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Chapter 29: Solid State Electronics Mike O’Connor Blake Barber Nick Keller Charlie Vana Nick Dotson.

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Presentation on theme: "Chapter 29: Solid State Electronics Mike O’Connor Blake Barber Nick Keller Charlie Vana Nick Dotson."— Presentation transcript:

1 Chapter 29: Solid State Electronics Mike O’Connor Blake Barber Nick Keller Charlie Vana Nick Dotson

2 Band Theory Because energy is quantitized, each atom has a specific energy level. As these atoms are brought together into solids, the many, interacting electric fields of the atoms change the energy levels of the atoms. No two electrons can have the same energy. The result is many differing energy levels, or energy bands. Energy bands are regions on a graph where the energy levels of the atoms in a solid are spread. These bands are separated by forbidden gaps which are values of energy electrons are not allowed to have. In insulators, the lowest band is fully filled because it forces electrons to jump to the next available band in order to move. This resists electron flow because the electrons must gain more enegry. In conductors, the lowest band is only partially filled because it allows electrons to move freely without gaining more energy.

3 Energy Band of Insulator

4 Energy Band of Conductor

5 Example Problem If a copper atom contributes one electron, how many free electrons exist in a cubic centimeter of copper? The density of copper is 8.96 g/cm 3 and it has an atomic mass of 63.54 g/mole. Don’t forget Avagadro’s magic number of 6.02 E 23 atoms/mole. Given: Copper 1 free electron per atom M = 63.54 g/mole ρ = 8.96 g/cm 3 Solution: Dimensional Analysis (Free e - )/(cm 3 Cu) = ((1 Free e - )/(1 atom))x(6.02 E 23 atoms/1 mole)x((1 mole Cu)/(63.54g Cu)) x(8.96g/1 cm 3 Cu) = 8.49 E 22 free e - / cm 3 Cu

6 Summary In solids, the allowed energy levels are spread into broad bands. The bands are separated by values of energies that electrons may not have called the forbidden gap.

7 What that means The levels are spread into broad bands. The bands are separated by values of energy that electrons are not allowed to have. These energies are called forbidden gaps. Electrical conduction in solids explained in terms of these bands, this is called the band theory.

8 MORE! In conductors, electrons can move because the band of allowed energy levels is only partially filled.

9 So….. Electricity can be possible because when a potential difference is placed across a material, the resulting electric field exerts a force on the charged particles. These accelerate, the field wont work on them, and they gain energy. They can only gain energy if there is a higher energy level into which they can move, this is true when the bands are only partly filled.

10 Example Find the fraction of atoms that has free electrons in silicon. At room temperature, thermal energy frees 1x 10^13 in pure silicon. The density of silicon is 2.33 g/m^3 and the atomic mass of silicon is 28.09 g/mol. A. How many silicon atoms are there in one cubic centimeter? B. What is the fraction of silicon atoms that have free electrons? That is, what is the ratio of free e^-/cm^3 to atoms/cm^3 in silicon?

11 Given Silicon: density, p = 2.33 g/cm^3 Atomic mass, M= 28.09 g/mol Free e^-/cm^3 = 1 x 10^13 Avogadro's number, Av = 6.02 x10^23 atoms/mol

12 Unknowns A. Si atoms/Cm^3 B. Free e^-/atoms Si

13 Solution- Part A A. By dimensional analysis, (2.33g/1cm^3)(1 mol/28.09 g Si)(6.02x10^23 atoms/1 mol) = 4.99 x 10^22 atoms/cm^3

14 Solution-Part B Free e^-/ atoms Si = (1x10^13 free e^- /cm^3)(1cm^3/4.99x1 0^22 atoms Si) = 2x 10^-10 Si atoms with free electrons

15 Semiconductors Who Occasionally Enjoy Partying Chapter 29

16 Key Points Conduction in semiconductors is usually the of result doping pure crystals with small numbers of impurity atoms. N- type semiconductors conduct by means of electrons while in p-type semiconductors, conductions is by means of holes.

17 Explained Conduction in semiconductors is usually the of result doping pure crystals with small numbers of impurity atoms. N- type semiconductors conduct by means of electrons while in p-type semiconductors, conductions is by means of holes. The dispersion of different atoms within a substance will effectively determine where and to what degree they conduct. The electrons within N-types have free flowing electrons that can transfer electricity to a degree. P-type conductors have positively charged holes in its valence shell that transfer electricity by vibrating motion.

18 History of Semiconductors Semiconductors are widely used in almost all electronic devices. Most widely used applications are transistors used in many electronics. The transistor was invented by John Bardeen, William Shockley, and Walter Brattain in 1947. Semiconductors were discovered in the 1800’s as poor conductors that exhibited photoelectric effects.

19 Practice Problem Silicon is doped with arsenic so that one in every 10^6 Si atoms is replaced by an As atom. Assume that each As atoms donates one electron to the conduction band. A. What is the density of free electrons in this extrinsic semiconductor? Given: As atom density= 1 As atom/ 10^6 Si atoms free e-/ As atom = 1 free e-/ As atom Si atom/ cm^3 = 4.99x 10^22 Si atoms/ cm^3 free e-/cm^3 in intrinsic Si= 1x10^13/ cm^3 Unknowns: As-donated free e-/cm^3 Solution: a. By dimensional analysis, Free e-/cm^3= (1 free/1 As atom) x (1 As atom/ 1x10^6 Si atoms) x ( 4.99x 10^22 Si atoms/ 1 cm^3)= 4.99 x10^3 free e-/cm^3

20 Diodes Diodes are the simplest semiconductor device. A diode consists of joined regions of p-type and n-type semiconductors. Holes are on the p-type side and electrons on the n-type side. The area in between is called the junction. This is also where the depletion layer occurs. The depletion layer is created through a loss of holes or free electrons around the junction. A Diode conducts charges in one direction only and can be used to produce current that flows in one direction only. Voltage Drop V=V d + IR

21 Diodes and Circuits When a diode is connected to a circuit as seen on the bottom left, the free electrons in the n-type semiconductor and the holes in the p-type semiconductor are attracted toward the battery the width of the depletion layer is increased and almost no current flows through the diode. However if the battery is connected in the opposite direction like in the bottom right electrons reach the p-end and fill the holes. The depletion layer is eliminated and current flows. Reverse-biased DiodeForward-biased Diode

22 Transistors A npn transistor consists of n-type semiconductors surrounding a thin p-type layer. If the transistor has a n-type in the center then it is a pnp transistor. The central layer is called the base and the two surrounding regions are the emitter and collector.

23 Transistors Transistors are used as amplifiers in almost every electronic instrument. The pn-junctions can be seen as back-to-back diodes. The base collector diode is reverse biased so no current flows, however the base-emitter diode is forward biased.

24 Practice Problem A diode has a voltage drop of.4V when 12 mA flows through it. If the same 470-Ω resistor is used, what battery voltage is needed? V=V d + IR V=.4V + (.012 A)(470 Ω) = 6.0 V

25 Conduction in Solids

26 Key Points 1.Electrons in metals have a very fast random motion. A potential difference across the metal causes a very slow drift of the electrons. 2.In insulators, electrons are bound to atoms. More energy is needed to move them than is available.

27 Key Points (In simpler terms) 1.The metallic bonds in metals allow the electrons of the metal atoms to move throughout the substance when little potential difference is added. 2.The potential energy needed to break electrons off insulator molecules is higher than metals. So insulators need a lot of energy to transfer a little current.

28 Example: A Diode in a Simple Circuit A silicon diode, whose I/V characteristics are shown, is connected to a battery through a 470Ω resistor. The battery forward-biases the diode and its voltage is adjusted until the diode current is 12mA. A. Draw a schematic diagram of the circuit B. What is the battery voltage?

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30 Solution Given: graph of V d versus I R=470Ω Needed: battery voltage, V V= V d + IR When I=12mA, V d = 0.7V (shown in graph) V= V d + IR 0.7V + (470Ω)(1.2x10 -2 A)=6.3V

31 The End


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