ENEE 313, Spr 09 Midterm II Solution

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1 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 N A (0) = /cm 3. At the x = L edge, the sample is doped at N A = /cm 3. 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 p 0 (x). Approximately calculate and roughly sketch the equilibrium electron concentration n 0 (x). (Hint: The doping concentration can be written in terms of x as N A (x) = x, where x is given in units of cm.) p 0 (x) N A (x) n 0 (x) = n2 i = p x n 0 (0) = ; n 0 (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 E bi (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 ˆx, 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 ˆx direction. 1

2 (d) (10 pts.) Calculate the magnitude of this electric field E bi (0) at x = 0. At equilibrium: J p,diff + J p,drift = 0 J p,diff = J p,drift ; at x = 0 dp 0 (x) [qd p x=0] = qµ p p 0 (0)E(0) kt q µ p = µ p E(0) E(0) = /10 17 = 0.65 V/cm (e) (BONUS: 5 pts.) Calculate the built-in electric field E bi (x) throughout the sample and sketch it. 2

3 PART II EXCESS CARRIER CONCENTRATIONS, 20 pts. 2. A uniformly-doped n-type silicon sample has N D = /cm 3. At room temperature, n i = /cm 3. Use µ n =1200 cm 2 /V.s, µ p =400 cm 2 /V.s. The carrier lifetimes are τ n = τ p = 10 7 sec. Let a fixed excess hole concentration p n = δp n (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 δp(x) = p n exp ( x ) (a) (10 pts.) Find the p n level required to maintain a hole diffusion current density J p,diff = 10 5 A/cm 2 =10 µa/cm 2 at the edge x = 0 at room temperature (T = 300 K, kt/q V). dp(x) J p,diff = qd p = qd dδp(x) p 1 = qd p p n e ( x/lp) = q D p p n = 10 5 A/cm 2 D p = V thermal µ p = = 10.4 cm 2 /sec = D p τ p = = cm J p,diff = p n = 10 5 p n = (since p 0 is uniform.) = /cm 3 (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) = p 0 + δp(x), where p 0 is the equilibrium hole concentration. Account for all hole current components.) Current continuity equation for holes: dp(x, t) dt = 1 q dj p (x, t) δp(x, t) τ p We have to define the hole current J p 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

4 J p (x, t) = J p,drift (x, t) + J p,diff (x, t) = qµ p p(x, t)e(x, t) qd p dp(x, t) = qµ p p(x, t)e(x, t) qd p dδp(x, t) since p(x, t) = p 0 + δp(x, t), and p 0 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) dt = 1 d q [qµ dδp(x, t) p(p 0 + δp(x, t))e(x, t) qd p ] = D p d 2 δp(x) 2 d [µ δp(x, t) pδp(x)e(x)] τ p δp(x, t) τ p 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) dt d 2 δp(x) dδp(x, t) δp(x, t) = D p µ 2 p E τ p 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 material 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 equilibrium, is the right magnitude to completely balance the diffusion current. Thus, at equilibrium there is no net current flowing through the junction. 4

5 For the rest of this question, consider a silicon pn-junction diode with the p-type bulk doped uniformly at N A = /cm 3 and n-type bulk doped uniformly at N D = /cm 3. n i = /cm 3 at room temperature. D n = 18 cm 2 /s in the p-side, and D p = 12 cm 2 /s in the n-side. τ n = τ p = 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 V 0 and the equilibrium depletion region width W D 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.) V 0 = kt q ln ( N AN D ) = ln ( ) = V n 2 i ɛSi V 0 N A + N D W D = = q N A N D = 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 J 0 expression comes from which current. Also, D n /L n will have the same order of magnitude as D p /.) J 0 = q( D p p n0 + D n L n n p 0) 5

6 Here, p n0 = n2 i N D = /cm 3 n p0 = n2 i = /cm 3 N A p n0 > n p0 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 (D n /L n )n p0 component. Then, with D p = 18 cm 2 /sec and = D p τ p = cm, D p / = cm/sec and J 0 = = A/cm 2 At V A = 0.5 V, J D = J 0 (exp (0.5/0.026) 1) = A/cm 2 At V A = 0.2 V, J D J 0 = A/cm 2. (f) (BONUS: 5 pts.) A diode to be used in a high-frequency circuit should respond quickly to any change in bias. This implies the excess minority carrier concentrations injected to either side of the junction should be built up or dispelled quickly. This can be achieved by having the injected minority concentrations, δp n and δn p, as small as possible for a given bias. Remembering that these injected concentrations are proportional to the equilibrium minority concentrations on either side, if you could alter the doping for the p-side, what would you choose for a faster-switching diode? A higher doping would mean smaller n p, which would help the diode switch faster. (2) 6

7 PART IV 15 pts. 4. Fill in the blanks, select the correct answer or indicate True or False for the statements below. Each blank or true/false answer is worth one point. Blanks might contain more than one word. You may choose 15 out of 20 to answer; if you get more than 15 answers right in this section your grade will be 15. (a) A non-uniform carrier distribution across a semiconductor sample creates a diffusion current. (b) (TRUE/FALSE) For a semiconductor sample with a uniform constant acceptor doping level N A, if a uniform donor concentration N D > N A is added to the sample, the hole diffusion current increases.: FALSE (c) Under reverse bias, the net current of the pn-junction depends on the generation, OR generation rate of minority carriers approximately within a diffusion length of the depletion region, OR junction. Therefore, this saturation current does not change with bias. (d) Zener breakdown in a pn-junction is based on the phenomenon of tunneling.. It occurs in reverse bias. (e) The depletion region width narrows with higher forward bias. This is why the diode responds to very high bias voltages as if it was a resistor. (f) Under reverse bias, the electric field across the depletion region of a pn-junction will be higher than the same field in a no-bias case. (g) It is possible to turn a rectifying metal-semiconductor junction to a tunneling OR conducting junction by doping the semiconductor much higher. (h) TRUE/FALSE: With the electron concentration increasing in the +x direction, the electron diffusion current is in the x direction: FALSE (i) If the donor doping of the n-side of a pn-junction is half of the acceptor doping of the p-side, the depletion region width in the n-side is... that in the p-side. (a) a quarter of (b) half of (c) equal to (d) twice (j) Under reverse bias, there is more charge stored in the capacitor formed by the depletion OR junction capacitance of a pn-junction. (k) The charge stored by the capacitor for the depletion (junction) capacitance of a pnjunction is the ionized dopants left behind in the depletion region after all the mobile carriers have been depleted from there. (l) Avalanche breakdown happens when electrons drifting across the depletion region can gain enough kinetic energy to knock out another electron-hole pair, which then also begin to accelerate. For this to be more likely, the diode should be designed with a longer depletion region, which is attained by using lower doping. 7