Please excuse the hand drawn
circuit presentation of schematics and
large file size.
Very Basic Balanced Shunt
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 analyze 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 k-ohms.
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 parallel 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 =
Distortion reduction factor , Drf
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____
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 ß 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
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 from 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 exercise 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.
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
at R3. The signal voltages 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
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.
Unfortunately 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 such circuits become de-facto phase shift oscillators.
correction in standard class A Ultralinear 30 watt
Considerable argument erupted 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
were all wrong. The class A amplifier consists 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 instantaneous
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
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 correction 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 correct
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 utilized 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 method 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.
Correction in fully balanced PP amp.
This schematic has a singe 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.
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 flawlessly. 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
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 flavor tube, the 6H30.
BTW, my client thought the 6CG7 sounded better. I
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.
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.