Please excuse the hand drawn circuit presentation of schematics and large file size.

Very Basic Balanced Shunt Feedback

The Balanced Shunt Feedback  amp was an idea dreamed up to exploit the idea of applying balanced loops of shunt NFB from the output stage tube anodes back to the output of a balanced drive stage using a long tail pair.
The feedback loops are the shortest loop available as they do not include the secondary of the output transformer, and avoids some of the instability and phase shift effects of leakage inductance in the OPT.
To make things slightly clearer, here is an even more basic simplified schematic circuit :-

The problem with anyone wanting to design a shunt FB using NFB between input and output and to calculate
the effect on distortion and output resistance is to work out R1 and R2 carefully.
Once that is done, all that is needed to work out the distortion reduction factor is R1, R2, and output tube gain
for a given load value. We do not need to know input tube voltage gain.
And the effective anode resistance after feedback has been added can also be calculated.
From these figures we can work out whether there enough NFB or too much.
So let us analyse the above PP amp with 6L6......

The input LTP is arranged using the pentode sections of a pair of 6U8A triode-pentodes. The triode sections
are configured so they make a high impedance load on the pentode at signal frequencies, but at low frequencies they act to supply the required amount of quiescent DC to the pentodes.
The input impedance to the triodes is high while the output impedance from the triode cathodes is nearly as low as a cathode follower.
There are feedback resistors of 470k between the 6L6 output tube anodes and the pentode driver tube anodes. When these are not connected, the gain of the driver stage is extremely high because pentode gain is about gm x RL, and the triodes above the pentodes have an equivalent very high loading dynamic impedance. In the case of the 6U8A triode section, µ = 40, so the effective impedance looking up into the 22k cathode resistor = [ ( µ + 1 ) x Rk ] + Ra = ( 41 x 21.2 k ) + 5 k = 874 kOhms. There is a 1M triode triode grid bias resistor but it is bootstrapped off the 1k2 and 22k triode cathode R network so the 1M is a load which is much higher than 1M so the triode plus its cathode R act as a dynamic load of approximately 800kOhms.
The feedback resistors between the output tube anodes and the pentode anodes provide the driver pentodes with their principle loads, equal in value to 470k divided by the output tube stage ( gain + 1 ).

In this case output tube gain = 80Va / 9.5Vg = 8.42. The voltage at the pentode anode = 9.8V since it always will be slightly higher than the output from the triode cathode.
The amp schematic shows that +80V and - 9.8V appear at each end of the 470k, so the official gain without follower losses
= 80Va  / 9.8Va = 8.16, so the effective RL appearing at the pentode anode is 470k / ( 8.16 + 1 ) = 51.3k.
Another way to look at it is to say there is a total of 89.8Vrms across the 470k, so I = 0.191mA.
If the voltage at the pentode anode is 9.8Vrms, then its load = Va / IRL = 9.8V / 0.191mA = 51.3k.

However the loading of the triode stage has to be included part of the load in paralell with the 470k effective load.
The triode loading = 800k, so the actual load seen by the pentode = 51.3 // 800 = 48.4k.

The gain of the input pentode = µ x RL / ( RL + Ra ) .
It is difficult to know exactly what the value of µ actually is because it varies because µ = gm x Ra and both the latter change for a given value of Ia. But we can say that after perusing the tube data sheets that gm = approx 2.8mA/V and Ra = approx 400k.
So µ = 1,120
So the actual gain of the pentode = 1,120 x 48.4 / ( 48.4 + 400 ) = 121.
This is close to the above measurements in the schematic.

The tetrode output stage gain is approximately inversely proportional to the RL. If the gain sags as a result of a lower RL then the output anode voltage will reduce and the effective load seen by the driver pentode stage will increase in value, since 470 k will be divided by a lower ( gain+1) total.
The amp has RLa-a = 6k, and if RL = 3k, then the anode signal voltage would fall from 80V to about 40V for a given
grid input signal so there would be 49.8V across 470k so the I flow in the 470k = 0.106mA, and the load appearing
at the pentode = 9.8 / 0.106 = 92.5k.
When this occurs, the driver stage gain rises because its gain = gm x RL, approximately, and more drive voltage is automatically applied to the output stage grids via the low output resistance of the triode's cathode above the pentode drivers.  The output impedance from the cathode of the triode is quite low, and the following 6L6 grid bias resistors have little effect on the pentode driver gain, due to the buffering effect of the triode.

This is important, since the circuit operation depends on having a highest possible
output resistance of the driver tube, which is the opposite of most conventional ideas which say one should only use low mu triodes to drive output stages. The arrangement of the circuit is negative voltage shunt feedback. To make the most from the
available NFB the value of R1 must be kept as high as possible and the value of R2 as low as possible without
causing the driver tubes to overload.
 
Beta, ß, the fraction of the output voltage fed back, is calculated by working out the R1 and R2 resistance arms of the basic shunt feed arrangement used in all shunt feedback circuits.
The R1 from input to the grids of the output stage include the effective anode resistance of the driver pentode in parallel with the triode current source impedance and any other loads on the pentode except the NFB resistance R2 which is 470k in this case.
The effective Ra' of the pentode = [ ( µ + 1 ) x Rk ] + Ra = [ ( 1,121 +1 x 2.2k ) + 400k ] = 2.86Mohms.

In this case, R1 is equal to the triode dynamic impedance of 800k // 3.86M = 625k.

Output stage gain = 80Va / 9.5Vg = 8.42 

Distortion reduction factor , Drf  =                1             =   0.172
                                                     1 + ( 8.42 x 0.57 ) 

Thus distortion of 3% without any NFB will be reduced to 0.517% with shunt NFB.

An even greater reduction of output resistance/impedance, ie, anode to anode resistance
of the 6L6 output tubes is available. Without NFB Ra for one 6L6 = approx 35k, and µ = 200 approx.

Ra', after FB,   =      Ra____         =           35,000           =  304ohms. (This is less than 6L6 strapped as a triode = 1.6k.)
                              1 + ( µ x ß )           1 + ( 200 x 0.57 )

Now the OPT impedance ratio is 6 k to 8 ohms, which is 750 to 1, so at the output we would see the
Rout = ( Ra-a / ZR ) + total P and S winding resistance for the OPT = ( 2 x 304 / 750 ) + Rw
= 0.81ohms + approx 0.6 = approx 1.4ohms.
The test amp had its total Rout = approx 1.3 ohms.

From the above calculations, any way we could increase the gain of the output stage, and increase the value of ß by having R1 as a higher value would improve the results considerably. In practice, getting B up to about 0.5 is about all we could manage,
and using higher gain output pentodes, such as EL34 or EL84, would make the circuit produce better measurements.

Basically the output stage operates as an anode follower. A fuller explanation of the principle is in the RDH4, pages 332 to 334, although they don't show the use of the mu-follower circuit anywhere. The circuit also works well when beam power output tubes are used in UL, but any output stage could be drafted into this way of applied feedback.
The amount of NFB applied = 20 log ( 1 / Drf ) =  is about 15 dB in this circuit. The input pentodes and driver triodes
can be separate tubes such as 6AU6, 6BX6, 6EJ7 plus 12AT7 or other suitable triodes. A separate heater supply biased at about 200 v+ for the follower part of the pentode & triode series arrangement.

To get the best fromj the circuit the OPT should have low winding resistance, the output stage should work in nearly all class A, and the CT of the OPT should be provided with a very well smoothed B+ voltage.

Where is any advantage to using balanced shunt NFB?  I just cannot see any for an output stage but it is an interesting exercize to consider.
Back in 1996 when I dreamed up this schematic and tried it out on an old amp chassis it was in response to
Allen Wright's and Joe Rasmussen's secret ideas of Forced Symmetry which used a combination of balanced shunt NFB
and balanced series voltage NFB from the OPT anode connections back to various points of the input/driver stage
which I suspect consisted of a long tail pair using two arms of j-fets and triodes in cascode which make
the equivalent of a pentode.
I recall discussing the whole idea of Forced Symmetry with Joe in the editions of the news letter of the Audiophile Society
of NSW, A.S.O.N and I wasn't in agreement with Joe at all. The Allen and Joe empire moved on to other
schemes to make their PP amps palatable to those who thought single ended amps were the best.

But please try things for yourself before deciding what is best.

    ______________________________________________________________________________________________

Balanced shunt FB for 100W amp.

I have not built this amp, and I doubt I ever will because it is too complex. I publish it here to show
where the idea for balanced shunt FB could be applied.
_______________________________________________________________________________________________

Balanced screen FB.

Automatic servo bias control SE amp.
This schematic shows a basic idea for active servo bias
control for a single output tube.

DC current in the tube generates a voltage at R3.
The signal volages are filtered out by R5/C2 LPF and the
DC voltage is applied to one the base of Q1.
The base of Q2 has a reference voltage of about 0.6V
applied from the voltage divider R8, R8, VR1.
When set up VR1 is adjusted to get the wanted bias current
in R3 10ohms. Once the DC idle current is set, and changes to the cathode current  in R3 will alter the voltage at R3 and the Q1 base voltage. If Ik rises, then the voltage at Q1
collector  will be driven more negative thus opposing any rise in cathode voltage. There is enough differential dc voltage gain in the Q1 / Q2 LTP to keep the idle current in the tube constant once it is set.

This type of circuit can be developed to be used in a PP
output stage by having a slightly more elaborate LTP arrangement which i found worked extremely well to control
the balance of dc current.
Unfortunatly when even a small amount of global NFB is applied the circuit became unstable at low frequencies and
I decided the complexity wasn't worth the benefits so this is another fine idea I have never actually used.
perhaps someone else might get it working better but I would prefer the simple RC cathode biasing circuit which regulates
the bias current very adequately in class A amps. Sure there is some wasted power dissipation in the cathode R
but the simplicity is worth the wasted heat.
This type of bias control is useless if used for class AB amps if it is used to try to set the actual bias levels because the cathode current changes and you don't want the grid bias to change in class AB / B amps.
But where the LTP is set up to just balance the bias which would otherwise stay fixed at the adjusted value then the LTP does try to balance the DC cathode currents. Trouble is when NFB is applied and such circuits become de-facto phase shift oscillators.
______________________________________________________________________________________________

Error correction in standard class A Ultralinear 30 watt amplifier.

Considerable argument errupted at rec.audio.tubes when I published this schematic at alt.binaries.schematics.electronic.

They all said it was just global NFB being applied in the same old way, but they are all wrong.
The class A amplifier consits of a standard UL amp with KT88 (etc) and driven by an LTP. All the ac signal voltages generated
are shown with 13.7Vrms into 6 ohms at the output giving 31 watts.
The differential input to V2 & V3 is +3.4Vrms, so the overall voltage gain = 13.7 / 3.4 = 4.029.
The ac voltage shown have their phase polarity shown as + or - for the instanteneous wave form condition.

The +input voltage is applied directly to the non inverting input port at the grid of V2.
It is also applied to the input grid of V1.
One would think that since there is a CCS dc supply to V1 anode, a healthy -voltage would be generated at the anode.
But V1 is set up so there can be as little as possible output signal from its anode. There is a 22k resistor, R9 cap coupled
to the output of the OPT secondary where the speaker connects. The Volatge at the output is +13.7V~,
so +0.622 mA~ of current flows from the output to the anode of V1.
The V has the characteristics of µ = 20, Ra = 5k for the 2 parallel sections, and gm = 4mA/V
So since the anode output voltage is 0.0V~, there must be a voltage required at Vg-k to generate a current of -0.622 mA
to oppose the current in R9 so no anode voltage signal appears.
This Vg-k required = I / gm = 0.622 / 4.0 = 0.155V~.
So where we have +3.4V~ applied to the grid there must be 3.4V - 0.155V = +3.244V~ at the cathode, and since the ac Ia = 0.622mA, then Rk must be V / I = 3.244 / 0.622 = 5.21k.
To bias the V1 properly for about 5mA of idle current and have a dc anode voltage of about +120Vdc,
Rk must be divided into 2 resistors to derive a biasing voltage of about -Vdc for the grid hence the Rk shown above
is divided into R6 = 1k and R7 = 4.2k.

Thus far we have calculated the values for the Rk to get zero signal voltage output from V1 anode.
In practice the R7 value would be adjusted for minimum signal voltage appearing at V1 anode when the load at the output is about the middle value of load to be connected, and I have chosen 6 ohms because it is about 1/2 way between 4 and 8 ohms.

So while there is 0.0V~ appearing at the V1 anode there is also no signal voltage appearing at V4 grid.
Therefore the 3.4V~ input signal causes the full output to appear.

The V1 anode output resistance = [ ( µ + 1 ) x Rk ] + Ra = [ ( 20 + 1 ) x 5.2k ] + 5k = 114.2k.

This resistance acts with the 22k R9 to form two arms of a shunt NFB network, and ß = 114k / ( 22k + 114k ) = 0.84.
If a distortion voltage +Vd~ appears at the output, then it is divided by V1 Rout and R9 to give +0.84Vd~
at the anode of v1 and hence is applied to the inverting port of the amp which is the grid of V3.
It is amplified by the gain of the amp so distortion correction voltage at the output is 0.84 x 4.029 = -3.38Vd~.
This subtracts from the distortion voltage of +4.438V~ which is would be present with no error correction connected
or if one simply grounded V3 grid. The distortion is thus reduced by a factor of Vd / 4.438Vd = 0.228
which is an applied error correction of 12.8 dB.

 In math terms, the distortion reduction factor  = 1 / [ ( 1 + ( A x ß ) ]  where A is the open loop gain without
error correction or NFB connected.
So in this case Drf = 1 / [1 + ( 4.029 x 0.84 ) ] = 0.228 which is the same as the ready reckoning we worked through above.

The effect of the error correction network on the output resistance of the amp is the same as with a loop of NFB.
In this case the Rout without NFB would be 8 ohms, and the transformed µ' of the output tubes at the output is approx 0.7,
so Rout' after error cirrection is applied = Rout  / [1 + ( A x µ' x ß ) ] where A is the gain of the driver stage.
In this case Rout' =  8 / [ 1 + ( 16 x 0.7 x 0.84 ) ] = 0.77 ohms. I have neglected winding resistance of the OPT
and in practice Rout would be about 1 ohm.

So the error correction works to corect distortion and lower output resistance in exactly the same way as NFB when considering the math of how it all happens. But there is no feedback signal voltage applied to the amp formed by
V2, V3, V4 and V5. By no feedback signal I mean a proportion of the wanted undistorted output signal.
All normal FB amplifiers have a portion of their output signal ß x Vout fed back and included in this signal
is a portion of the distortion at the output = Vdn x ß. 

In a NFB amp the total applied input signal must be ( Vout / open loop gain ) plus Vout x ß.
If the above amplifier were to be set up as a normal NFB amp and not use V1 at all then
to get the same distortion reduction factor we would have to apply 11.5V~ of feedback signal to V3 grid
and 14.9V~ to 2 grid.
Of course in practice this would not be done and V1 would be utilised as a traditional gain tube ahead
of V2&V3 and so a smaller feedback voltage and smaller open loop voltage is used since the open loop gain would be
much higher for the amp.
The error correction methode involves less gain tubes between input and output and employs the error correction amp
as an additional active amp to provide an error correction voltage to be applied without any wanted signal voltage to the
main amp. It cannot be any worse than having the extra gain tube in a conventional tube line up.

The above amp is a bit insensitive with 3.4Vrms needed for full power. But V2&V3 could be 12AT7,
or a couple of  6BX6 or 6EJ7 strapped as triodes for some remarkable performance.
V1 could be a variety of triodes such as 6DJ8, 12AT7 etc.
V1 does not have to ever produce much output signal voltage and the values of Rk ensure whatever voltage is
produced is at low thd because of the local current NFB acting with the Rk involved.

Time constants and stability issues may need additional phase shift and gain tailoring networks to be applied as in a conventional
amp with global NFB 

Perhaps it sounds better with a test signal by Beethoven, but to really find out, build the circuit.
_______________________________________________________________

Error Correction in fully balanced PP amp.

This schematic has a singele ended input converted to two oppositely phased signals at the anode and cathode outputs
of V1. Signal voltages are shown as +V or -V to indicate the phases, and instead of having a balanced arangement of the
single V1 in the 30W UL amp above there is an arrangement with the cross coupled four triodes to two balanced shunt feedback networks shown as R1A & R2A, and R2B & R2B to give very little signal output voltage at the resistor junctions.
At the junctions there is only the correction voltage, ie, ß x Vout where Vout is the anode signal at each side of the output PP circuit.
The schematic is a challenge for anyone to build, as well as to understand. Rather than say exactly how this works and
put everyone to sleep, I leave it to the few wide awake mortals with enquiring minds to work it all out like I did.

There are many circuits for amplifiers which could be employed but nearly all those like this last one are more difficult to understand and are more complex and have a few more triodes, R and C components than simpler circuit topologies which
have already been proven to work flawlessy. So thus it would be difficult to justify the extra complexity and the cost.
Some makers such as McIntosh and ARC do have more complexity in their circuits than I would care to use.
These makers with very long established reputations are under no threat from minor designers like myself who would challenge the validity of complexity where I see no need for it; I achieve splendid subjective fidelity and quite good enough
thd/imd measurements.
_____________________________________________________________________________________________

Please excuse the copy of the handwritten work book page I did back in about 2004 when I measured the THD for
a line stage. The conclusions were:-
All triode types and configurations were acceptable.
6CG7 had less THD than the present fashionable flavour tube, the 6H30.

And BTW, my client thought the 6CG7 sounded better. I like 6CG7.
_____________________________________________________________________________________________

More simple preamps

Here are a couple of simple line stages. Both 1 and 2 are inverting amplifiers, ie, the phase of the output is 180degrees
to the input. Thousands of  preamps have been made with the above recipes.

 
3 more line stages for examination. No3 is inverting, but types 4 and 5 are non inverting.
_____________________________________________________________________________________________

This inverting SET preamp has constant current sources for the dc supply to both gain triode and cathode follower.
There is also shunt NFB around the input gain stage to reduce its gain  from being too high.
The gain stage can be switched out of the circuit when no gain is needed.
12AU7, 6DJ8, 6SN7 are all other suitable twin triodes.
______________________________________________________________________________________________