Lecture 16. Equilibrium and Chemical Potential. Free Energy and Chemical Potential Simple defects in solids Intrinsic semiconductors

Lecture 16 Equilibrium and Chemical Potential Free Energy and Chemical Potential Simple defects in solids Intrinsic semiconductors Reference for this Lecture: Elements Ch 11 Reference for Lecture 12: Elements Ch 12 Lecture 16, p 1

Free Energy, Equilibrium and Chemical Potential Last time: Free energy F sys = U sys T reservoir S sys This is the maximum available work we can get from a system that is connected to a reservoir (environment) at temperature T reservoir. Equilibrium corresponds to maximum S tot = S reservoir + S small system. We saw that minimizing F is equivalent to maximizing S tot, but with the advantage that we don t have to deal explicitly with S reservoir. Consider now two small systems in equilibrium with a reservoir (not shown) at temperature T. Thermal equilibrium at temperature T is given by minimizing total free energy, F = F 1 + F 2 : df 1 df 2 F = N1 + N2 = 0 dn dn 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: N 1 N 2 F = F 1 +F 2 N 1 µ i dfi dn i The chemical potential of subsystem i Equilibrium condition: F = µ N + µ N = 1 1 2 2 0 Lecture 16, p 2

The Path Ahead... Having considered thermal equilibrium when volume and energy exchanged, now we ll consider systems in which particles can be exchanged (or created ). Minimization of total free energy will allow us to understand a wide variety of different physical processes. Some examples: Particles can move from place to place. Particles can combine into new types (e.g., chemical reactions). This will lead to the concept of chemical equilibrium. And lots and lots of applications... Let s start with a concrete example. Lecture 16, p 3

F-minimum example: Defects in Crystal Lattices interstitial vacancy In a perfect crystal at low temperatures, the atoms are arranged on a lattice like the one shown at the left. Consider M atoms on lattice sites, where M is a very large number, about 10 22 for a mm-sized crystal. As the crystal is heated up the atoms jiggle around, and some atoms will jump to interstitial sites, leaving a vacancy behind. There is an energy cost to form each interstitial-vacancy pair ( I-V pair ) like the one shown above, i.e., there is an energy difference between an I-V pair and a normally occupied site. By minimizing the Free energy of N I-V pairs, we will calculate the average number of defects that form at a temperature T. Lecture 16, p 4

ACT 1 As we let the temperature of the solid 0, what fraction of the atoms will sit at the interstitial sites? a) none b) half c) all Lecture 16, p 5

Solution As we let the temperature of the solid 0, what fraction of the atoms will sit at the interstitial sites? a) none b) half c) all Because it costs energy to create an interstitial-vacancy pair, at low temperature the decrease in F due to entropy gain (increasing the number of available sites) will be smaller than the increase in F due to the energy cost. Therefore, the free energy will be minimized by staying at home. F = U - TS As T 0, the TS term becomes unimportant Lecture 16, p 6

Defects in Crystal Lattices (2) Suppose we have M possible vacancy sites, and M possible interstitial sites (essentially one per atom). We want to know N, the number of interstitial-vacancy pairs at temperature T. We want to minimize F(N) as a function of N. Call F(0)=0 for convenience. F( N) = U( N) TS( N) U( N) = N How to calculate S(N)? Need energy for each institialvacancy pair Assume the crystal s vibrational entropy is not much changed by making an interstitial. S is then due to the number of places each vacancy could be, and the number of places each interstitial could be. Lecture 16, p 7

Defects in Crystal Lattices (3) Perfect lattice Lattice with a Defect interstitial Entropy of I-V pair: S( N) = k ln Ω vacancy # ways to put N identical particles in M cells = M N /N! (single occupancy, dilute limit) But there is no correlation between the position of a vacancy and the position of an interstitial. Therefore, the total number of accessible states Ω = Ω I Ω V =(M N /N!) 2. Entropy: Free energy: N M S( N) = k ln Ω = k ln = 2k N lnm ln N! N! ( ) ( ) F( N) = U( N) TS( N) = N 2kT N lnm ln N! 2 typo! In equilibrium: df( N) dn = 0 Lecture 16, p 8

Stirling s Approximation It will often be necessary to calculate d(ln N!)/dN. We ll use a well known approximation for N!, known as Stirling s Approximation*: Try some numbers: ln N! N lnn - N N ln N! N lnn - N 10 15.1 23.0-10 = 13.0 50 148.5 195.6-50 = 145.6 1000? 6908-1000 = 5908 d(ln N!) d N ( N ln N N ) = ln N + - 1 = ln N dn dn N The derivative is only defined for large N. *This is not Robert Stirling ( Stirling engine ) but James Stirling, Scottish mathematician. Lecture 16, p 9

Defects in Crystal Lattices (4) By minimizing the Free energy of N interstitial-vacancy pairs, we determine the average number of defects that form at temperature T: (from a previous slide) ( ) F( N) = U( N) TS( N) = N 2kT N lnm ln N! Minimize F: df dn M = 2kT lnm + 2kT ln N = 2 kt(ln ) = 0 N Solve for the fraction N/M = # defects lattice sites: N 2 kt(ln ) M = N M = e / 2kT This looks like a Boltzmann factor: an exponential temperature dependence. As we predicted before, as T 0 the fraction of interstitial-vacancy pairs is exponentially suppressed. n n c = e / 2kT with N n = = pair density V M nc = = cell density V Notice the 2 in the Boltzmann factor. It came from squaring the (M N /N!) number of positional states, because there are 2 movable objects, vacancy and interstitial. Lecture 16, p 10

Act 2 We just saw the fraction of interstitial-vacancy pairs is given by 1. Suppose the energy cost to create such a pair is 1 ev. If we want to keep the fraction of vacancies less than 1%, what is the maximum temperature T 1% we should heat the material to? a) 100 C b) 1000 C c) 10,000 C N M 2kT = e 2. Suppose that for some reason the vacancy and interstitial sites were always right next to each other. How would this safe temperature T 1% change? a) T 1% will decrease b) T 1% will increase c) T 1% will stay the same Lecture 16, p 11

Solution We just saw the fraction of interstitial-vacancy pairs is given by N M 2kT = e e 2kT 1. Suppose the energy cost to create such a pair is 1 ev. If we want to keep the fraction of vacancies less than 1%, what is the maximum temperature T 1% we should heat the material to? 1% a) 100 C b) 1000 C c) 10,000 C = 0.01 = ln(0.01) = 4.6 2kT 1% / 2 0.5eV T1% = = = 1264 K 5 4.6k 4.6(8.6 10 ev/k) Interpretation: As we raise the temperature higher, the material is literally coming apart. 2. Suppose that for some reason the vacancy and interstitial sites were always right next to each other. How would this safe temperature T 1% change? a) T 1% will decrease b) T 1% will increase c) T 1% will stay the same Lecture 16, p 12

Related Example: Solid Solutions A B Important for real devices e.g., silicon-gold, tin-lead In equilibrium, some A atoms are in the B crystal and vice versa. Assume: There are M A sites, N of which are occupied by B atoms. N << M. The only entropy is due to site counting (ignore vibrations, etc.) The energy increase when a B goes to an A site is. Let s call the chemical potential of the B atoms in their own crystal 0, by choosing a convenient zero for energy. Then, in equilibrium, the chemical potential of the B s in A must also be zero: F U S M N N kt µ = 0 = = T = kt ln. So, if N M : = e N N N N M V, T V, T V, T M! S = k ln N!( M N)! Lecture 16, p 14

Electrons in Semiconductors Energy gap, At T = 0: Conduction band has no electrons Valence band totally filled with electrons In many materials, electrons cannot have every conceivable energy. There is a low energy range (the valence band ) and a high energy range (the conduction band ). A gap of disallowed energies separates them. (The reason for the gap is a Physics 214 topic.) At T = 0, every valence band state is occupied. (S=0. Why?) At T 0, electrons are thermally excited from the valence band to the conduction band. How many determines the electrical conductivity. The activated free electrons and the holes (unfilled states) left behind act as two ideal gases. We can compute the density of thermally excited electrons (and holes) by minimizing F electron + F hole. We simplify the problem by assuming that excitation from valence to conduction band always requires the same energy, i.e., every conduction state has energy more than every valence state. This avoids having to do integrals. Lecture 16, p 15

Electrons in Semiconductors (2) Conduction electron Hole This is Shockley s* cartoon of an intrinsic semiconductor. At T = 0, the cars (electrons) can t move. If some are raised to the upper level (the conduction band) then motion becomes possible. The vacant spaces on the lower level are holes. Motion of the cars on the lower level is more simply described by pretending that the holes are the objects that move. An intrinsic semiconductor is one in which the number of electrons equals the number of valence band states, so that at T = 0 every state is filled, and no electrons are left over. *John Bardeen, Walter Brattain, and William Shockley invented the transistor in 1947. Lecture 16, p 16

Intrinsic Semiconductors We have N electrons N valence states (call their energy zero) N conduction states (energy ) Conduction electron We want to know N e and N h, the numbers of conduction electrons and holes at temperature T hole Our analysis will be similar to the Defects in a Lattice problem. Conduction electrons and holes are created in pairs, so: N e = N h (and dn e = dn h ) Method: Minimize F = F e + F h : df dfe dfh dne = dn df h e df = + + dn dn dn dn dn e e e e h h = 0 µ + µ = h e 0 One big difference: electrons and holes actually behave like ideal gases --they are free to rapidly move around in the crystal, with an effective mass m e, m h. So, what are S, F, and µ for an Ideal Gas of Particles? Lecture 16, p 17

How Do S, F, µ Depend on N/V? In general, this is a complicated function of particle properties and the environment. To keep things simple, we ll only work with ideal gases. For an ideal gas the internal energy per particle, u U/N, depends on T, but not on N. For many problems we can set this to 0. How does S depend on N? Let s count microstates and see. N N M M Ω = so S = k ln( ) = k N lnm ln N! N! N! Thus, ( ) S M d(ln N!) = k ln since = ln N N M N dn The number of bins, M, is proportional to V. It also depends on T (see next slide). and the chemical potential for an ideal (monatomic) gas is: µ F S M N n = u T = u kt ln = u + kt ln = u + kt ln N V, T N M N VnQ nq n N/V = particle density, n Q M/V = number of states per unit volume. n Q can only be calculated using quantum mechanics, so we will treat it as an empirical quantity (i.e., we ll tell you the numerical value when necessary). n Q is a function of T, but not of the particle density, n. For ideal gases, µ ~ logarithm of particle density. Lecture 16, p 18

n nv Q Q = = n N What is n Q? n Q = total no. states (per unit volume) available to a particle at temp. T number of states in volume V number of particles Considering both position and momentum. QM: particle with momentum p has wavelength λ = h/p. (h = Plank s constant) Taking λ as a characteristic length, the effective cell volume is ~ V λ 3 = (h/p) 3. Using p 2 /2m = (3/2)kT, the quantum-mechanical cell density has a T-dependence: n Q 1/λ 3 = (3mkT/h 2 ) 3/2 If we do the problem more carefully (see Elements for details) we find* n = (2 πmkt / h ) = (10 meter )( m / m ) ( T / 300 K ) Q 2 3 / 2 30 3 3/ 2 3/ 2 p Examples: n Q (H @ 300K) = (10 meter )( m / m ) ( T / 300 K) = 10 meter 30 3 3/ 2 3/ 2 30 3 p n (e @ 300K) = (10 meter )( m / m ) (300 / 300 K) = 1.27 10 meter 30 3 3/ 2 3/ 2 25 3 Q e p n (e @ 3K) = (10 meter )( m / m ) (3/ 300 K) = 1.27 10 meter 30 3 3/ 2 3/ 2 22 3 Q e p *For a monatomic gas, the n T (T) we had before is n Q ; otherwise there are modifications. Lecture 16, p 19

Electrons and holes in 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 n Q = (geometric mean of n Qe n Qh ) (n Qe n Qh ) 1/2. 2kT Compare to I-V result (p. 11). Lecture 16, p 20

E-h h pair density vs. T In a pure semiconductor, there are equal numbers of conduction-band electrons and valence-band holes: n = n n = n e e h i Q - /2kT This intrinsic density n i clearly varies strongly with energy gap. Some numerical values (at T = 300 K): material (ev) n Q (meter -3 ) n i (meter -3 ) Si 1.14 1.72 x 10 25 5.2 x 10 15 Ge 0.67 7.21 x 10 24 - GaAs 1.43 2.63 x 10 24 - Question: Why aren t the n Q all equal? In particular, if you set m = m e in the formula, n Q = 1.27 10 25. The answer is that the effective mass of a hole is not equal to the effective mass of an electron (and even m e,effective m e 9.11 x 10-31 kg!), due to interactions with the lattice. Therefore, we will treat n Q as an empirical quantity. Lecture 16, p 21

Act 3: Intrinsic Silicon At 300 K the intrinsic carrier density in Si is n i = 5.2 x 10 15 /m 3, with a bandgap of = 1.14 ev, and quantum density n Q = 1.72 x 10 25 /m 3. What is the carrier density at 150 K? a. 6.3 10 5 /m 3 b. 1.8 10 6 /m 3 c. 2.6 10 15 /m 3 Lecture 16, p 22

Solution At 300 K the intrinsic carrier density in Si is n i = 5.2 x 10 15 /m 3, with a bandgap of = 1.14 ev, and quantum density n Q = 1.72 x 10 25 /m 3. What is the carrier density at 150 K? n = n = n e e i Q a. 6.3 10 5 /m 3 b. 1.8 10 6 /m 3 c. 2.6 10 15 /m 3 /2kT T T ( ) = ( ) n n e e 2 Q 2 /2 k( T /2) T ( ) ( ) 1 3/2 = ( ) = 0.35 ( ) n n T n T Q 2 Q 2 Q /2 k( T /2) 1.14/2 k(150) 44.1 20 e = e = e = 7 10 n 0.35(1.72 10 )(7 10 ) 4.3 10 /m e 25 20 5 3 = = Exponential sensitivity to temperature! Change to n Q (T) relatively minor This is very rapid variation. To get a feeling for it, consider T 301 K (an 0.3% change). Then 5.2 5.6 (a 7.6% change!) Lecture 16, p 23

Digital Thermometers The rapid (exponential) variation of resistance with temperature is used to measure temperature. Many modern digital thermometers use a thermistor, a semiconductor device whose resistance depends on temperature. Lecture 16, p 24

Exercise: Other Semiconductors What is the number of intrinsic carriers in the conduction band, n i, for two other common semiconductors, Ge ( = 0.67 ev) and GaAs ( = 1.43 ev) at 300 K? Note: kt = 0.026eV at T = 300K. Lecture 16, p 25

Solution What is the number of intrinsic carriers in the conduction band, n i, for two other common semiconductors, Ge ( = 0.67 ev) and GaAs ( = 1.43 ev) at 300 K? Note: kt = 0.026eV at T = 300K. Ge: n = n e = 1.8 10 /m i Q 0.67 ev / 2 0.026eV 19 3 using n = 7.2 10 /m Q 24 3 The difference is almost entirely due to the band gap. GaAs: n = n e = 3.0 10 /m i Q 1.43 ev / 2 0.026eV 12 3 using n = 2.6 10 /m Q 24 3 Lecture 16, p 26

Next Time Applications of free energy Doped Semiconductors Law of atmospheres, revisited Lecture 16, p 27