EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri Second Edition ( 2001 McGraw-Hill) Chapter 5 5.6 Doped GaAs Consider the GaAs crystal at 300 K. a. Calculate the intrinsic conductivity and resistivity. b. In a sample containing only 10 15 cm -3 ionized donors, where is the Fermi level? What is the conductivity of the sample? c. In a sample containing 10 15 cm -3 ionized donors and 9 × 10 14 cm -3 ionized acceptors, what is the free hole concentration? Solution a Given temperature, T = 300 K, and intrinsic GaAs. From Table 5.1 (in the textbook), n i = 1.8 × 10 6 cm -3 , µ e ≈ 8500 cm 2 V -1 s -1 and µ h ≈ 400 cm 2 V -1 s -1 . Thus, σ = en i ( µ e + µ h ) ∴ σ = (1.602 × 10 -19 C)(1.8 × 10 6 cm -3 )(8500 cm 2 V -1 s -1 + 400 cm 2 V -1 s -1 ) ∴ σ = 2.57 × 10 -9 Ω -1 cm -1 ∴ ρ = 1/ σ = 3.89 × 10 8 Ω cm b Donors are now introduced. At room temperature, n = N d = 10 15 cm -3 >> n i >> p. σ n = eN d µ e ≈ (1.602 × 10 -19 C)(10 15 cm -3 )(8500 cm 2 V -1 s -1 ) = 1.36 Ω -1 cm -1 ∴ ρ n = 1/ σ n = 0.735 Ω cm In the intrinsic sample, E F = E Fi , n i = N c exp[−(E c − E Fi )/kT] (1) In the doped sample, n = N d , E F = E Fn , n = N d = N c exp[−(E c − E Fn )/kT] (2) Eqn. (2) divided by Eqn. (1) gives, N d n i = exp E Fn − E Fi kT     (3) ∴ ∆E F = E Fn − E Fi = kT ln(N d /n i ) (4) Substituting we find, ∆E F = (8.617 × 10 -5 eV/K)(300 K)ln[(10 15 cm -3 )/(1.8 × 10 6 cm -3 )] ∴ ∆E F = 0.521 eV above E Fi (intrinsic Fermi level) 5.1 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri c The sample is further doped with N a = 9 × 10 14 cm -3 = 0.9 × 10 15 cm -3 acceptors. Due to compensation, the net effect is still an n-type semiconductor but with an electron concentration given by, n = N d − N a = 10 15 cm -3 − 0.9 × 10 15 cm -3 = 1 × 10 14 cm -3 (>> n i ) The sample is still n-type though there are less electrons than before due to the compensation effect. From the mass action law, the hole concentration is: p = n i 2 / n = (1.8 × 10 6 cm -3 ) 2 / (1 × 10 14 cm -3 ) = 0.0324 cm -3 On average there are virtually no holes in 1 cm 3 of sample. We can also calculate the new conductivity. We note that electron scattering now occurs from N a + N d number of ionized centers though we will assume that µ e ≈ 8500 cm 2 V -1 s -1 . σ = en µ e ≈ (1.602 × 10 -19 C)(10 14 cm -3 )(8500 cm 2 V -1 s -1 ) = 0.136 Ω -1 cm -1 5.7 Degenerate semiconductor Consider the general exponential expression for the concentration of electrons in the CB, n = N c exp − ( E c − E F ) kT     and the mass action law, np = n i 2 . What happens when the doping level is such that n approaches N c and exceeds it? Can you still use the above expressions for n and p? Consider an n-type Si that has been heavily doped and the electron concentration in the CB is 10 20 cm -3 . Where is the Fermi level? Can you use np = n i 2 to find the hole concentration? What is its resistivity? How does this compare with a typical metal? What use is such a semiconductor? Solution Consider n = N c exp[−(E c − E F )/kT] (1) and np = n i 2 (2) These expressions have been derived using the Boltzmann tail (E > E F + a few kT) to the Fermi − Dirac (FD) function f(E) as in Section 5.1.4 (in the textbook). Therefore the expressions are NOT valid when the Fermi level is within a few kT of E c . In these cases, we need to consider the behavior of the FD function f(E) rather than its tail and the expressions for n and p are complicated. It is helpful to put the 10 20 cm -3 doping level into perspective by considering the number of atoms per unit volume (atomic concentration, n Si ) in the Si crystal: n at = (Density)N A M at = (2.33 × 10 3 kg m -3 )(6.022 × 10 23 mol − 1 ) −3 −1 (28.09 ×10 kg mol ) i.e. n at = 4.995 × 10 28 m -3 or 4.995 × 10 22 cm -3 Given that the electron concentration n = 10 20 cm -3 (not necessarily the donor concentration!), we see that 5.2 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri n/n at = (10 20 cm -3 ) / (4.995 × 10 22 cm -3 ) = 0.00200 which means that if all donors could be ionized we would need 1 in 500 doping or 0.2% donor doping in the semiconductor (n is not exactly N d for degenerate semiconductors). We cannot use Equation (1) to find the position of E F . The Fermi level will be in the conduction band. The semiconductor is degenerate (see Figure 5Q7-1). (a) (b) E Fp E v E c E Fn E v E c CB VB CB E g(E) Impurities forming a band Figure 5Q7-1 (a) Degenerate n-type semiconductor. Large number of donors form a band that overlaps the CB. (b) Degenerate p-type semiconductor. 50 100 1000 2000 10 15 10 16 10 17 10 18 10 19 10 20 Electrons Holes Dopant Concentration, cm -3 Figure 5Q7-2 The variation of the drift mobility with dopant concentration in Si for electrons and holes at 300 K. Take T = 300 K, and µ e ≈ 900 cm 2 V -1 s -1 from Figure 5Q7-2. The resistivity is ρ = 1/(en µ e ) = 1/[(1.602 × 10 -19 C)(10 20 cm -3 )(900 cm 2 V -1 s -1 )] ∴ ρ = 6.94 × 10 -5 Ω cm or 694 × 10 -7 Ω m Compare this with a metal alloy such as nichrome which has ρ = 1000 nΩ m = 10 × 10 -7 Ω m. The difference is only about a factor of 70. 5.3 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri This degenerate semiconductor behaves almost like a “metal”. Heavily doped degenerate semiconductors are used in various MOS (metal- oxide- semiconductor) devices where they serve as the gate electrode (substituting for a metal) or interconnect lines. 5.8 Photoconductivity and speed Solution t A B t = 0 Time t ′ Time G ph Laser on t = 0 Laser off t = 10 µ s τ A τ B τ A τ B τ A G ph τ B G ph t ′ = 0 Consider two p -type Si samples both doped with 10 15 B atoms cm -3 . Both have identical dimensions of length L (1 mm), width W (1 mm), and depth (thickness) D (0.1 mm). One sample, labeled A , has an electron lifetime of 1 µ s whereas the other, labeled B , has an electron lifetime of 5 µ s. a . At time t = 0, a laser light of wavelength 750 nm is switched on to illuminate the surface ( L × W ) of both the samples. The incident laser light intensity on bot h samples is 10 mW cm -2 . At time t = 50 µ s, the laser is switched off. Sketch the time evolution of the minority carrier concentration for both samples on the same axes. b . What is the photocurrent (current due to illumination alone) if each sample is connected to a 1 V battery? a Figure 5Q8-1 Schematic sketch of the excess carrier concentration in the samples A and B as a function of time from the laser switch-on to beyond laser switch-off. 5.4 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri b From the mobility vs. dopant graph (Figure 5Q8-2), µ h = 450 × 10 -4 m 2 V -1 s -1 and µ e = 1300 × 10 -4 m 2 V -1 s -1 . Given are the wavelength of illumination λ = 750 × 10 -9 m, light intensity I = 100 W/m 2 , length L = 1 mm, width W = 1 mm, and depth (thickness) D = 0.1 mm. 50 100 1000 2000 10 15 10 16 10 17 10 18 10 19 10 20 Dopant Concentration, cm -3 Electrons Holes Figure 5Q8-2 The variation of the drift mobility with dopant concentration in Si for electrons and holes at 300 K. The photoconductivity is given by (see Example 5.11 in the textbook): ∆ σ = e η I λ τ µ e + µ h () hcD where η = 1 is the quantum efficiency. Assume all light intensity is absorbed (correct assumption as the absorption coefficient at this wavelength is large). The photocurrent density and hence the photo current is given by: ∆J = ∆I/A = E∆ σ Substitute: ∆I = W × D() V L     e η I λ τ µ e + µ h () hcD ∴ ∆I = WVe η I λ τ µ e + µ h () Lhc The photocurrent will travel perpendicular to the W × D direction, while the electric field E will be directed along L. Substituting the given values for sample A (electron lifetime τ A = 10 -6 s, voltage V = 1 V): ∆I A = 0.001 m ()1 V()1.602 × 10 −19 C () 1()100 W/m 2 () 750 ×10 −9 m () 10 −6 s () 0.13 m 2 V s + 0.045 m 2 V s      0.001 m ()6.626 ×10 −34 J s () 3.0 ×10 8 m/s () ∴ ∆I A = 1.06 × 10 -5 A The photocurrent in sample B can be calculated with the same equation, using the given value of τ B = 5 × 10 -6 s. After calculation: ∆I B = 5.29 × 10 -5 A 5.5 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri 5.6 . N d n i = exp E Fn − E Fi kT     (3) ∴ ∆E F = E Fn − E Fi = kT ln(N d /n i ) (4) Substituting we find, ∆E F = (8.617 × 10 -5 eV/K)(300 K)ln[(10 15 cm -3 )/(1.8 × 10 6