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. Voltage Dividers    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.Cathode Biasing    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
 regulate the output voltage -- all of which would work, albeit at great cost, effort, and weight.    Or we could interject some of the power supply noise into the output stage to null the noise from the output. This sometimes happens by accident, when the input signal is cascaded from one stage to another to the output, the power supply noise is inverted and attenuated to the just right amount that results in the plate experiencing the same amount of noise as the output transformer's B+ connection, which means the transformer never sees any noise voltage across its primary; hence, no noise at the output.   Our goal then is to design the countervailing power supply noise into the circuit. Because the grounded cathode operation of the output tube inverts the grid signal on its plate, the noise would have to be phase inverted, if it were to be added at the grid. Otherwise, the noise would increase at the output, although it may decrease or even disappear at the plate.    Here lies the problem of attaching a short feedback loop from the plate of the output tube to the driver circuit in an SE amplifier. The feedback will strive to eliminate any noise on the plate, which means the plate would resemble the ground in AC terms. This means the output transformer will see the whole of the power supply noise developed across its primary winding. The amplifier's PSRR will be a sad 0 dB, in spite of having zero noise on the plate of the output tube.  (Remember, the speaker is not attached to the plate, but to the secondary winding of the output transformer.)   On the other hand, signals presented to the cathode remain in phase at the plate. If the required amount of noise were to be interjected at the cathode of the output tube, the plate would track the B+ noise; thus, the transformer would never see any noise voltage developed across its primary; therefore, no noise at the secondary. Because the tube amplifies the cathode signal, the amount of noise needed to match the power supply connection would the reciprocal of the gain realized by the tube. For example, if the gain from cathode to plate were 10, then the amount of noise needed at the cathode would be 1/10 of the power supply noise. An analogy    An analogy of what is going on here would be to imagine a movie that is being displayed on a screen on a large passenger ship, from a projector on the shore.     If the water is very still, the image looks good; but the movie is ruined by any waves in the water or movement aboard ship. Extreme and elaborate measures could be made to try to hold the ship steady, such as making the ship truly massive, damming up the waterways, and pouring tons of oatmeal into the water - all ultimately fruitless.  But if we step back and realize that the movie is being shown not for the benefit of those on the shore, but for those on the ship, a better approach presents itself:  simply move the projector