NAME:______________________________________ PUID: ______________________________________ SOLUTIONS ECE 606 Exam 4: Spring 2013 March 28, 2013 Mark Lundstrom Purdue University This is a closed book exam. You may use a calculator and the formula sheet at the end of this exam. There are four equally weighted questions. To receive full credit, you must show your work (scratch paper is attached). The exam is designed to be taken in 60 minutes, but you may use the full, 75 minute class period. Be sure to fill in your name and Purdue student ID at the top of the page. DO NOT open the exam until told to do so, and stop working immediately when time is called. 40 points possible, 10 per question 1) 2 points for each part – 10 points total 2) 2 points for each part – 10 points total 3a) 4 points 3b) 2points 3c) 2 points 3d) 2 points 4) 2 points for each part – 10 points total ECE-­‐606 1 Spring 2013 Exam 4: ECE 606 Spring 2013 Answer the five multiple choice questions below by choosing the one, best answer. 1) What can we determine from a plot of one over junction capacitance squared vs. reverse bias voltage? a) The minority carrier lifetime. b) The minority carrier diffusion coefficient. c) The semiconductor bandgap. d) The doping density. e) The series resistance. 2) When is the diffusion capacitance important? a) For low DC forward bias and for frequencies that are well below 1 2!" n . ( )
b) For low DC forward bias and for frequencies that are well above 1 ( 2!" ) . c) For high DC forward bias and for frequencies that are well below 1 ( 2!" ) d) For high DC forward bias and for frequencies that are above 1 ( 2!" ) . e) For low DC forward bias and for frequencies that are well below 1 ( 2!" ) . n
n
n
n
3)
4) There are two PN junctions in a bipolar transistor. Each one can be either forward biased (FB) or reverse biased (RB). In the “saturation” region of operation, how are these two junctions biased? a) Emitter-­‐base is FB and base-­‐collector is FB. b) Emitter-­‐base is FB and base-­‐collector is RB. c) Emitter-­‐base is RB and base-­‐collector is FB. d) Emitter-­‐base is RB and base-­‐collector is RB. e) Emitter-­‐base is zero-­‐biased and base-­‐collector is zero-­‐biased. The increase in collector current as the magnitude collector voltage increases is known as the Early effect and is due to what physics? a) Quantum mechanical tunneling. b) Impact ionization. c) SRH recombination d) The decreasing width of the quasi-­‐neutral base. e) Generation in the space charge region. 5) To achieve a high common emitter current gain, ! DC , the base of an NPN silicon BJT must be doped much less heavily than the emitter. Why? a) To minimize recombination in the emitter-­‐base junction. b) The minimize recombination in the base. c) To minimize current crowding in the base. d) To put the base in high-­‐level injection. e) To increase the ratio of the electron current injected into the base and the hole current injected into the emitter. ECE-­‐606 2 Spring 2013 Exam 4: ECE 606 2) Consider an NPN BJT with the following parameters: I Ep = 0.01 mA I En = 2 mA I Cn = 1.98 mA Spring 2013 I Cp = 0.0001 mA 2a) Determine the base transport factor, ! T . Solution: I
1.98
! T = 0.99 ! T = Cn =
= 0.99 I En 2.00
2b) Determine the emitter injection efficiency, ! Solution: I En
2.00
! = 0.995 ! =
=
= 0.995 I En + I Ep 2.00 + 0.01
2c) Determine the current, I E Solution: I E = I En + I Ep = 2.00 + 0.01 = 2.01 I E = 2.01 mA 2d) Determine ! DC . Solution: ! DC = " ! T = 0.995 # 0.99 = 0.985 ! DC = 0.985 2e) Determine ! DC . Solution: ! DC =
ECE-­‐606 " DC
0.985
=
= 65.9 1# " DC 1# 0.985
! DC = 65.9 3 Spring 2013 Exam 4: ECE 606 3) Spring 2013 Sketch the minority hole concentration in the base of a PNP BJT for the following conditions. In each case, indicate the magnitude of !p at x = xn and at x = xn + WB . 3a) Forward active region. 3b) Inverted active region. ECE-­‐606 4 Spring 2013 Exam 4: ECE 606 3c) Saturation Spring 2013 3d) Cut-­‐off ECE-­‐606 5 Spring 2013 4) Exam 4: ECE 606 Spring 2013 Consider the impact of changing some key BJT device parameters on some key figures of merit. For each of the 5 changes below, explain in words whether the change increases, decreases, or has no effect on the current gain, ! DC . Assume negligible recombination in the base and emitter. 6a) Double the width of the quasi-­‐neutral base. Solution: I
N D W
! = C = DE n E I B N AB D p WB
Double the base width, the beta drops by a factor of 2. This is because the diffusion gradient of electrons in the base drops by a factor of 2, so the collector current drops by the same factor, but the base current is unaffected. 6b) Double the base lifetime. Solution: I
N D W
! = C = DE n E I B N AB D p WB
No effect. Since the assumption is no recombination, the base lifetime has no effect. Making it bigger will have no effect. 6c) Double the base doping. Solution: I
N D W
! = C = DE n E I B N AB D p WB
Double the base doping, the beta drops by a factor of 2. This is because the electron density at the beginning of the base, as given by the Law of the Junction, will drop by a factor of two, because it goes as ni2 N AB . ECE-­‐606 6 Spring 2013 Exam 4: ECE 606 Spring 2013 6d) Double the width of the quasi-­‐neutral emitter. Solution: I
N D W
! = C = DE n E I B N AB D p WB
Double the emitter width, the beta increases by a factor of 2. This is because the gradient of minority carrier holes in the emitter, which determines the back-­‐
injected hole current (the base current), is reduced by a factor of 2. 6e) Double the emitter doping. Solution: I
N D W
! = C = DE n E I B N AB D p WB
Double the emitter doping, the beta increases by a factor of 2. This is because the hole density at the beginning of the emitter, as given by the Law of the Junction, will drop by a factor of two, because it goes as ni2 N DE . ECE-­‐606 7 Spring 2013 ECE-­‐606 SCRATCH PAPER 8 Spring 2013 ECE-­‐606 SCRATCH PAPER 9 Spring 2013