[Patrick Turner]

To calculate effective Ra when shunt or series VOLTAGE NFB is applied
there is a simple
equation,
Effective Ra' = Ra / ( 1 + [ µ x ß ] ) .

But another handy one for a pentode is

Ra' = 1 / ( [ gm x ß ] + 1/Ra )

So a pentode such as 6AU6 with Ra = 1M, and gm = 0.003A/V at the
operating point
and with a shunt NFB network using a pair of high resistances of 900k
and 100k
will act with Ra effectively =
1 / ( [ 0.003 x 0.1 ] + 1 / 1,000,000 ) = 1 / ( 0.0003 + 0.000001 ) =
3.32 k.
Gain will be nearly 9.

If the tube was a triode, does the formula work?

Let's have the same high resistance shunt NFB network to avoid loading
effects
by the NFB network and let us use a 1/2 12AX7 with Ra = 65 k, and Gm =
0.0016A/V .

Ra' = 1 / ( [ 0.0016 x 0.1] + 1/65,000 ) = 1 / ( 0.00016 + 0.0000153 ) =
5,701 ohms.

Using the first formula at the top of this post I get

Ra' = 65,000 / ( 1 + [ 104 x 0.1 ] ) = 65,000 / 11.4 = 5,701 ohms.

Note that the 12AX7's µ is 104, because µ = gm x Ra = 0.0016 x 65,000 in
this case.

It seems like both the formulas all work wth all triodes and pentodes.

Where Ra becomes really high in the case of pentodes, then approximately

Ra' = 1 / [ gm x ß ].
Including the 1/Ra figure always makes the calculated Ra' lower, which
is better for most purposes,
so the real Ra' is a bonus for where we said we have a very high Ra for
the pentode to begin with.

In the case of the triode internal NFB there is an electrostatic NFB
shunt network of very high resistance
so there is a figure for ß due to electrode distances and spacings and
wire pitches on the grid.

The Ra of a triode would be very high, like a pentode, if its NFB wasn't
operational.
So approximately the Ra of a triode is the Ra', ie, effective Ra with
NFB applied,
and since Ra is so darn high without NFB the Ra without the triode NFB
needn't be included in the
FB equation and nor does it need to be following the above reasonings.

So say we have triode Ra = 1,000 ohms, and gm = 0.006A/V as in the case
of a 6550 in triode,
then Ra = 800 = 1 / [ 0.006 x ß ] approximately.
So ß in the triode = 1 / [ 0.006 x 800 ] = 1 / 4.8 = 0.208.

For shunt FB, ß = d1 / ( d1 + d2 ), where d1 could be R1, d2 could be
R2;
d1 & 2 are distances, R1 & 2 are resistances....

So if d1 = 1mm, 0.208 = 1 / ( 1 + d2 ) so d2 must be 3.807mm, and this
last
distance is between space charge and screen, which does the job of the
anode
were the tube a real triode with its anode positioned where the grid is.

Perhaps I am being too simplistic in my logic; hell, its only my logic,
and I am sure some expert has done a far better job than I have.

But the second formula for effective Ra with NFB is useful to show
how a pure current source becomes a finite voltage source of finite Ra'
when NFB is applied.

Patrick Turner.