onto the ship. This way, no matter how much the ship sways, screen, projector, and passengers are in sync. From the shore, the movie would seem to roll and wave; but that would be irrelevant to the viewers.
To carry this analogy over to the SE amplifier, the screen is the primary of the output transformer; the rolling water, the power supply noise; the output transformer's secondary winding and loudspeaker, the passengers; the projector, the triode's plate; and, at the risk of stretching the analogy too far, the test bench oscilloscope attached to ground and the triode's plate, the onlookers on the shore.
Most voltage dividers consist of two resistors in series with each other, but other circuit elements and combinations are possible: triodes, pentodes, and photocells. If the voltage division is to be preformed primarily on an AC signal, then two triodes, two pentodes, two capacitors, or two inductors could be used.
When capacitors are used to divide an AC signal, the division is the inverse of resistors. Here the smaller valued capacitor develops the greatest signal voltage across its plates. For example, if a 30 µf and 10 µf capacitors are wired in series and an AC signal of 400 volts is applied across the series, the voltage ratio will be 3:1, with 100 volts across the bigger capacitor and 300 volts across the smaller. If you find this too paradoxical, then think in terms of impedances rather than capacitance values. In this example, the 30 µf capacitor will at any given frequency have one third the reactive impedance of the 10 µf capacitor. The math is simple enough:
Xc = 1 / (2pFC)
And, as one capacitor is three times bigger than the other one,
Xc2 = 1 / (2pF3C)
Xc2 = Xc / 3.
This trick will only work with cathode biased amplifiers, as what is needed is some degree of separation between cathode and ground. A cathode biased amplifier uses the current flowing through the tube to develop a voltage across the cathode resistor and to push the cathode up to a more positive voltage than the grid, which presumably is at ground potential. The value for the cathode resistor is determined by the needed grid-to-cathode voltage divided by the desired idle current for the tube:
Rk = Vgk / I.
For example, a 300B might require a grid-to-cathode voltage of 100 volts in order to maintain an idle current of 50 mA with a cathode-to-plate voltage 440 volts. The cathode resistor would then be equal to 100 / 0.05, or 2,000 ohms. This resistor would then be bypassed with a large