About NFB within triodes,

Testing a 6550 to measure Ra, Gm and µ.

Fig 1. Schematic A and B for testing 6550.

Schematics A and B explained,

How to determine Beam Tetrode and Triode Ra, µ, and Gm.

Simple method to find Beam Tetrode and Triode Ra, µ, and Gm.

Equivalent Model, Beam Tetrode, UL, Triode, 6550

Fig 2. Equivalent Model for calculation gain of UL connected tubes.

Fig 3. Graph of Ra vs UL tap % for SE 6550.

SEUL 6550 and SEUL 6550 with CFB.

Fig 4. Schematics for SEUL and SEUL+CFB for 6550.

CFB with UL taps and without UL taps explained with maths.

Other related pages which should be read first :- Basic tube 1

Triodes
have NFB inside them already....

Triodes or beam & pentode tubes which are triode
connected have

internal electrostatic negative feedback voltage acting between
the anode

and the electron stream and grid voltage. In a triode, when a
+going

voltage is applied to the grid, there is a -going anode voltage
with

increased load current.

The +V grid change produces a
more positive electrostatic field which

increases electron flow from the cathode to anode. The -V anode

voltage produces a more negative electrostatic field through out
the

triode and even between cathode and grid, so the anode voltage
reduces

the electron flow to the anode.

The
electron flow in a triode is controlled by TWO voltage field

effects, one from the grid voltage and the other from
the anode.

And the grid voltage field effect is opposite to the anode
voltage field effect.

If the grid +voltage change is trying to increase electron flow,
the -anode

voltage is trying to reduce electron flow.

So two voltages are together
working intimately and instantaneously and

at all frequencies to control the electron flow and the anode
voltage action

opposes the action of the grid voltage. As frequencies rise the
capacitance

between anode and grid and between grid and cathode become low
reactances

which have a gain reducing effect, but for the audio band we may
neglect

capacitance effects in this page in order to understand the
simple operation

at low frequencies.

The voltage field effects in a
real triode such as a 300B are effectively

the same to a 6550 or other beam tetrode or pentode which has
its screen

grid connected to its anode so that it works like a triode. In
the case of the

"triode strapped" 6550, it is the screen grid which propagates
the voltage

field to alter the electron stream to itself and to the anode.

If any distortion voltage
appears at the triode anode which is not present

at the input grid this voltage is applied to the electron stream
in a manner

which opposes its own production. If there was a negative going
distortion

voltage at the anode it indicates increasing current in the RL.
This -going

voltage lessens the anode current and opposes the increasing
current.

In effect, the anode voltage field effect is the reason why
triodes have such

low dynamic anode resistance.

Consider that you have a 300B
with Ea = 400V, Ia = 70mAdc, biased with

Eg = -80Vdc.

You
could be forgiven for thinking the 300B acts like a resistance
=

V / I = 400 / 0.07 = 5,714 ohms, because that's Ohm's Law at
work. And if

you raised Ea by +10V, you might expect to see Ia change = 10V /
5,714

= 1.75mAdc. You'd be wrong. Because of the NFB within the
triode, the

increase in Ea causes much more Ia to flow because the dynamic

Ra = 800 ohms. So a rise of Ea by +10V produces an Ia change =
10 / 800

= 12.5mA.

In the case of a 6550 with
screen connected to anode, the effect is the

same as a real triode. But the fact there is a separate screen
connection

allows any fraction of anode voltage field effect to be applied
by the screen

to the electron flow. This can be done using taps on an OPT
winding as it

is done with an "Ultralinear" OPT.

One may ask more questions about
this and you need to consider a 6550

more closely. The screen grid is a second control grid using a
helical wire

coil mounted on support rods but placed between the main control
grid and

anode. Screen grids were designed to allow a fairly unimpeded
anode

current flow of electrons from cathode to anode, and yet the
screen blocks

roughly 95% of the the electrostatic field effect due to anode
voltage changes.

The screen needs to be at a high positive dc voltage to maintain
a nearby

positive voltage field to maintain velocity of electrons on
their way to whatever

is positive. About 10% of tube current from the cathode strike
the screen

wires and are absorbed and form the screen supply current. The
other

90% of electrons miss hitting the screen but continue on to the
next positively

charged element, the anode, where they are absorbed and form the
anode

load current. The anode voltage can change over a wide range but
anode

current is determined by the load. The dynamic anode resistance,
Ra, of a

6550 beam tetrode may be 32,000 ohms when there is a fixed
screen voltage.

If the screen is tied to the anode to make a triode the 6550 Ra
becomes

about 1,000 ohms.

If the screen has 50% of the
anode signal voltage taken from a tap on

an OPT the tube is intermediary between a beam tetrode and
triode,

and the name 'Ultralinear', or UL has been given to this form of
use

of a multigrid power tube. With "50% UL taps" the dynamic anode

resistance could be about 2,240 ohms which means this simple
application

of local NFB gives a very worthwhile reduction of the high beam
tetrode

anode resistance. The tube gain between control grid and anode
will be

roughly twice that of the triode connection depending on load.
The beam

tetrode capability of producing up to 45% efficiency in class A
is largely

maintained and 40% efficiency is typical with UL. The 50% UL
connection

reduces THD spectra to being very similar to triode connection
and with

a huge reduction of odd order products such as 3H and 5H. UL
operation

of beam tetrodes and pentodes originated soon after WW2 and
became

the the most popular way to connect multigrid power tubes to
give

approximately twice the power of triode connection yet with
triode

characteristics of low Ra and low THD. A huge number of amps use

the UL connection.

WARNING.
FOLLOWING TESTS ON TUBES INVOLVE HIGH VOLTAGES

AND
TEMPORARY CONNECTIONS OF ELECTROLYTIC CAPACITORS.

DO NOT
ATTEMPT TESTING UNLESS YOU ARE FULLY EXPERIENCED.

Testing a 6550 to measure Ra, Gm and µ.

Fig 1.

Fig 1 shows TWO schematics, A
and B, for testing a 6550 Beam Tetrode to

determine Ra, Gm and µ.

For determination of Triode Ra,
Gm, µ, connect Screen g2 to anode,

and use the same method steps.

Schematic
A explained......

An air gapped choke is used to pass Idc to the anode of
the tube under test.

It should have a minimum of 50 Henrys at 60mA dc and at low Vac
of say

10Vrms across the choke. There is nothing to stop anyone using
more

than 50H, but to achieve 100H at low Bac max and at 60mAdc will
require

a core weighing perhaps a kilogram, and thousands of turns of
0.4 mm

dia wire. Winding resistance should be less than 300 ohms.

Loads on the anode may be varied between 2k0 and 10k0 and rated
for

at least 10W. These may be connected between the output side of
the

100uF x 450Vrated electrolytic capacitor and 0V.

Care must be taken to
avoid a bad electrical shock when connecting

a load from 0V to the cap and the load should be only
connected or

disconnected by a switch and without touching any part of the
circuit.

The choke may be difficult to
source, but you may use 2 or 3 chokes of 20H

each connected in series. If the choke winding resistance total
exceeds

200 ohms, the B+ supply voltage should be adjusted upwards to
ensure the

Ea remains at +400Vdc. To make a choke, see my pages on choke
design.

A 50H choke has XL reactance of
314k ohms at 1 kHz. This reactance is in

parallel to RL, and will have an extremely small effect on the
actual load on

the anode.

At 1kHz, For RL = 8k, Z R+L = 8 x 314 / sq.rt ( 64 + 98,596 ) = 2,512 / 314.1

= 7.997k ohms. The presence of the choke with reactance = 39 times RL has

negligible effects on test results.

Schematic B explained.

Schematic B shows a constant current source using three HV rated NPN

mosfets, Q1, Q2, Q3. There are many types around, but 2SK1461 should

work OK. Maximum possible heat generated in each device, Pd, with B+

= +850Vdc, and say Idc = 100mA and with anode shorted to 0V will be

28 Watts, so it is essential to mount the 3 devices on a heatsink capable of

liberating 85W for a 30C rise which means the surface area of the aluminium

should total around 3,400 sq.cm. During normal use in the schematic shown,

each device generates about 7.7 Watts of heat.

For a test circuit to be reliable, you MUST allow for MUCH higher Idc and

Vdc than used during testing, lest the poor little mosfets fail to become a

short circuit within seconds of an accidental short circuit of anode to 0V.

One could use a vacuum tube to function as a CCS, say another 6550,

but the circuit complexity increases and the heater supply must be biased

at the cathode Vdc, and probably you'll never get a finite resistance value

of say 5 Megohms which is so easy to get with solid state devices.

There are THREE
mosfets used so that the voltage between the B+ rail

of +850Vdc and the anode voltage is divided across each mosfet
nearly

equally even if the anode of the tube should become shorted to
0V. If the

6550 anode is shorted to 0V, then the voltage across each
mosfet, Vds,

is only 283Vdc, and if Iadc = 100mA, Pd = 28 Watts, and OK for
most

TO3P type mosfets. The maximum possible Idc is controlled by the
Q3

500r source resistance and the 38V zener diode. The maximum Vdc
across

the 500r source resistance is 38V, so Idc is limited to 76mAdc.

But as anode voltage changes at
audio frequencies the current flow will

remain nearly constant. The low output resistance of Q3 source
drives the

series 100k drive resistors to Q2 and Q1 gates and also
bootstraps the

junction of the two series 100k bias resistances for Q3 gate.
Thus none of

the biasing 100k resistances load the anode significantly. The
anode load

of the CCS is the effective bootstrapped value of Q3 source
500r. Mosfets

have very high drain resistance and Gm of perhaps 1A/V, so that
their

amplification factor µ, ( like a vacuum tube ) is also
very high. Data for mosfets

don't include the dynamic drain resistance, Rd, or the µ,
which is the product

of Gm x Rd, which may be a figure of several thousand. But the
500r appears

to the anode as a resistance of perhaps 10,000 x 500r, ie,
5Meg ohms.

A signal change of 260Vrms at the anode may produce a current
change of

only 0.052mArms and voltage change across the 500r may be only
0.026mV,

and not measurable with most voltmeters.

The mosfet CCS will definitely be easier and cheaper to make
than a

choke, but will be more easily damaged in the hands of inexperienced

and careless persons.

For testing tubes with higher or
lower Ea, and for higher Idc, the same

schematic may be used but with care taken to make sure smoke
will never

be generated.

For both schematics A and B....

Voltage rail supplies should be prepared using
information elsewhere at this

website. If the schematics A&B are used to test SE tube
operation up to

maximum SE output voltage of perhaps 270Vrms, A signal generator
with low

output impedance is needed to apply low distortion signals of up
to say

50Vrms at 1kHz. This is usually unavailable from most signal
generators and

a low THD voltage preamp must be prepared using say a 12AU7 plus
6BQ5

in triode and with a global NFB loop to reduce the open loop
gain of about

250x to about 20x, thus reducing driver THD to less than 0.3% at
50Vrms.

The schematics allow for
measurements of triode operation with g2

connected to the anode.

If taps are placed along the
choke used in A, UL operation may be examined.

Wherever possible, anode signal voltage applied to RL should be
kept to less

than 40Vrms to keep well way from the high THD levels when
output voltage

swing becomes high.

An oscilloscope SHOULD be used
to monitor all signal voltages and currents

to ensure THD is below 2% at all times.

Beam Tetrode.

(1)
Connect RL = 4k, apply 1.00Vrms to g1,
measure output signal across RL.

Suppose VRL = 19.55Vrms. Ia = 19.55V / 4k = 4.887mA.

Gain from g1 to anode = VRL / Vin = 19.55.

(2)
Connect RL = 8k and without
changing input of 1.00Vrms, measure VRL.

Suppose VRL = 35.2Vrms. Ia = 35.2V / 8k = 4.40mA

Gain from g1 to anode = 35.2.

The Ra may be calculated by dividing the difference in anode voltages by

the difference in anode currents for two different loads.

In this example, the anode voltages have been measured for 1.0Vrms of

input, so Va is the same figure for tube gain. But Vin could be any

convenient voltage and the Ra maybe calculated thus...

Ra = difference in Va for 2 different loads / difference in Ia for 2

different loads.

For this example, difference in Va = 35.2V - 19.55V = 15.65Vrms.

Difference in Ia = 4.887mA - 4.400mA = 0.487mA

Ra = V / I = 15.65 / 0.487mA = 32.13k.

(4) Calculation of beam tetrode amplification factor, µ.

Now for all tubes, grid 1 to anode Gain, A, = µ x RL / ( RL + Ra ),

Thus µ = A x ( RL + Ra ) / RL.

Gain A for 4k = 19.55 from step (1),

Thus µ = 19.55 x ( 4k + Ra ) / 4k.

µ may be calculated by substituting Ra from step (3)

For RL = 4k, µ = 19.55 x ( 4k +32.13k ) / 4k,

= ( 19.55 x 36.13k ) / 4k = 706.34k / 4k

= 176.58.

(5) Calculate Beam Tetrode g1 transconductance,

Gm
= µ / Ra = 176.58 / 32.13 = 5.495 mA/V (
0.005495A/V ).

(6)
Connect RL = 4k, apply 1.00Vrms to g1,
measure output signal

across RL.

Suppose VRL = 4.95Vrms. Ia = 4.95V / 4k = 1.2375mA.

Gain from g1 to anode = VRL / Vin = 4.95.

(7)
Connect RL = 8k and without
changing input of 1.00Vrms,

measure VRL.

Suppose VRL = 5.57Vrms. Ia = 5.57V / 8k = 0.6963mA

Gain from g1 to anode = VRL / Vin = 5.57.

The Ra may be calculated by dividing the difference in anode voltages

by the difference in anode currents for two different loads.

In this example, the anode voltages have been measured for 1.0Vrms of

input, so Va is the same figure for tube gain. But Vin could be any

convenient voltage and the Ra maybe calculated thus...

Ra = difference in Va for 2 different loads / difference in Ia for 2

different loads.

For this example, difference in Va = 5.57 - 4.95 = 0.62Vrms,

Difference in Ia = 1.2375mA - 0.6962mA = 0.541mA,

Triode Ra, RaT = V / I = 0.62V / 0.541mA = 1.145k.

(9) Calculation of Triode amplification factor, µT.

Now for all tubes, grid 1 to anode Gain, A, = µ x RL / ( RL + Ra ),

Thus µT = A x ( RL + RaT ) / RL, where µT and RaT are for Triode.

Thus µT = 4.95 x ( 4k + RaT ) / 4k.

µ may be calculated by substituting RaT from step (8)

For RL = 4k, µ = 4.95 x ( 4k + 1.145k ) / 4k,

= 4.95 x ( 5.145k ) / 4k = 25.468k / 4k

= 6.367

(10) Calculate Triode g1 transconductance, Gm = µT / RaT

This example, Gm = 6.367 / 1.145k = 5.56mA/V ( 0.0056A/V ).

(11) Triode Gm should be equal to Gm for beam tetrode?

Expect accuracy of calculations to be within 2%. Error %

= 100 x ( [GmT- GMBT] / GmT )

This example. Error = 100 x [5.56 - 5.495] / 5.56 = 1.169%, OK.

-------------------------------------------------------------------------------------------------

NOTE. For all tubes, grid 1 to anode Gain, A, = µ x RL / ( RL + Ra ),

Thus µ = A x ( RL + Ra ) / RL.

Where gain for two different RL are known, let A1 = gain for RL1 and

A2 = gain for RL2.

You can derive the following equation...

Ra = ( A2 - A1 ) / ( [A1 / RL1] - [A2 / RL2] )

You should notice that this equation is giving the same calculation as for

calculating Ra by dividing the difference in anode voltages by the difference

in anode currents for two different loads.

NOTE. There will be slight variations between tube samples because of

slight tube manufacturing differences.

In previous tests I have done on
EH6550 I have found average values at

the above nominated idle conditions of Ea = +400Vdc, Ia =
55mAdc,

typical in many amplifiers :-

Beam Tetrode Ra = 32,000 ohms,
Gm = 5.5mA/V, µ = 176.

Triode Ra = 1,160 ohms, Gm = 5.5mA/V, µ = 6.38.

The
tube data sheets will show gm = 11mA/V and
at Ea = 250Vdc and

Ia = 140mAdc and Ra maybe 15k ohms. Nobody in their right mind
would

ever set up a 6550 or in such conditions for class A operation.

NOTE.
The above method for determination tube measuring and testing

to find µ, Ra and Gm is accurate enough depending on the
accuracy of

the resistance loads and the voltmeter.

-------------------------------------------------------------------------------------------------

(12)
Calculation of Ultralinear Ra, µ and Gm.

NOTE.
µ for ANY mode of operation between pure beam tetrode and
triode

may be calculated. The screen may be connected to a tap on the
anode

primary winding of the OPT and the UL tap position is often
quoted in OPT

specifications as a percentage of winding turns between B+ and
anode.

A common figure is 43%, an old Mullard figure, or 50% for Leak
amps, but

UL% may be anywhere from 20% to 75%. The UL% needs to be
converted

to a FRACTION for use in equations. UL% as fraction, ULFr = UL%
/ 100.

NOTE.
To calculate µ for ANY mode we need to know Ra for beam
tetrode,

Gm for either beam tetrode or triode and µ for Triode. The
above figures in

steps (1) to (10) provide the basic information.

This example, Ra BT = 32k, Gm =
5.5mA/V, µBT = 176, µT = 6.38, let .

Any mode µ =
Gm x Ra
BT

( ULFr x [ µBT
- µT ] ) + 1

µT

Example, Beam Tetrode µ,
ULFr = 0.0,

( 0.0 x [ 176 - 6.38 ] ) + 1

6.38

agrees with data above.

For triode, ULFr = 1.0,

µT = 5.5 x 32k = 176 / 27.58 = 6.38,

( 1.0 x [ 176 - 6.38 ] ) + 1

6.38

For 50% UL tap, ULFr = 0.50

µUL = 5.5 x 32k = 176 / 14.29 = 12.31,

( 0.5 x [ 176 - 6.38 ] ) + 1

6.38

agrees with Fig 3 graph below.

(13) Determination of Screen Grid g2 Gm, Gmg2.

This may be calculated without measuring Gmg2.

Triode Ra = 1/Gmg2 in parallel with RaBT.

= [ 1/Gmg2 ] x RaBT / ( [1/Gmg2] + RaBT ).

From this we can derive screen Gmg2 = ( RaBT - RaT ) / ( RaBT x RaT )

This example, Gmg2, mA/V = ( 32k - 1.16k ) / ( 32k x 1.16k ) = 30.84 / 37.12

= 0.83mA/V

If full ohms were inserted, Gmg2 A/V = (32,000 - 1,160) / (32,000 x 1,160)

= 30,840 / 37,120,000

= 0.00083A/V = 0.83mA/V.

(14) Calculation of Screen grid amplification factor, µg2.

µg2 = Gmg2 x RaBT.

This example, µg2 = 0.83mA/V x 32k = 26.56.

Simple method to find Beam Tetrode and Triode Ra, µ, and Gm :-

This method does not need a
signal generator and may use any accurately

adjustable voltage source including a variac or transformer
winding at

mains frequency.

(1)
To find grid Gmg1 for any mode of operation :-

Shunt the TP1 to 0V with 470uF
cap, 450Vdc rated. Apply 1.00Vrms

signal to g1. Measure voltage across 10r between TP1 and TP2.

Suppose we measure 0.055Vrms. Ia = 0.055 / 10 = 0.0055Arms =
5.5mArms.

This occurs as a result of a 1Vrms grid change, so Gm = 5.5mA/V.

To measure the change in Gm with
Iadc idle current, the tube should be

biased for a range of Ia at say 10, 20, 30, 40, 50, 60, 70,
80mAdc and

careful Vin & Ia measurements taken.

(2)
To Find Beam Tetrode RaBT :-

Set up schematic A or B as
shown. Disconnect any RL. Connect 0.47uF

between grid 1 and 0V, and no g1 signal is applied.
Connect output side

of 100uF to a low impedance signal source.

Apply say 25Vrms signal from a
low impedance signal source and measure

signal voltage across the 10r between TP1 and TP2. Calculate
signal current

flow in 10r = V10r / 10. Suppose you measure 0.0078Vrms.

Ra = Va / Ia.

Ia = 0.78mA. Va = 25Vrms,

Ra
= 25 / 0.0078 = 32,000 ohms.

(3)
To find grid 1 µg1, for beam Tetrode :-

µBT
= Gm x Ra = 0.0055 x 32,000 = 176.

NOTE,
it is only necessary to know Gm and Ra, from which µ

may be easily calculated.

NOTE.
If the Idc was supplied to tetrode with a constant current
source

instead of a choke, Voltage gain g1 to anode = µ, because
there is negligible

load current occuring in the load which has almost an infinite
number of ohms.

The g1 to anode gain with an infinite load value would also be
infinite but is

limited to Gm x Ra, because the Ra is the internal anode to
cathode dynamic

resistance, acts just like a resistance strapped across a
current generator in

Fig 2 below.

If Gm = 5.5mA/V, and Ra = 32,000
ohms, the voltage gain with no load should

be Gm x Ra = 176. Schematic B in Fig 1 shows the CCS which in
fact has a

finite R of say 5 Megohms, and if Ra was 32k, then with the 5Meg
the gain

with no other load becomes Gm x (Ra + RL) in parallel. 32k
+ 5M = 31.8k,

and µ = 174.88, close enough to 176.

internal dynamic resistance. This will vary with Ia and for UL or triode

connection. In triode, the 6550 may have Ra = 1,000

ohms, and if Gm = 5.5mA/V, µ = 5.5. The presence of the finite high

resistance of a CCS or a choke reactance will have no significant effect

on triode measurements.

Now Gain, A = µ x RL / (
RL + Ra ). But µ = Gm x Ra for all tubes,

so A = Gm x Ra x RL / ( RL + Ra ), which removes µ
from the equation.

In fact the expression Ra x RL / ( RL + Ra ) is the result of RL
and Ra

being parallel load resistances. For where there is a choke used
instead

of CCS, and no other RL, gain A = Gm x Ra in parallel with choke
reactance,

XL.

At 1 kHz, Z of Ra // XL = 32k x 314k / sq.rt ( [32k x 32k] + [314k x 314k] )

= 10,048 / 315.56 = 31.94k ohms.

Gain A = Gm x ( Ra
in parallel to RL ) = 5.5 x 31.94 = 175.68. This is
very

close to 176.0, the value of gain we should see without the
reactance load

of inductance present.

In practice, the 50H choke will
have much lower real L at low voltage levels

and hence lower XL and so for 1Vrms grid 1 input, we would
always measure

a lower anode voltage output with a "50H" choke. The choke shunt

capacitance could be 400pF at this is a reactance load of 397k
at 1kHz.

However, where L and C reactances are equal, the L becomes a
parallel

tuned circuit and the load impedance of a parallel tuned circuit
is higher

than either reactance value at the frequency involved.

NOTE.
For the idle conditions in Fig1 above Ra = 32k, Gm =
5.5mA/V,

hence µ = 176. Data sheets for 6550 have Ea = 250V, Ia =
140mAdc

which gives Ra = 12k and Gm = 11mA/V so µ = 132. The
conditions of

Ea = 400V and Ia = 55mA in Fig 1 above are far more common and

typical of many amps made since the 6550 was first produced.

Gm and Ra vary considerably depending on Ia and the relationship
is

not a very linear.

(4)
To measure and find screen grid 2 µ :-

Use A or B schematics in Fig 1
above. Then ground input grid with

0.47uF cap. Ground anode with 470uF cap from TP1 to 0V.

Connect 1,000 ohms between screen B+ supply and screen.

Connect 100uF electrolytic cap from screen to low resistance
signal supply.

Apply 10Vrms to the screen and
measure voltage across 10r between

TP1 and TP2. Suppose the voltage reading is 0.083Vrms. Then
current

= 0.083V / 10r = 8.3mArms, and screen Gm = I change / Vin change

= 0.0083A / 10V = 0.83mA/V.

With a grounded grid, the 6550
may be operated as a triode with screen

as the control grid. The Ra of this triode is approximately the
same as

the Ra in beam tetrode mode. The µ of the screen grid is
calculated

the same way as µ for grid 1.

µ
g2 = Gm g2 x Ra = 0.83 x 32,000 = 26.5.

NOTE.
Just remember that the Gmg2 and Gmg2 vary with the value
of Ia.

NOTE.
The above calculations ignore the slight signal input
current to

g2 which then has a very small loading effect on the anode
circuit.

The equations would become far too complex if screen current
input

was included. But from the equations given, predictions of gain
for any

load and effective Ra may be determined with enough accuracy to

allow the overall amp design to be well described without any
trial

and error procedures.

Equivalent
Model, Beam Tetrode, UL, Triode, 6550 :-

Fig
2.

Fig
2 shows an Equivalent Model
for a single 6550 beam tetrode set up

with anode load of 6,200 ohms.
Those not used to using an equivalent

model of an electronic device need to realize that
clever minds invent a

diagrammatic model like Fig 2 to reduce the device function to
simple

known circuit units which allow mathematical equations to be
derived

which can then predict the device behavior during the design of
an amplifier.

Symbols for voltage source and constant current generator are
used

universally by electronics engineers.

In Fig 2, the screen is shown to be connected at a tapping point
arrow along

RL from where a variable amount of the anode signal may be
applied to

the screen. Beam tetrode will have the arrow at the top of RL,
Triode

operation will have arrow at the bottom of RL and Ultralinear
operation

will have the arrow somewhere along the anode RL as it is shown
in Fig 2.

Other places on this website
describe Equivalent models for tubes being a

low Rout voltage generator producing Vout = µ x Vin, and
with Ra as a

series resistance in series with V generator and the anode. The
use of a

current generator and shunting Ra as I have here in Fig2 is just
as valid

a method to describe what happens inside the tube in useful
mathematical

terms.

For
Beam Tetrode operation, screen g2 is connected to top
of RL, ie,

to B+, which is well bypassed to 0V. Therefore the current
generator2 acts

as an infinite value of resistance between output anode and 0V
and it has

no loading effect upon the current changing ability of g1.

the circuit and there is no need to consider the dc flow conditions

which enable the signal operation.

The signal
generator "voltage source" symbol is a circle with sine
wave

within, and it is assumed the output resistance of the voltage
source is

very low, less than 600 ohms. The mains supply at the wall
socket is also

a "voltage source" with a very low output resistance, often less
than 5ohms.

The
constant current generator symbol with two circles
produces an

unchanging current output which equals Vin input x Gm of the
grid of the

device. The output resistance of the Constant Current Generator,
CCG

is infinitely high, so whatever load ohms are connected to the
CCG

will determine the voltage produced by the current applied
because

V = I x R.

Fig 2 shows the operating Idle
conditions with Ea = +400Vdc,

Iadc = 55mAdc, Grid g1 bias = -50Vdc, and it is assumed the 6550
has

been tested and found to have the following characteristics :-

Ra
for pure beam tetrode operation with fixed Eg2 and grounded

cathode = 32k ohms. Gm
g1 = 5.5mA/V, and g1 µ = Gm g1 x Ra = 176.

screen voltage,

A
= µ x RL / ( RL + Ra ) = 176 x 6.2k / ( 6.2k +
32k ) = 28.56.

Beam tetrode gain may also be
expressed as

A = Gmg1 x ( RL in parallel with Ra)

This is easy to visualize when Fig is inspected.

Generator 1 produces current = 5.5mA/V which acts on Ra
// RL thus giving...

A
= Gmg1 x RL x Ra / ( RL + Ra ) = 5.5mA/V x 6.2k x 32k /
( 6.2k + 32k )

= 28.56

NOTE.
With any two paralleled R, the resultant R = 1 / ( 1/R1 + 1/R2 )

This may be more conveniently expressed as R = ( R1 x R2 ) / (
R1 + R2 ),

therefore load on which the
current generator works is

Ra and RL in parallel = ( RL x Ra ) / ( RL + Ra ).

drive only,

Ra = 32,000 ohms and Gm g2 = 0.83mA/V. g2 µ = Gm g2 x Ra = 26.5

Gain from screen drive, Ag2 = µg2 x RL / ( RL + Ra )

= 26.5 x 6.2k / ( 6.2k + 32k ) = 4.31.

Gain is also Ag2 = Gm g2 x ( RL in parallel with Ra )

= 0.83 x 6.2k x 32k / ( 6.2k + 32k ) = 4.31.

You may wonder what significance these equations may have, but to

understand how the beam tetrode works you need to understand the maths.

Ultralinear and Triode.

Fig 2 shows Ultra-Linear operation with a tap on RL to feed the screen

with 50% of the anode signal. As soon as one begins to consider the UL

operation of the beam tetrode, then the screen grid acts to oppose the

action of grid g1. The signal applied to the screen grid can be any fraction

of the anode signal between 0% which is pure beam tetrode operation and

100% which is triode operation.

Consider Fig 2 above with a 50% UL tap on an OPT, and consider what

happens when +1Vrms is applied to g1. The +1Vrms generator 1 current flow

= Gm g1 x input voltage = 5.5mA. This current acts on Ra and RL and also

upon the resulting resistance created by generator 2. With no signal between

screen grid generator input and and cathode, the generator 2 has infinite

resistance between anode and cathode. But this resistance becomes a much

lower finite resistance when a signal exists between g2 and cathode, and the

value of the resistance = 1 / (ULFr x gmg2).

The ULFr is the UL fraction of anode signal fed to the screen, and officially

may be calculated as UL Fraction = ( UL tap Vg2 - Vk ) / ( Va - Vk ).

This allows for any mode of operation including where CFB is used.

This may be difficult to understand but consider the anode voltage rises +1Vrms.

If the UL tap delivers 0.5V to g2, then the change in current in generator 2

= 0.5V x Gmg2 = 0.5 x 0.83mA = 0.415mA, and so the generator 2 set up

acts like a resistance of Vout / Iout = 1V / 0.415mA = 2.41k ohms.

Thus the current of 5.5mA from generator 1 acts upon THREE

resistances, Ra, RL, and 1 / (ULFr x gmg2) all in parallel. These three R

are 32k, 6.2k and 2.41k, and total = 1.646k.

So the anode voltage = 5.5mA x 1.646k = 9.053V, as shown on Fig 2.

Now if there 6550 acts in beam tetrode mode, for +1.0Vrms g1 input,

Va = -28.56Vrms. But with 50%UL, we have -9.053Vrms. There is 50% UL,

so the screen signal input = -4.527Vrms. The gain of the screen to anode

= 4.31 with RL = 6.2k, so the -4.527V is amplified x 4.31 to become + 19.511V

at anode which sums with what the g1 is trying to produce so that resultant signal

= -28.56V + 19.511V = -9.053Vrms. The UL connection reduces gain by a factor

of 9.053 / 28.56 = 0.317.

Should any THD voltage appear at anode in beam tetrode mode, then it will

be reduced in theory by the factor of 0.316.

However, the action of g1 and g2 upon the electron stream is not perfectly

linear and Ia is varied by a factor of cube root of Vin change squared x constant,

and because 2H is the main H generated by such a transfer function, the 2H for

UL will remain high. But the reduction of 3H and 5H which is produced by the

pure beam tetrode mode is MUCH reduced by the screen UL tap.

Maximum THD reduction because of UL screen tap achieved when ULFr = 100%,

ie, the 6550 is triode connected. In real triodes, such as a 300B without a screen,

the anode has Gm and acts just like the screen upon what might be the Ra with

a screen the resistance = 1 / Gmg2. A 300B may be considered to have anode

Gm = 1 / Ra = 1 / 800 ohms = 1.25mA/V.

In Fig 2 above, the gain g1 to anode for the beam tetrode ( or pentode)

with any value of UL tap% may be calculated....

Gain g1 to anode = Gm g1 x ( RL // Ra // [ 1 / { UL fraction x Gm g2 } ] )

For UL gain calculations, UL.

(1) Calculate RL // Ra, where Ra is pure beam tetrode Ra.

This example, RL // Ra = 6k2 // 32k = 6k2 x 32k / ( 6k2 +32k ) = 5.197k.

(2)

Consider UL% = 50%, ULFr = 0.5,

Calculate 1 / { UL fraction x Gm g2 } = 1 / { 0.5 x 0.83mA/V } = 2.409k.

(3) Calculate UL gain = 5.5mA/V x 5.197k // 2.409k

= 5.5mA/V x 5.197k x 2.409k / ( 5.197k + 2.409k )

= 5.5 x 1.646 = 9.053.

(3) Let us check. Fig 2 shows +I Vrms is applied to g1, -9.05V appears at

the anode.

-4.53Vrms is applied to g2. Without the screen FB, Va = -28.56V would appear

at the anode. But with -4.53Vrms applied to g2, there is a FB voltage generated

at the anode which = screen gain x screen signal = 4.31 x 4.53V = +19.55V,

and this reduces the -anode voltage swing without NFB leaving a residual voltage

= -28.6 + 19.55V = -9.05V at anode. The analysis of currents generated will

show all this to be true.

(4) Calculate UL Ra for any value of UL% tap.

Ra for any UL% = Ra // [ 1 / { UL fraction x Gm g2 } ], where Ra is pure

beam tetrode Ra.

This example, UL tap at 50%,

Calculate UL Ra

= RaBT // 1 / { UL fraction x Gm g2 } = 32k x 2.409k / ( 32k + 2.409 )

= 2.24k.

(5) Calculate UL µ = Gmg1 x UL Ra.

UL µ = Gm g1 x ( RL // Ra // [ 1 / { UL fraction x Gm g2 } ] )

= Gm x UL Ra.

This example, UL tap at 50%, µ = 0.0055 x 2.24k = 12.32.

(6) Triode Ra

When tetrode is triode connected, Triode RaT = RaBT // 1/Gm g2.

In this case, Ra triode

= 32k // 1/0.83ma/V = 32k // 1.204k

= 32k x 1.204k / ( 32k + 1.204k ) = 1.16k.

(7) Triode µ = Gm g1 x RaT.

This example, Triode µ = 0.0055ma/V x 1.16k = 6.38.

NOTE, UL and triode Gm will always stay the same as beam tetrode

Gm because it is measured when there is no change of anode or screen

voltages when a signal is applied to grid g1.

It can be seen that the 50% UL tap reduces beam tetrode gain from 28.6

to 9.05, ie -10dB, but reduces Ra from 32k to 2.24k, -23dB, a very useful amount.

Fig 3.

Fig 3 shows the reduction of Ra
with a given UL tap % of anode signal applied

to the screen grid 2. There is seldom any point of having UL%
more than 60%

where Ra is less than 2 x triode Ra.

UL operation has two main
benefits to amplifier performance. The first and most

important is that the maximum class A1 or AB1 output power of
the 6550 is kept

close the maximum achievable with pure beam tetrode where Eg2 to
Ek is kept

constant at all times. As the UL tap % approaches 100%, or
triode connection,

maximum power is limited by grid current required to swing the
anode voltage

towards 0V. The extra circuitry needed for Class A2 or AB2 is
seldom justified

and creates extra distortion and reliability problems.

The second main benefit of UL
operation is the lessening of odd numbered

harmonics produced by beam tetrodes in addition to their even
numbered

harmonics. Usually, 50% UL amps will give very much lower THD
than pure

beam tetrode operation especially in PP operation. Nobody can
usually

tell any difference in the sound between a well designed triode
amp or a

UL amp. But the UL amp will make nearly twice the maximum power
of the

triode amp.

Fig 3 graph is only valid for
the operating conditions shown for a 6550.

Usually such conditions are for each tube in a class AB1 PP amp.

Pda is only 22 Watts. And for pure class A1 SE power the ideal
RL for one

tube is 6k2 for the given Ea and Ia. With UL tap at 50%, about
10.4 Watts is

available at approximately 5% mainly 2H at clipping. For two
tubes in PP,

the pure class AB1 anode to anode load would 12k4 and PO max =
21W

at about 1% THD, mainly 3H at clipping. But many people will
insist on

more power and therefore RLa-a would be about 6k1 for the two
tubes with

PO perhaps 40W max. But at normal levels of audio where average

PO = 1/40 of the maximum at clipping, ie, maybe 1W, the class
AB1 amp is

working in pure class A1 PP and load for each 6550 is 3.05k.
Tube gain will

be highest while the AB1 amp PO is low and the gain at low PO

should be used for working out the gain of the driver amp and
the amount

of global NFB to be applied because instability is most likely
where gain is

highest.

If you wish to work out the gain
of the output stage:-

From Fig 3 you can see 50% UL
Ra = 2.24k.

From the chart on RHS side Fig
3, for 50% UL µ = 12.3.

Class
A1 UL gain between g1 and anode = UL µ x RL / ( UL Ra +
RL )

= 12.3 x 3.05 / ( 2.24 + 3.05 )
= 7.09.

The maximum grid drive required
depends on the bias voltage and

= ( Bias Vdc x 0.707 ) This will be the maximum drive Vrms
needed for

1 x 6550 and for all classes of A1 and AB1 conditions and all
loads.

SEUL
6550 and SEUL 6550 with CFB.

Fig 4 shows a 6550 used in two ways, SEUL, and SEUL + CFB, ie,

as an SE amplifier with Ea and Ia conditions being the same as used in

the test circuits A&B in Fig1 and in the Equivalent Model in Fig 2.

Fig 4 shows the plain UL amp stage with 50% UL tap and the gain equations

have already been explained above.

The use of a Cathode Feedback winding on the OPT allows the 6550 or

many other beam tetrodes or pentodes to have their performance enhanced

to give lower Ra and less THD than possible with pure UL or triode operation.

To cut a long story short, it may be proven that the CFB schematic as shown

on Fig 4 works like a UL configured tube but with some portion of the total

anode to cathode signal allowed to appear at the cathode as a form of

applied series voltage NFB.

The CFB schematic requires some expert explanation.

When the CFB stage in Fig 4 was designed, it was intended that there be 20%

of Va-k signal appearing at the cathode. This is because it happens to be an

optimal amount of CFB to use so that there is a fair amount of FB applied so

that it will give a significant reduction of THD and Ra but without needing

a grid 1 drive voltage which is so high that the driver stage distortion does

not add more distortion than is reduced by the use of the CFB in the output

tube.

It was also decided that the tube would also have the same signal operation

within in it as for the SEUL stage also shown on Fig 4, so there is a 50%

portion of Va-k applied between g2 and cathode, Vg2-k.

Fig 4 shows that the Va-k = -9.05Vrms, same as for the SEUL stage, and

that the OPT primary has an anode section with +7.24Vrms and cathode

section with -1.81Vrms. -1.81Vrms is 20% of the 9.05Vrms of Vak. These

voltages are kept in in their 4:1 ratio by the turn ratio of the two windings.

To get the tube to work with a 50% UL tap, Vg2-k = 50% of Vak = 0.5 x 9.05V

= 4.525V.

We already have -1.81V at the cathode, so the voltage at the tap on the

anode winding is at a point = -4.525 + 1.81 = -2.71Vrms as you can see

in Fig 4. The + or - sign used in front of Vrms signals indicates the phase

of the voltages.

The relationship of UL + CFB voltages can be expressed mathematically as

UL fraction with or without CFB = ( Vg2 - Vk ) / ( Va - Vk ), which becomes

UL% / 100 = UL Fraction = ( UL tap Vg2 - Vk ) / ( Va - Vk )

Where all voltages are measured between 0V and anode

or cathode or g2.

For the example above, what is the UL Fraction?

UL Fraction = ( -2.715 - {+1.81} ) / ( -7.24 - {+1.81} ) = - 4.525 / -9.05

= 0.5 = CORRECT.

Its easy to get muddled with signs and phases, but don't blame me,

I didn't invent maths, and you need to practice the calculations with some

idea in mind for what may be the likely result which may prevent you

proceeding further if you get an absurdly wrong calculated parameter.

What we now know about the UL+CFB example is that the 6550 with

CFB and screen tap % will have exactly the same operating internal gain

conditions as the plain SEUL without CFB. To generate the whole Va-k

signal there must be 1Vrms between input grid and cathode, Vg1-k.

For all the working voltages to occur as we seen them, the grid g1

signal with respect to 0V must be 1Vrms more than Vk and have the

same phase.

So Vg1-0V = +Vg-k + Vk = 1V + 1.81V = +2.81V.

Now the lessening of overall stage gain is due to the Cathode Feed Back

at cathode.

Engineers call the UL gain = Va-k / Vg1-k, open loop gain, or OLG.

The gain with CFB applied by the cathode is called the closed loop gain,

or CLG, and is Va-k / Vg1-0V.

This can all be expressed more simply in maths if we introduce ß, Greek

letter 'beta' which is the fraction of the output signal fed back in series

with the input signal.

In this case, ß = 20% of total Va-k so ß fraction = 0.20

We also need to simplify OLG Gain to be A, and CLG to be A',

as in RDH4 which has huge amount of very useful information about all

sorts of feed back.

In this example, A = 9.05 / 1 = 9.05, and ß = 0.2.

For all series voltage NFB applications, A' = A / ( 1 + [A x ß] ).

In this example, A' = 9.05 / ( 1 + [ 9.05 x 0.2 ]) = 9.05 / 2.81 = 3.22.

The amount of applied NFB is expressed in dB and

dB applied FB = 20 x log ( OLG / CLG ),

OR = 20 x log ( Vin with NFB / Vin without NFB ),

OR = 20 log ( A / A' ), and for where signal change is small,

class A, and where Vout of the amp is the same with NFB

and without NFB.

In this case, applied CFB = 20 log ( 9.05V / 3.22V ) = 8.98dB.

With real world amplifier using the UL+CFB arrangement of Fig 4,

what would the maximum grid input voltage Vg1-0V need to be?

Consider 10W is produced in class A1 into 6k2 load.

Va = sq.rt ( P x RL ) Vrms = sq.rt ( 10 x 6,200 )

= 249Vrms. A' = 3.22,

so max grid Vg1-0V input needs to be Va-k max / A' = 249 / 3.22

= 77.32Vrms.

This is about twice the input voltage needed for the plain SEUL stage,

so the driver stage must be carefully designed, so what are the benefits

of CFB?

The CFB leads to a further useful reduction of THD and Ra.

The "open loop" THD is reduced by series voltage NFB by the same

amount as gain is reduced by CFB to become "closed loop" THD,

or THD'.

THD' = THD / ( 1 + [A x ß] ), and if THD at 10W = 5%, we might expect

THD at 1W = 1.5%.

THD' at 1W = 1.5 / ( 1 + [9.05 x 0.2] ) = 0.53%, and THD is reduced by

-8.98dB.

Now Ra is reduced according to the formula Ra' = Ra / ( 1 + [µ x ß] ).

Notice that the amplification factor µ or product of Gm x Ra replaces Gain A

in the feedback equation. This is because Gain may change with RL,

but the Ra without NFB and Gm of the tube remains constant.

In the example for 6550 with 50% UL, the UL Ra and UL µ may be

found from the graph in Fig 3. UL Ra = 2.24k, µ = 12.3.

UL with CFB Ra' = 2,240 ohms / ( 1 + [12.3 x 0.2] ) = 647 ohms.

Ra of plain UL connection with no CFB = 2.24k, and with CFB added

the Ra is again reduced x 1/4. To achieve this the grid 1 input signal is

only doubled, so the CFB is very effective.

The CFB + UL Ra is close to 1/2 the value of triode connection.

Usually the THD is much lower than triode connection.

Figure 4 shows an OPT with 6,200 : 5.6 load impedance ratio.

This means that any source resistance of a device powering the primary

winding is reduced by the OPT impedance ratio.

So the output resistance of the amplifier stage at the OPT sec

= Ra' x sec load / primary load.

In this case Rout = 647 x 5.6 / 6,200 = 0.585 ohms.

To this we must add secondary winding resistance which may be 5%

of the load, or 0.28 ohms, so Rout = 0.865 ohms.

Damping factor,

DF = Secondary load / (calculated Rout + Sec Rw).

In this case, DF = 5.6 / ( 0.585 + 0.28 ) = 6.47, and good enough for

many people.

Some GLOBAL NFB could be applied from the OPT secondary back

to an input stage to further reduce the Rout and THD of the output stage

and the THD of the input and driver stages. Usually never more than 10dB

is needed, but with such GNFB, the DF may become 20, and the THD at

1W perhaps 0.17%.

This assumes the usual 2H of the driver tube is negligible, but usually the

driver 2H is as high as the output stage with CFB, and it is not uncommon

to measure such SE amps with THD < 0.1% at a Watt, mainly all 2H, and

benign, and the amp gives excellent sound.

Using CFB without a tap for UL screen voltage, Acoustical.

The most famous use of "local OPT CFB" in amplifiers was probably in

Quad-II amplifiers from the 1950s.

Quad's invention was called the Acoustical Connection.

The CFB winding had 10% of the total turns in the OPT primary and

had the screens taken directly to a steady and filtered B+ supply rail.

The Quad-II design enabled a pair of KT66 to make 22 Watts but the

Ra' with CFB was about the same as if the KT66 were triode strapped.

THD was as low as triode. Driving the KT66 was no more difficult

than driving KT66 strapped as triodes.

How does one take best advantage of the Acoustical?

The Vdc screen voltage does not have to be equal to the B+ anode

supply voltage and for many tubes such as 6L6, 807, EL34, E:84, 6V6,

KT66 KT88, 6550 etc, the Eg2 may be only +200Vdc to 400Vdc where

the B+ anode supply might be as high as +900V without exceeding tube

ratings.

A pair of 807 in class AB2 could be used to make a PA amp giving 80

Watts with Eg2 = +300Vdc and Ea = +600Vdc. THD was rather high at

about 13% 3H for where there was no CFB and where pure beam tetrode

was used. For hi-fi, and reliability, I might use 6550 with Ea = +480Vdc,

and Eg2 = +300Vdc, and I might have 20% CFB. My 8585 amp is an

example with 12.5% CFB.

The heat generated in the screen wires is much less if Eg2 is kept low.

And the lower Eg2 means that the bias voltage for grid g1 may be a lot less

and if the amp is a class A type with cathode biasing we don't need to have

such a high Ek which means less heat is wasted in Rk, and the B+ may also be

lower to accommodate the Ea + Ek required.

How do we predict the outcome using a fixed g2 voltage but with CFB?

With a fixed Eg2 supply and a CFB cathode signal voltage, there is in effect

some screen FB being applied very much like the UL tap from the anode

winding of an OPT. So in effect, to calculate final Ra' and gain we need to

calculate the UL Ra and µ for the effective amount of UL%.

This is easy because UL% = CFB%.

Let us consider the 6550 used in Fig 4 where there is 20% CFB, but where

we might have the screen taken to a fixed Eg2 = +400Vdc.

This ensures the model in Fig 2 and graph in Fig 3 will be valid.

With 20% CFB, UL% = 20%, so from graph in Fig3,

UL Ra = 5.07k and UL µ = 27.9.

Ra' for the CFB = Ra' = Ra / ( 1 + [µ x ß] ),

In this case, CFB Ra' = 5.07k / ( 1 + [27.9 x 0.2] ) = 770 ohms.

With RL = 6k2, open loop UL gain = UL µ x RL / ( RL + UL Ra ),

For 20% UL, A = 27.9 x 6.2k / ( 6.2k + 5.07k ) = 15.35.

CLG with CFB = UL A / ( 1 + [UL A x ß] )

= 15.35 / ( 1 + [15.35 x 0.2] ) = 3.77.

Applied CFB = 20 log ( 15.35 / 3.77 ) = 12.2dB NFB.

Assume 20% UL THD with no CFB = 7% at 10W,

then at 1W THD = 2.2%.

With CFB, expect THD to be -12,2dB or 0.55% at 1W.

The required grid g1 drive signal required will be about 66Vrms.

The results are in theory very close to the same as for the use of a

UL tap as in Fig 4 above.

In fact, if the CFB % was slightly increased to about 22%,

and g2 taken to a fixed Eg2, the function would be nearly

identical in terms of Ra, THD and DF.

Using CFB with screen bypassed to cathode.

One may use a beam tetrode with CFB windings and bypass the screens

to cathode. In a single ended output this may be done using a high value

electrolytic cap between g2 and cathode and the DC feed to g2 is via a

suitable resistance or high value choke from a B+ rail.

In PP output stages, the screens maybe taken to taps on the anode windings

which have the same phase as the cathode signals and the same % of turns

as for the CFB. Capacitors of 2uF should be used to bypass these taps to the

cathodes. Such taps then make the operation of the tube equal to pure

beam tetrode with CFB.

Effective Ra' = Ra / ( 1 + [µ x ß] ), where Ra and µ are for pure beam

tetrode, ß is fraction of Va-k fed back.

This example, Ra' = 32,000 / ( 1 + [176 x 0.2] ) = 883 ohms.

Open loop Gain, A = µ x RL / ( RL + Ra )

This example, RL = 6k2, A = 176 x 6.2 ( 32 + 6.2 ) = 28.56.

Closed loop gain A' = A / ( 1 + [ A x ß ] ),

This example, A' = 28.56 / ( 1 + [ 28.56 x 0.2 ] ) = 4.25.

Amount of applied CFB = 20 log ( A/A' ), this example = 16.5dB

Open loop distortion of an SE class A beam tetrode at 10W may be 10%

with conditions shown.

At 1W open loop THD will about 3%. With CFB, THD' should be 2% at 10W

and 0.45% at 1W. THD will mainly be a mix of 2H and 3H. PP tubes at 20W

class A should give open loop 3% THD, and at 1W 0.67%. With CFB the

THD at 20W should be 0.5% and at 1W 0.15%. THD will be mainly 3H and

other odd H.

Va-k at 10W into 6k2 = 249Vrms.

Vin max, Vg1 - 0V = Va-k / A' = 249 / 4.25 = 58.6 Vrms.

NOTE. This connection is best if the circuit works in pure class A

where least switching harmonics are generated, but the odd numbered H

of the pure class A beam tetrode will remain, although lowered in quantity.

I believe the Acoustical or UL + CFB to be superior to where pure beam

tetrode with FB is used. Some audio enthusiasts of the 1950s used screen

taps from the OPT with the same phase of signal but at a higher % tap than

the CFB. This made the operation of the beam tetrodes subject to positive

screen feedback, thus boosting Ra by a moderate factor of 2 and µ by a

factor of perhaps 3, and thus achieving open loop gain much higher. So was

the THD, But where there was CFB, final measured results gave marginally

lower effective Ra' and THD' while keeping overall closed loop gain about the

same. Such PFB schemes are very likely to be unstable and cause HF

oscillations, and I am not aware of any manufacturer who has applied

positive FB in the way described. McIntosh should be mentioned because

their amps have 50% CFB which means anode and cathode windings have

equal number of turns which total the turns used for normal amp with no CFB.

The screens of the 6550 are taken to anode connections of tubes on opposite

sides of the PP circuit thus allowing pure beam tetrode operation with CFB.

Possibly the operation may be improved if screens were taken to the fixed

B+ supply which results in 50% UL operation with 50% CFB. But McIntosh

amps require a very high grid drive voltage of up to 150Vrms per grid1.

In practice the use of a UL screen tap in class A SE amp with CFB winding

gives the least complex mix of harmonics in the measured amount of THD,

and people tell me this gives the best sound.

For most people, it is enough to know CFB windings and a fixed Eg2 voltage

are more effective than having only UL taps to reduce Ra and THD while

keeping voltage gain high enough to avoid a grid drive signal of more than

75Vrms.

The screen of a beam tetrode may be used to be the main control grid for

anode voltage change so the tube works as a triode with µ = 26.5, Ra = 32k,

and gm = 0.8mA/V. But it is rarely ever done because the gain tends to be low,

the Ra remains high as beam tetrode and the screen current requires a direct

coupled cathode follower to apply the drive signal. So if Va was say 250Vrms

for RL = 6k2, Vg2 = 58Vrms and to the CF drive it would be 61Vrms.

It would become much higher for class AB2 operation into a lower load value

of 3k2.

The negative feedback in triodes or multigrid tubes due to the electrostatic

voltage effects is not a perfectly linear application of NFB because the

current/voltage relationships in vacuum tubes is not linear. The anode current

change for a grid voltage or anode voltage change is proportional to the

square root of a number cubed. Professor Child described triode "self regulation"

better than I can in Terman's 1937 book, Radio Engineering.

Any tube will have its highest gain when the anode load is the highest possible

number of ohms. The highest load will be a constant current source.

The internal feedback effect gives maximum linearity, ie, least THD when the

load is a CCS and no anode current change occurs.

In a power amp one must allow RL to be low enough to get sufficient current

change and hence sufficient output power and anode NFB becomes less

effective as RL is lowered so even with triodes there is unavoidable distortion.

But in preamp and driver stages RL can be made to be quite high so THD

will be low. Therefore I like to use constant current source loads in signal

preamp triode circuits and chokes plus resistances for the RL delivering DC

to the anode.

Any tube is said to be a voltage
device, and it is true true because there can

be changes to the anode voltages even when there is no current
change to the

idle current flow between anode and cathode. The voltage gain in
triodes is

determined by the relative distances between the cathode grid
and anode.

These distances determine the effects of voltage field
intensities and the

effect on the electron stream. Triodes are said to be voltage
sources because

their Ra is usually lower than the RL they power. Pentodes and
tetrodes are

said to be current sources because their Ra can be much greater
than the RL

they power.

My pages on load matching loads
to power tubes emphasize the need for

careful loading of tubes.