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How the Junction Works When the N-type material is negative with respect to the P type, as in Fig 19-4A, electrons flow easily from N to P The N-type semiconductor, which already has an excess of electrons, receives more; the P-type semiconductor, with a shortage of electrons, has some more taken away The N-type material constantly feeds electrons to the P type in an attempt to create an electron balance, and the battery or power supply keeps robbing electrons from the P-type material This condition is illustrated in Fig 19-5A, and is known as forward bias Current can flow through the diode easily under these circumstances When the battery or dc power-supply polarity is switched so the N-type material is positive with respect to the P type, the situation is called reverse bias Electrons in the N-type material are pulled toward the positive charge pole, away from the P-N junction.

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Prefix piconanomicromillikilomegagigateraFraction 0.000000000001 (one-trillionth) 0.000000001 (one-billionth) 0.000001 (one-millionth) 0.001 (one-thousandth) 1000 1,000,000 1,000,000,000 (one billion) 1,000,000,000,000 (one trillion)

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In the P-type material, holes are pulled toward the negative charge pole, also away from the P-N junction The electrons are the majority carriers in the N-type material, and the holes are the majority carriers in the P-type material The charge therefore becomes depleted in the vicinity of the P-N junction, and on both sides of it, as shown in Fig 19-5B This zone, where majority carriers are deficient, is called the depletion region A shortage of majority carriers in any semiconductor substance means that the substance cannot conduct well Thus, the depletion region acts like an electrical insulator This is why a semiconductor diode will not normally conduct when it is reverse-biased A diode is, in effect, a one-way current gate usually! Junction Capacitance Some P-N junctions can alternate between conduction (in forward bias) and nonconduction (in reverse bias) millions or billions of times per second.

Sometimes you need to use the power equation to find currents or voltages. Then you should use I P/E to find current, or E P/I to find power. It s easiest to remember that P EI (watts equal volt-amperes), and derive the other equations from this by dividing through either by E (to get I) or by I (to get E).

This results in less end-to-end energy loss and avoids the dispersion problem that occurs in multimode installations. Chromatic Dispersion The speed at which an optical signal travels down a fiber is absolutely dependent upon its wavelength. If the signal comprises multiple wavelengths, then the different wavelengths will travel at different speeds, resulting in an overall spreading or smearing of the signal. As discussed earlier, chromatic dispersion comprises two subcategories: material dispersion and waveguide dispersion.

Other junctions are slower The main limiting.

There is an important difference between energy and power. You ve probably heard the two terms used interchangeably, as if they mean the same thing. But they don t. Energy is power dissipated over a length of time. Power is the rate at which energy is expended. Physicists measure energy in joules. One joule is the equivalent of one watt of power, dissipated for one second of time. In electricity, you ll more often encounter the watt hour or the kilowatt hour. As their names imply, a watt hour, abbreviated Wh, is the equivalent of 1 W dissipated for an hour (1 h), and 1 kilowatt hour (kWh) is the equivalent of 1 kW of power dissipated for 1 h. An energy of 1 Wh can be dissipated in an infinite number of different ways. A 60-watt bulb will burn 60 Wh in an hour, or 1 Wh per minute. A 100-W bulb would burn 1 Wh in 1/100 hour, or 36 seconds. A 6-watt Christmas tree bulb would require 10 minutes (1/6 hour) to burn 1 Wh. And the rate of power dissipation need not be constant; it could be constantly changing.

Material Dispersion Simply put, material dispersion occurs because dif-

P-N junction. At B, reverse bias of the same junction. Solid black dots represent electrons. White dots represent holes. Arrows indicate direction of charge-carrier movement.

32 Electrical units Figure 2-6 illustrates two hypothetical devices that burn up 1 Wh of energy. Device A uses its power at a constant rate of 60 watts, so it consumes 1 Wh in a minute. The power consumption rate of device B varies, starting at zero and ending up at quite a lot more than 60 W. How do you know that this second device really burns up 1 Wh of energy You determine the area under the graph. This example has been chosen because figuring out this area is rather easy. Remember that the area of a triangle is equal to half the product of the base length and the height. This second device is on for 72 seconds, or 1.2 minute; this is 1.2/60 0.02 hour. Then the area under the curve is 1/2 100 0.02 1 Wh.

2-6 Two devices that burn 1 Wh of energy. Device A dissipates a constant power; device B dissipates a changing amount of power.

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