at clipping for a pair of 6550
triodes.
Graph 2. Class AB triode Po vs RLa-a.
Graph 2 gives anode Po for values of RLa-a where THD < 2% for
sine wave input.
There are graphs for 2 x 6550 and KT120, and we need only
concentrate on 6550 because a pair
will provide excellent music, with KT120 offering only marginal
increase in maximum output power.
However, KT120 have Pda max rating = 65W, and so can have idle
Iadc = 70mA, thus nearly doubling
the initial maximum pure class A for RLa-a below 11k6. Thus
KT120 may not give much more Po,
but may sound different, neither better or worse, but just as
enjoyable.
The graph shows there is no point ever considering the lowest
minimum RLa-a to be less than 2k5.
The Pda increases for clipping Po as load decreases, and IMHO,
all amplifiers should be able to
sustain clipping Po with a continuous sine wave for a load which
produces Pda = rated limit from
data sheets, for at least 30 minutes.
My formula for Pda for class AB amps where there is
substantial idle current is based on
Pda = [ Power input from PSU ] - [ Audio power output ].
Pay great attention to the equation brackets or you will make
terrible mistakes !!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Pda for BOTH tubes =
Ea x [[ ( 0.364 x Iadc ) + ( 1.8 x Va-a / RLa-a ) + ( 0.364 x
Iadc squared / { [ 2.83 x Va-a / RLa-a ] - Iadc } ) ]] - Po.
Pda for RLa-a = 2k5 at 40W. Va-a = 316Vrms. Iadc per tube at
idle = 50mA = 0.05A.
Pda = 500V x [[ ( 0.364 x 0.05A ) + ( 1.8 x 316Vrms / 2,500r ) +
( 0.364 x 0.05A x 0.05A / [{ 2.83 x 316Vrms / 2,500r } - 0.05A ]
) ]] - 40W
= 500V x [[ 0.0182A + 0.2275A + ( 0.00295A ) ]] - 40W.
= 500V x [[ 0.2486A ]] - 40W = 124.3W - 40W = 84.3W. Therefore
Pda for each 6550 = 42.15W.
It is co-incidental that I picked the load where Pda has risen
to exactly the rated maximum Pda.
For all RLa-a less than 2k5, Pda max will be above 42W. For all
RLa-a above 2k5 the Pda will be below 42W.
The calculation for KT120 with Iadc 50mA would give the same max
Pda, but because they have rating of
65W, they will not fail when 2k5 load is connected.
The sensible lowest RLa-a should be where Po = 90% of the
maximum possible Po. In this case it is
40W x 90% = 36W, and from graph 1, lowest RLa-a load = 4k0, ie,
1/2 of the Nominal RLa-a = 8k0.
(12) Conclusions for class AB1, 2 x 6550, Ea 500V, Iadc = 50mAdc
each tube.
Minimum AB RLa-a = 4k0.
Nominal or Middle AB RLa-a = 8k0.
Pure class A RLa-a = 16k0.
(13). Define OPT wanted for 2 x 6550 or KT88 in class AB triode
with Ea = 500V :-
Nominal load ratio :- 8k0 : 4r0, 8r0, 16r0.
------------------------------------------------------------------------------------------------------------------------------
Some related issues.
(A) What is the best Ea for PP TRIODE amps?
Usually the tube data gives a reasonable guide. For small tubes
like EL84, 6V6, class AB or A PP triode
operation does not give enough Po, and such tubes are best in UL
mode with Ea max of +325V, or with 20% CFB.
2A3 can give a very nice 10W class AB1 with Ea < 350V.
For 6L6GC, KT66, 807, EL34 in triode, or 300B, Ea may be between
350V and 450V.
For 6550, KT88, KT90, KT120, Ea may be between 350V and 500V.
For 6550 with Ea = 350V, Iadc = 80mA, RLa-a can be 2k5 to get
maximum 18W class AB1.
THD, IMD and damping factor are all not as good as using Ea
above 400V.
Minimum Ea for class AB1 may be calculated = square root of ( 4
x Ra x Pda max ) where = Ra is average
Ra at Eg1 = 0V or data value for Ra. For 6550, with Ra = 900r,
Pda max = 42W,
Ea min = sq.rt ( 4 x 900 x 42 ) = sq.rt 151,200 = 388Vdc.
For 2A3, Ra = 800r, Pda = 12W, Ea min = 195V.
For 6BQ5, Ra = 2k0, Pda = 12W, Ea min = 309V.
(B) Harmonic Distortion for 100% maximum possible class A1 power
= 1% approximately, mainly 3H.
Harmonic Distortion for this example at maximum class AB1 power
for minimum RLa-a =
4% approximately, mainly 3H, 5H.
(C) 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, with one appearing normal with
the other upside down and reverse direction and combined that
looks very interesting but which is impossible
for anyone to produce themselves. A classic example is on page
575, Radiotron Designer's Handbook, 4th Ed,
1955. I don't know anyone who understands the sketch on that
page or anyone who has ever bothered
to draw a page up like that one.
(D) The above calculations are all based on having well
Regulated B+ power supplies. In practice, class
AB amps may draw up to 3 times the idle Idc at full Po with the
minimum RLa at clipping. The PSU B+ rail
should have Rout < 100r, so that if Idc changes from 100mAdc,
250mAdc, the 150mAdc change produces a
B+ drop of only 15Vdc. To achieve this, tube rectifiers should
not be used, and S1 diodes used.
Reservoir capacitor should be at least 235uF and choke = 4H with
Rw = 50r.
Following C at CT to 0V = 235uF, which gives LC resonance 5.2Hz.
In practice, most tubed class AB hi-fi
amps have very stable B+ because average Po rarely moves above
the pure class A levels.
Fig 5. A good schematic for a 30W integrated class AB
triode amp.
This schematic suits KT88 or 6550 in triode. But KT90 or KT120
could be used with higher Iadc at
idle to increase the amount of possible class A. KT120 with Ia =
70mAdc and Ea 450Vdc could produce
22W max for RLa-a = 9k2, and I bet this 22W will sound just as
well and having an 845 or anything else
able to make 22W. The power supply required for two such
channels may be similar
to PSU in 5050 integrated amp at 5050 integrated amplifier.
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