[Alex]

Hello,

What is the *official* definition of "screen grid mu" or "mu g1g2" of a
pentode?

Obviously it is - dVg2/dVg1, but under which conditions:
- triode connected (Va=Vg2);
- Ig2=const; Va=const (as Ian Bell suggested);
- Ic=const; Va=const;
- anything else?

Measured under these different conditions, screen grid mu values will be
most likely quite close. (The question is rather of an academic then
practical interest.)

I think mu g1g2 is a triode-connected mu, but why then it is usually not
called so?

Regards,
Alex

[Ian Iveson]

Alex wrote:

What is the *official* definition of "screen grid mu" or
"mu g1g2" of a pentode?

Obviously it is - dVg2/dVg1, but under which conditions:
- triode connected (Va=Vg2);
- Ig2=const; Va=const (as Ian Bell suggested);
- Ic=const; Va=const;
- anything else?

Measured under these different conditions, screen grid mu
values will be most likely quite close. (The question is
rather of an academic then practical interest.)

I think mu g1g2 is a triode-connected mu, but why then it
is usually not called so?

Seems the last guru switched the lights out.

Where's John? He was the only old bugger, er, guru, left :-(

If it's measured with the anode at constant voltage, which
would be sensible, then there may be some difference between
it and triode mu, where the anode voltage varies. The anode
voltage of a pentode has not much effect on current beyond
the knee, but it does have some, so I would expect screen mu
to be slightly different from triode mu.

In what direction would the difference be?

But this is guesswork. Maybe if we keep asking...

Ian

[glenbadd]

On Sun, 10 Jan 2010 21:18:47 -0800, "Alex"
wrote:

Hello,

What is the *official* definition of "screen grid mu" or "mu g1g2" of a
pentode?

Obviously it is - dVg2/dVg1, but under which conditions:
- triode connected (Va=Vg2);
- Ig2=const; Va=const (as Ian Bell suggested);
- Ic=const; Va=const;
- anything else?
I think mu g1g2 is a triode-connected mu,
but why then it is usually not
called so?

http://www.r-type.org/addtext/add002.htm
contains
"Tetrodes and pentodes do not really have definable ì values, except
when operated as triodes. Attempts to measure ì in pentodes tend to
yield very high values (eg. 1000s) which vary widely according to the
method of measurement. Nevertheless, occasionally one sees ì values
quoted for certain tetrode and pentode types. More often quoted,
however, is the inner amplification factor (ì g1-g2) which relates to
the grid/screen spacing but not to the anode. Typical values for mu g1-
g2 lie between 5 and 10 for the majority of common valve types. As
with triode ì values, ì g1-g2 remains pretty constant for a given
valve under all conditions of use. "

HTH, Glenn.

[Ian Iveson]

Alex

Per RDH4, (any) mu "is the ratio of the incremental
change in any one
electrode voltage to the incremental change in any other
electrode
voltage, under the conditions that a specified current
remains
unchanged and that all other electrode voltages are
maintained
constant."

For screen mu the "specified current" is Ia.

So screen mu = dVg2/dVg1 @ Va = const and Ia = const.

Is it what you mean?

No, Va constant and Iscreen constant. A proof-reading
slipup, I guess. Only one electrode is held at constant
current according to RDH4. Clearly if the screen were not
that one, then it would be at constant voltage, which
wouldn't be any good.

Ian

[Ian Iveson]

Come to think of it, if you have mu g1.a, and mu g1.g2, you
should know everything, surely?

Erm, what about:

Let mug1a = x
Let mug1s = y
Then musa = x/y
and muas = y/x

Let triodemu = w

Then w = y + y.w/x

So w = x.y/(x-y)

That is:

Triode mu = anode mu * screen mu / (anode mu - screen mu)

Wadya think?

Ian

[Ian Iveson]

"flipper" wrote in message
...
On Wed, 13 Jan 2010 11:47:10 -0000, "Ian Iveson"
wrote:

Alex

Per RDH4, (any) mu "is the ratio of the incremental
change in any one
electrode voltage to the incremental change in any
other
electrode
voltage, under the conditions that a specified current
remains
unchanged and that all other electrode voltages are
maintained
constant."

For screen mu the "specified current" is Ia.

So screen mu = dVg2/dVg1 @ Va = const and Ia = const.

Is it what you mean?

No, Va constant and Iscreen constant. A proof-reading
slipup, I guess. Only one electrode is held at constant
current according to RDH4.

The specified current is Ia.

Clearly if the screen were not
that one, then it would be at constant voltage,

That isn't clear at all, just as when measuring g1 mu g1
current can
change, as in positive grid drive being an obvious case.
That in no
way implies it's at a "constant voltage.'.

***Only*** one electrode is at constant current, it says.
****Others are at constant voltage**** it says.

I'm afraid you've lost the plot entirely. For measuring
screen mu, the grid voltage is varied, and the consequent
change in screen voltage is measured. Constant screen
voltage is no good for that. Unfortunately RDH uses the word
"other" ambiguously. It's up to us to find the sensible
interpretation. For measuring mu from a to b, the voltage at
a is varied, b is held constant current, and other
electrodes are held at constant voltage.

That last part...other electrodes are held constant at
voltage...answers Alex's question. There is only one
plausible interpretation.

Ian

[Alex]

"flipper" wrote in message
...
On Sun, 10 Jan 2010 21:18:47 -0800, "Alex"
wrote:

Hello,

What is the *official* definition of "screen grid mu" or "mu g1g2" of a
pentode?

Obviously it is - dVg2/dVg1, but under which conditions:
- triode connected (Va=Vg2);
- Ig2=const; Va=const (as Ian Bell suggested);
- Ic=const; Va=const;
- anything else?

Per RDH4, (any) mu "is the ratio of the incremental change in any one
electrode voltage to the incremental change in any other electrode
voltage, under the conditions that a specified current remains
unchanged and that all other electrode voltages are maintained
constant."

For screen mu the "specified current" is Ia.

So screen mu = dVg2/dVg1 @ Va = const and Ia = const.

Is it what you mean?

[Ian Iveson]

Alex wrote:

Come to think of it, if you have mu g1.a, and mu g1.g2,
you should know everything, surely?

Erm, what about:

Let mug1a = x
Let mug1s = y
Then musa = x/y
and muas = y/x

Let triodemu = w

Then w = y + y.w/x

So w = x.y/(x-y)

That is:

Triode mu = anode mu * screen mu / (anode mu - screen mu)

Wadya think?

Perhaps
triode mu = x * y / (x + y) ?

And it will only be true (or rather might be true)
ignoring current redistribution effect, in other words
assuming that the ratio Ig2/Ia = constant and not subject
to modulation with Va and or Vg2 changing.

That assumption isn't made anywhere in my maths, so if it's
a problem it would be because it's necessary in order for
the principle of superposition to apply. I don't see how
though, if the two mus are measured using a very small
incremental change in Vg1, and if electrode starting
voltages are the same for each of the two measurements. As
the increment approaches zero, can that current distribution
be taken to be constant? And if not, how would it's
variation effect the measured voltages such that the maths
becomes suspect?

I was quite proud of my bit of algebra :-(

My derivation seems correct to me...on the assumption that
the given values of mu apply to the same operating
conditions.

Take your pentode with anode mu = 300 and screen mu = 36.
Anode mu isn't quoted in the datasheet but can be roughly
estimated from given values of transconductance and load.

Then triode mu would be 10800/264 = 41, which ties in nicely
with your original estimate of 40.

Your equation would make triode mu less than screen mu. That
can't be true, surely, because the anode voltage change,
from constant voltage for measuring screen mu, to triode
mode for measuring triode mu, is in the same direction as
the screen voltage change, and so I can't see how mu could
be diminished. My maths simply uses the anode/screen mu to
calculate the effect on the screen voltage of that variation
in anode voltage.

Er, actually my incremental change in anode voltage is much
greater than very small, I see, because it's multiplied by
the triode mu. Neither anode nor screen is subject to either
constant current or constant voltage, but rather share the
same constant current, meaning that the current split is
free to change. I see what you mean and you could be right
but I still don't see exactly how. It's one step too far for
the pentode in my head.

The only people I ever trusted on this kind of stuff were
Alan Douglas and Duncan Munro, and they disappeared years
ago. It's a shame that Duncan's latest triode model
principles never got applied to the pentode because existing
screen models aren't very good so simulation won't prove
anything.

Perhaps we should ask Menno van der Veen? He's quite good at
maths.

Ian

[Alex]

"flipper" wrote in message
...
On Wed, 13 Jan 2010 20:32:04 -0800, "Alex"
wrote:


"flipper" wrote in message
. ..
On Sun, 10 Jan 2010 21:18:47 -0800, "Alex"
wrote:

Hello,

What is the *official* definition of "screen grid mu" or "mu g1g2" of a
pentode?

Obviously it is - dVg2/dVg1, but under which conditions:
- triode connected (Va=Vg2);
- Ig2=const; Va=const (as Ian Bell suggested);
- Ic=const; Va=const;
- anything else?

Per RDH4, (any) mu "is the ratio of the incremental change in any one
electrode voltage to the incremental change in any other electrode
voltage, under the conditions that a specified current remains
unchanged and that all other electrode voltages are maintained
constant."

For screen mu the "specified current" is Ia.

So screen mu = dVg2/dVg1 @ Va = const and Ia = const.

Is it what you mean?


Mathematically, yes.

Well, it is possible to measure screen mu this way. The test circuit will
look like that:
1. A small AC signal is applied to grid 1;
2. A (fixed) resistor is connected from (fixed) +B supply rail to the plate;
3. Negative feedback (may be AC coupled) is applied from the plate to the
screen grid g2 (Unusual, isn't it?) via a non-inverting amplifier with very
high gain.
4. To calculate the mu, measure AC component on g2 and divide by the AC
component on g1.

Now because of the deep NFB, plate voltage will be maintained almost
constant, and because of the fixed plate load resistor, Ia will be constant
as well.

Cathode current will not be constant varying exactly the same way as the g2
current.

A strange test circuit, but theoretically and practically possible.

[Alex]

"Ian Iveson" wrote in message
...
Come to think of it, if you have mu g1.a, and mu g1.g2, you should know
everything, surely?

Erm, what about:

Let mug1a = x
Let mug1s = y
Then musa = x/y
and muas = y/x

Let triodemu = w

Then w = y + y.w/x

So w = x.y/(x-y)

That is:

Triode mu = anode mu * screen mu / (anode mu - screen mu)

Wadya think?

Perhaps
triode mu = x * y / (x + y) ?

And it will only be true (or rather might be true) ignoring current
redistribution effect, in other words assuming that the ratio Ig2/Ia =
constant and not subject to modulation with Va and or Vg2 changing.


Ian

[Ian Iveson]

Alex

I got signs wrong in my original working. Sreen mu and anode
mu are actually both negative, such that my derivation for
mu.s.a should have been of opposite sign to mu.g.a and
mu.g.s, and so our results come out pretty much the same.

i.e. as far as we know, the approximation:

Triode mu = anode mu * screen mu / (anode mu + screen mu)

Is better than the assumption that screen mu and anode mu
are the same!

Checking with a simulated EL84 @ 330V and around 53mA I get:

Anode mu = - 900
Screen mu = - 19
Triode mu = 18.62
Screen/anode mu = 47.5 (notably close to - anode mu / screen
mu)

I recognise that my method makes some dubious assumptions,
but so does yours...more, in fact, one of which at least is
false, and mine's quicker, and needs hardly any data. I
didn't need to know how a valve works to derive it, either.
Most of all, I didn't need any notion of feedback. But then
neither did you, really.

It could be that your's is better, because, taking your A to
be around 0.1 and using your equation gives a better
approximation to the simulation. But then one or more of
your assumptions are wrong, and the simulation could also be
wrong.

However, I think we are on the right track, with you a
little further along, at the expense of much extra work,
more assumptions, and extra data requirements. So it
goes...half the error costs twice as much.

Is it useless knowledge I wonder. How many datasheets give
anode mu for a pentode? How many give both anode mu and
screen mu?

And, btw, you initial figure of 40 for the triode mu of the
valve you originally asked about was miles out :-b

We need someone to measure a real pentode!


Your equation would make triode mu less than screen mu.
That can't be true, surely, because the anode voltage
change, from constant voltage for measuring screen mu,
to triode mode for measuring triode mu, is in the same
direction as the screen voltage change, and so I can't
see how mu could be diminished.

Er, inverting amp, silly me...

In my mind it is quite natural that triode mu is less
than screen mu. Screen mu is defermined by the field ffom
g2 penetrating through g1. In a triode connection both
the field from g2 and the field from the plate (the later
is additionally reduced by g3) penetrate g1. Therefore
there is more "field negative feedback" in triode
connection as opposed to using g2 as the plate and
connectind the real plate to a fixed voltage rail...

Anyway, now when your have changed your definition, I
will try to check the maths as well. Let us agree that:
Plate mu = dVa/dVg1 @ Ia = const; Vg2 = const;
Screen mu = dVg2/dVg1 @ Ig2 = const; Va = const;
Triode mu = dVa/dVg1 @ Ia+Ig2 = const.

Our task is to (try to) express triode mu through Screen
mu and Plate mu.

If we agree on this srarting point, let us redo our
derivations and compare results. My gut feel is, when
unfolding full derivatives into partial derivatives,
there will be undefined term(s) related to current
redistribution. We shall see... Others are welcome to
join too.

So it goes:
No one would deny that very generally plate current
(increment thereof) depends on Vg1, Vg2 and Va. [Indexi
are as follows: 1 -control grid; 2 - screen grid; 3 -
plate (anode))]

dIa = S31 * dVg1 + S32 * dVg2 + S33 * dVa (1)
Here you can recognise:
S31 as a normal transconductance;
S33 -- output conductance (reciprocal of Rp, which bloody
reduces Q of any LC tank in the IF amplifier stage,
etc...);
S32 is a strange beast. It is sort of transconductance if
a screen grid is used as control grid. (I remember some
folks were trying to use horizontal deflection output
tubes in such mode in audio...)

Similarly:
dIg2 = S21 * dVg1 + S22 * dVg2 + S23 * dVa (2)
Here you can recognise:
S21 is a transconductance of the screen grid (normally the
same fraction of transconductance as the fraction of
screen current to plate current);
S22 output (shunting) conductance of screen grid (strange
beast);
S23 - shows how plate voltage affects screen current
(!!!). A strange animal too.

Now let us express triode mu. Note that dIa + dIg2 = const
and dVa = dVg2, so:
0 = dVg1 * (S31 + S21) + dVa * (S32 + S33 + S22 + S23)
Obviously:
screen mu = (S31 + S21) / (S32 + S33 + S22 + S23)
(3)

To calculate screen mu from (3) we need to know six
S-parameters. Can we get all of them from plate mu and
screen mu values? Let us try.

From plate mu definition (dIa = 0 and dVg2 = 0) and (1) we
derive:
S33 = S31 / (plate mu) (4)
(This is a variation of the classical: S * Rp = mu
equation; Rp = 1 / S33, as I mentioned earlier.)

Similarly from screen mu definition (dIg2 = 0 and dVa = 0)
and (2) we derive;
S22 = S21 / (screen mu) (5)

Now substituting (4) and (5) into (3) we get:
Triode mu = (S31 + S21) / (S32 + S31 / (1/(plate mu)) +
S21 / (1/(screen mu)) + S23) (6)

Having eliminated two S-parameters we still need four to
get the answer. Obviously it is NOT POSSIBLE to work out
triode mu only from screen mu and plate mu. However, we
might try to make some assumpltions.
a) Normally we know plate transconductance from datasheets
and can calculate screen transconductance taling the same
pro rata as screen current to plate current. So let us
introduce
A = S21 / S31 or S21 = S31 * A
(6)

Substituting (6) into (5):
Triode mu = (1 + A) / ((S32 / S31) + (1/(plate mu)) +
(1/(screen mu)) + (S23 / S31)) (7)

Still three unknowns, actually, two, because we can easily
get S31 (plate transconductance from the datasheet). So
more assumptions are needed.
b) "S32 / S31" indicates how screen grid affects plate
current compared to control grid. WITH GREAT RESERVATION
we can assume that it is 1 / (screen mu);
c) "S23" indicates how plate voltage affects grid current.
Let us think. If plate voltage goes up... what happens
with screen current? Screen current may go up because
simply cathode current will slightly increase because some
field from the plate manages to penetrate through all
three grids and reach cathode space charge. On the other
hand, screen current may go down, because higher plate
voltage will suck more electrons from the space charge
under the suppressor grid, and less electrons will be
returning to the screen grid. The former effect I would
think is more typical for beam tetrodes, the later -- for
pentodes, especially for "dual control pentodes" with
closely wound suppressor grid, such as 6AS6, for example.
So we can not guess, and let us agree on the middle and
just put S23 = 0.

Assuming screen voltage is constant, screen current falls as
anode voltage rises in every datasheet I've ever seen, of
pentode or beam tetrode. Anode voltage appears to control
the share of current between anode and screen, whereas
screen voltage determines the total current, from a casual
inspection.

Now with these two rough wet finger estimations:
triode mu = (screen mu) * (plate mu) * (1 + A) / ( (plate
mu) * (1 + A) + (screen mu)) (8)

It appears that triode mu is lower than screen mu defined
as -- dVg2 / dVg1 @ dVa = 0; dIg2 = 0, again the it is NOT
CONCLUSIVE because of very wild assumptions made.

So it is not possible to say who was right, me or Ian
Everson, we are both wrong. One thing is definite -- it is
not possible to fully describe triode mu from the screen
mu and the plate mu alone.

Later I might present similar derivations for screen mu
defined as -- dVg2 / dVg1 @ dVa = 0; dVa = 0, as
originally suggested by "Flipper".

That's not what he suggested, and what he suggested was
wrong anyway, so what's the point?

Ian

[Alex]

"Ian Iveson" wrote in message
...
Alex wrote:

Come to think of it, if you have mu g1.a, and mu g1.g2, you should know
everything, surely?

Erm, what about:

Let mug1a = x
Let mug1s = y
Then musa = x/y
and muas = y/x

Let triodemu = w

Then w = y + y.w/x

So w = x.y/(x-y)

That is:

Triode mu = anode mu * screen mu / (anode mu - screen mu)

Wadya think?

Perhaps
triode mu = x * y / (x + y) ?

Your equation would make triode mu less than screen mu. That can't be
true, surely, because the anode voltage change, from constant voltage for
measuring screen mu, to triode mode for measuring triode mu, is in the
same direction as the screen voltage change, and so I can't see how mu
could be diminished.

In my mind it is quite natural that triode mu is less than screen mu. Screen
mu is defermined by the field ffom g2 penetrating through g1. In a triode
connection both the field from g2 and the field from the plate (the later is
additionally reduced by g3) penetrate g1. Therefore there is more "field
negative feedback" in triode connection as opposed to using g2 as the plate
and connectind the real plate to a fixed voltage rail...

Anyway, now when your have changed your definition, I will try to check the
maths as well. Let us agree that:
Plate mu = dVa/dVg1 @ Ia = const; Vg2 = const;
Screen mu = dVg2/dVg1 @ Ig2 = const; Va = const;
Triode mu = dVa/dVg1 @ Ia+Ig2 = const.

Our task is to (try to) express triode mu through Screen mu and Plate mu.

If we agree on this srarting point, let us redo our derivations and compare
results. My gut feel is, when unfolding full derivatives into partial
derivatives, there will be undefined term(s) related to current
redistribution. We shall see... Others are welcome to join too.

Regards,
Alex

[Alex]

"Alex" wrote in message
...

"Ian Iveson" wrote in message
...
Alex wrote:

Come to think of it, if you have mu g1.a, and mu g1.g2, you should know
everything, surely?

Erm, what about:

Let mug1a = x
Let mug1s = y
Then musa = x/y
and muas = y/x

Let triodemu = w

Then w = y + y.w/x

So w = x.y/(x-y)

That is:

Triode mu = anode mu * screen mu / (anode mu - screen mu)

Wadya think?

Perhaps
triode mu = x * y / (x + y) ?

Your equation would make triode mu less than screen mu. That can't be
true, surely, because the anode voltage change, from constant voltage for
measuring screen mu, to triode mode for measuring triode mu, is in the
same direction as the screen voltage change, and so I can't see how mu
could be diminished.

In my mind it is quite natural that triode mu is less than screen mu.
Screen mu is defermined by the field ffom g2 penetrating through g1. In a
triode connection both the field from g2 and the field from the plate (the
later is additionally reduced by g3) penetrate g1. Therefore there is more
"field negative feedback" in triode connection as opposed to using g2 as
the plate and connectind the real plate to a fixed voltage rail...

Anyway, now when your have changed your definition, I will try to check
the maths as well. Let us agree that:
Plate mu = dVa/dVg1 @ Ia = const; Vg2 = const;
Screen mu = dVg2/dVg1 @ Ig2 = const; Va = const;
Triode mu = dVa/dVg1 @ Ia+Ig2 = const.

Our task is to (try to) express triode mu through Screen mu and Plate mu.

If we agree on this srarting point, let us redo our derivations and
compare results. My gut feel is, when unfolding full derivatives into
partial derivatives, there will be undefined term(s) related to current
redistribution. We shall see... Others are welcome to join too.

So it goes:
No one would deny that very generally plate current (increment thereof)
depends on Vg1, Vg2 and Va. [Indexi are as follows: 1 -control grid; 2 -
screen grid; 3 - plate (anode))]

dIa = S31 * dVg1 + S32 * dVg2 + S33 * dVa (1)
Here you can recognise:
S31 as a normal transconductance;
S33 -- output conductance (reciprocal of Rp, which bloody reduces Q of any
LC tank in the IF amplifier stage, etc...);
S32 is a strange beast. It is sort of transconductance if a screen grid is
used as control grid. (I remember some folks were trying to use horizontal
deflection output tubes in such mode in audio...)

Similarly:
dIg2 = S21 * dVg1 + S22 * dVg2 + S23 * dVa (2)
Here you can recognise:
S21 is a transconductance of the screen grid (normally the same fraction of
transconductance as the fraction of screen current to plate current);
S22 output (shunting) conductance of screen grid (strange beast);
S23 - shows how plate voltage affects screen current (!!!). A strange animal
too.

Now let us express triode mu. Note that dIa + dIg2 = const and dVa = dVg2,
so:
0 = dVg1 * (S31 + S21) + dVa * (S32 + S33 + S22 + S23)
Obviously:
screen mu = (S31 + S21) / (S32 + S33 + S22 + S23) (3)

To calculate screen mu from (3) we need to know six S-parameters. Can we get
all of them from plate mu and screen mu values? Let us try.

From plate mu definition (dIa = 0 and dVg2 = 0) and (1) we derive:
S33 = S31 / (plate mu)
(4)
(This is a variation of the classical: S * Rp = mu equation; Rp = 1 / S33,
as I mentioned earlier.)

Similarly from screen mu definition (dIg2 = 0 and dVa = 0) and (2) we
derive;
S22 = S21 / (screen mu)
(5)

Now substituting (4) and (5) into (3) we get:
Triode mu = (S31 + S21) / (S32 + S31 / (1/(plate mu)) + S21 / (1/(screen
mu)) + S23) (6)

Having eliminated two S-parameters we still need four to get the answer.
Obviously it is NOT POSSIBLE to work out triode mu only from screen mu and
plate mu. However, we might try to make some assumpltions.
a) Normally we know plate transconductance from datasheets and can calculate
screen transconductance taling the same pro rata as screen current to plate
current. So let us introduce
A = S21 / S31 or S21 = S31 * A (6)

Substituting (6) into (5):
Triode mu = (1 + A) / ((S32 / S31) + (1/(plate mu)) + (1/(screen mu)) + (S23
/ S31)) (7)

Still three unknowns, actually, two, because we can easily get S31 (plate
transconductance from the datasheet). So more assumptions are needed.
b) "S32 / S31" indicates how screen grid affects plate current compared to
control grid. WITH GREAT RESERVATION we can assume that it is 1 / (screen
mu);
c) "S23" indicates how plate voltage affects grid current. Let us think. If
plate voltage goes up... what happens with screen current? Screen current
may go up because simply cathode current will slightly increase because some
field from the plate manages to penetrate through all three grids and reach
cathode space charge. On the other hand, screen current may go down, because
higher plate voltage will suck more electrons from the space charge under
the suppressor grid, and less electrons will be returning to the screen
grid. The former effect I would think is more typical for beam tetrodes, the
later -- for pentodes, especially for "dual control pentodes" with closely
wound suppressor grid, such as 6AS6, for example.
So we can not guess, and let us agree on the middle and just put S23 = 0.

Now with these two rough wet finger estimations:
triode mu = (screen mu) * (plate mu) * (1 + A) / ( (plate mu) * (1 + A) +
(screen mu)) (8)

It appears that triode mu is lower than screen mu defined as -- dVg2 / dVg1
@ dVa = 0; dIg2 = 0, again the it is NOT CONCLUSIVE because of very wild
assumptions made.

So it is not possible to say who was right, me or Ian Everson, we are both
wrong. One thing is definite -- it is not possible to fully describe triode
mu from the screen mu and the plate mu alone.

Later I might present similar derivations for screen mu defined as -- dVg2 /
dVg1 @ dVa = 0; dVa = 0, as originally suggested by "Flipper".

Regards,
Alex

[Alex]

"Ian Iveson" wrote in message
...
Is it useless knowledge I wonder. How many datasheets give anode mu for a
pentode? How many give both anode mu and screen mu?

Useless. Pentode mu is not usually given in datasheets and, unlike triode
mu, greatly depends on the operating point.

Muxh more practical would be discussing the following question:
How to estimate mug1g2 in a pentode if it is not given in a datasheet?

Someone alredady suggested finding a plate curve close to cutoff and then
mug1g2 = Vg2/Vg1.
And what if the curves are not given in a datasheet?
Then I use the following estimation:
mug1g2 = Vg2 / (|Vg1| + (Va / S)) .
The second term in the denominator virtually brings the tube to cutoff from
a specified class "A" operation point, then the propblem reduced to the
above principle. This estimation is very rough.

What would you do?

Later I might present similar derivations for screen mu defined as --
dVg2 / dVg1 @ dVa = 0; dVa = 0, as originally suggested by "Flipper".

That's not what he suggested, and what he suggested was wrong anyway, so
what's the point?

Not much point, but I found I forgot another condition which can
mathematically make another assumption redundant. This condition is the
follows. Suppressor grid does not steal electrons. So if plate current
increases while triode mu measurement, screen current has to reduce by the
same amount.

Thus: triode mu = dVg2 / dVg1 @ dIg2 + dIa = 0 and dIg2 = --dIa.
I will try to review calculations with the later condition. It will throw
another equation in a system of concurrent equations, and one of the
unknowns will be eliminated.

Alex

[Patrick Turner]

On Jan 18, 7:44*am, "Alex" wrote:
"Ian Iveson" wrote in message

...

Is it useless knowledge I wonder. How many datasheets give anode mu for a
pentode? How many give both anode mu and screen mu?

Useless. Pentode mu is not usually given in datasheets and, unlike triode
mu, greatly depends on the operating point.

Muxh more practical would be discussing the following question:
How to estimate mug1g2 in a pentode if it is not given in a datasheet?

Someone alredady suggested finding a plate curve close to cutoff and then
mug1g2 = Vg2/Vg1.
And what if the curves are not given in a datasheet?
Then I use the following estimation:
mug1g2 = Vg2 / (|Vg1| + (Va / S)) .
The second term in the denominator virtually brings the tube to cutoff from
a specified class "A" operation point, then the propblem reduced to the
above principle. This estimation is very rough.

What would you do?

Later I might present similar derivations for screen mu defined as -- *
dVg2 / dVg1 @ dVa = 0; dVa = 0, as originally suggested by "Flipper".

That's not what he suggested, and what he suggested was wrong anyway, so
what's the point?

Not much point, but I found I forgot another condition which can
mathematically make another assumption redundant. This condition is the
follows. Suppressor grid does not steal electrons. So if plate current
increases while triode mu measurement, screen current has to reduce by the
same amount.

Thus: triode mu = dVg2 / dVg1 @ dIg2 + dIa = 0 and dIg2 = --dIa.
I will try to review calculations with the later condition. It will throw
another equation in a system of concurrent equations, and one of the
unknowns will be eliminated.

Alex

Hi All, I have come a little late to this thread but some time ago I
gave a lot of thought to making up formulas which one could apply to
the design of output stages using power pentodes or beam tetrodes
where there is cathode FB tertiary windings.

Nothing in RDH4 adequately discusses the issues about screen Gm in
regard to Ultralinear or Acoustical (CFB from OPT) operations.

I did post extensively a few months ago on my findings and with
formulas but mainly everyone ignored me, and the shots of information
went over everyone's head it seemed to me. So I ain't gonna re-post;
go search the archives for postings under my name.

I extensively tested 6550 to measure the Gm and Ra of both normal
tetrode operation with triode connection using G2 connected to anode
and with G1 taken to a fixed bias and using G2 as the control grid.

As everyone should know, the G1 Gm varies hugely with Ia and Ea so to
know how to design or understand a 6550, you must completely ignore
the data sheets where the G1 Gm and Ra is stated for Ea and Ia
conditions that nobody in their right mind would ever use.
Typically data sheets give G1 Gm = 11mA/V, and Ra = 18k, so this gives
µ = 198. This is with Ea at about 250V and Ia at 140mA. But modern
6550 samples may have lower Gm, and to find outabout YOU MUST MEASURE
the tubes at the exact operating conditions, and forget the data
sheets which at best are a vague guide only.

I found that with Ia = 50mA and Ea = 450V and Eg2 at about 300V, G1 Gm
5.5mA/V, and Ra was 33k, so µ = 181, showing yet again that µ is the
most constant parameter. G2 gm is simply measured with Eg2 at say
300V, but with no RL in the anode circuit except a 100 ohm resistor
and with G1 at 0V with cathode biasing. A signal is applied to G2, and
the anode current measured in the 100r and I got G2 Gm = approx 0.83mA/
V. Ra was found to be slightly lower than the Ra when the tube is in
beam tetrode, at about 30k I recall, so the µG2 becomes 0.00083 x
30,000 = 25 approx.

From this info one may draw up an equivalent circuit of the beam
tetrode (or pentode) with a current source producing current between
cathode and anode terminals = Vg1 x G1 Gm and with a shunt resistance
equal to Ra also between a and k.
There is also a current source of VG2 x G2Gm also strapped across a
and k but in beam tetrode mode the G2 voltage is constant so this
current source has infinite resistance.
But where you have some voltage from the anode fed back in Ulralinear
or triode or CFB then the screen grid current source becomes active
and the fed back voltage changes the total Ia outcome. I won't
elaborate by providing the formulas I have derived before. You can all
get busy and do it yourselves like I did, but I found that I could
predict what the Ra would be for UL, Triode, or CFB and what gain I
would get armed only with the measured G1 Gm and Ra which gave me beam/
pentode µ, and the triode µ when the tube was tested in triode for
gain with with a CCS load.

After all this investigation I concluded that the Acoustical
connection was the most effective way to set up any beam/pentode in an
OP stage to get reasonably good gain at least as high as using a 300B,
and to get Ra lower than triode connection and lower even than the Ra
of a 300B. Quad used 10% of the P winding as a CFB tertiary winding
but 15% to 20% is far better if you employ methods to keep the driver
stage distortion nice and low by never ever using Quad-II's gutless
wonder EF86 driver stage.

One could follow one's nose of basic logic and apply it to signal
pentodes after one has taken appropriate measurements and applied the
relevant equivalent modelling.

There was also a paper titled "Amplifiers and Superlatives" published
in Wireless World in about 1955 which described the operation of beam
and pentodes with varying amounts of FB from the anode winding or
simply by means of having a CFB winding with fixed Eg2, like Quad-II.
The paper is now online somewhere and can be googled. Unfortunately, I
defy anyone to be able to understand the maths because there is a
mystery factor "m" included in the accompanying maths, and it isn't
clearly explained or why it is there. So I derived my own formulas
which work AFAIAC.

I now rarely need to explore further; I just set up the tubes for
acoustical CFB and adjust the fixed Eg2 for lowest THD and that's
that.
When I first used CFB in a large 8585 amp in 1996 I had Ea only 400V,
and the screens were fed from taps from the anode coils, about 25% of
anode to cathode total voltage. So Eg2 = Ea, which is fine for where
Ea is low. Then I had 12.5% of FB in cathode winding coils and at 50W
in class A from 4 x 6550 I had THD = 0.5% without any global loop FB
and the amp sounded magnificent.
with some Eg2. Rout was much lower than using straight ordinary 37.5%
UL taps.
One only needs 10dB of GFB to reduce the Rout of the CFB of about 1.5
ohms down to 0.5 ohms. Distortion spectra is better than pure class A
triode.

Patrick Turner.