Re: Kathode folower question



Ian Iveson wrote:

RealInfo wrote

"Ian Bell" <ruffrecords@xxxxxxxxx> ???
??????:gm7cfh$bep$1@xxxxxxxxxxxxxxxxxxxxxxxx
RealInfo wrote:
Hi all

How do I calculate the cathode follower cathode resistor
in case of DC coupling between the 1st and the 2nd
stage, the 2nd tube grid is directly
connected to the 1st tube anode .


Depends on the tube type and what sort of load you want
to drive and the its stage tube and its anode voltage.
But in simple terms - you know the anode voltage of the
preceding stage so you know the CF grid voltage. Then CF
needs to be a few volts above this - how many we don't
know yet, but it is only a few volts so it is usually
safe to say that the operating point of the CF is with a
plate voltage of the CF plate supply minus its grid
voltage. On the CF tube plate characteristics draw a
vertical line at this voltage.

Now comes the hard part. You need to decide what static
current the CF is to draw. To find this you need to know
the load impedance the CF will drive and the voltage you
want to drive in it - this will give you the load
current. As a rough guide you should set the static
current to four times this value. Now is the time to
check you have chosen a tube that is meaty enough by
calculating its dissipation as static current times plate
to cathode volts.

Assuming that's OK, draw a horizontal line at this
current on the plate tube characteristics. Where it
intersects the vertical line is your desired operating
point and you can see what the grid bias voltage needs to
be so add this to the grid voltage and this is the
desired cathode voltage. You know the standing current
and the cathode voltage so working out the cathode
resistor is just a matter of dividing the latter by the
former.


In practice it is not quite that easy because it is a
good idea for the cathode resistor be be at least twice
the tube's plate resistance and this method takes no
account of that.

This seems to involve such a common misconception that
sometimes I wonder if I've got the wrong end of the stick.

For me, it is the *total load* on a triode that should, as a
rule of thumb, be somewhere around 2.5 times the anode
resistance. Why should you, and many others I have read,
consider only the cathode resistor's contribution to the
load?

Consider a typical 6SN7 line stage amp with gain triode with 47k dc RL,
then direct coupled CF with 47k cathode dc load. Ia for gain triode and
CF are nearly the same if VRLdc = Ea.

Suppose we couple the cathode of the CF to a power amp input or some
other gear and the input resistance of that amp or gear is 47k.
The the CF has to drive 47k dc load in parallel with 47k ac coupled
load. The load becomes 23.5k. RL is more than 2.5 x Ra, distortion is
low.

Everything is fine until a certain voltage is reached where suddenly the
bottom of the sine waves cut off sharply. The lower the ac coupled load
is, the lower the voltage threshhold at which this severe asymetrical
clipping takes place.

But if you only want a few volts from your preamp, the cut off behaviour
won't be seen.


Now, if the total load met that condition, then the CF
wouldn't be able to serve it's common purpose of driving low
impedance loads. Consequently it might be said that the CF
has no place in high fidelity, because it abuses a triode
and then tries to cover up with loads of feedback. May as
well use SS, which has the advantage of not needing heater
current.

The CF only becomes a fudge when you allow the ac coupled load to be too
low. if a following power amp has input resistance >50k, then there is
no problem with the use of a CF. The low output resistance of the
follower means that bandwidth will be maintained despite the cable and
amp input capacitance or perhaps the inductance of an input transformer.

Anyway, in answer to the OP's question, surely calculation
of the cathode resistance needed to establish a particular
operating point is the same regardles of the voltage on the
grid? That is, your choice of operating point gives you Vgk
and Ia. Simple ohms law returns Rk, bearing in mind that Vgk
= Vg - Vk. The main reason a CF is more difficult to design
than an AF is that Vk is a much more significant part of
Vak, so you don't know the necessary Va until the end of the
calculation. Often you will be designing for a given HT
voltage, so you end up with an iterative process of
rethought and recalculation. Also, the signal Vk is a much
more significant part of the signal Vgk so it's not so easy
to choose the best operating point in the first place.

CF are easy to design if you simply make the cathode dc load suit the Ea
and Ia of the tube based upon the normal load line analysis.

OTOH, perhaps if I were to use a CF, I shouldn't bother too
much about choosing an operating point...just choose Va and
Ia, mess with Rk until it fits, and let Vgk and Vak be
whatever they fancy. Feedback willing, it'll be alright on
the night.

Or look for typical configurations in data sheets.

Yes, there is the option of copying he who has gone before thee.

Patrick Turner.

Ian
.

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