F. H. Richardson's bluebook of projection (1935)

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THEORY AND WIRING OF AMPLIFIERS 563 Switch D-2 is evidently extraordinary in its wiring. A two-blade, double-throw switch, its lower blade switches 110-volts a. c, and its upper blade switches 12 volts d. c. In physical fact it is a rotary switch, having the positions "off," "operate" and "charge." In the "off" position the switch is open, as it is shown in the drawing. The "operate" connections are those which the switch in the drawing would make if thrown to the right, closing the circuit of the line a. c. to the power transformers and connecting the positive leg of the 12volt d. c. to the internal wiring of the amplifier. The "charge" position is that of the same switch thrown to the left in the drawing, in which both line a. c. and 12-volt d. c. are disconnected from the amplifier and connected to the battery charger. D. C. Filament Circuit of Figure 140 Assuming that switch D-2 has been thrown to the right, or "operate" position, let us trace the 12-volt circuit through the wiring of this amplifier. From the 12-volt battery terminal (the upper of the two 12-volt positive terminals) the line runs right through switch D-2, then upward through R-21 rheostat, and thence upward to the filament of V-2. Through this filament, then down, left and up to the filament of V-l. Through the filament of V-l and down through the two bias resistors, R-l and R-24; then down and right to the 92-AL key switch, K-2. Through the two right-hand contacts of this switch to the milliammeter M-3, and thence left to the negative 12-volt terminal. Switch K-2 needs a word of explanation. As shown in the drawing, it places the milliammeter in series with the d. c. filament circuit, using that instrument to show whether filament rheostat R-21 has been correctly adjusted. But when this key switch is thrown, the two central prongs are spread apart. The upper arrowheads then make contact with the central prongs, and the lower arrowheads are isolated. In that position of the switch the battery circuit we have just traced runs through the left-central prong to the left-outside prong, and thus