If you have no cathode bypass capacitor in a pentode voltage amplifier,
should the screen bypass capacitor be connected to the ground or to the
cathode? If the capacitor is connected to ground, would there be
resulting negative or positive feedback? Thanks

Not realy an answer , but .....
http://dddac.de/at07.htm might be a nice read .


"gojamo" schreef in bericht
oups.com...
If you have no cathode bypass capacitor in a pentode voltage amplifier,
should the screen bypass capacitor be connected to the ground or to the
cathode? If the capacitor is connected to ground, would there be
resulting negative or positive feedback? Thanks

gojamo wrote:

If you have no cathode bypass capacitor in a pentode voltage amplifier,
should the screen bypass capacitor be connected to the ground or to the
cathode? If the capacitor is connected to ground, would there be
resulting negative or positive feedback? Thanks

It is the usual practice to bypass the screen to the cathode,
so that the highest possible gain is achieved since bypassing to
0V reduces gain due to the NFB applied between cathode and screen
since there is a voltage developed across Rk.

So consider the extreme case where one used a pentode
with screen bypassed to 0V, but with equal RL at the anode and
cathode, as a concertina phase inverter.
I have never sen this done but
there is no reason why it couldn't be done; it will have its pros
and cons.

Say the Va = +10v, the Vk = -10, and
let us suppose Vg1 = -11v
Then Vg2 goes 10v positive in respect to Vk, so this would
reduce the gain at the anode, so more g1 voltage would have to be applied
than if the tube was connected as a pentode.
Connected with G2 bypassed to 0V the tube is intermediary between
pentode and triode, sort of like a UL stage.
Consider if +D volts of distortion should appear at the anode
with G2 bypassed to.
Because RLa = RLk, -D volts of distortion must appear at the cathode.
So there is effectively +D volts between G1 and k applied,
and this gets amplified say 10 times to make -10D volts appear
at the anode, and the only way this situation can exist is if
the anode voltage at the anode distortion was +11D volts before NFB was
present.
Also there is +D volts of distortion applied between k and G2, so the
screen circuit amplifies this distortion to make a certain value of -D
voltage at the anode
and this further acts to cancel the original distortion.

So yes there is NFB.

Just to explain further, just consider a normal tube amp with global
FB from the speaker connection back to V1.
V1 acts like a differential amp with a low impedance drive to one input
port, the cathode,
and the other input port, the grid1, a high impedance drive.
The anode output is the third terminal and its signal is
the V1 tube gain x the difference between Vk and Vg1.

So, let us suppose you have a Williamson which is a
non inverting amp like most amps using traditional
global series voltage negative feedback.

Let us suppose the open loop gain, or gain without GNFB
is say 200 without any FB so to make say 20v at the output,
some 0.1 v is needed at the input.
Suppose we apply the GNFB so that 1.1 v is applied to the
input for the 20v output, and 1v is the feedback voltage at the cathode of
V1.
This must come from a resistance divider so the fraction of the output
voltage, ß, fed back to the input
is 1/20 = 0.05.

Let us suppose that with FB applied, there is 0.1% of thd at the output
with 20v output,
and that ths 20 v is a few dB less than clipping.
Then the output distortion voltage is 0.1% of 20v = 0.02v.
We should indicate the phase of this voltage as +0.02v.

There must be distortion voltage of ß x 0.02 applied to the V1 cathode
so we have - ( 0.05 x 0.02 )v of thd applied to V1k = +0.001.

But no distortion voltages generated by the amp exist at the grid,
to the +thd applied to the cathode act to cause a -ve thd voltage at the
output.
This voltage *must be* open loop gain x thd voltage = 200 x -0.001v =
-0.2v.
But hey, we said there was only +0.02 to begin with.
Well in fact there must have been +0.22v of thd without NFB applied, and
the
application of NFB causes -0.2v of NFB generated "error"
voltage to be simultaneously applied and the THD error voltage subtracts
from the
open loop THD and you get a resultant +0.02v remaining after FB is applied.

Anywhere NFB is applied this scenario is acted out
and same applies to a humble pentode used at the input
of some amp with its screen bypassed to 0V instead of to its cathode.
The amount of NFB applied to the screen where the Rk merely acts to bias
the tube
with the g1 at 0V bias potential is quite low, almost insignificant, unless

Rk was large, which it could be depending on how we arrange the g1 bias
to make Rk somewhat larger than a normal R of say 470 ohms, typical
for a 6AU6 input tube with a load of 100k.

A typical 6AU6 input pentode, ( or one of many in the family )
can have its gain reduced in the manner described, and its up to
**your** ingenuity to play around with a tube tomorrow on a breadboard to
find out
more about tube properties. It is sunday, and sundays are your day off, so
get busy to learn because if you don't do things and observe closely and
precisely,
them all I am saying will go in one ear and out the other.

You should know how to calculate the Ro of a pentode
by applying two different loads with the same input voltage by carefully
measuring the two gains, the effective Ra of the tube can be calculated
because gain, A, = µ x RL / ( RL + Ra ).
This formula applies to all tubes.

Therefore µ = A x ( RL + Ra ) / RL.
µ is a fixed tube parameter for the same tube with different loads.

So for two different values of A and RL,

A1 x ( RL1 + Ra ) / RL1 = A2 x ( RL2 + Ra ) / RL2

From this you can work out Ra, since it is the only unknown
value in this equation.

Series negative current FB usually causes effective Ra to rise,
and series negative voltage FB causes effective Ra to fall.

Now I will leave you with enough information to
do the bread board experiment with your pentode, and answer my question,
What type of series NFB is applied by bypassing the screen to 0V
instead of to the cathode, voltage or current NFB??

Patrick Turner.

"gojamo" wrote

If you have no cathode bypass capacitor in a pentode voltage
amplifier,
should the screen bypass capacitor be connected to the ground or
to the
cathode? If the capacitor is connected to ground, would there be
resulting negative or positive feedback? Thanks

As the valve passes more current, the screen in your case would
become less positive with respect to the cathode, tending to reduce
the current. Hence it is negative feedback.

I think you will have a high output impedance though.

cheers, Ian