ECE 305 Spring 2015 ECE 305 Homework SOLUTIONS: Week 8 Mark Lundstrom Purdue University 1) The sketch below shows the carrier concentrations in a PN junction at room temperature. Answer the following questions. 1a) Is the diode forward or reverse biased? Explain your answer. Solution: Forward biased because there are excess electrons on the P-side and excess holes on the N-side. 1b) What is the acceptor concentration on the P-side? 16 -3 Solution: N A = 10 cm 1c) What is the donor concentration on the N-side? 14 -3 Solution: N D = 10 cm 1d) What is the intrinsic carrier concentration? Solution: n p n2 0 0 = i On the P-side: n p = 1016 ×107 = 1023 n = 1023 = 3.16 ×1011 cm-3 0 0 i On the N-side: n p = 1014 ×109 = 1023 n = 1023 = 3.16 ×1011 cm-3 0 0 i n = 3.16 ×1011 cm-3 i ECE-305 1 Spring 2015 ECE 305 Spring 2015 HW8 solutions (continued): 1e) Do low level injection conditions apply? Solution: YES. On the P-side: Δn −x = 1010 << p = 1016 ( p ) 0 On the N-side: Δp x = 1012 << n = 1014 ( n ) 0 1f) What bias (in volts) is applied to the diode? Solution: According the the law of the junction: 2 n qV k T qV k T Δn −x = i e A B = n e A B ( p ) N p0 A 10 k T Δn −xp ⎛ 10 ⎞ V = B ln ( ) = 0.026ln = 0.18 V A q n ⎜ 107 ⎟ p0 ⎝ ⎠ V = 0.18 V A 1g) Which is longer: the electron diffusion length on the P-side of the junction or the hole diffusion length on the N-side of the junction. Solution: The electron diffusion length on the P-side. (The diffusion length is the distance it takes for the minority carrier concentration to return to its value in the bulk. Question: What would the figure look like for a reverse biased diode? 2) This problem concerns the energy band diagram shown below for a diode under bias. ECE-305 2 Spring 2015 ECE 305 Spring 2015 HW8 solutions (continued): 2a) Is the diode forward or reverse biased? Solution: Forward biased because F > F . n p 2b) What is the value of the applied bias? Solution: qV = F − F A n p V = +0.5 V A 2c) What is the bandgap of the semiconductor? Solution: Reading from the graph: E − E = 1.25 eV C V 2d) What is the built-in potential of the junction. Solution: From the plot: V = V −V = 0.25 V j bi A Since: V 0.5 V A = + V = V +V = 0.75 V V = 0.75 V bi j A bi 3) A silicon diode is asymmetrically doped at N 1019 cm-3 and N 1016 cm-3. (Note D = A = that at N 1019 the semiconductor is on the edge of degeneracy, but we can assume D = that non-degenerate carrier statistics are close enough for this problem.) Answer the following questions assuming room temperature. Assume that the minority electron and hole lifetimes are τ = τ = 10−6 s. The lengths of the N and P regions are n p L = 500 µm and L >> x ,x . p n ECE-305 3 Spring 2015 ECE 305 Spring 2015 HW8 Solutions (continued): 3a) Find the zero-bias depletion region capacitance per cm2 of diode area. Solution: The junction capacitance per unit area is: K ε C = s 0 J W V ( A ) The depletion region width for a one-sided junction is: 1/2 ⎡2KSε0 ⎤ W (VA = 0) = ⎢ Vbi ⎥ ⎣ qN A ⎦ The built-in potential is k T ⎛ N N ⎞ V B ln D A bi = ⎜ 2 ⎟ q ⎝ ni ⎠ Putting in numbers, we find: k T ⎛ N N ⎞ ⎛ 10191016 ⎞ V B ln D A 0.026ln 0.90 V bi = ⎜ 2 ⎟ = ⎜ 20 ⎟ = q ⎝ ni ⎠ ⎝ 10 ⎠ 1/2 1/2 ⎡2K ε ⎤ ⎡2 ×11.8 × 8.854 ×10−14 ⎤ W V 0 S 0 V 0.90 3.43 10−5 cm ( A = ) = ⎢ bi ⎥ = ⎢ −19 16 × ⎥ = × ⎣ qN A ⎦ ⎣ 1.6 ×10 ×10 ⎦ (Note that we used in F/cm and N in cm-3 so that the result would come out ε0 A in cm not in meters.) K ε 11.8 × 8.845×10−14 C = s 0 = = 3.05×10−8 F/cm2 J 0 W V = 0 2.8 ×10−5 ( A ) C = 3.05×10−8 F/cm2 J 0 3b) Find the depletion capacitance at V 5 V (reverse biased). A = − Solution: W V 0 Ksε0 Ksε0 ( A = ) Vbi CJ 0 CJ = = × = CJ 0 = W V W V = 0 W V Vbi −VA 1−V V ( A ) ( A ) ( A ) A bi −8 3.05×10 −8 2 CJ = = 1.19 ×10 F/cm < CJ 0 1+ 5 0.9 C V = −5 V = 1.19 ×10−8 F/cm2 J ( A ) Reverse bias decreases the junction capacitance. ECE-305 4 Spring 2015 ECE 305 Spring 2015 HW8 Solutions (continued): 3c) Find the depletion capacitance at V 0.5 V (forward biased). A = + Solution: −8 CJ 0 3.73×10 −8 2 CJ = = = 5.6 ×10 F/cm > CJ 0 1−V V 1− 0.5 0.9 A bi C V = −5 V = 4.58 ×10−8 F/cm2 Forward bias increases the junction J ( A ) capacitance. 4) A silicon diode is asymmetrically doped at N 1019 cm-3 and N 1016 cm-3. (Note D = A = that at N 1019 the semiconductor is on the edge of degeneracy, but we can assume D = that non-degenerate carrier statistics are close enough for this problem.) Assume that the minority electron and hole lifetimes are τ = τ = 10−6 s. The lengths of the N and n p P regions are L = 500 µm and L >> x ,x . p n 4a) Estimate the applied forward bias at which the P-region enters high-level injection. Solution: High injection will occur first on the lightly doped side, the P-side. The maximum excess electron concentration occurs at the beginning of the P-side. From the law of the junction: 2 n qV k T Δn 0 = i e A B −1 ( ) N ( ) A Low level injection means: Δn 0 << N ( ) A Let’s estimate the voltage needed to go into high-level injection, this way 2 n qV k T Δn 0 = i e HL B −1 = N (*) ( ) N ( ) A A A more conservative estimate would be Δn 0 = 0.1N , but (*) is close enough ( ) A for us now. Using (*) as the criteria: 2 k T ⎛ N ⎞ k T ⎛ N ⎞ V = B ln A = 2 B ln A HL q ⎜ n ⎟ q ⎜ n ⎟ ⎝ i ⎠ ⎝ i ⎠ ⎛ 1016 ⎞ V = 2 × 0.026 × ln = 0.781 V HL ⎜ 1010 ⎟ ⎝ ⎠ V = 0.718 V HL ECE-305 5 Spring 2015 ECE 305 Spring 2015 HW8 Solutions (continued): 4b) Compute the current density at the onset of high-injection. Solution: The current density is given by: qV k T J = J e A B −1 0 ( ) From HW8, problem 1): J = 9.1×10−12 A/cm2 0 qV k T J = J e A B −1 = 9.1×10−12 e0.718/0.026 −1 = 9 A/cm2 0 ( ) ( ) J = 9 A/cm2 5) A silicon diode is asymmetrically doped at N 1019 cm-3 and N 1016 cm-3. (Note D = A = that at N 1019 the semiconductor is on the edge of degeneracy, but we can assume D = that non-degenerate carrier statistics are close enough for this problem.) Answer the following questions assuming room temperature. Assume that the minority electron and hole lifetimes are τ = τ = 10−6 s. The lengths of the N and P regions are n p L = 500 µm and L >> x ,x . What is the reverse breakdown voltage of this diode? p n Assume a critical field for breakdown of E = 3×105 V/cm. cr Solution: 1/2 2Vbi 2(Vbi +VR ) ⎡2qN A (Vbi +VR )⎤ E (0) = = 1/2 = W ⎢ K ε ⎥ ⎡2KSε0 ⎤ ⎣ S 0 ⎦ ⎢ (Vbi +VR )⎥ ⎣ qN A ⎦ 1/2 ⎡2qN A (Vbi +VR )⎤ E cr =E (0) = ⎢ K ε ⎥ ⎣ S 0 ⎦ K ε E 2 V = S 0 cr −V BR 2qN bi A k T ⎛ N N ⎞ ⎛ 10191016 ⎞ V B ln D A 0.026ln 0.90 bi = ⎜ 2 ⎟ = ⎜ 20 ⎟ = q ⎝ ni ⎠ ⎝ 10 ⎠ 2 2 11.8 8.854 10−14 3 105 KSε0E cr × × × ( × ) VBR = − 0.90 = −19 16 − 0.90 = 29.4 − 0.90 = 28.5 V 2qN A 2 ×1.6 ×10 ×10 VBR = 28.5 V This assumes a planar junction. If there is junction curvature at the edge of the diode, the breakdown voltage will be lower. ECE-305 6 Spring 2015 .