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) Lecture 17, p 1
Lecture 17 Applications of Free Energy Minimum Semiconductors, doped Law of Atmospheres, revisited Reference for this Lecture: Elements Ch 12 Reference for Lecture 19: Elements Ch 13 Lecture 17, p 2
Last time: Intrinsic Semiconductors Energy gap, E = E = 0 At T = 0: Conduction band has no electrons Valence band totally filled with electrons 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 n µ h = kt ln n h Qh n e µ e = + kt ln nqe 2. Electrons and holes are created in pairs total free energy is minimized when n h ne nhne µ h + µ e = 0 = kt ln + + kt ln = + kt ln nqh nqe nqhnqe 3. For a pure semiconductor, n e = n h = n i ( intrinsic pair density ) 2 n hne nhne ni n kt i ln = = = e = e nqhnqe kt nqhnqe nqhnqe nq The quantum density for an e-h pair is n Q (n Qe n Qh ) 1/2. 2kT Compare to I-V result (p. 11). Lecture 17, p 3
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 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. Silicon (Group IV) Phosphorous (Group V) Si Si Si Si Si Si Si P Si Si Si An extra (free) electron. No extra hole. 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 N e = N h. Instead, N e = N h + N d, where N d is the number of donor atoms. More of Shockley s cartoon Lecture 17, p 4
ACT 1 We can also replace Si atoms with Al. Aluminum has one fewer electron than silicon. What is the relation between N e and N h in this case? Silicon (Group IV) Aluminum (Group III) Si Si Si Si Si Si Si Al Si Si Si a) N e = N h - N Al b) N e = N h c) N e = N h + N Al Lecture 17, p 5
Solution We can also replace Si atoms with Al. Aluminum has one fewer electron than silicon. What is the relation between N e and N h in this case? Silicon (Group IV) Aluminum (Group III) Si Si Si Si Si Si Si Al Si Si Si a) N e = N h - N Al b) N e = N h c) N e = N h + N Al We are missing some valence electrons, so even when N e (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 N d. Lecture 17, p 6
Doping of Semiconductors (2) We can follow the same procedure to minimize the free energy. Assume n d <<n Q. (This is called light doping.) F = 0 implies that µ e + µ h = 0. This is again like the vacancy-interstitial problem. ne nh µ e + µ h = + kt ln( ) + kt ln( ) = 0 n n Qe Qh n e = n h + n d Assume: n e and n h << n Q. n n Solution: or n e n h = n i 2 e h kt = e 2 Q n Typically, n d >> n i. Therefore n e = n i + n d n d. Almost all of the free (conduction) electrons come from the dopant atoms. For Si at T = 300 K: nenh 10 2 n Q This is an example of the law of mass action. intrinsic pair density n i = n e = n h n Q = (n Qe n Qh ) 1/2 If there were no doping, both n e /n Q and n h /n Q would be ~3 10-9. Even a small amount of doping: n d /n Q = 10-7, increases n e dramatically and suppresses n h. n e /n Q ~ 10-7, n h /n Q ~ 10-12. In the I-V problem, adding interstitial atoms to a crystal reduces the number of vacancies. 19 Lecture 17, p 7
Summary: Particle equilibrium in Semiconductors µ µ = + = = 2 / kt F 0 e h 0 nenh nqe if n e and n h << n Q. Pure (intrinsic) semiconductor n e = n h = n i = n Q e - /2kT Doped semiconductor n i is called the intrinsic carrier concentration = intrinsic e-h pair concentration n e = n h + n d n e n h = n i 2 The law of mass action. This is valid for intrinsic and doped semiconductors. Lecture 17, p 8
Example: Law of Mass Action The addition of impurities increases the crystal s conductivity. Let s add n d = 10 24 /m 3 phosphorous atoms to Si produces. Using the intrinsic carrier density, n I, that we calculated for silicon, compute the density of holes, n h, in this doped crystal. Lecture 17, p 9
Solution The addition of impurities increases the crystal s conductivity. Let s add n d = 10 24 /m 3 phosphorous atoms to Si produces. Using the intrinsic carrier density, n I, that we calculated for silicon, compute the density of holes, n h, in this doped crystal. n I = 5.2 10 15 << n d, so n e n d. This doping overwhelms the intrinsic carrier density. n e n h, but n e n h = n i2 still holds. Law of Mass Action ( 5. 2 10 15 ) 2 2 ni n h = = = 2. 7 10 /m 24 n 1 10 7 3 The addition of electrons has depressed the density of holes by over 8 orders of magnitude! e Question: Is 10 24 atoms/m 3 a lot or a little? Hint: There are 5 10 28 Si atoms/m 3. Lecture 17, p 10
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 Lecture 17, p 11
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 n e n h = constant, increasing the number of holes decreases the number of free electrons. This technique is called compensation. Lecture 17, p 12
Last Time Thumbnail review of free energy: Equilibrium corresponds to maximum S tot = S reservoir + S small system. When we calculate S, we only need to know the temperature of the reservoir. In minimizing F (equivalent to maximizing S tot ) we don t have to deal explicitly with S reservoir. 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/dn 1 = 0: df df df df df dn dn dn dn dn df dn 1 2 1 2 = + = = 1 1 1 1 2 df = dn 1 2 1 2 The derivative of free energy with respect to particle number is so important that we define a special name and symbol for it: 0 N 1 N 2 F = F 1 +F 2 N 1 µ i dfi dn i The chemical potential of subsystem i For two subsystems exchanging particles, the equilibrium condition is: µ 1 = µ 2 Maximum Total Entropy Minimum Free Energy Equal chemical potentials Lecture 17, p 13
Why Bother with Yet Another Definition? Answer: It makes the various equilibrium conditions look the same: Exchange of: σ d 1 d 2 Volume: = p 1 = p 2 dv dv σ 1 2 d 1 d 2 Energy: = T 1 = T 2 du du df1 df2 Particles: = µ 1 = µ 2 dn dn σ σ 1 2 1 2 System 1 System 2 The two systems can exchange volume, energy, or particles. Why does the last equation use df/dn, instead of dσ/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. Lecture 17, p 14
Physical Significance of µ Last time we looked at the chemical potential for an ideal (monatomic) gas: Consider the case of no external potential (u is a constant): µ = kt ln(n) - constant. Apply it to the free expansion problem: µ high µ low µ F n = u + kt ln N V, T nq Remove the piston: Diffusion: Particles move from high µ to low µ Some time later: In equilibrium, µ is the same everywhere. Lecture 17, p 15
Equilibrium and Chemical Potential Recall the situation when systems can exchange energy. The definition of temperature: 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 already seen that in thermal and particle equilibrium, the chemical potentials are equal: µ 1 = µ 2. Out of equilibrium (µ 1 > µ 2 ): The larger µ system has a larger df/dn, so particles flow from high µ to low µ. Note that dµ/dn (= d 2 F/dN 2 ) must be positive, or equilibrium isn t stable. F µ N 1 F V, T energy particles 2 N Lecture 17, p 16
Build Your Intuition 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. F A local minimum of F unburned burned Lecture 17, p 17
Next Monday Chemical equilibria - Law of mass action again Surface chemistry Phase equilibria and chemical potentials Lecture 17, p 18