BASIC TUBE 4. Created March 2011.
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.
schem-test-circuit-se-6550.GIF

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.

Impedance Z, of parallel R+L = R x XL / sq.rt ( R squared + XL squared ),
 
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.

How to determine Beam Tetrode or Triode Ra, µ, and Gm :- 

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.

(3) Calculate Ra for beam tetrode operation.

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 ).

Triode.

(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.

(8) Calculate Ra for Triode operation.

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,

µBT =              5.5 x 32k                      = 176, OK,
            ( 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.

In fact, the Ra of any tube is the highest possible ohm value for the tube
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.

Z of Ra // L = Ra x XL / sq.rt ( Ra squared + XL squared )

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.
Equivalent-model-current-generator-6550-KT88.GIF

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 model focuses purely on the signal voltages and currents within
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.

With pure beam tetrode, anode voltage gain, A from Vg1 to Va with fixed
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 ).

Screen drive. With a grounded cathode, grounded g1 and with screen
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.
graph-Ra-vs-UL%25-CFB%25-se6550.GIF

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.
schem-ULmodel-SE-6550.GIF

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.

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