Consider the 12SK7 vacuum tube: gm of 0.002, plate resistance of 0.8MegOhms, grid capacitance of 6pF, output(plate) capacitance of 7pF.

Predict bandwidth by gm/C. Assume nodal C is 6p + 7p + 7p parasitic = 20pF.

Bandwidth is 0.002 / 20e-12 = 0.0001 * e+12 = 1e+8 = 100MHz, providing 10 nanosecond (10 feet one way, 5 feet 2-way) resolution.

http://www.r-type.org/pdfs/6sk7.pdf

Consider the 12SK7 vacuum tube: gm of 0.002, plate resistance of 0.8MegOhms, grid capacitance of 6pF, output(plate) capacitance of 7pF.

Predict bandwidth by gm/C. Assume nodal C is 6p + 7p + 7p parasitic = 20pF.

Bandwidth is 0.002 / 20e-12 = 0.0001 * e+12 = 1e+8 = 100MegaRadians/second or 16MHz; using the Tektronix rule-of-thumb of 0.35/bandwidth for the response of multi-stage systems, or 0.35/16MHz, the Trise is 20nanoseconds; 20nS providing 20 feet one way, 10 feet 2-way, resolution.

http://www.r-type.org/pdfs/6sk7.pdf