On Sep 14, 8:48*am, (Scott Dorsey) wrote:
wrote:
On Sep 13, 3:36=A0pm, (Scott Dorsey) wrote:
wrote:
The 33 Meg resistor is being bootstrapped to at least 330 Meg. =A0The
bootstrap is direct coupled. =A0The output Z of the capsules I'm using
is ~50 Meg at 40 Hz.
How did you get that number? =A0What is the capacitance of the capsule wh=
en
idle?
Capsule capacitance is about 80 pF. *So its output Z it just its
capacitive reactance, right?
I'm not sure if you can assume that, but it would make sense. *80 pF is
a pretty high capacitance capsule and you can probably get away with a higher
Z than most with it.
As long as the leak resistance is low enough to permit all the spurious
charge appearing on the grid to be grounded out, you're fine. =A0The amou=
nt
of spurious charge that appears there is dependant on the kind of tube yo=
u
use and it's one of the more important things to select when you pick a
tube for microphone applications.
What do you mean by "spurious charge"? *Do you mean input current?
No, the whole point of the leak resistor is to drain off the charge that
appears spuriously on the grid..... as electron flow occurs inside the tube,
the grid slowly picks up a charge. *As long as there is a path to ground
for that charge, everything is fine... but if the grid is completely
disconnected it will gradually charge up toward the plate value.
The value of the leak resistor required is determined by the degree to
which this happens. *There is a good discussion in the Radiotron Handbook.
Don't think about input current... it's really not a useful concept here
anyway.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."
Apparently you're right about enlarging the grid resistor, but for a
different reason. A bigger grid/charge resistor results in LOWER
noise. How's that for counter-intuitive? My beef with my prototype
is noise, so hopefully this is the answer. I'll let y'all know.
Check out the paper below, page 2:
http://www.etymotic.com/publications/erl-0053-1967.pdf
Sean B