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MISCELLANEOUS  SCHEMATICS
2

DIY
and other interested armchair experts might well wonder what

sort of schematics I have tried just for fun or to see if they would hold

promise because of their clever novelty or special sonic
characteristics.
WARNING!

TUBE
AMP CIRCUITRY REQUIRES EXTREME CARE

WITH
POTENTIALLY LETHAL VOLTAGES.

YOU HAVE BEEN WARNED.

Please
excuse the hand drawn circuit presentation of schematics and

large file size.

Basic
Balanced Shunt
FB
2 x 6L6GC
Balanced Shunt FB  100W

4 x 6550 per channel
Automatic servo bias
control
1 x KT88 etc.
Error correction standard UL amp
2 x KT88 etc.

Error correction fully balanced
2 x KT88 etc.
Simple
line preamps 1 and
2.
2 triodes.
Simple line preamps 3, 4 and 5.
2 and 3
triodes.
Line preamp with switched gain and
CCS      2 triodes.

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 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.

and
ß, the fraction fed back
=
625          = 0.57

625 + 470

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
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
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 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.

________________________________________________________________

Balanced
screen FB.

This
schematic was used for awhile for an 8585 prototype but abandoned for

the circuit now shown at 8585
Integrated Amp page.
But
the balanced screen FB schematic worked fabulously well technically but
just
didn't sound as later pages for the 8585.
Maybe
someone else could get it to sound better.
The principle of operation
is simple enough.
The OPT has cathode FB windings where 12.5% of the
primary windings are devoted to
the cathode circuit of the
output tubes. So there is about 8dB of local NFB applied in the

output stage.
The driver input stage is a pair of EL84 and are
working in an LTP and in pentode mode and
their dc supply is
via a choke with CT to allow a high anode current without loading down
the
EL84 with a low value anode resistance load.
The EL84
screens are not each bypassed to the cathodes of the EL84 but bypassed to
the
cathode feedback windings of the OPT so that the voltage applied
to the EL84 screens is
about half the anode  voltage
output and in effect there is approximately 12dB of series

voltage NFB applied back to the EL84 but to the screens  instead of
FB to the cathodes
via a low impedance resistance divider
network.
Hence the total applied FB is about 20dB and thd and Rout is
nice and low.
But I later rigged the same amp with the
schematic shown on the 2006 8585 page and
I thought the
detail and fidelity was better. The customer who owned the amp I fiddled

with at the time agreed with me that my more conservative
modification with global NFB
from OPT secondary to an SET input
triode gave far better imaging which he said
was 3
dimensional. The use of EL84 with choke loading in an LTP is a feature of
all
such amplifiers I now make and in triode the EL84 is a very
accurate driver.

__________________________________________________________________________

Automatic
servo bias control SE amp.
Thisschematic
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
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 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.

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.

______________________________________________________________________________

Error correction in standard
class A Ultralinear 30 watt amplifier.

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
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
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.

_________________________________________________________________________________

Error 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.

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 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

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 flavor
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.

______________________________________________________________

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