[Patrick Turner]

In some recent articles I posted here I showed that the formulas I
thought were OK simply
were not OK for use with shunt FB around an inverting amp.

After a night's sleep, I speny an hour untangling the mystery
of shunt NFB formulas that don'r work properly, and came up with this
one....


Gain with shunt feedback applied, or closed loop gain,


A' = ( A - Axß ) / ( Axß + 1 )


Where ß = R1 / ( R1 + R2 ),
A is open loop gain measured when FB network is connected.
A' is closed loop gain.

Let's see if it works, to prove this is right.

Example 1, using triode with A = 20.

Let us suppose this is a 1/2 x 6DJ8 Ia = 5mA, so Ra = 5k, µ =33, RL = 8k
approx.
R1 = R2 = 210k , so ß = R1 / ( R1 + R2 ) = 0.5.

This set up is an "anode follower".

A' = ( 20 - 20x0.5 ) / ( 20x0.5 +1 )
= 10 / ( 10 + 1 )
= 10 / 11, which is what we would measure,
ie, -11V input gives +10V output.


Example 2, using triode with gain = 20.
R1 = 70k, R2 = 210k, so ß = 70 / ( 70 + 210 ) = 0.25.

A' = ( 20 - 20x0.25 ) / ( 20x0.25 +1 )
= 15 / 6
= 2.5, which agrees with what we would measure,
because for -4V input, we would get +10V output.

How I derived the correct formula is my secret, and involves several
pages of trying to epress gain in terms of A, R1 and R2.

But since ß = R1 / ( R1 + R2 ), then if we let R1 = 1, then ß = 1 / 1 +
R2.

So finally by a process of schoolboy agebraic elimination of unwanted
terms I ended
up with an equation for A' that involved ONLY ß, A and 1.0, and without
having to know values of R1 or R2, but just knowing ß.

I am not sure of the equation for the output resistance, ie, effective
Ra, or Ra'
after NFB has been applied.

But Ra' depends on µ, so perhaps we might say

Ra' = Ra / ( 1 + µxß ).

Consider example 1, the classic anode follower.

This would mean Ra' = 5k / 1 + 33 x 0.5

= 5k / 17.5 = 284 ohms approx.

To check this you need to adjust the value of RL to
change the gain.

Rout = change in anode output voltage / change in load current.

The A for varying loads can be easily we worked out from A = µ x RL / (
RL + Ra ),
correct for all tubes.


Patrick Turner.

[Patrick Turner]

In Addition to what I said below yesterday,
some precautions need to be given when measuring the voltages outcome of
shunt NFB when tried say between an output tube and input tube.

There are two arms of the NFB resistance divider, R1, and R2.
R2 is anode to grid, and R1 is grid to input.

It is important that the output resistance of the input tube be taken
into account.
This R is the Ra of the driving tube, its dc carrying RL, and a grid
bias R
for the output tube grid straight after the coupling cap between two
tubes.
There will also be a cap between output anode and R2.

Suppose you have 1/2 a 6SN7 with Ia = 4 mA, then dc carrying 50k, and
output tube biasing R of
200k after a 2.2uF coupling cap. Assume the 6SN7 is fully bypassed.
Ra of the 6SN7 will be about 10k, so Rout of the whole input stage
is 10k // 200k // 50k //, or about 8k.

This 8k acts in series with R1, and IS REALLY PART OF R1.

But say one has 32k between input tube and output grid, then R1 is 40k.

Suppose we would want ß = 0.1.

Therefore 0.1 = 40k / ( 40k + R2 ), so R2 = 9 x 40k = 360k.

Now the 8k od Rout for V1 is like an equivalent resistance in series
with an imaginery
voltage source.

The model of the input tube needs to be considered, a voltage generator
producing an output voltage of µ x Vg, with a series Ra = 10k, and at
the output of the 10k
there is the 50k and 200k shunting the output to fixed voltage points.
Hence Rout = 8k.

We do not need to know the gain conditions of V1, although its load
should not be too low.
We have 50k, 200k and the 32k to power into, and of course the 32k
doesn't
appear as 32k, it appears as a higher R because there is similarly
phased voltages at each
end of this R.

But we do have V1 RL = approx 20k, rather low if we wish to swing about
30Vrms
to drive an output tube with low THD.

Now when we observe the working voltages in a real amp with two cascaded
tubes
the measurement of input voltage is impossible because its
actually at the "front end" of ther imaginery 8k series Rout we
calculated above.

Therefore, for all **accurate** observations of the NFB voltages that we
might wish to
view, don't use a driver tube at all for experiments, use a low Rout
signal gene,
or something with negligible Rout of less than 100 ohms preferably.

Then what you measure will adhere to what you calculate in the formulas
I have given below.

However, in the real world, we'd know the concept, and would not worry
too much about
exact measuring, and just apply the shunt FB with approximately the
right
R values and put up with the result of much lower THD of the amp,
wider bandwidth, and lower Rout.
In fact, with ß = 0.1 approximately, the Ra of 12k for an EL34 in
pentode or 18k of Ra of a
KT88 in beam tetrode mode is reduced to below 1k in both cases, and
below triode Ra.
Class A THD will have more artifacts than triode connection of the
multigrids, but the
levels of THD will be low enough. THD will be reduced by factors between
1/3 and 1/5.

There will be those who'll say the multigrids with such FB won't sound
as well as the triodes.

The argument has gone on since 1933, and still no consensus.
The multigrids will produce nearly twice the PO though, where the THD
will be still under what the triode does at half the multigrid.
In othe words, a KT88 with shunt FB where ß = 0.1 will have
less THD than a triode at 5 watts

Where a driver tube is used to power the low input resistance of a
typical shunt
FB arrangement, careful considerations need to be given, like using a
µ-follower
drive stage.

R1 can be increased in value to make ß effectively much higher
by using a CCS dc load for V1 and not bypassing V1 anode,
But you'll then need more input voltage to V1, so there are
no free lunches with shunt NFB.

I investigated shunt FB for serious power amps and the schematics are
shown at

http://www.turneraudio.com.au/miscel...hematics2.html

See the full explantions there.

I have not stuck with shunt FB for power amps.
The overload behaviour is awful! But the real reason for not using
the shunt FB was because I found conventional series voltage global NFB
sounded better
when you have series voltage NFB in the output stage with ß = between
0.1 and up to 0.33,
using dedicated transformer widings al la the Acoustical Quad-II method,
only all done better than Quad ever would.

However, I have not tried all possibilities, so anyone is free to try.

Patrick Turner.




Patrick Turner wrote:

In some recent articles I posted here I showed that the formulas I
thought were OK simply
were not OK for use with shunt FB around an inverting amp.

After a night's sleep, I speny an hour untangling the mystery
of shunt NFB formulas that don'r work properly, and came up with this
one....

Gain with shunt feedback applied, or closed loop gain,

A' = ( A - Axß ) / ( Axß + 1 )

Where ß = R1 / ( R1 + R2 ),
A is open loop gain measured when FB network is connected.
A' is closed loop gain.

Let's see if it works, to prove this is right.

Example 1, using triode with A = 20.

Let us suppose this is a 1/2 x 6DJ8 Ia = 5mA, so Ra = 5k, µ =33, RL = 8k
approx.
R1 = R2 = 210k , so ß = R1 / ( R1 + R2 ) = 0.5.

This set up is an "anode follower".

A' = ( 20 - 20x0.5 ) / ( 20x0.5 +1 )
= 10 / ( 10 + 1 )
= 10 / 11, which is what we would measure,
ie, -11V input gives +10V output.

Example 2, using triode with gain = 20.
R1 = 70k, R2 = 210k, so ß = 70 / ( 70 + 210 ) = 0.25.

A' = ( 20 - 20x0.25 ) / ( 20x0.25 +1 )
= 15 / 6
= 2.5, which agrees with what we would measure,
because for -4V input, we would get +10V output.

How I derived the correct formula is my secret, and involves several
pages of trying to epress gain in terms of A, R1 and R2.

But since ß = R1 / ( R1 + R2 ), then if we let R1 = 1, then ß = 1 / 1 +
R2.

So finally by a process of schoolboy agebraic elimination of unwanted
terms I ended
up with an equation for A' that involved ONLY ß, A and 1.0, and without
having to know values of R1 or R2, but just knowing ß.

I am not sure of the equation for the output resistance, ie, effective
Ra, or Ra'
after NFB has been applied.

But Ra' depends on µ, so perhaps we might say

Ra' = Ra / ( 1 + µxß ).

Consider example 1, the classic anode follower.

This would mean Ra' = 5k / 1 + 33 x 0.5

= 5k / 17.5 = 284 ohms approx.

To check this you need to adjust the value of RL to
change the gain.

Rout = change in anode output voltage / change in load current.

The A for varying loads can be easily we worked out from
A = µ x RL / ( RL + Ra ),
correct for all tubes.

Patrick Turner.

[Ian Thompson-Bell]

Patrick Turner wrote:
In Addition to what I said below yesterday,
some precautions need to be given when measuring the voltages outcome of
shunt NFB when tried say between an output tube and input tube.

There are two arms of the NFB resistance divider, R1, and R2.
R2 is anode to grid, and R1 is grid to input.

It is important that the output resistance of the input tube be taken
into account.
This R is the Ra of the driving tube, its dc carrying RL, and a grid
bias R
for the output tube grid straight after the coupling cap between two
tubes.
There will also be a cap between output anode and R2.

Suppose you have 1/2 a 6SN7 with Ia = 4 mA, then dc carrying 50k, and
output tube biasing R of
200k after a 2.2uF coupling cap. Assume the 6SN7 is fully bypassed.
Ra of the 6SN7 will be about 10k, so Rout of the whole input stage
is 10k // 200k // 50k //, or about 8k.

This 8k acts in series with R1, and IS REALLY PART OF R1.


This is basically what is says in RDH i.e that the source resistance of
the signal driving the input arm needs to be included in R1.

An alternate way of looking at it is that R1 is the input resistance of
the shunt/shunt NFB stage and thus the input signal is reduced by
Rout/(Rout +R1) where Rout is the source resistance of the driving
stage. This is something you should do anyway when calcualting overall gain.


But say one has 32k between input tube and output grid, then R1 is 40k.

Suppose we would want ß = 0.1.

Therefore 0.1 = 40k / ( 40k + R2 ), so R2 = 9 x 40k = 360k.

Now the 8k od Rout for V1 is like an equivalent resistance in series
with an imaginery
voltage source.

The model of the input tube needs to be considered, a voltage generator
producing an output voltage of µ x Vg, with a series Ra = 10k, and at
the output of the 10k
there is the 50k and 200k shunting the output to fixed voltage points.
Hence Rout = 8k.

We do not need to know the gain conditions of V1, although its load
should not be too low.
We have 50k, 200k and the 32k to power into, and of course the 32k
doesn't
appear as 32k, it appears as a higher R because there is similarly
phased voltages at each
end of this R.

But we do have V1 RL = approx 20k, rather low if we wish to swing about
30Vrms
to drive an output tube with low THD.

Now when we observe the working voltages in a real amp with two cascaded
tubes
the measurement of input voltage is impossible because its
actually at the "front end" of ther imaginery 8k series Rout we
calculated above.

Therefore, for all **accurate** observations of the NFB voltages that we
might wish to
view, don't use a driver tube at all for experiments, use a low Rout
signal gene,
or something with negligible Rout of less than 100 ohms preferably.


Good advice.

Cheers

Ian