LOAD MATCHING 1
SINGLE 6550 TRIODE
OPERATION.
The use
of the 6550 beam
tetrode as a single triode can make a superlative amplifier. My pages on the use
of beam tetrodes give much information on the load matching outcomes from using
the 6550 as a single amplifying device but many people would prefer to use
a multi grid
output tube as a triode by connecting the screen grid to the anode.
In fact
most people will want to use the 6550 in triode mode for an SE ( single ended )
amplifier rather than use the 6550
in beam tetrode mode, or have a quad of
them in parallel to make enough pure class A1 power.
This page has the
following content :-
* Fig1. Anode resistance curves for GE6550A in
triode from the 1950s.
Explanations of how the curves were obtained and what
the curves mean.
* Fig 2. Schematic
for testing power triodes.
Triode voltage generator model is included in
explanations.
* Fig 3. GE6550 triode curves with 2.5k RL and
tangent to
calculate Ra, µ and gm for any chosen working point.
Explanation of
what load lines are.
How to plot them graphically to read the graphs for
gain, calculate power output, and 2H distortion.
* Fig 4. Measured power output vs distortion
for 3 load values used with EH6550 in triode.
* Fig 5. Measured power output vs
distortion for 3 load values used with GE6550A in triode.
* Fig 6. Measured power output vs
distortion for 3 load values used with KT88jj Tesla in triode.
* Fig 7. Measured power output vs
distortion for 3 load values used with KT90EH in triode.
* Fig 8. My corrected Ra curves for EH6550 in
triode with 3 values of RL plotted.
Comments on loads recommended for EH6550
in triode.
* Fig 9. My anode curves
for EH6550 in triode with no load lines so you can download them for
use.
* Fig 10. Ra curves for 300B measured after
1990 with 4.5k loadline details.
Comparison of THD with beam tetrodes
strapped as triodes.
* Fig 11. Ra
curves for trioded GE6550 measured after 1990 with 4.5k load line
details.
List of conclusions about beam tetrodes used as SE
triodes.
* Fig 12. Intermodulation test rig schematic
for measurement.
* Fig 13. Measuring
the intermodulation distortion using an oscilliscope wave form.
Comments
about THD and IMD significance.
---------------------------------------------------------------------------------------------
Fig1.

This set of anode
resistance curves is taken from a scan of an ancient GE data sheet and tidied up
in MS Paint.
The positions of the curves have been carefully
preserved.
To be able to
establish these curves in the 1950s a curve plotter machine was used to draw the
curves on paper
while someone worked a calibrated analog electronic circuit
for testing the tubes.
Another way it may have been done was by
photographing an oscilliscope screen and making a black ink on white paper
drawing by tracing over the photo. All these methods were prone to errors,
but despite errors there was enough information in the curves to design any
circuit using the tube.
The process of viewing
the curves shown required that the tube be set up in a test circuit with its
cathode grounded,
and its anode taken to a ac signal source of low source
impedance less than 50ohms, with a 10 ohm current sensing
resistance.
Current wave forms and applied anode voltages are fed to a dual trace
oscilliscope.
The oscilliscope is capable of X-Y function and transfer
curve for Ea vs Ia is displayed as a curve as plotted above.
The grid of the
tube under test has a range of fixed bias voltages applied while the anode has a
large ac voltage swing applied
of up to say 800V peak for many output tubes.
There is a separate curve generated for each Eg value.
Each curve represents
the dynamic anode resistance known as Ra, for the tube.
Each Ra line is a
curve rather than a straight line. This is because where you change the Ea, the
Ia changes non linearly
so that Ia = a constant x cube root of Ea squared. To
fully understand electrostatic behaviour in tubes one has to study
the old
books from the 1920s and 1930s, and read up on Child's Law.
Today hardly anyone
can buy a tube electronic curve tracing circuit because of the high voltages
involved and lack of demand.
But a few have succeded in using a PC to very
easily create a ditital file of the curves which can be printed out,
there
are some programs available now which automatically plot loadlines across the
tube characteristic curves and calculate
the distortions.
Considering the Ra
line in the above set of curves for Eg = 0V, when Ea is raised to 100V, Ia =
100mA, and when Ea is further increased to 200V, Ia rises to 275mA, so the rate
of Ia increase isn't linear with applied Ea. Ra varies as explained above but we
really only want to know Ra at one very small range of Ea variations where Ra
won't vary much.
All resistances can be measured in ohms from Ohm's Law where
R = E / I.
The Ra is the dynamic output
resistance of the tube and has no strict relationship to the quiescent
fixed operating Ea divided by the fixed idle supply current. So because a
typical operating point could be with Ia = 100mA, and Ea = 330V, it does not
mean Ra = 330 / 0.1A, or 3,300 ohms. The DC operation of the tube should be
considered separately to the ac operation.
The equivalant ac model of a triode can be
considered a voltage generator which has an extremely low source resistance for
its output signal voltage which is the product of the amplification factor, µ,
multiplied by input voltage between grid and cathode. There is then a model
resistor between the voltage gene output and the anode terminal. This resistor
is the Ra, or anode resistance or impedance as it is known. Every amplifying
device has its own particular value of "generator resistance"
or "source
resistance" or "anode impedance". Even the mains power supply at the wall has
source impedance.
The mains supply can
be considered to be two elements. The first is a perefct 240Vrms supply with
zero output resistance so that any load connected will not affect the voltage.
The second is some amount of resistance in series bnetween the perfect
voltage supply and the wall sockets in your house. The voltage at the wall
socket will change depending on whether you plug a lamp in which draws little
current, or a 2 kilowatt heater because of the copper wire resistance between
your appliances and the perfect voltage supply away from your house. The source
resistance in ohms can be measured by dividing the voltage change by the current
change which occurs for two different known load values. The mains source
resistance or a triode's ac anode resistance can simply be measured by recording
the voltage change between the unloaded condition and the loaded condition and
generator resistance Rg = Vchange / I change.
Suppose a room heater
is rated for 2,400 watts. With 240Vrms the current is 10 amps rms.
The resistance of the
heater = V / I = 240 / 10 = 24 ohms. Suppose the voltage drop when the heater is
turned on is 10Vrms. The resistance at the connection point and
measurement point of the room heater is thus 10V / 10A = 1 ohm.
This
resistance includes the wall wiring and wires to the supply pole and back to the
generator and all the
other users also connected across the the supply.
Power = I squared x Resistance so where the wire resistance = 1 ohm, there
will be a power loss
in that wire = 10 x 10 x 1ohm = 100 watts. This explains
why heat is generated in the leads to appliances.
In a triode in a test
circuit with a large value choke the load can be made to be such a high value
that it is a negligible load.
With a choke of 20H, at 1kHz the choke ac
impedance will be F x 6.28 x L = 1,000 x 6.28 x 20 = 125,600ohms,
and well
above any value which will seriosly affect the measurement of most power
triodes.
So when such a choke is connected there is little anode signal
current and the amplification we see is at its maximum
value which is µ, or
the amplification factor, since no voltage drop is occuring across the internal
generator resistance
of the triode.
Here is a schematic of
my power tube test rig which includes the triode shown as an "imaginery voltage
generator"
with a series resistance to represent the dynamic anode resistance
of the tube....
Fig 2.

This
actual schematic was used for load testing in Fig 8 below. It includes the dc
set up for the triode, the cathode biasing,
and the triode is shown modelled
as a voltage generator with a series resistance of the value of Ra also
calculated below.
The ac voltage shown at the voltage gene output does not
actually appear anywhere really; it is an imaginery signal
created for
modelling purposes. But while Vg stays at 29.63Vrms, the voltage at the gene
output will always
be 217.57Vrms for the model. if we had 3k for the anode
load, there would be a circuit from the voltage gene
to 0V with Ra of
837ohms V in series with 3k, so the signal Ia = 217.57 / 3,837 ohms = 57mA, so
the signal at the 3k load, ie, anode output voltage = 0.057 x 3,000 = 170Vrms.
The coupling capacitor C2 has an impedance of only 3.2 ohms at 1 kHz.
This
test circuit is called a "parafeed" circuit and is actually used for some SET
amps where there is a non gapped normal
ac output transformer located in the
place of R2 anode load.
Now let us look at some lines
plotted on the above curves for the GE6550A....
Fig 3.

These
curves are the same as for Fig 1 above.
Let us find out what is the anode
resistance, Ra, for this tube.
We have decided to try to use the quiescent
idle condition of Ea = 330V, Ia = 100mA which gives
a combined anode and
screen dissipation, Pda, of 33 watts and about 3/4 of the maximum rated
Pda = 42 watts.
We would never set up a triode at the rated maximum Pda
because the tube may only last a short time. The operating point is called Q,
and it is on the Ra curve for Eg = -30V. The Ra we want to know is for a tiny
swing voltage on the Ra curve so to find out Ra for point Q a tangent line to
the Ra curve is drawn from N to M using a ruler, or mouse with a line draw
facility in an image program.
The Ia difference of N to M read off
vertically = 370mA.
The Ea difference of N to M read off horizontally = 560V
- 250V = 310V.
The resistance value of any sloped line on the graph is
determined using Ohm's Law, R = E / I.
Ra = E / I = 310V / 0.37A = 837
ohms.
As you can see by the slope of possible tangent lines, Ra could be
1.8k at Ia = 25mA, and 400 ohms at Ia = 300mA.
The amplification factor,
µ, is the voltage gain of the tube achieved with a load that does not cause any
current change.
This over simplifies the idea of µ, but its all I have room
for now. µ is determined by the designer by dimensioning the inter-electrode
distances.
A load where no current change occurs is called a constant current
source. Such a load is easy to provide to the tube and can be a high value choke
to supply the dc idle current which may have 200,000 ohms of ac impedance at 1
kHz. A 32 Henry
choke has 200,000 ohms of ac impedance at 1 kHz, and such a
load with say 250vrms applied across it results in only
1.25mA of ac anode
current. If you draw the load line for 200k, it is virtually a horizontal
line on the graph.
The constant current source, CCS, could be another tube or
transistor connected in such a way to act as an
"active load" and equivalent
to many megohms of resistance. We would need to then apply an input signal and
measure
the output signal to calculate the voltage gain which in this case be
equal to µ.
But we don't have to do that if the curves are accurately drawn
for the tube. To plot a loadline on the tube curves where no current change
occurs, the line will be a horizontal one at the dc current level.
So let us
look at the horizontal line at Ia = 100mA. This line is intersected by all the
Ra curves and the Ea change caused by an Eg change can easily be read off.
So
for Eg = 0V, Ea = 100V, and for Eg = -60V, Ea = 540V.
Therefore a grid
voltage change of -60V gives an anode change of +440V.
The gain = µ = Ea
change / Eg change = 440V / -60V = - 7.33. The sign for µ is accurate, but for
general purposes
nobody bothers to worry about it and just refers to µ as a
positive number. But in other gain equations it may be important to
consider
µ with the correct -ve sign.
For all tubes, transconductance, gm,
in amps per volt, is the ability of the grid voltage to cause anode current
change and
gm = µ / Ra.
In
this case gm = 7.33 / 837 = 0.00876A/V = 8.8mA/V
So by working with
curves we know the 3 basic parameters for the tube for the working Q point. For
all tubes,
Signal voltage gain, A, = µ x
RL / ( RL + Ra ) This is the very useful universally applicable equation
for tube voltage gain.
Now we want to see what outcome we may get
with a load resistance. Once we have an operating point, we can calculate a
reasonable "centre value" for RL for any power triode.
The method to establish the triode load follows :-
(1) Triode RL = ( Ea / Ia ) -
Ra.
In this case RL = ( 330 / 0.1 ) - 837 = 2.46k .
Let us round that up to 2.5k. The damping
factor, DF = RL / Ra.
In this case DF = 2,500 / 837 = 2.98, which is
fair for a power triode where we want RL to be at least 2 x Ra.
(2) To draw the line representing RL =
2,500 ohms, first work out Ea / RL = 330V / 2,500 = 132 mA.
(3) Add Ia quiescent of 100mA, 132 + 100
= 232 mA.
(4) Plot point A on the
Ia axis at 232 mA.
(5) Draw the RL
straight line from A through Q and on to point D if the RL line will intersect
the Ea axis.
I had to extend the axis out to Ea = 580V to get point
D.
(6) Check that the line is
correct for RL. Ea = 580V / Ia = 232 mA = 2,500 ohms, OK.
The line A to D
crosses the Ra line for Eg = 0V at point B, and for twice the grid swing at -60V
Ra line at point C.
BQC represents the maximum anode voltage changes with
2.5k RL connected.
(7) Drop
vertical lines from B and C to the Ea axis and read off the Ea minimum, and Ea
maximum. These are at 147V
and 472V. Therefore the -ve anode v swing = 330V -
147V = 183V. The +ve anode v swing = 472V - 330V = 142V.
(8) Total peak to peak load swing = 183V + 142V
= 325V, so load ac V = 325V / 2.82 = 115.2 Vrms.
(9) Power output = ( V x V ) / RL = (
115.2 x 115.2 ) / 2,500 = 5.3 watts.
(10) Second harmonic distortion % =
100 x 0.5 x ( difference in peak +ve
and -ve load swings )
sum of peak load swings
In this case 2H % = 100 x 0.5 x (
183 - 142 ) / (
183 + 142 ) = 6.3%.
The smart people amoung you will notice that
the Ra lines for the GE6550 are not spaced along any horizontal
line at even
spaces; the Ra curves tend to crowd closer as you move to the right side of the
graph.
If you measured the +ve and -ve anode swings for -/+ 30V grid change,
you find that with a CCS load at Ia = 100mA
that there is about 2.5% 2H
distortion which isn't very good because many power triodes are capable of
2H less than 0.5% even with a large anode V swing.
So were the curves
drawn in 1955 correct? I decided to measure a couple of slightly used samples of
EH6550 to see if there was much difference between today's Russian made EH6550
and the tubes made all those years ago in the US.
My test rig circuit has
a high value choke to supply DC to the tube, switched load resistances of
9.7k, 7.2k, 4.5k, 3k, 2k, and 1k across the choke. I have switched cathode
resistors and a massive cathode bypass cap to set the bias and a regulated
B+ supply. The basic schematic is in Fig 2 above .There is a low
distortion triode signal amp for grid signals producing less than 0.5% 2H at
50Vrms.
The test signal from the ultra low THD oscillator is at 1 kHz. The
conditions for the test were Ea = 420V, Ia = 73mA,
Eg = -47V.
Here are
three following graph sheets each with the same 3 loads showing the THD for each
load vs output voltage.
I measured samples EH6550, GE6550A, KT88jj, and KT90
:-
Fig 4.

The
2H is lower when no load is connected at < 1.6%. at 235vrms output. The old
GE6550A triode curves
are not correct for the EH6550.
Fig 5.

The
sample of GE6550A tested has 2,000 hrs of use and is middle aged, yet it managed
to measure better than EH6550
for a 3k RL, but only slightly worse for 4.5k
and 9.7k. Ra and µ were about the same.
The results indicate that the
original GE anode curves are slightly fictitious.
Fig 6.

This
KT88 measures slightly better.....
Fig
7.

The
KT90 THD is less than each of the 3 previous tests. Part of the reason is the
slightly higher Ia = 78mA, placing the operating region further "up the curves"
and thus further away from the bottoms of the curves which swing round to the
left
and thus contribute to the distortion. If the Ra lines were straight
lines and all exactly parallel, there would be no distortion,
but amoung the
laws of nature there is seldom any linearity.
Here are my corrected Ra
curves for EH6550 with load lines plotted which must be correct because the
above
measurements must co-relate with the data curves :-
Fig 8.

You
can see that the Ra lines are all spaced more evenly along the horizontal
Ia line for Ia = 50mA.
If we go from Eg1 = 0V to -50V, then from -50V
to -100V, we get Ea variations of 355V and 345V.
2H % = 100 x 5 / 700 =
0.71%. When tested with no load at Ea = 72mA, the triode gave
approximately
1% at 235vrms.
When the actual the Eg grid bias voltage
used in a test does not coincide with an Eg value for an Ra line, it is more
difficult to
accurately calculate the 2H at anode voltage load swings less
than maximum. I found it tedious and confusing. Its faster for me to go out to
my workshop and measure the tube, because then I am certain of the results. But
if the results fit the curves, then the curves can be handy to use later for any
other amplifier including a push pull type.
I believe the curves above are
closer to the real characteristics of the EH6550 than the GE 6550 are to the
curves drawn for it 50 years ago.
If I were to apply the formula
for centre value of RL we get
RL = ( 420 / 0.073 ) - 866 = 4.89k. By
measuring Ra with a tangent line got Ra = 866 ohms for the above working
point.
RL could be rounded to 5k.
The output transformer ratio
chosen should have the nominal speaker value matched to the above calculation of
RL.
So if we have 6 ohm speakers which are common today, then the OPT
impedance ratio = 5k : 6 ohms.
This is a 833 : 1 Z ratio, ( thus the
turn ratio = 28.9 : 1 since the turn ratio is the square root of the impedance
ratio ) .
The Ra of the output tube is transformed by the OPT
impedance ratio
so that the Ra
measured at the secondary speaker connection is Ra / ZR = 866 / 833 = 1 ohm
approximately. To this we must add the winding resistances, Rw, of both
primary and secondary because both appear in series with the load and secondary
load x 7% should be allowed if the Rw is unknown.
So typically an amp with
6550 would have Rout = 1.0 + 0.42 = 1.42 ohms. This gives a damping factor of 6
/ 1.42 = 4.2.
Bass speaker impedance is usually higher than nominal so
usually a triode without any global NFB will control a bass
speaker quite
adequately and sound well.
But cross over networks and the use of 3 way
speakers can lead to speaker impedances falling well below
the nominal value
of 6 ohms and perhaps down to say 4 ohms, and the load match then approaches the
case above where
the 3k load is drawn in place. The damping factor is then
much worse, and since speakers are designed to work where
Rout from
amplifiers < 1 ohm, then the SE triode without global NFB will cause a 1db
level drop below the 6 ohm level
where load = 4 ohms.
Thus when we add
12dB of global NFB the Ra seen at the output and the Rw is reduced from 1.42
ohms to
about 0.5 ohms and quite satisfactory. THD is also reduced about
10dB.
Here is a copy of the EH6550 anode curves which
readers are free to download and copy for load matching purposes :-
Fig 9.

How
good is the EH6550 in triode compared to the "gold standard" of power triodes,
the 300B?
I
don't have a WE 300B to test but here are some curves I found from
www.audiomatica.com, measured on a Sofia
tube curve tracer during the last
few years...
Fig 10.

Notice the Ra curves are only about 7%
closer together along the Ia = 50mA level, so that at far left,
between 0V
and -15V we a distance about 7% more than between 135V and -150V.
At Ia =
50mA, we go from Eg1 = 0V to -75V, then from -75V to -150V, we get
Ea
variations of 305V and 295V.
2H % = 100 x 5 / 600 = 0.83% for no load at
50mA.
What other amplifying device produces 212Vrms output at less than 1%
thd without any external loop
of NFB?
With a load of 4.5k plotted on the
curves the +ve and -ve anode voltage swings for -/+75V grid swing
are 268V
and 233V, so 2H = 3.5% for 7.0 watts. For 7.0 watts from each of the 4 beam
tubes tested above,
EH6550 gave 4.2%, GE6550 4.6%, KT88jj 3.3%, KT90EH 3.0%.
So there really isn't much difference between the
2H distortion.
Here
are Ra curves for GE6550A taken with the same gadgets by audiomatica, and with
my loadline of 4.5k
superimposed to read off the expected power and 2H
distortion :-
Fig 11.

Notice
the Ra curves are about 22% closer together at the far right than at the left
side along the Ia = 50mA level.
At Ia = 50mA, we go from Eg1 = 0V to -40,
then from -40V to - 80V, we get
Ea variations of 290V and 270V.
2H % =
100 x 10 / 560 = 1.79% for no load at 50mA.
With a load of 4.5k, the maximum
anode voltage swings are 297V and 220V, so maximum power = 7.5 watts
and 2H =
7.4%.
But my measurements in the graph above for GE6550A, THD
vs output voltage, the THD = 4.9% at 7.5 watts/4.5k.
My conclusions :-
Don't trust all the data curves you read.
Unless the testing method was extremely accurate,
errors of several % can
occur.
Always measure a circuit
before you know the facts about the distortions.
EH6550 in triode probably do measure nearly as well as
a 300B and could be used to give the same sonic performance.
KT90EH in triode does measure better than EH6550,
GE550A, KT88.
All the 4 tested
beam tetrodes have Ra = approximately 850 ohms to 1,100 ohms for the SE
triode test conditions.
I hear all the audiophiles groaning with
disbelief, but then there are also other brands of 6550 such as Svetlana
and
some chinese brands which could be as good. The EH KT90 has a 55 watt Pda
rating, it will make a nice amount of power, and could be run at Pda = 40
watts.
I have a pair of KR audio 300B which have Pda = 65watts, and I
expect them to be excellent sounding with measurements
at least as good as
the above.
At present I am struggling to build a couple of 50 watt class A
SE amps with a pair of
KR Audio 845 tubes in parallel. These do have nice
anode curve data, but how they actually measure is unknown.
The benefit
of using a pair of 845 is the 50 watt ceiling at a nominal load 4% THD, using a
load with a similar RL / Ra ratio as the case for the 6550 above using 4.5k. At
an average level of 1 watt the output voltage will be 1/7 of the 50 watt level
and THD should be about 0.5%. The same 1 watt developed by an 8 watt SET amp
would have about 1.4%, as indicated in my measurements above.
The
harmonic distortion is of interest in that we should use the measurement of it
as a guide. There is some interesting reading
in the Radiotron Designer's
Handbook, 4th Edition, 1955.
For most tube amps RDH4 says IMD will be
several times the value of THD when tested in the following way.
The IMD is
measured with a bass signal of say 80Hz and a treble signal of 5 kHz which has
1/4 of the amplitude of the bass signal. When high levels of bass signal occur
in a single tube output stage one half of the sine wave has more gain than the
other due to variations in gm and Ra during the bass wave cycle. This causes the
5 kHz tone to be amplitude modulated 80 times per second so that "side band" IMD
products exist at 4,920 Hz and 5,080 Hz, and neither are harmoniously related to
the 5kHz. If the main THD product was 3H as with a PP amp, there is a change of
gain twice for each crest of the wave, so the amplitude of the HF wave is varied
by 160 times a second, so products exist at 4,840 Hz and 5160Hz.
The
test rig to determine IMD is :-
Fig
12.

The high pass
filter can be a simple cascaded CRCRCRCRCR type with each section having a -3dB
pole at 1kHz.
The series sections should allow a 5kHz signal to pass without
much attenuation but at 80Hz the
attenuation will be over 100dB and the
ultimate slope of the filter will ensure rejection of the 80Hz signal or any
other low frequencies such as mains related hum signals.
The cathode ray
oscilliscope wave forms will be :-
Fig
13.

Wave
form 1 shows a LF and HF signal viewed separately on a dual trace CRO with their
amplitude
set so that the LF wave is 4 times the HF wave.
With both
signals present in the amp under test they look like wave form 2; the HF appears
to be "riding on" the larger amplitude LF signal.
Wave form 3 shows the
HF signal "envelope" shape where the LF signal has been filtered away.
If
there is any intermodulation distortion the amplitude of the HF signal wave will
be seen to vary at a rate related to the frequency of the LF wave
signal.
The measurements of IMD are usually taken before any part of the
combined wave is clipping.
During actual measurements in a good amp the peak
to peak modulation voltage, a - b, is usually a tiny amount,
and IMD could
typically be 3% when the THD of the bass
signal was 1% at a normal listening level.
This may be dificult to see on a
CRO, and difficult to measure by reading the graticules.
I use a peak
detector circuit with a diode plus R&C which converts the LF rate of
amplitude changes of the HF wave form
to a LF signal voltage which can be
measured by a millivolt meter or shown on a second trace on the CRO.
Its the
same sort of cirucuit used for detecting AM radio waves but the RC time constant
is such that it
filters the HF away leaving only LF below about
250Hz.
The 80Hz modulation expected is shown here as a LF wave which is
shown drawn of scale to what will actually be seen on the CRO because for 1
cycle of 80hz, there really are 62.5 waves of 5kHz sine waves. The actual
LF tone is not critical but about 80 Hz is away from mains related hums,
and is also where there is a fair amount of LF content in music. However, should
you be a perfectionist, then use 30Hz, and this will test the IMD effect of
output transformer saturation at high levels.
The amount of IMD will increase
very suddenly when clipping occurs or output transformer saturation or grid
current occurs in the amp.
Intermodulation distortion is generated by the
non linearities of the amp and the artifacts are created by all the
frequencies present, so where hundreds of music frequencies exist as the
same time, there are thousands of harmonic products generated. If we were to
separate the IMD products from the music and play it through the speakers by
itself,
we would here a kind of rustling noise in time with the music. Its
not a nice noise to listen to. All amplifiers and speakers produce IMD.
As
long as the IMD is kept to low enough levels the music will sound quite clear
and free of perceptible IMD.
IMD is reduced by global NFB along with THD and
phase shift.
Triode amplifiers with global loop NFB tend to have less IMD
than beam tetrodes or pentodes with a similar amount of
NFB where the power
maximums the amps concerned are equal. So to make 28 watts in SE triode class A,
4 x 6550 strapped as triodes are needed but to make the same 28 watts in SE
class A with tetrodes,
only 2 x 6550 may be required strapped as tetrodes.
The triode amp would sound better.
In the case with an SET single triode
with just two test frequencies, when the triode is conducting more Ia
current
during peaks in the current wave for LF, the gm of the tube
increases, thus the smaller amplitude HF signal present
is more greatly
amplified during the increase in gm. Similarly when the Ia reduces in the
troughs of the LF Ia waves,
the gm reduces from its value at the zero
crossing point. So the HF wave is not amplified as much and its amplitude
reduces.
The wave form3 of the above figure13 can be diplayed on a dual
trace CRO with the bass frequency wave
and the relationship of the bass wave
to IMD can be seen clearly if the IMD level is high enough.
The slight
difference in amplitude of the HF wave is actuall as a result of additional
frequencies being created by the
intermodulation phenomena. Those
frequencies in the case of a single ended amp = HF + LF, and HF -LF,
so that
for 80Hz and 5,000Hz the addititional intermodulation products are 5,080Hz and
4,920Hz.
Niether of these additioanl frequencies which are actually present
will sound musical and related to the
fundementals of the LF and HF
tones.
In music, many tones are harmonically related so that many of the IMD
products are at a frequency which is
related to the harmonic overtones in
instruments, so that where there are say two string instrument tones with one at
F and another at 4F then IMD products are at 3F and 5F, both of which are at
harmonics of the lower note string of 1F.
But 1F + 8F would give IMD products
at 7F and 9F, and either could sound bad if at high level.
It is possible to
filter out these F for viewing/measururing with high Q filters.
The way our
ears react to THD and IMD is a very much argued topic.
It is also
possible to use 5kHz and 8kHz as the two test frequencies of equal
amplitude.
The sum and difference products are then at 3kHz and 13kHz,
neither of which is harmonically related to either test F,
and a high Q
filters can easily be made to measure either IMD product to then calculate the
IMD %, without the measurement
being muddled by the presence of any harmonic
product which would be the case if tones of 5kHz and 10kHz were used.
In
tube amps the worst amounts of IMD are caused by LF modulation of higher
frequencies hence the
use of the standard 4:1 ratio of LF:HF
test signals has been a fair indicator of amplifier performance
for
the last 80 years. The reason is that a tube amp tends to suffer iron caused
distortion more as F reduces.
And also because bass frequencies have a much
higher amplitude than mid-treble frequencies.
Push pull amplifiers have 3H
as the dominant harmonic and the rate at which gain changes occur is twice that
of the SE amp
where the harmonic is mainly 2H. The PP amp will therefore
produce substantially different spectra of IMD products.
Some may say that PP
amps can therefore never sound as well as a SE amp of the same power
and THD
measured level. But usually the PP amp will have perhaps 4 times less thd than
the SE amp
because of the 2H current cancelling in the output stage and
because the THD is lower the IMD will also be lower,
thus offsetting the
perception that PP IMD is worse sounding than SE amps.
Many fine sounding
amplifiers of the past used very ordinary iron which promoted IMD
generation.
With today's top quality grain oriented silicon steel and with a
sufficient ratio of turns per volt
the tube amp can avoid the problems of the
past for even better sound.
At the end of the day, how the
music sounds is all that matters.
The lower the IMD, the better the
music.
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