# collisions of electrons. In semiconductor, in certain temperature ranges the conductivity increases rapidly by increasing temperature

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1 1.9. Temperature Dependence of Semiconductor Conductivity Such dependence is one most important in semiconductor. In metals, Conductivity decreases by increasing temperature due to greater frequency of RESISTIVITY Intrinsic Material Extrinsic Material Conductivity T ( o K) Figure1.23 Temperature Dependence of Semiconductor Conductivity collisions of electrons. In semiconductor, in certain temperature ranges the conductivity increases rapidly by increasing temperature 1

2 In figure 1.23, at low temperature, the charge carriers are frozen and the resistivity is extremely high, as the temperature raises, increasing fraction of carriers to ionize and the resistivity decreases rapidly because of increasing the ionized charges. When temperature is sufficiently high, most of dopants are completely ionized. The Conductivity begins to decrease and the resistivity is increased again just as in metals. At still higher temperature, there is further sharp decrease in resistivity due to appreciable excitation of all carriers and crossing the energy gap. 2

3 Example (31) Band gap of Si depends on the temperature as Eg = 1.17 ev T 2 / T Find a concentration of electrons in the conduction band of intrinsic (undoped) Si at T = 77 K if at 300 K n i = cm 3. Solution: Therefore, n 1 T 2 = n 1 T 1 T 2 T 1 n i 2 = N c N v exp E g KT T 3 exp E g KT 3 2 exp E gt 2 + E gt 1 2 K T 2 2 K T 1 Putting the proper values in the formula we obtain that n i (77K) cm 3. Solved Examples Example (32) Calculate the intrinsic carrier density in germanium, silicon and gallium arsenide at 300, 400, 500 and 600 K. Solution The intrinsic carrier density in silicon at 300 K equals: n i 300K = N c N v exp E g 2 KT = x x x exp x = x 10 9 m 3 Similarly, one finds the intrinsic carrier density for germanium and gallium arsenide at different temperatures, yielding: Germanium Silicon Gallium Arsenide 300 K 2.02 x x x K 1.38 x x x K 1.91 x x x K 1.18 x x x

4 When an electron introduced without subjected external electric field at absolute temperature in a perfect semiconductor sample. The electron will move freely random through the crystal as shown in figure The Concept of Mobility Figure1.24. Random Electron Path at no Field Figure1.25. Directed Electron Path under Field 4

5 This is a general property associated with perfect periodic structures. At room temperature, valance electrons are liberated by random motion through the crystal. The random velocity of an electron at room temperature is about 10 7 cm/sec. The mean free path of the electrons between two collisions is in the order of 10-5 about 500 times the distance between neighboring atoms. The average time between collisions is about second. Under thermal equilibrium condition, the random motion of electrons leads to zero current in any direction. Thermal vibration introduced treated as a particle (phonons). Its collisions with electrons and holes called scattering. The scattering phenomena increased as temperature increases for a maximum scattering velocity. 5

6 6 If we apply electric field to the crystal, the electrons gain a force q toward the positive pole as shown in fig.1.25, and moves with acceleration given by: q ε a = Eq m After time ( ), the electrons suffer of collisions decreases velocity (θ) between each two consecutive collisions as θ = a, produces what is called drift velocity (θ d ). θ d = θ = μ ε Eq The term (q /m) gives the mobility of the charge carriers, then (Eq.1.51) becomes: θ d = μ ε Eq Normally, n > p for any material. For example, in silicon n = 1800, p = 400 cm 2 /V.sec, and in germanium we find that n = 3800, p = 1800 cm 2 /V.sec.

7 FIELD DEPENDENCE ON MOBILITY By increasing the subjected field, the drift velocity increases, due to the increasing of kinetic energy of the electrons. When electric field is higher increased, a critical value developed and reaches maximum scattering velocity; this velocity is not more increased by increasing electric field, figure (M/S) 5x10 4 Maximum Scattering Velocity For Si Figure 1.26 Field Dependence on Mobility 7

8 Example (36) Electrons in silicon carbide have a mobility of 1400 cm 2 /V-sec. At what value of the electric field do the electrons reach a velocity of 3 x 10 7 cm/s? Assume that the mobility is constant and independent of the electric field. What voltage is required to obtain this field in a 5 micron thick region? How much time do the electrons need to cross the 5 micron thick region? Solution: The electric field obtained from the mobility and the velocity: ε = μ ν = 1400 = 30 kv/cm 3 x 107 Combined with the length one finds the applied voltage. V = εl = 30,000 x 5 x 10-6 = 0.15 V The transit time equals the length divided by the velocity: t r = L/v = 5 x 10-6 /3 x 10 7 = 16.7 ps Solved Examples Example (37) The resistivity of a silicon wafer at room temperature is 5 Ωcm. What is the doping density? Find all possible Solution: starting with a initial guess that the conductivity is due to electrons with a mobility of 1400 cm 2 /V-s, the corresponding doping density equals: N D = n = 1 q μ n ρ = x x 1400 x 5 = 8. 9 x cm 3 The mobility corresponding to this doping density equals μ n = μ max + μ max μ min 1 + N D N ρ a = 1366 cm 2 /vs Since the calculated mobility is not the same as the initial guess, this process must be repeated until the assumed mobility is the same as the mobility corresponding to the calculated doping density, yielding: N d = 9.12 x cm -3 and µ n = 1365 cm 2 /V-s For p-type material one finds: 8

9 RECOMBINATION CENTERS MECHANISM It is an electronic state in the energy gap of semiconductor materials. These states considered as imperfections in the crystal. Metallic impurities are capable of introducing such energy states in the energy gap. The recombination rate is affected by volume of impurities, and surface imperfections. There are three main factors affecting the mobility of charge carriers in semiconductors, they are: Temperature: As temperature increases, the thermal kinetic energy increases the vibration of atoms and the charge carriers suffer from Collisions, the dependence of mobility in temperature given by: μ L = KT 3Τ2 Eq

10 Impurities: The scattering of charge carriers results from the presence of ionized donors or acceptors or impurities. This charged centers will deflect the motion of carriers by the electrostatic forces between two bodies, so the density of such centers affect the velocity; it is also being noted that, the impurity scattering decreases as temperature increases. μ I = KT 3 Τ 2 Eq N I Where N I, is the density of ionized centers. Appreciable reduction of mobility results. For example, in germanium the hole mobility falls to 900 cm 2 /V.S. (Half its maximum) when the resistivity is Materials in this order of impurity are actually employing in semiconductor devices. 10

11 Dislocations: Dislocation is atomic misfit, where atoms not probably arranged, so it has a considerable role of scattering carriers. For example, in germanium, the dislocations behave as acceptors, and the mobility affects by: μ D = KT Eq Now if we combine these three parameters, we have a general expression to determine such effects in mobility of charge carriers. 1 μ = μ L μ I 1 μ D = α L T Τ α I T Τ α D T 1 Eq Only the first two terms are normally important, because they are depending more on temperatures. 11

12 Vacancies (missed) atom in Covalent Bonding in a Semiconductor Crystal Intertitles atom in the Covalent Bonding in a Semiconductor Crystal Missed Atom One atom more 12

13 Example (38) Calculate the electron and hole densities in an n-type silicon wafer (N d = cm -3 ) illuminated uniformly with 10 mw/cm 2 of red light (E ph = 1.8 ev). The absorption coefficient of red light in silicon is 10-3 cm -1. The minority carrier lifetime is 10 ms. Solution The generation rate of electrons and holes equals: G n = G p = P OPT E ph q = x 1. 6 x = 3. 5 x cm 3 s 1 Where, α is absorption coefficient, P opt illumination power, E pt is the red light, where the photon energy was converted into Joules. The excess carrier densities then obtained from: δ n = δ p = τ p G p = 10 x 10 3 x 3. 5 x = 3. 5 x cm 3 The excess carrier densities then obtained from: So that the electron and hole densities equal: n = n o + δ n = x = 3. 5 x cm 3 Solved Examples Example (39) What are the approximate thermal velocities of electrons and holes in silicon at room temperature? SOLUTION: Assume T = 300 K and recall m n = 0.26 m 0. V th = Kinetic energy = 1 2 m n V 2 th = 3 2 KT 3 KT m = 3 x x J 300 K K x = 2. 3 x m 5 mτ s = 2. 3 x 10 7 x 9. 1 x 10 cmτ s Note that 1 J = 1 kg m 2 /s 2. Using m p = 0.39 m 0 instead of m n, one would find the hole thermal velocity to be cm/s. Therefore, the typical thermal velocity of electrons and holes is cm/s, which is about 1000 times slower than the speed of light and 100 times faster than the sonic speed. 13

14 1.11. THE DRIFT CURRENT Transport of charge carriers under electric field produces a drift current. The current flow in a sample of figure (1.27), having electron concentration (n) given by: I n = q n θ d A Eq Substitute in the drift velocity, gives: I n = q n μ n ε A Eq L Cross Sectional Area Drift Electron Current I n V.Figure Current Conduction in a Semiconductor Bar 14

15 we know, ε = V L, R = V, σ = qn μ I n We can find the resistance of the sample due to the electron component as: ε L R n = q A n μ n ε = L L = ρ q A n μ n n A Eq The same steps can be done to determine the drift current due to the hole component in p-type material, I p = q p μ np ε A ρ p = σ p = qp μ p 1 Eq qpμ 15

16 In semiconductor, both carriers are included, so: σ T = σ n + σ p = q n μ n + p μ p I T = q A ε n μ n + p μ p Eq In intrinsic, n = p = n i, then, σ j = q n i n μ n + p μ p Eq Note that n > p, so in an intrinsic material the electrons contribute more to conductivity than holes. It seems that the conduction is due to electrons, so the conduction in intrinsic material considered as in n-type material. For extrinsic material, the conductivity given by: σ n = q n i μ n where n n > p n & σ p = q p μ p here p p > n p 16

17 1.12. THE DIFFUSION CURRENT Transport of charges in semiconductors called diffusion. In a semiconductor bar, the concentration of charge carrier is not uniform. Diffusion of electrons or holes results from their movement from higher concentration to lower concentration with gradient d/dx where concentration of carriers is not distributed uniform and varies with distance x. sectional area A is, N/m 3 Diffusion Figure 1.28 Electron Diffusion Gradient Electron Concentration Of Electron Current I n (m 2 ) 17

18 Charges (electrons or holes) are in random motion because of their thermal energy. Their motion gives rise to a current flow known as diffusion current as shown in figure 1.28, diffusion flux obeys Fick s first law: N F = D Eq x F is flux of carriers and defined as the number of carriers passing through m 2 /sec. Net transport of charges across the surface constitutes current, it is proportional to the concentration gradient. Current passing through cross I = q A F Eq So that, the diffusion current due to electrons given by: N I n = q A D n Eq

19 19 And, the diffusion current due to holes is given by: P I p = q A D p Eq x Where D n, D p are the diffusion constants for electrons and holes respectively The negative sign in (Eq.1.66) indicates that the hole current flows in the direction opposite to the gradient of the holes. The diffusion constants related with the mobility by Einstein relationship as: D p = D n = V μ p μ T = KT n q = T Eq For lightly doped Si at room temperature = 39D. The total current of electrons and holes components given by the summing of its diffusion and drift currents as: I n = qa n μ n E + D n n x Eq I p = qa p μ p E + D n p x Eq

20 Example (43) An abrupt silicon p-n junction (N a = cm -3 and N d = 4 x cm -3 ) is biased with V a = 0.6 V. Calculate the ideal diode current assuming that the n-type region is much smaller than the diffusion length with W n = 1 µm and assuming a "long" p-type region. Use µ n = 1000 cm 2 /V-s and µ p = 300 cm 2 /V-s. The minority carrier lifetime is 10 µs and the diode area is 100 µm by 100 µm. Solution: The current calculated from: I = q A D nn po + D pp no V o e Vt 1 L n W n With D n = µ n V t = 1000 x = 25.8 cm 2 /V-s D P = µ P V t = 300 x = 7.75 cm 2 /V-s n p0 = n 2 i / N a = /10 16 = 10 4 cm -3 p n0 = n 2 i / N d = /4 x = 2.5 x 10 3 cm -3 L n = D n τ n = x 10 5 = 161 μm Yielding I = 40.7 µα Solved Examples Note that the hole diffusion current occurs in the "short" n-type region and therefore depends on the quasi-neutral width in that region. The electron diffusion current occurs in the "long" p-type region and therefore depends on the electron diffusion length in that region. Example (44) For a p + -n Si junction the reverse current at room temperature is 0.9 na/cm 2. Calculate the minority-carrier lifetime if N d = cm 3, n i = cm 3, and μ p = 450 cm 2 V 1 s 1. Solution: For a p + -n junction J S = e p D p L p 2 D p = e n i N d L p = e n 2 i N d D p τ p 1 2 Taking into account that μ = ed/kt, we finally get τ p = s. 20

21 LIFE TIME OF CHARGE CARRIERS In intrinsic material n = p, and due to thermal excitation new hole - electron pairs are produced and others are disappeared as a result of recombination after a time n, p (mean lifetime). The lifetime of carriers is in the range from nanosecond to some hundreds of microseconds. Consider a Si bar of N-type illuminated by light of the proper frequency, as a result, n and p concentration will increase by the same amount, so: p n0 p n0 = n n0 n no Eq Where p no, n no are the equilibrium concentration of holes and electrons, and p no-, n - no represents the carrier concentration during the process. If the source of light is turned off, the carrier concentration will return to its equilibrium values exponentially and with time constant = n = p, and we can write:

22 p n p n0 = p n0 p n0 e t Τ τ Eq n n n n0 = n n0 n n0 e t Τ τ Eq From the equations above, the rate of concentration change for hole is: pn t = p n p n0 = p n p n0 Eq τ t And, the rate of concentration change for electrons is: n n t = n n n n0 = n n n n0 Eq τ t The quantity p n - p no or n n - n no represents the excess carrier density, and the rate of change of excess density is proportional to the density itself. The (-) sign indicates that the change is decreases in case of recombination. 22

23 1-14 CONTINUITY EQUATIONS carrier concentration is a function of both time and distance, upon fact that charge created or destroyed. Consider a volume of area A and length dx, fig.1.29, in which the hole concentration is P, and P/ p is the decreasing of hole concentration by recombination, so, the change in hole P Holes/m 3 + Cross Section Area A x Figure 1.29 Carriers concentration in volume dx 23

24 24 concentration by recombination within volume (Adx) is: q A x p τ p Eq And if g is the thermal rate for generation hole - electron pair/unit volume, then the increase of concentration by generation is, q A x g Eq In general, current vary with distance, current entering volume at x is I and leaving at x+dx is I+dI, then change in current within volume is di. Let the increase in current due to diffusion within volume is q A dx p/ t, since. charge tends to equilibrium, then, p q A x t = q A x p τ p + q A g x I Eq In addition, since the hole- current is the sum of the diffusion current and the drift current components, as:

25 25 p I p = q A D p x + q A p μ p ε Eq If the semiconductor is in thermal equilibrium and no electric field is biased, then the hole density will attain a constant value p o, under this condition, I = 0 and p/ t = 0, = 0, so that: g = Eq p 0 τp Equation 1.79 indicates those rate at which holes generated thermally equal the rate of recombination under equilibrium condition. Combining equations 77,78,79, we obtain the continuity equation or equation of conservation of charges: p t = p p 0 2 p + D τ p p x 2 μ p ε p Eq x If we consider holes in n-type, Eq.1.80 becomes: p n t = p n p n0 τ p + D p 2 p n x 2 μ p p n ε x Eq

26 The general continuity equation 1.81 be considered for three special cases. 1.When concentration is independent of x at zero bias: If = 0 and concentration is independent of x, Eq.1.81 will be rewrite as: p n t = p n p n0 τ p The solution of (Eq.1.82) is: p n p n0 = Eq p no e Τ p n0 t τ Eq

27 2.When concentration is independent of t at zero bias: If = 0, and a steady state has been reached at no time, so that p n / t = 0, Eq.1.81 will be rewritten again as: 2 p n D p x 2 = p n p n0 = 2 p n τ p x 2 = p n p n0 Eq D p τ p The solution of Eq.1.84 is: p n p n0 = K 1 e xτl p + K 2 e x Τ L p Eq Where K 1, K 2 are integration constants, L p = D p τ p and it represents the distance into the semiconductor at which the injected concentration falls to 1/e of its value at distance x = 0. 27

28 3. When concentration varies sinusoidal with t at zero bias : If = 0, and the injected concentration varies sinusoidal with angular frequency ( ), p n x, t = p n x e j ω t Eq If Eq.1.86 substituted in to the continuity Eq.1.81, results: j ω p n x = p n x 2 p n + D p x 2 Eq τ p 2 p n x 2 = 1 + j ω τ p p 2 n Eq L p And at F = 0, Eq.1.88 rewritten again as: 2 p n x 2 = p n Eq L p 28

29 Example (48) Consider n-type silicon with N d = cm 3 at T = 300 K. light source is turned on at t = 0. The source illuminates the semiconductor uniformly, generating carriers at the rate of G n = G p = cm 3 s 1. There is no applied field. (a) Write down the continuity equation and solve it to get the expression for the excess minority carrier concentration, δp t, as a function of time for t 0. Solution: When there is no applied electric field, the carrier distribution is diffusion driven. The continuity equation for the minority carrier p x, t = 1 J p x, t + G t q x p x, t R p x, t Then reduces to, δp t = G p δ p τ p With the general solution δ p t = A exp t π p + G p τ p Using the initial condition (before the light was turned on) that δp t = 0, then we found that A = G p τ. The full solution then is, Solved Example δ p t = G p τ p 1 exp t τ p (b) As t >, the system will approach steady state. When the steady state excess carrier concentration is cm 3, get the minority carrier lifetime, τ p. Solution: The system will approach steady state as t >, Evidently, the steady state carrier density is given δp t t = G p τ p. The τ p must then take on the value τ p = 5 x G p. With G p = cm 3 s 1, then the minority (hole) carried lifetime must be τ p = s. (c) Determine the time at which the excess carrier concentration becomes half of the steady state value, δp t t that you calculated in (b). Solution: The value at which exp t τ = 1 2 so, t = ln 2 τ p = x 5 x 10 6 s 29

30 1-15. Flux of Charged Spices To determine the formula of flux (F), F = D P o T E n T I A l N x (a) (W) F 1 F 2 F 3 F 4 (x-a/2) x x+a/2 (x-a) Figure (b) x+a consider the motion of positively charged impurities in a crystal. The atom of the crystal forms a series of potential hills, which hinder the motion of the charged impurities as shown. The height of potential barrier (W) is in the order of some of electron volt. In most material, the distance between successive potential barrier (a) is in the order of the lattice spacing which is in some angstroms (1 o A = meter). Figure (a). Distance Fig. (1-30) Model of ionic motion within crystal, Potential distribution with and without bias 30

31 31 If a constant field is applied, the potential distribution as a function of distance is shown in figure b, this make flow of positively charged particles to the right easier than left. To calculate the flux (F) at a position (x), it is average of fluxes at positions (x a/2) and at (x + a/2), these two fluxes given by (f 1 f 2 ), and (f 3 f 4 ).Consider component (F 1 ), is product of: 1. Intensity per unit area for impurities (charges) at the potential valley at (x a). 2. The probability of a jump of any of these impurities (charges) to the next valley at position (x) 3. The frequency of attempted jump (ⱱ)

32 32 Thus, we can write: F 1 = a C x a exp q K T W 1 a ε (ⱱ) Eq (1 90) 2 Where, a C x a is the density per unit area of the particles situated in the valley at (x a). The exponential factor is the probability of a jump from the valley at (x a) to the valley at position (x), and (ⱱ) is frequency of attempted jump note that, the lowering of the barrier due to the electric field (ε). Similar formulas can be written for (F 2 ), (F 3 ), and (F 4 ). By combined them to give a formula for the flux (F) at position (x), with the concentration (C(x ± a)) approximated by ( C(x) ± a CΤ x, We obtain,

33 33 F x = - ⱱ a 2 e qwτkt sinh - q a ε 2 KT Eq (1 91) C x cosh q a ε 2 KT + 2 a ⱱ e Τ qw Kt C As extremely important, limiting from of this equation is obtained for the case when the electric field is relatively small, i.e., ε «KT/q a. In this case, we can expand the cosh and the sinh terms in the equation. Noting that cosh (x) = 1 and sinh (x) = x for x 0, this results in the limiting form of the flux equation for a positively charged spices. F x = D C x + μ ε C Eq. (1 92), Where: D = va 2 e qw Τ KT μ = va2 e qw Τ KT KT q

34 34 Note that, mobility (µ) and the diffusivity (D) related by: D = KT q μ This is the well Known Einstein relationship. A similar derivation made for the motion of negatively charged spices. In addition, we can derive that, D = D o e E a Τ KT Where, (D o ) is the diffusivity of impurities in oxide, (E a ) is the activation energy. Thus, the activation energy corresponds to the energy required to form a silicon space rather than to the energy required to move the impurity. Since silicon to silicon, bunds must be broken to form a space. In fact, activation energy of the diffusivity of acceptor and donor type impurities, and it is in the range of two and three electron volt in germanium.

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