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Miscellaneous Notes

Miscellaneous Notes. The end is near – don’t get behind. All Excuses must be taken to 233 Loomis before 4:15, Monday, May 1. The PHYS 213 final exam times are * 8-10 AM, Monday, May 7

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Miscellaneous Notes

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  1. Miscellaneous Notes • The end is near – don’t get behind. • All Excuses must be taken to 233 Loomis before 4:15, Monday, May 1. • The PHYS 213 final exam times are * 8-10 AM, Monday, May 7 • * 8-10 AM, Tuesday, May 8 and* 1:30-3:30 PM, Friday, May 11. The deadline for changing your final exam time is 10pm, Monday, May 1. • Homework 6 is due Tuesday, May 2 at 8 am. (NO late turnin). • Course Survey = 2 bonus points (soon to accessible in SmartPhysics)

  2. Lecture 17Applications of Free Energy Minimum Semiconductors, doped Law of Atmospheres, revisited Reference for this Lecture: Elements Ch 12 Reference for Lecture 19: Elements Ch 13

  3. Last time: Intrinsic Semiconductors At T = 0: Conduction band has no electrons Valence band totally filled with electrons E =D Energy gap, D E =0 We now have the tools to solve for the equilibrium density of e-h pairs: 1. Since they act like ideal gases, the chemical potentials are 2. Electrons and holes are created in pairs total free energy is minimized when 3. For a pure semiconductor, ne = nh = ni (“intrinsic pair density”) Compare to I-V result (p. 11). The quantum density for an e-h pair is nQ  (nQenQh)1/2.

  4. Silicon (Group IV) Phosphorous (Group V) An extra (free) electron.No extra hole. Si Si Si Si Si Si P Si Si Si Si Doping of Semiconductors At T = 0, in a pure semiconductor the valence band is completely filled with electrons, and there are none in the conduction band. Suppose we replace one Si atom with a P atom.Phosphorous has one more electron, which is very weakly bound (its D is very small), and will almost certainly end up in the conduction band. Think of Shockley’s garage. If we add one car, it will have to go into the upper level, and there are still no vacancies (holes) in the lower level. Phosphorus is called a “donor atom”. The addition of impurities that have a different number (more or fewer) of valence electrons is called doping. It can have a dramatic effect on the material’s electrical properties. With doping, we no longer have Ne = Nh.Instead, Ne= Nh + Nd,where Nd is the number of donor atoms. More of Shockley’s cartoon

  5. Silicon (Group IV) Aluminum (Group III) Si Si Si Si Si Si Al Si Si Si Si ACT 1 We can also replace Si atoms with Al. Aluminum has one fewer electron than silicon. What is the relation between Ne and Nh in this case? a) Ne = Nh - NAl b) Ne = Nhc) Ne = Nh + NAl

  6. Silicon (Group IV) Aluminum (Group III) Si Si Si Si Si Si Al Si Si Si Si Solution We can also replace Si atoms with Al. Aluminum has one fewer electron than silicon. What is the relation between Ne and Nh in this case? a) Ne = Nh - NAl b) Ne = Nhc) Ne = Nh + NAl We are missing some valence electrons, so even when Ne (electrons in the conduction band) is zero, we have one hole per Aluminum atom. Look at the “p-type” part of Shockley’s cartoon. Aluminum is called an “acceptor atom”. In the equations, acceptors act like negative Nd.

  7. We can follow the same procedure to minimize the free energy. Assume nd<<nQ. (This is called “light doping”.) DF = 0 implies that me + mh = 0. This is again like the vacancy-interstitial problem. Solution: or nenh = ni2 Typically, nd >> ni. Therefore ne = ni + nd nd.Almost all of the “free” (conduction) electrons come from the dopant atoms. For Si at T = 300 K: If there were no doping, both ne/nQ and nh/nQ would be ~310-9.Even a small amount of doping: nd/nQ = 10-7, increases ne dramatically and suppresses nh. ne/nQ ~ 10-7, nh/nQ ~ 10-12. In the I-V problem, adding interstitial atoms to a crystal reduces the number of vacancies. This is an example of “the law of mass action”. Doping of Semiconductors (2) ne = nh + ndAssume: ne and nh << nQ. “intrinsic pair density” ni = ne = nh nQ= (nQenQh)1/2

  8. Summary:Particle equilibrium in Semiconductors if ne and nh << nQ. Pure (intrinsic) semiconductor ne = nh = ni = nQ e-D/2kT ni is called the “intrinsic carrier concentration” = intrinsic e-h pair concentration Doped semiconductor ne= nh + nd nenh = ni2The law of mass action. This is valid for intrinsic and doped semiconductors.

  9. Example: Law of Mass Action The addition of impurities increases the crystal’s conductivity. Let’s addnd = 1024/m3 phosphorous atoms to Si produces. Using the intrinsic carrier density, nI, that we calculated for silicon, compute the density of holes, nh, in this “doped” crystal.

  10. Solution The addition of impurities increases the crystal’s conductivity. Let’s addnd = 1024/m3 phosphorous atoms to Si produces. Using the intrinsic carrier density, nI, that we calculated for silicon, compute the density of holes, nh, in this “doped” crystal. nI = 5.21015 << nd, so ne  nd. This doping overwhelms the intrinsic carrier density. ne nh, but nenh = ni2 still holds. The addition of electrons has depressed the density of holes by over 8 orders of magnitude! Law of Mass Action  Question: Is 1024 atoms/m3 a lot or a little? Hint: There are 51028 Si atoms/m3.

  11. Act 2 Suppose you have a piece of “dirty” silicon, with lots of unwanted phosphorous impurities (making the conductivity too high because there are too many free electrons). What might you do to fix this? a) dope with more phosphorous b) dope with a different type of donor c) dope with an acceptor atom, like boron

  12. Solution Suppose you have a piece of “dirty” silicon, with lots of unwanted phosphorous impurities (making the conductivity too high because there are too many free electrons). What might you do to fix this? a) dope with more phosphorous b) dope with a different type of donor c) dope with an acceptor atom, like boron Adding more donors will only make the problem worse. Adding more acceptor atoms, we increase the number of holes. Since nenh = constant, increasing the number of holes decreases the number of free electrons. This technique is called “compensation”.

  13. Last Time N 1 F = F1+F2 N N 1 2 Maximum Total Entropy Minimum Free Energy Equal chemical potentials Thumbnail review of free energy:Equilibrium corresponds to maximum Stot = Sreservoir + Ssmall system. When we calculate DS, we only need to know the temperature of the reservoir. In minimizing F (equivalent to maximizing Stot) we don’t have to deal explicitly with Sreservoir . Consider exchange of material (particles) between two containers.These are two small systems in equilibrium with a reservoir (not shown) at temperature T. In equilibrium, dF/dN1 = 0: The derivative of free energy with respect to particle number is so important that we define a special name and symbol for it: For two subsystems exchanging particles, the equilibrium condition is: The chemical potential of subsystem “i”

  14. Why Bother with Yet Another Definition? System 1 System 2 Answer: It makes the various equilibrium conditions look the same: Exchange of:  Volume: p1 = p2  Energy: T1 = T2  Particles: m1 = m2 Why does the last equation use dF/dn, instead of ds/dN? Remember that there is a thermal reservoir (not shown). When particles are exchanged, the reservoir’s entropy might change. (It might gain or lose energy.) That’s what F takes care of. The two systems can exchangevolume, energy, or particles.

  15. Physical Significance of m high m low m Last time we looked at the chemical potential for an ideal (monatomic) gas: Consider the case of no external potential (u is a constant): m = kT ln(n) - constant . Apply it to the free expansion problem: Diffusion:Particles move from high m to low m Remove the piston: In equilibrium, m is the same everywhere. Some time later:

  16. energy particles 2 1 Equilibrium and Chemical Potential Recall the situation when systems can exchange energy. The definition oftemperature: 1/T = dS/dU (holding V and N fixed) tells us that temperatures are equal in thermal equilibrium. Otherwise we could increase S by exchanging some energy. We also know what happens when the systems are out of equilibrium (unequal T). Because high T means a small derivative, energy flows from the hot system to the cold one. Let’s look at the situation when we have particle exchange. From the definition of chemical potential, we have alreadyseen that in thermal and particle equilibrium, the chemicalpotentials are equal: m1 = m2. Out of equilibrium (m1 > m2):The larger m system has a larger dF/dN, soparticles flow from high m to low m. Note that dm/dN (= d2F/dN2) must be positive, or equilibrium isn’t stable. F N

  17. A process will happen spontaneously if the free energy would be decreased. This is just another way of saying that, left on their own, systems will tend to thermal equilibrium, i.e., minimum free energy. If you have to do work on a system (the opposite of a spontaneous process), you are increasing the free energy of the system. Why do matches burn (after you strike them)? The burned state has less free energy than the unburned state. Why must you strike the match? There is an energy barrier - it needs help getting started. Build Your Intuition F A local minimum of F unburned burned

  18. Next Monday Chemical equilibria - Law of mass action again Surface chemistry Phase equilibria and chemical potentials

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