John said:
Which is reasonable.
It is *not* the end of story.
It is as far as internal shot noise goes.
If the metal-film resistor jams zero shot noise current into the
emitter,
This statement makes absolutely no sense. Internal shot noise has
absolutely nothing to do with any ability "to jam" noise free current in
from an
external source. The simplified model of shot noise assumes from the outset
that external sources are noise free. So, its simply irrelevant whether this
is via a metal film resister or teletubbies.
There is a shot noise generator associated with an ideal, otherwise noise
free, DC current source feeding the emitter, irrespective of whence the
current originates. This generator is due to the statistical nature of
charge carriers crossing a junction. This noise generator is directly across
the dynamic impedance re. This drops a voltage right at the base-emitter
junction.
Well, actually, rather than stating that there is a collector noise dropped
across re as one simplified convenient model, it is better to use the output
noise equivalent model, that is:
icn_shotout = (icn_in.re).gm = icn_in
That is, the collecter shot noise is directly at the internal ce nodes. This
internal noise current is a constant set by ICDC, irespective of any
external circuitary. Now..if there is an emitter resister, a simple
calculation on the equivelent circuit will show that the noise actually
generated into the external collecter load is:
in_load = icn_shotout/(1+Re/re), because the controlled source swips some
of the current away.
So, in an actual circuit, the output noise current, dropped across the load
resister, is reduced by Re feedback. However, again, this noise is not a
function of the nature of the driving source or resistance, in addition this
reduced output noise does not imply a better S/N ratio, as the signal also
gets reduced by Re. Of course, if Re generates additional noise, then this
will have to be added into the calculation.
how can full shot noise emerge from the collector?
I hope the above explains what Re does, but draw the equivalent circuit and
do the algebra by all means.