It can be alternatively fixed by adding a large pull-down resistor from the gate to the ground, to dissipate the deposited charge when the switch is opened, but at the expense of lowering input impedance. All wires running inside this shielding layer will be to a large extent decoupled from external electrical fields, particularly if the shield is connected to a point of constant voltage, such as earth or ground. The drawback of the transistor switches shown earlier is that they work in a manner similar to a single-pole switch: they can connect the load to the ground (NPN, n-channel MOSFET) or to the supply rail (PNP, p-channel MOSFET), or simply leave it in open circuit state. Every word of this can be written at the present moment, that is, ten years later, with exactly the same significance. A simple way to fix it is to have the input signal drive two complementary transistors at the same time, placed on both sides of the "out" node; this is known as a push-pull amplifier.

The second circuit - a bridge or full-wave rectifier - is a bit more clever, but also easy to follow: opposing pairs of diodes are used to select the more positive or negative out of two input leads, and always produce a particular output polarity. Heavy cables should be laid out to depths of 400 fathoms, where there are tide-ways. The system is constructed as follows: Eight wrought-iron pipes, three inches in diameter, are laid side by side in two rows, about four feet below the surface. Although neither expense nor pains were spared in the construction of this line, the cost being comparable with that of the Prussian system, two years had not elapsed before some of the wires ceased to work, and, though these were replaced and workmen kept constantly busy on the line, at the end of seven years the line was wholly abandoned in favor of overhead wires. If you see two dropouts that were N minutes apart then you will probably continue to see dropouts every N minutes. When the voltage is close to the positive rail, the situation will reverse; in effect, this arrangement is an inverting switch.

This arrangement still suffers from the 0.6V bias - but this time, it can be solved more neatly than with AC coupling: by biasing both transistors into symmetrical, slight conduction when Vin is at the mid-point - and simply relying on the input signal to swing the ratio. For starters, there is a voltage offset present between the input and the output; for inputs between 0 and 0.6V, the output will be simply clipped at 0V. Connecting the emitter to a negative voltage at least 0.6V higher than the lowest signal voltage is a potential solution to the clipping problem - but it may be impractical in some settings. In all cases, the driving voltage applied to the base (or gate) must be high enough to trigger the transistor; that is, at least 0.6V in BJT, and at least 1-2V for most MOSFETs. The base-emitter voltage is kept near the diode conduction threshold, so the current flowing through this path is also very low and self-limiting (and naturally, avoided entirely in MOSFETs). Naturally, a wide range of refinements to these basic designs is known today - say, to improve linearity, bandwidth, or to compensate for temperature variations.

While very useful for controlling high-impedance signals, the diode simply serves as a "crowbar" across the supply terminals - and therefore, for input voltage sources that can source a significant current, this arrangement gets dangerously inefficient; a resistor can be used to limit supply current, of course - but this simply takes you back to the high-impedance scenario - not very useful for, say, driving motors. The problem with this is that all real-world loads will develop some voltage across them in normal operation; this raises the emitter or source voltage accordingly - perhaps close to, or even above, the driving base / gate voltage. Alas, when Vin later drops to 2V, Vout will stay at 5.4V - and because emitter voltage is now higher than base voltage, the transistor will not conduct. R1 / R2. In other words, Vout will be amplified by R1/R2 - with no dependency on hfe! When Vin2 is higher than Vin1, the right transistor will insist on getting the emitter voltage to a point where the left one no longer conducts - and so, the current flowing through the right R1 (and the associated voltage drop) will increase. The conducting path for the current in the cable is provided by the conductor.

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