30 Aug
2008
30 Aug
'08
2:45 a.m.
der Mouse wrote:
Standard Disclaimer: It's been 30 years since I've done such calculations, so,
if I still have the concepts right:
Given a B+ (cathode-plate differential) of 200V, an electron accelerated from
velocity of 0 at the cathode, to the plate, will have acquired 200 electron-Volts
(eV) of energy when it hits the plate.
Circumventing force/acceleration calculations and simply going backwards from the
energy/mass/velocity relationship:
Given the mass of an electron and now knowing it's kinetic energy, we can calculate
the final velocity of the electron when it hit the plate (with a conversion factor
inserted to convert eV to Joules (m^2Kg/s^2):
E = m v^2 / 2
v = sqrt( 2E / m )
= sqrt( 2 (200 eV) (1.6e-19 (m^2Kg/s^2)/eV) / (9.1e-31 Kg) )
= ~ 8,400,000 m/S
Supposing the plate-cathode distance is 3 mm, and given the average velocity over
that distance is half the final velocity since the electron started at 0:
t = d / (v/2)
= 0.003 m / ( (8.4e6 m/S) / 2 )
= ~ 0.7 nS
.. seems valid perhaps .. Of course, it's not really the cathode-to-plate time
that matters, but the grid-to-plate time (for amplifying tubes), but anyways..
Ballpark example, take a 12AU7: [...3pF
capacitance...RC time
constant...]
(Not to say there weren't other reasons they
were slow..)
That brings up something I've occasionally wondered about: how fast do
electrons move in a vacuum tube? (I'm talking about the free-flight
path between cathode and plate.) In particular, what's the time delay
between emission from the cathode and reception by the plate? I
imagine it depends on the plate voltage; does that make enough of a
difference to care about? I don't remember enough of the constants to
figure out what sort of acceleration a gradient of, say, 200 V/cm
imparts to a free electron....