The Semiconductor in Equilibrium

Lecture 6 Semiconductor physics IV The Semiconductor in Equilibrium

Equilibrium, or thermal equilibrium No external forces such as voltages, electric fields. Magnetic fields, or temperature gradients are acting on the semiconductor. All properties of the semiconductor will be independent of time in this case.

Goal The concentration of electrons and holes in the conduction and valence bands with the Fermi-Dirac probability function and the density of quantum states. The properties of an intrinsic semiconductor. The properties of an semiconductor with impurities (dopants). 3

CHARGE CARRIERS IN SEMICONDUCTORS Two types of charge carrier, the electron and the hole. The current in a semiconductor is determined largely by the number of electrons in the conduction band and the number of holes in the valence hand. The distribution (with respect to energy) of electrons in the conduction band is n( E) = gc( E) ff( E) the density of quantum states in the conduction band the probability that a state is occupied by an electron 4 The total electron concentration per unit volume in the conduction band is n n( E) de =

The distribution (with respect to energy) of holes in the valence band is p( E) = g ( E)[1 ff( E)] v the density of allowed quantum states in the valence hand the probability that a state is not occupied by an electron The total hole concentration per unit volume is found by integrating this function over the entire valance-band energy. p = p( E) de 5

The location of Fermi energy E F At T=K, valence band is full and the conduction band is empty in an intrinsic semiconductor (no impurities and no lattice damage in crystal). Ev < EF < Ec At T>K, the valence electrons gain energy and a few move to conduction band and leave empty states. Electrons and holes are created in pairs by the thermal energy. The number of electrons in the conduction band is equal to the number of holes in the valence band. Fig (b) The splitting of the 3s and 3p states of silicon into the allowed and forbidden energy bands.

* 3/2 4 π (2 mn) gc( E) = E E 3 h * 3/2 4 π(2 mp) gv( E) = E 3 v E h c If we assume that the electron and hole effective masses are equal, then g c (E) and g v ( E ) are symmetrical functions about the midgap energy. The function f F (E) for E > E F is symmetrical to 1 - f F (E) for E < E F about the energy E = E F. 1 ff ( E) = E EF 1+ exp( ) The areas representing electron and hole concentrations are equal => E F in the middle of bandgap energy 7 Fig Density of states functions, Fermi-Dirac probability function, and areas representing electron and hole concentrations for the case when E, is near the midgap energy

The n andp Equations The thermal-equilibrium concentration of electrons = n g ( E) f ( E) de c F The lower limit of integration is E c The upper limit of integration should be the top of the allowed conduction band energy. However, since the Fermi probability function rapidly approaches zero with increasing energy we can take the upper limit of integration to be infinity. 8

* 3/2 4 π(2 mn) gc( E) = E E 3 h If (E c - E F ) >> k T, then (E - E F ) >> 1 [ ( E EF )] ff ( E) = exp ( E EF ) 1 + exp * 3/2 4 π(2 mn) ( E EF) no = E Ec exp[ ] de 3 h EC c Boltzmann approximation Let η = n E Ec * 3/2 4 π(2 mn) ( Ec EF) 1/2 exp[ ] exp( ) 3 h = η η dη 1 2 1/2 η exp( η) dη = π n π 2( ) exp[ ] h * 2 m n 3/2 ( Ec EF) = 2 9

The thermal-equilibrium electron concentration in the conduction band ( Ec EF ) N c is called the effective density states function in the * conduction band. 2πm 3/2 N c n = N n = 2( ) 2 h c exp[ ] The thermal-equilibrium concentration of holes in the valence band is p = gv( E)[1 ff ( E)] de * 3/2 4 π (2 mp ) gv( E) = E 3 v E h 1 1 ff( E) = ( EF E) 1 + exp 1

11 If ( EF Ev) >> 1 ( EF E) 1 ff( E) = exp[ ] ( EF E) 1 + exp E v * 3/2 4 π (2 mp ) ( EF E) = 3 v h p E E exp[ ] de The lower limit of integration is taken as minus infinity instead of the bottom of the valence band since the probability function of holes approach zero when energy is minus infinite. p ' ( Ev E) η = 4 π (2 m ) ( E E ) = exp[ ] ( ) exp( ) d * 3/2 p F v 3 h + * 3/2 π p F p 2 exp 3 2 m ( E E) = h ' 1/2 ' ' η η η

The thermal-equilibrium concentration of holes in the valence band p = N v ( Ev E) exp[ ] The effective density of states function in the valence band is * 2πm p Nv = 2 2 h The magnitude of N v is also on the order of 1 19 cm -3 at T = 3 K for most semiconductors. 3/2 12

The effective density of states functions, N c and N v, are constant for a given semiconductor material at a fixed temperature. Table Effective density of states function and effective mass values N c * 2πmn = 2 2 h 3/2 N v * 2πm p = 2 2 h 3/2 13

Example: Calculate the probability that a state in the conduction band is occupied by an electron and calculate the thermal equilibrium electron concentration in silicon at T= 1 K. Assume the Fermi energy is.25 ev below the conduction band. The value of N c for silicon at T = 1 K is N c = 2.8 x 1 19 cm -3. Solution: The probability that an energy state at E = E c is occupied by an electron is given by 1 ( Ec EF ).25 f ( E ) exp[ ] exp( ) 6.43 1 F c = = = Ec EF 1 + exp( ).259 5 14 ( Ec EF ).25 n = Nc exp[ ] = (2.8 1 ) exp( ) = 1.8 1 cm.259 19 15 3 The probability of a state being occupied can be quite small, but the fact that there are a large number of states means that the electron concentration is a reasonable value.

Example Calculate the thermal equilibrium hole concentration in silicon at T= 4 K. Assume that the Fermi energy is.27 ev above the valence band energy. The value of N v for silicon at T = 3 K is N v = 1.4 1 19 cm -3. Solution Nv The hole concentration is p 4 = (1.4 1 )( ) = 1.6 1 cm 3 19 19 3 ( E E ).27 = Nv =.3453 15 3 = 6.43 1 cm F v 19 exp[ ] (1.6 1 ) exp( ) 15

The Intrinsic Carrier Concentration For an intrinsic semiconductor, The concentration of electrons in the conduction band n i is equal to the concentration of holes in the valence band p i. The Fermi energy level is called the intrinsic Fermi energy, or E F = E Fi. EFi Ec Ev EFi n = ni = Nc exp[ ] p = pi = ni = Nv exp[ ] n = N N 2 i c v EFi Ec Ev EFi exp[ ]exp[ ] 2 Ev E E c g ni = NcNv exp[ ] = NcNv exp[ ] E g is the bandgap energy 16

For E g = 1.12 ev, n i = 6.95 x 1 9 cm -3 from the equation for silicon at T = 3 K. The commonly accepted value of n i for silicon at T = 3 K is approximately 1.5 1 9 cm -3 This theoretical function does not agree exactly with experiment. Fig The intrinsic carrier concentration of Ge, Si, and GaAs as a function of temperature. 17

The Intrinsic Fermi-Level Position Since the electron and hole concentrations are equal N c ( ) ( ) exp[ E c E Fi Fi v ] Nv exp[ E = E ] If we take the natural log of both sides of this equation 1 1 Nv EFi = ( Ec + Ev) + ln( ) 2 2 N * 1 3 mp EFi = ( Ec + Ev) + ln( ) * 2 4 mn 1 ( E + E ) = E 2 The midgap energy c N c v midgap v * 2πmp = 2 2 h 3/2 N c * 2πmn = 2 2 h 3/2 18

3 m EFi Emidgap = ln( ) 4 m * p * n The intrinsic Fermi level is m m m = m * * p n < m * * p n > m * * p n exactly in the center of the bandgap below the center of the bandgap. above the center of the bandgap. 19

Example To calculate the position of the intrinsic Fermi level with respect to the center of the bandgap in silicon at T = 3 K. The density of states effective carrier masses in silicon are Solution m = 1.8 m m =.56m * * n p The intrinsic Fermi level with respect to the center of the * bandgap is 3 m 3.56 E E = ln( ) = (.259)ln( ) Fi midgap p 4 * mn 4 1.8 E E =.128eV = 12.8meV Fi midgap 2 If we compare 12.8 mev to 56 mev, which is one-half of the bandgap energy of silicon, we can, in many applications, simply approximate the intrinsic Fermi level to be in the center of the bandgap.

DOPANT ATOMS AND ENERGY LEVELS The real power of semiconductors is realized by adding small, controlled amounts of specific dopant, or impurity, atoms. The doped semiconductor, called an extrinsic material adding a group V element, such as phosphorus 21 The phosphorus atom without the donor electron is positively charged. At very low temperatures, the donor electron is bound to the phosphorus atom.

The donor electrons jump to the conduction band with thermal energy Fig The energy-band diagram showing (a) the discrete donor energy state and (b) the effect of a donor state being ionized. 22 The electron in the conduction band can now move through the crystal generating a current, while the positively charged ion is fixed in the crystal. The donor impurity atoms add electrons to the conduction band without creating holes in the valence band. The resulting material is referred to as an n-type semiconductor.

Adding a group III element, such as boron, as a substitution impurity purity to silicon. One covalent bonding position appears to be empty Fig Valence electrons may gain a small amount of thermal energy and move about in the crystal. 23 The "empty" position associated with the boron atom becomes occupied, and other valence electron positions become vacated. These other vacated electron positions can he thought of as holes in the semiconductor material.

Fig The energy-band diagram showing (a) the discrete acceptor energy state and (b) the effect of a acceptor state being ionized. Acceptor atom gets electrons from the valence band with thermal energy. If an electron were to occupy this "empty" position, its energy would have to be greater than that of the valence electrons. The acceptor atom can generate holes in the valence hand without generating electrons in the conduction band. This type of semiconductor material is referred to as a p-type material 24

Ionization Energy Energy required to elevate the donor electron into the conduction band. Bohr theory The most probable distance of an electron in a hydrogen atom from the nucleus from quantum mechanics is the same as Bohr radius. The coulomb force of attraction between the electron and ion equal to the centripetal force of the orbiting electron. This condition give a steady orbit. 25 e m v = πεr r 2 * 2 2 4 n n

26 If we assume the angular momentum is also quantized, then we can * write m rnv = nħ n is a positive integer nh 2 * 2 2 2 Substitute v = e m v nħ 4πε * m r into = 2 r n 4πεrn r n = * 2 n m e The assumption of the angular momentum being quantized leads to the radius being quantized The Bohr radius is 4πεħ = =.53A 2 2 me normalize the radius of the donor orbital to that of the Bohr r radius n 2 m = n ε ( ) rest mass of an electron r * a m the relative dielectric constant of the semiconductor material a effective mass of the electron in the semiconductor.

If we consider the lowest energy state in which n = 1, and if we consider silicon in which ε r = 11.7 and the conductivity effective mass is m*/m =.26. then we have that r 1 a = 45 r 1 = 23.9 A. This radius corresponds to approximately four lattice constants silicon. Recall that one unit cell in silicon effectively contains eight atoms, so the radius of the orbiting donor electron encompasses many silicon atoms. => The donor electron is not tightly bound to the donor atom. 27

28 The total energy of the orbiting electron is given by E = T + V The kinetic energy is 1 * 2 T = m v 2 2 2 * n ħ 4 πε Since m rnv = nħ rn = * 2 m e * 4 m e T = 2 2 2( nħ) (4 πε) 2 * 4 e m e The potential energy is V = = 4 πεrn ( nħ) (4 πε) * 4 m e The total energy is E = T + V = 2 2 2( nħ) (4 πε) 2 2

In silicon the ionization energy is E = -25.8 mev, much less than the bandgap energy of silicon. This energy is the approximate ionization energy of the donor atom. Table Impurity ionization energies in silicon and germanium 29 Germanium and silicon have different relative dielectric constants and effective masses, resulting in different ionization energy.

THE EXTRINSIC SEMICONDUCTOR A material has impurity atoms. One type of carrier will predominate. The Fermi energy will change as dopant atoms are added. n type: the density of electrons is greater than the density of holes n >p majority carrier: electrons; minority carrier: holes the Fermi energy is above the intrinsic Fermi energy p type: the density of holes is greater than the density of electrons n <p majority carrier: holes; minority carrier: electrons the Fermi energy is below the intrinsic Fermi energy 3

n type Fig Density of states functions. Fermi-Dirac probability function, and areas representing electron and hole concentrations for the case when E F is above the intrinsic Fermi energy. 31

p type Fig Density of states functions, Fermi-Dirac probability function, and areas representing electron and hole concentrations for the case when E F is below the intrinsic Fermi energy. 32

The electron concentration in extrinsic semiconductor n = N i The intrinsic carrier concentration n = N i c ( Ec EFi) + ( EF EFi) exp[ ] c ( Ec EFi) exp[ ] EF EFi n = ni exp[ ] ( EF EFi) p = ni exp[ ] 33

The n and p Product n p = N N c v ( Ec EF ) ( EF Ev) exp[ ]exp[ ] Eg np = NcNv exp[ ] n p = n 2 i 34 The product of n and p is always a constant for a given semiconductor material at a given temperature. The equation is invalid if the Boltzmann approximation is not valid since it is derived by the Boltzmann approximation E E >> and E E >> c F F v

The Fermi-Dirac Integral If the Boltzmann approximation (E-E F >>) does not hold. 1/2 4 π * 3/2 ( E E ) 3 (2 ) c de n = m n h E E E F c 1+ exp( ) If we again make a change of variable and let E E η c EF Ec = ηf = * 1/2 2mp 3/2 η dη n = 4 π( ) 2 h 1+ exp( η η ) The integral is defined as F 35 Fermi-Dirac integral

F ( η ) 1 2 η dη 1/2 F ( F ) 1 e η + η The Fermi-Dirac integral, is a tabulated function. if η F > The Fermi energy is actually in the conduction band. ( ) / EF EC ηf Fig The Fermi-Dirac integral as a function of the Fermi energy 36

p 2 m ( ) 1 exp( η ) * ' 1/2 ' p 3/2 η dη = 4 π( ) 2 ' ' h + η F ' η = Ev E η = E E ' v F F if η > ' F the Fermi level is in the valence band. 37

Example To calculate the electron concentration using the Fermi-Dirac integral.let η F = 2 so that the Fermi energy is above the conduction hand by approximately 52 mev at T = 3 K. 38 Solution * 1/2 2m η dη 2 n = 4 π( ) n = N F ( η ) n 3/2 2 t 1/2 h 1+ exp( η η ) F π For silicon at 3K, N c = 2.8 1 19 cm -3 The Femi-Dirac integral has a value of F 1/2 (2) = 2.3 2 n 19 19 3 (2.8 1 )(2.3) 7.27 1 cm = = π With the Boltzmann approximation n = N c ( Ec EF ) exp( ) n o = 2.8 1 2 cm -3 F

STATISTICS OF DONORS AND ACCEPTORS The probability function of electrons occupying the donor state is Nd nd = 1 Ed EF 1+ exp( ) 2 n d is the density of electrons occupying the donor level 39 N d is the donor concentration E d is the energy of the donor level The factor 1/2 in this equation is a direct result of the spin factor. Each donor level has two possible quantum states (spin orientations). The insertion of an electron into one quantum state, however, precludes putting an electron into the second quantum state.

The density of electrons occupying the donor level is equal to the donor concentration minus the concentration of ionized donors nd = Na N a The concentration of holes in the acceptor states p N a 1 N N a = = a a EF Ea 1+ exp( ) g N a is the concentration of acceptors N a- is the concentration of ionized acceptors g is a degeneracy factor g=4 in silicon and gallium arsenide. 4

41 Complete Ionization and Freeze-Out If Ed EF >> Nd Ed EF nd = 2Nd exp( ) 1 E exp( d EF ) 2 The Boltzmann approximation is also valid ( Ec EF ) n = Nc exp( ) The ratio of electron in the donor state to the total number of electrons in the conduction band plus donor state. ( Ed EF ) 2Nd exp( ) nd = n ( E ) ( ) d + n d EF Ec Ed 2Nd exp( ) + Nc exp( )

n d nd + n 1 = Nc ( Ec Ed ) 1+ exp( ) N d Small value The factor (E c - E d ) is just the ionization energy of the donor electrons. At room temperature, the donor states are essentially completely ionized. => All donor impurity atoms have donated an electron to the conduction band. 42 At room temperature, there is also essentially complete ionization of the acceptor atoms. =>each acceptor atom has accepted an electron from the valence band.

Complete Ionization at T=3 K Fig Energy-band diagrams showing complete ionization of (a) donor states and (b) acceptor states. Partial ionization of donor or acceptor atoms when K<T<3K 43

Freeze-Out at T=K Conduction band Conduction band Valence band Valence band Fig Energy-band diagram at T = K for (a) n-type and (b) p-type semiconductors. N d+ = =>Each donor state must contain an electron. => E F >E d N a- = =>Each acceptor state does not contain electron. => E F >E v 44

CHARGE NEUTRALITY In thermal equilibrium, the semiconductor crystal is electrically neutral. =>The electrons are distributed among the various energy states, creating negative and positive charges, but the net charge density is zero. A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region. An n-type compensated semiconductor occurs when N d > N a, A p-type compensated semiconductor occurs when N a > N d. If N d = N a, it is an intrinsic material. 45

Equilibrium Electron and Hole Concentrations The charge neutrality condition is n + N = p + N + a d n + ( N p ) = p + ( N n ) a a d d n : total electron concentration = thermal electrons + donor electrons p : total hole concentration = thermal holes + acceptor holes p a : the hole concentration in acceptor states n d : the electron concentration in donor states 46 Fig Energy-band diagram of a compensated semiconductor showing ionized and un-ionized donors and acceptors

If we assume complete ionization, n d and p a are both zero n + N = p + N a 2 ni n + Na = + N n Nd Na Nd Na n = + ( ) + n 2 2 d d 2 2 i Two factor to impact n --- the concentration of impurity atoms --- the intrinsic carrier concentration 47

48 Fig Energy-band diagram showing the redistribution of electrons when donors are added

Complete ionization n = N i c ( Ec EFi) exp[ ] Fig Electron concentration versus temperature showing the three regions: partial ionization, extrinsic, and intrinsic. 49 As the temperature increases, additional electron-hole pairs are thermally generated so that the n j term may begin to dominate.

Similarly n 2 i Na p + = + p N d If N a -N d >>n i, then Na Nd Na Nd p = + ( ) + n 2 2 p = Na Nd 2 2 i n 2 2 ni ni = = p ( N N ) a d 5

POSITION OF FERMI ENERGY LEVEL The position of the Fermi energy level is a function of the doping concentrations and as a function of temperature n = N c ( Ec EF ) exp[ ] n c N N N N E c EF = ln d a d a 2 2 where n = + ( ) + n n i 2 2 For an n-type semiconductor, N d >>n i then n Nd 51 n c Ec EF = ln nd As the donor concentration increases, the Fermi level moves closer to the conduction band. EF EFi n = ni exp[ ] E E F n Fi = ln ni

For a p-type semiconductor N v EF Ev = ln p If we assume that Na>>Ni N v EF Ev = ln Na As the acceptor concentration increases, the Fermi level moves closer to the valence band. The difference between the intrinsic Fermi level and the Fermi energy in terms of the acceptor concentration E E p Fi F = ln ni 52

Fig Position of Fermi level for an (a) n-type and (b) p-type semiconductor. Fig Position of Fermi level as a function of donor concentration (n type) and acceptor concentration (p type). 53

E E F n Fi = ln ni p E E = ln Fi F n i n = N i c ( Ec EFi) exp[ ] Fig Position of Fermi level as a function of temperature for various doping concentrations. As T increases, n i F Fi E > E 54

Summary The concentration of electrons and holes n = gc( E) ff ( E) de p = gv( E)[1 ff ( E)] de Using the Maxwell-Boltzmann approximation n ( Ec EF ) ( EF Ev) exp[ ] p = Nv exp[ ] = Nc The intrinsic carrier concentration is n = N N 2 i c v Eg exp[ ] 55

The concept of doping the semiconductor with donor atoms and acceptor atoms to form n-type or p-type material. The fundamental relationship of the hole and electron 2 concentration is n p = n i The electron and hole concentrations is a function of impurity doping concentrations Nd Na Nd Na n = + ( ) + n 2 2 2 2 i Na Nd Na Nd p = + ( ) + n 2 2 2 2 i 56 The position of the Fermi energy level is a function of impurity doping concentration E E F n Fi = ln ni E E p Fi F = ln ni