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 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.

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
oscilloscope

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 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
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

behavior 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
succeeded in

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
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 equivalent 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 perfect

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
between

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 = 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.
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 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

maximum
value which is µ, or the amplification factor, since no
voltage drop

is occurring 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 "imaginary 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 for Fig 8 below.

It
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
voltage

shown at the voltage gene output does not actually appear
anywhere

really; it is an imaginary 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
837

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
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.

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 or 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 SE triode load follows :-

The example tube is 6550 strapped as a triode with loadlines drawn in

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 = 2.98,

which is 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.

(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 downwards to the right

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 point D.

(6) Check that the line is correct for RL. RL = Ea at point D / Ia at point A.

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 below 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

Ea minimum, and point C is at the Ea maximum. These are at 147V and 472V.

Therefore the 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 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 among you will notice that the Ra lines for the GE6550

are not evenly spaced along any horizontal line. The Ra curves tend to

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 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

USA.

My test rig circuit has a high value choke to supply DC to the tube.

This simulates a constant current source. There are switched load resistances

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 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 among 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.

KT120 EH could be a very fine choice of tube and will definitely run at

Pda = 40Watts and give slightly more PO than KT90.

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 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 unknown.

Ah, but as I edit now in 2011, see my pages on the SE55 with 845.

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 oscilloscope 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 difficult 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 circuit 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 displayed 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 actual 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 additional intermodulation products are

5,080Hz and 4,920Hz. Neither of these additional frequencies which are

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 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/measuring 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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