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ENEE 313, Spr '09 Midterm II Solution

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ENEE 313, Spr '09 Midterm II Solution ENEE 313, Spr ’09 Midterm II Solution PART I—DRIFT AND DIFFUSION, 30 pts 1. We have a silicon sample with non-uniform doping. The sample is 200 µm long: In the figure, L = 200 µm= 0.02 cm. At the x = 0 edge, the sample is doped at an acceptor density of 17 3 16 3 NA(0) = 10 1/cm . At the x = L edge, the sample is doped at NA = 5 × 10 1/cm . In between, the doping density varies linearly, as shown in the figure (y-axis not to scale). (a) (2 pts.) What type is the material? The material is p-type. (b) (10 pts.) Sketch the equilibrium hole concentration p0(x). Approximately calculate and roughly sketch the equilibrium electron concentration n0(x). (Hint: The doping 17 18 concentration can be written in terms of x as NA(x) = 10 − 2.5 × 10 x, where x is given in units of cm.) p0(x) ≈ NA(x) 2 20 ni 2.1 × 10 n0(x) = = 17 18 p0 10 − 2.5 × 10 x 3 ⇒ n0(0) = 2.1 × 10 ; n0(L) = 4200 (See next page for the sketch.) (c) (8 pts.) Assume this sample is at equilibrium. Mark the directions of diffusion for the majority and for the minority carriers. Mark the direction of the electric field Ebi(x) which must be built-in to oppose this diffusion current at equilibrium. The majority carriers are the holes, and they will diffuse towards +ˆx, as the h+ concentration decreases in that direction. The minority carriers are the electrons, and they will diffuse towards −xˆ, as the e− concentration decreases in that direction. This will create a hole diffusion current and an electron diffusion current both in the +ˆx direction, and an opposing drift current must balance this for the sample to be at equilibrium. Therefore the built-in electric field will point in the −xˆ direction. 1 (d) (10 pts.) Calculate the magnitude of this electric field Ebi(0) at x = 0. At equilibrium: Jp,diff + Jp,drift = 0 ⇒ |Jp,diff | = |Jp,drift| ; at x = 0 dp (x) ⇒ |[qD 0 | ]| = qµ p (0)E(0) p dx x=0 p 0 kT ⇒ µ × 2.5 × 1018 = 1017µ E(0) q p p ⇒ E(0) = 0.026 × 2.5 × 1018/1017 = 0.65 V/cm (e) (BONUS: 5 pts.) Calculate the built-in electric field Ebi(x) throughout the sample and sketch it. 2 PART II—EXCESS CARRIER CONCENTRATIONS, 20 pts. 18 3 2. A uniformly-doped n-type silicon sample has ND = 10 1/cm . At room temperature, 10 3 2 2 ni = 1.45 × 10 1/cm . Use µn=1200 cm /V.s, µp=400 cm /V.s. The carrier lifetimes are −7 τn = τp = 10 sec. Let a fixed excess hole concentration ∆pn = δpn(x = 0) be injected and maintained at one edge of the sample, at x = 0. Then the excess hole concentration profile throughout the sample will be given by −x δp(x) = ∆pn exp ( ) Lp (a) (10 pts.) Find the ∆pn level required to maintain a hole diffusion current density −5 2 2 Jp,diff = 10 A/cm =10 µA/cm at the edge x = 0 at room temperature (T = 300 K, kT/q ≈ 0.026 V). dp(x) dδp(x) J = −qD = −qD (since p is uniform.) p,diff p dx p dx 0 −1 (−x/Lp) = −qDp∆pn e Lp Dp −5 2 = q ∆pn = 10 A/cm Lp 2 Dp = Vthermalµp = 0.026 × 400 = 10.4 cm /sec q √ −6 −3 Lp = Dpτp = 10.4 × 10 = 1.02 × 10 cm 10.4 ⇒ J = 1.6 × 10−19 ∆p = 10−5 p,diff 0.00102 n 10−5 × 0.00102 ⇒ ∆p = = 6.13 × 109 1/cm3 n 1.6 × 10−19 × 10.4 (b) (10 pts.) While maintaining this excess injection level, a constant electric field E is applied across the sample. Starting from the hole current continuity equation, derive the differential equation that will govern δp(x) throughout the sample. (Hint: You do not need to insert numbers in this derivation, so simply use δp(x) to show the excess electron concentration. Remember that p(x) = p0 + δp(x), where p0 is the equilibrium hole concentration. Account for all hole current components.) Current continuity equation for holes: dp(x, t) 1 dJ (x, t) δp(x, t) = − p − dt q dx τp We have to define the hole current Jp to use in this equation. Since there is both a non-uniform hole distribution and an electric field in question here, both diffusion and drift current components must be present: 3 Jp(x, t) = Jp,drift(x, t) + Jp,diff (x, t) dp(x, t) = qµ p(x, t)E(x, t) − qD p p dx dδp(x, t) = qµ p(x, t)E(x, t) − qD , p p dx since p(x, t) = p0 + δp(x, t), and p0 is constant in x and in t for a uniformly-doped sample. This also implies dp(x, t)/dt = dδp(x, t)/dt in the left-hand side of the current continuity equation; therefore that equation becomes: dδp(x, t) 1 d dδp(x, t) δp(x, t) = − [qµp(p0 + δp(x, t))E(x, t) − qDp ] − dt q dx dx τp d2δp(x) d δp(x, t) = Dp 2 − [µpδp(x)E(x)] − dx dx τp The electric field is specified to be constant throughout the sample; therefore the full differential equation which governs δp(x, t) in this system is dδp(x, t) d2δp(x) dδp(x, t) δp(x, t) = Dp 2 − µpE − dt dx dx τp PART III—PN-JUNCTION DIODES, 35 pts. 3. Consider pn-junction structures. (a) (7 pts.) Sketch an example junction, built-in electric field (indicating its direction) and band energy diagram at equilibrium. Indicate conduction and valence bands, intrinsic Fermi level and Fermi level in your sketch. (b) (7 pts.) Qualitatively explain the formation of the depletion region and the built-in potential/electric field in the pn-junction. Use terms such as ”diffusion,” ”dopants” and ”recombination” in your explanation. At the junction of the two materials, holes diffuse from the p-type material towards the n-type material, and recombine with the electrons diffusing from the n-type ma- terial towards the p-type material, leaving behind uncompensated ionized dopants in a depletion region (“depleted” of mobile charges): Positively-charged donors on the n-side and negatively-charged acceptors on the p-side. These revealed charges set up an electric field between them, pointing from the n-side of the depletion region to the p-side. This electric field direction sets up a drift current for carriers of either type which opposes their diffusion current, and its magnitude, at equilib- rium, is the right magnitude to completely balance the diffusion current. Thus, at equilibrium there is no net current flowing through the junction. 4 For the rest of this question, consider a silicon pn-junction diode with the p-type bulk 18 3 doped uniformly at NA = 5×10 1/cm and n-type bulk doped uniformly at ND = 5× 15 3 10 3 2 10 1/cm . ni = 1.45×10 1/cm at room temperature. Dn = 18 cm /s in the p-side, 2 and Dp = 12 cm /s in the n-side. τn = τp = 1 µs on either side. Assume the bulk regions are much longer compared to the diffusion lengths. (c) (7 pts.) Calculate the built-in potential V0 and the equilibrium depletion region width WD at room temperature. Which side of the junction is the wider part of the depletion region? Why? (You do not need to calculate the depletion region sizes on the two sides of the junction, just comment on the relative size.) 18 15 kT NAND 5 × 10 × 5 × 10 V0 = ln ( 2 ) = 0.026 ln ( 20 ) = 0.843 V q ni 2.1 × 10 s s −12 2SiV0 NA + ND 1 × 2.07 × 10 × 0.843 −16 WD = = −19 × 2.002 × 10 q NAND 1.6 × 10 = 4.67 × 10−5 cm = 46.7 µm (1) The n-side is doped three orders of magnitude lighter than the p-side; therefore the much greater part of this depletion region width is on the n-side. (d) (4 pts.) Remember that we calculated the ideal diode current by capturing the minority diffusion currents at the depletion region edges. Show which is more significant in size for this particular diode—electron diffusion current at the p-side depletion region edge, or hole diffusion current at the n-side depletion region edge? (Hint: Identify which term in the J0 expression comes from which current. Also, Dn/Ln will have the same order of magnitude as Dp/Lp.) Dp Dn J0 = q( pn0 + np0) Lp Ln 5 2 ni 5 3 Here, pn0 = = 0.42 × 10 1/cm ND 2 ni 2 3 np0 = = 0.42 × 10 1/cm NA ⇒ pn0 > np0 ⇒ The hole injection into the n-side will yield a higher diffusion current than the electron injection into the p-side. (e) (10 pts.) What is the current density at 0.5 V applied bias? At -0.2 V applied bias? (Hint: You may ignore the component that you found to be less significant in part (c).) From the discussion above, we can ignore the (Dn/Ln)np0 component. Then, with 2 q − Dp = 18 cm /sec and Lp = Dpτp = 3.46 × 10 3 cm, Dp/Lp = 3468.21 cm/sec and −19 5 −11 2 J0 = 1.6 × 10 × 3468.21 × 0.42 × 10 = 2.33 × 10 A/cm 2 ⇒ At VA = 0.5 V,JD = J0(exp (0.5/0.026) − 1) = 0.0052 A/cm −11 2 At VA = −0.2 V,JD ≈ −J0 = −2.33 × 10 A/cm .
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