# 6.012 Electronic Devices and Circuits

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2 Problem 1 - (34 points) This problem contains 4 independent short problems that can be worked in any order. Page 2 of 1 a) [6 pts] You find a sample of silicon, but the accompanying data sheet has been partially destroyed so you only know that it is n-type, its resistivity is 1 Ohm-cm, and its electron mobility is 16 cm 2 /V-s. You also remember that n i (Si) = 1 1 cm -3. i) What is the approximate equilibrium electron concentration, n o, in this sample? n o = cm -3 ii) What is the approximate equilibrium hole concentration, p o, in this sample? p o = cm -3 b) [1 pts] The excess hole population illustrated on the right is established in an n-type silicon sample, N D = 1 17 cm -3. The minority carrier lifetime in this sample is 1-4 s, the hole mobility is 64 cm 2 /V-s, and the cross-section of the sample is 1-4 cm 2. 5 x 1 14 p (x) [cm-3] i) What is the hole current, i h = A J h, in this sample? 1 2 Note: 1 µm = 1-4 cm i h = Amps ii) What is the total number of excess holes in this sample? Excess holes = Problem 1 continues on the next page

3 Problem 1 continued Page 3 of 1 iii) What is the total rate at which excess holes are recombining with excess electrons in this sample, and what is the corresponding hole current, i h,recomb? Total recombination occurring in sample = holes/s Hole recombination current, i h,recomb = Amps c) [9 pts] Consider two p + -n diodes which are identical except for the fact that in Diode A the minority carrier lifetime is infinite making L h >> w N, while in Diode B the minority carrier lifetime is finite and small enough that L h < w N. Note: w N is the width of the n-type side, and L h is the minority carrier diffusion length. i) Which of these two diodes has the larger saturation current, I S? Explain your answer. Diode A Diode B They are similar Explanation: ii) Which of these two diodes has the larger total number of excess holes in the n- side quasi-neutral region when a forward biased V AB is applied? Diode A Diode B They are similar Explanation: iii) Which of these two diodes has the wider space charge layer (depletion region) in thermal equilibrium? Diode A Diode B They are similar Explanation: Problem 1 continues on the next page

4 Problem 1 continued Page 4 of 1 d) [9 pts] This question concerns the npn bipolar transistor shown below. There is negligible minority carrier recombination throughout except at the ohmic contacts. E Emitter n-si, N DE = 5 x117 cm -3 Base p-si, N AB = 1 x117 cm -3 Collector n-si N DC = 5 x116 cm -3 C B i) On the axes provided, sketch and label the excess minority carrier profiles when both junctions are forward biased with v BE = v BC =.6 V. Notice that this is not a bias in the forward active region. n, p [cm-3] ii) Calculate the forward current gain, β f ( 1/δ E ), for this device when it is biased in the forward active region, i.e. v BE > and v CE. D e = 4 cm 2 /s, D h = 15 cm 2 /s. iii) Redesign this transistor to increase the forward current gain, β f, to 1 by increasing the doping level of one of the three regions in this device. Indicate which region should be changed and to what the new doping level should be. β f = End of Problem 1 Region: N =

5 Problem 2 - (32 points) Page 5 of 1 Consider the silicon diode pictured below. It is 4 µm long, with ohmic contacts at each end, and it is uniform p-type with N A = 1 x 1 17 cm -3 for 1 µm on its far left end and uniform n-type with N D = 1 x 1 17 cm -3 on its far right end. In between these two uniformly doped regions, the net concentration, N d (x) N a (x), slowly grades linearly over a distance of 2 µm from -1 x 1 17 cm -3 on the left to 1 x 1 17 cm -3 on the right, as shown in the lower figure. A p-si, N A = 1x1 17 cm -3 p-si linearly graded n-si linearly graded n-si, N D = 1x1 17 cm -3 B N d (x)-n a (x) [cm-3] 1 x x 117 Note: 1 µm = 1-4 cm a) [6 pts] In thermal equilibrium, what is the electrostatic potential, φ(x), in the lefthand quasi-neutral region at x = -1.5 µm, and what is the electrostatic potential, φ(x), in the right-hand quasi-neutral region at x = +1.5 µm, and what is the built-in potential step, Δφ b, seen transiting from x = -1.5 µm to x = +1.5 µm? Use the 6 mv rule, and log 2 =.3. φ(-1.5 µm) = φ(1.5 µm) = V V Problem 2 continues on the next page Δφ b = V

6 Problem 2 continued Page 6 of 1 b) [4 pts] For the rest of this problem, a bias voltage, V AB, is applied to this diode resulting in a total depletion region width of 1 µm, and x N = x P =.5 µm. On the axes provided below, plot and label the net charge density, ρ(x), for x in the range - 2 µm < x < 2 µm with this bias voltage applied. Use the depletion approximation and assume that the regions outside the depletion region are quasi-neutral, i.e. ρ(x).!(x) [coul/cm3] c) [4 pts] On the axes provided below, plot and label the electric field, E(x), for x in the range - 2 µm < x < 2 µm with the bias voltage V AB applied. E(x) [V/cm] d) [4 pts] What is the change in potential, Δφ, transiting the depletion region when the bias is the same as in Part b, i.e. what is φ(.5 µm) φ(-.5 µm)? Δφ Depl.Reg. = φ(.5 µm) φ(-.5 µm) = Volts Problem 2 continues on the next page

7 Problem 2 continued Page 7 of 1 e) [4 pts] What is the change in potential, Δφ, in transiting the quasi-neutral n-type graded region between x =.5 µm and x = 1. µm, i.e. what is φ(1. µm) φ(.5 µm)? Δφ nqnr. = φ(1. µm) φ(.5 µm) = Volts f) [4 pts] On the axes provided below, plot and label the electrostatic potential, φ(x), for x in the range -2 µm < x < 2 µm with the same bias voltage applied as in Part c. Use the depletion approximation, and assume that the regions outside the depletion region are quasi-neutral. In your plot make φ() = Volts.!(x) [Volts] g) [2 pts] What is the applied bias voltage, V AB? V AB = Volts End of Problem 2

8 Problem 3 (34 points) Page 8 of 1 A p-type semiconductor sample with acceptor concentration N A and length L, illustrated below, has ohmic contacts at both its ends. A light source generates M A electron-hole pairs/cm 2 -s in the plane at x = X A, i.e. g L (x) = M A δ(x A ). Assume low-level injection and quasi-neutrality everywhere in the bar. g L (x) = M A!(X A ) p-type N A X A L x The general equation governing the excess minority carriers in a uniformly doped material is d 2 n'(x) " n'(x) = " 1 g dx 2 2 L (x) L e D e a) [4 pts] What boundary condition is imposed on the excess minority carriers n at x = and x = L? Boundary condition at x = : Boundary condition at x = L: b) [4 pts] We now make the assumption that the minority carrier lifetime is very long, which simplifies the general equation to: d 2 n'(x) dx 2 " # 1 D e g L (x) What quantitative restriction is placed on the minority carrier lifetime, τ e, for this assumption to be valid? τ e >> << (circle one) Problem 3 continues on the next page (fill in the blank)

9 Problem 3 continued Page 9 of 1 c) [6 pts] Using the long-lifetime approximation in part (b), determine two constraints (i.e. boundary conditions) on the excess minority carriers at x = X A, i.e. relating n (X A ) to n (X A + ). Constraint on n at x = X A : (Hint: Finite currents imply finite dn /dx.) Constraint on dn /dx in the vicinity of x = X A : (Hint: What goes in must come out.) d) [6 pts] On the axes provided sketch the excess minority carrier concentration, n (x), everywhere inside the semiconductor. Label your sketch with the relevant equations for n (x). n (x) X A L x e) [6 pts] On the axes provided sketch the minority carrier diffusion current, J e,diff (x), everywhere inside the semiconductor. Label your sketch with the relevant equations for J e,diff (x). J e,diff (x) X A L x Problem 3 continues on the next page

10 Problem 3 continued Page 1 of 1 f) [4 pts] In the space below, briefly explain (approx. 25 words or less) why the minority carrier diffusion is the dominant minority carrier current. g) [4 pts] A second light source is added illuminating a single spot along the semiconductor at x = X B, where x B > X A, and generating electron-hole pairs at a rate M B, so that g L (x) is now g L (x) = M A "(X A ) + M B "(X B ). Find n (x) and J e,diff (x) everywhere inside the semiconductor under this new illumination condition. If you could not do Parts d and e, indicate how you would use the results of those parts to answer this question. n (x) = J e,diff (x) = End of Problem 3 End of Exam One

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