SINGLE 6550 TRIODE
The use of the 6550 beam tetrode
as a single triode can make a
amplifier. My pages on the use of beam tetrodes give much
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
the 6550 in beam tetrode mode, or have a quad of them in
parallel to make
enough pure class A1
This page has the following
Anode resistance curves for GE6550A in triode from the
Explanations of how the curves were obtained and what the curves
Fig 2. Schematic for
testing power triodes.
Triode voltage generator model is included in explanations.
3. GE6550 triode curves with 2.5k RL and tangent to
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,
output, and 2H distortion.
4. Measured PO vs Dn for 3 load values
used with EH6550 in triode.
Fig 5. Measured PO
vs Dn for 3 load values used
with GE6550A in triode.
Fig 6. Measured PO
vs Dn for 3 load values used
with KT88JJ Tesla in triode.
Fig 7. Measured PO
vs Dn for 3 load values used
with KT90EH in triode.
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.
10. Ra curves for 300B measured after 1990 with 4.5k
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.
12. Intermodulation test rig schematic for measurement.
Fig 13. Measuring the
intermodulation distortion using an
Comments about THD and IMD significance.
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
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
circuit for testing the tubes. Another way it may have been done
by photographing an oscilloscope
screen and making a black ink on white paper
drawing by tracing over the photo. All these
methods were prone to errors, but
despite the 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 oscilloscope. The oscilloscope is capable of
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.
represents the dynamic anode resistance known as Ra, for the
Each Ra line is a curve rather
than a straight line. This is
you change the Ea, the Ia changes non linearly so that
Ia = a constant x cube root of Ea squared.
To fully understand electrostatic
behavior in tubes one has to study the old books from the 1920s
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
using a PC to very
easily create a digital 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
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,
rate of Ia increase isn't linear with applied Ea. Ra varies as
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
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 equivalent ac model of a triode can be
considered a voltage generator
which has an extremely low source resistance for its output
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 perfect
240Vrms supply with zero output resistance so that any load
connected will not
the voltage. The second is some amount of resistance in series
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
wire resistance between your appliances and the perfect voltage
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
resistance can simply be measured by recording the voltage
between the unloaded condition
and the loaded condition and generator
resistance Rg = V change / 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.
voltage drop when the heater is turned on is 10Vrms. The
resistance at the
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
connected across the the supply.
Power = I squared x Resistance so where the wire resistance = 1
there will be a power loss in that wire = 10 x 10 x 1ohm = 100
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 seriously 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
value which is µ, or the amplification factor, since no
is occurring across the
internal generator resistance of the triode.
Here is a schematic of my power
tube test rig which includes
Now let us look at some lines
plotted on the above curves for the
shown as an "imaginary voltage generator" with a series
represent the dynamic
anode resistance of the tube....
This actual schematic was used for load testing for Fig 8 below.
includes the dc set up for the triode, the cathode biasing, and
the triode is shown modeled as a
voltage generator with a series
resistance of the value of Ra also calculated below. The ac
shown at the voltage gene output does not actually appear
really; it is an imaginary signal created for modelling
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
there would be a circuit from the voltage gene to 0V with Ra of
ohms 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
of only 3.2 ohms at 1 kHz. This test circuit is called a
and is actually used for some SET amps where there is a non
normal ac output
transformer located in the place of R2 anode load.
These curves are the same as for Fig 1 above.
Let us find out what is the anode resistance, Ra, for this tube.
have decided to try to use the quiescent idle condition of Ea =
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
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
find out Ra for point Q a tangent line to the Ra curve is drawn
from N to M
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
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 or now. µ is determined by the
designer by dimensioning
inter-electrode distances. A load where no current change occurs
is called a
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,
another tube or transistor connected in such a way to act as an
equivalent to many megohms of resistance. We would need to then
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
So let us look at the horizontal line at Ia = 100mA. This line is
all the Ra curves and the Ea change caused by an Eg change can
So for Eg = 0V, Ea = 100V, and for Eg = -60V, Ea = 540V. Therefore
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
it may be important to consider µ with the correct -ve sign.
For all tubes, transconductance, gm, in amps per volt, is
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
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.
we have an operating point, we can calculate a reasonable "centre
for any power triode.
The method to establish the SE triode load follows :-
The example tube is 6550 strapped as a triode with loadlines drawn
Fig 3 above.
(1) Triode RL = ( Ea / Ia ) - Ra,
where Ra is anode resistance at the Q point.
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 =
fair for a power triode where we want RL to be at least 2 x Ra.
(2) Calculate Ea /
RL = +330Vdc / 2,500 ohms = 132 mA.
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 downwards to the
side of graph. The line will intersect the Ea axis at point D.
I had to extend the Ea axis out to Ea = +580V to get
(6) Check that the line is
correct for RL. RL = Ea at point D / Ia at
RL = 580V / 232 mA =
2,500 ohms, OK.
The load line A to D crosses the Ra line for Eg = 0V at point B.
The grid voltage change is +30V.
Find the point C on the load line where the grid voltage is 30V
the Q bias
point of -30Vdc, ie, at - 60V.
(7) Drop vertical lines
from B where Ea = +147V. A
horizontal line left from B
gives Ia = 175mA.
Similarly point C occurs
where Ea = +472V and Ia
= 38mA. Point B is at the
minimum, and point C is at the Ea maximum. These are at 147V and
negative Ea swing = 330V - 147V = 183V.
The positive Ea swing = 472V - 330V = 142V.
(8) Calculate total peak
to peak load swing = Ea maximum - Ea minimum
472V - 147V = 325V.
Calculate load Vrms =
V pk-pk / 2.82 = 325V / 2.82 = 115.2 Vrms.
(9) Power output = ( Load Vrms
squared ) / RL in ohms
= ( 115.2 x 115.2 ) / 2,500 = 5.3
(10) Second harmonic distortion
= 100 x 0.5 x ( difference in
peak +ve and -ve load swings )
In this case 2H % = 100 x 0.5 x ( 183 - 142 ) / ( 183 + 142 )
The smart people among you will notice that the Ra lines for the
are not evenly spaced along any horizontal line. The Ra curves
crowd closer as you
move to the right side of the graph. If you measured the
positive going and negative going anode swings for
-/+ 30V grid change, you
find that with even with a CCS load at Ia = 100mA there is about
distortion which isn't very good because many power triodes are
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
slightly used samples of EH6550 to see if there was much
made EH6550 and the tubes made all those years ago in the
My test rig circuit has a high value choke to supply DC to the
This simulates a constant current source. There are switched load
across the choke = 9.7k,
7.2k, 4.5k, 3k, 2k, and 1k. 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
triode signal amp for grid signals producing less than 0.5% 2H at
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
THD for each load vs output voltage. I measured samples EH6550,
KT88JJ, and KT90 :-
The 2H is lower when no load is connected at < 1.6%. at 235vrms
The old GE6550A triode curves are not correct for the EH6550.
The sample of GE6550A tested has 2,000 hrs of use and is middle
yet it managed to measure better than EH6550 for a 3k RL, but only
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
This KT88 measures slightly better.....
The KT90 THD is less than each of the 3 previous tests. Part of
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 among
nature there is seldom any linearity.
Here are my corrected Ra curves for EH6550 with load lines plotted
be correct because the above measurements must co-relate with the
data curves :-
You can see that the Ra lines are all spaced more evenly along the
Ia line for Ia = 50mA. If we go from Eg1 = 0V to -50V, then
from -50V to -100V,
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
an Eg value for an Ra line, it is more difficult to accurately
calculate the 2H at
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
believe the curves above are closer to the real characteristics of
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
working point. RL could be rounded to 5k.
The output transformer ratio chosen should have the nominal
matched to the above calculation of RL. So if we have 6 ohm
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
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
typically an amp with 6550 would have Rout = 1.0 + 0.42 = 1.42
This gives a damping factor
of 6 / 1.42 = 4.2.
Bass speaker impedance is usually higher than nominal so usually a
without any global NFB will control a bass speaker quite
sound well. But cross over networks and the use of 3 way speakers
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
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
Rw is reduced from 1.42 ohms to about 0.5 ohms and quite
THD is also reduced
Here is a copy of the EH6550 anode curves which readers are
download and copy for load matching purposes :-
How good is the EH6550 in triode compared to the "gold standard"
power triodes, the 300B?
I don't have a WE 300B to test but here are some curves I found
www.audiomatica.com, measured on a Sofia tube curve tracer during
the last few years...
Notice the Ra curves are only about 7% closer together along the
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
of 305V and 295V. 2H % = 100 x 5 / 600 = 0.83% for no load at
What other amplifying
device produces 212Vrms output at less than
1% thd without any external loop of NFB?
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,
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 :-
Notice the Ra curves are about 22% closer together at the far
at the left side along the Ia = 50mA level. At Ia = 50mA, we go
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
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
read. Unless the testing method
was extremely accurate, errors
of several % can occur.
Always measure a circuit before
know the facts about the distortions.
EH6550 in triode probably do measure nearly as well as a 300B and
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
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.
KT120 EH could be a very fine choice of tube and will definitely
Pda = 40Watts and give slightly more PO than KT90.
I have a pair of KR audio 300B which have Pda = 65watts, and I
them to be excellent sounding with measurements at least as good
above. At present in
2006 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
Ah, but as I edit now in 2011, see my pages on the SE55 with
The benefit of using a pair of 845 is the 50 watt ceiling at a
4% THD, using a load with a similar RL / Ra ratio as the case for
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
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
and a treble signal of 5 kHz which has 1/4 of the amplitude of the
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
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
so the amplitude of the HF wave is varied by 160 times a second,
so products exist at 4,840 Hz and
The test rig to determine IMD is
The high pass filter can be a simple cascaded CRCRCRCRCR type with
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
The cathode ray oscilloscope wave forms will be :-
Wave form 1 shows a LF and HF signal viewed separately on a
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
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
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
1% at a normal listening level. This may be difficult to see on a
difficult to measure by reading the graticules.
I use a peak detector circuit with a diode plus R&C which
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 circuit
used for detecting
AM radio waves but the RC time constant is such that it filters
away leaving only LF below about 250Hz.
The 80Hz modulation expected is shown here as a LF wave which is
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
music. However, should you be a perfectionist, then use 30Hz, and
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
Intermodulation distortion is generated by the non linearities of
and the artifacts are created by all the frequencies present, so
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
a kind of rustling noise in time with the music. Its not a nice
noise to listen
All amplifiers and speakers produce IMD. As long as the IMD is
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
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
required strapped as tetrodes. The triode amp would sound better.
In the case with an SET single triode with just two test
when the triode is conducting more Ia current during peaks in the
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,
reduces from its value at the zero crossing point. So the HF wave
amplified as much and
its amplitude reduces.
The wave form3 of the above figure13 can be displayed on a dual
CRO with the bass frequency wave and the relationship of the bass
to IMD can be seen clearly
if the IMD level is high enough.
The slight difference in amplitude of the HF wave is actual as a
of additional frequencies being created by the intermodulation
Those frequencies in the case
of a single ended amp = HF + LF, and HF - LF,
so that for 80Hz and 5,000Hz the additional
intermodulation products are
5,080Hz and 4,920Hz. Neither of these additional frequencies which
actually present will sound musical and related to the
fundamentals 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
tones with one at F and another at 4F then IMD products are at 3F
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/measuring with
high Q filters. The way our ears react to THD and IMD is a very
It is also possible to use 5kHz and 8kHz as the two test
amplitude. The sum and difference products are then at 3kHz and
which is harmonically related to either test F, and a high Q
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
In tube amps the worst amounts of IMD are caused by LF modulation
frequencies hence the use of the standard 4:1 ratio of
has been a fair indicator of amplifier performance for the last 80
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
the dominant harmonic and the rate at which gain changes occur is
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
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
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.
Educational and DIY
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