Download EE 105, Midterm 2, Fall 1996 Answers: Electrical Engineering Exam and more Exams Analysis and Design of Digital Integrated Circuits in PDF only on Docsity! EE 105, Midterm 2, Fall 1996 Answers Ground Rules: โ Close book; one 8.5x11 crib sheet (both sides) โ Do all work on exam pages โ Default bipolar transistor parameters: โ npn: รn=100, VAn=50 V, VCE-sat=0.2 V โ pnp: รp=50, VAp=25 V, VEC-sat=0.2 V โ Default MOS transistor parameters: note LAMBDA depends on L! โ NMOS: MUn Cox=100e-6 A/V2, LAMBDAn=[0.1/L] V-1 (L in micrometers) VTp=1 V โ PMOS: MUp Cox=50e-6 A/V2, LAMBDAp=[0.1/L] V-1 (L in micrometers) VTp=-1 V Problem #1: Small-Signal Amplifier [24 points] file:///C|/Documents%20and%20Settings/Jason%20Raft...20-%20Fall%201996%20-%20Howe%20-%20Midterm%202.htm (1 of 6)1/27/2007 4:43:32 PM EE 105, Midterm 2, Fall 1996 Answers a) [4 pts.] What is width of transistor M2 such that the DC output voltage Vout=0 V for VIN=0 V. Given: the length of M2 is L2=2e-6 m. W=100 micrometers b) [4 pts.] What is the numerical value of the input resistance Rin of this amplifier? Your answer should be correct to within +/- 5%. If you couldn't solve (a) you can assume for this part that W2=25e-6 m. Of course, this isn't the correct answer to part (a). Rin=14.31 mega-ohms c) [4 pts.] What is the numerical value of the output resistance Rout of this amplifier? Your answer should be correct to within +/- 5%. If you couldn't solve (a) you can assume for this part that W2=25e-6 m. Of course, this isn't the correct answer to part (a). Rout=260 ohms d) [6 pts.] What is the numerical value of the overall voltage gain vout/vs, with Rs=100 kilo-ohms and RL=20 kilo-ohms? Your answer should be correct to within +/- 5%. Again, If you couldn't solve (a) you can assume for this part that W2=25e-6 m. Of course, this isn't the correct answer to part (a). Av-overall=0.9804 e) [6 pts.] Sketch the transfer curve VOUT versus VIN for -2.5 <= VIN <= +2.5 V on the graph below. For this part, RL is infinity and RS=0 V. Hint: you should note that the current supplies each require at least VSUP(min)=0.5 V in order to function. file:///C|/Documents%20and%20Settings/Jason%20Raft...20-%20Fall%201996%20-%20Howe%20-%20Midterm%202.htm (2 of 6)1/27/2007 4:43:32 PM EE 105, Midterm 2, Fall 1996 Answers NOTE: The default npn transistors do not apply for this problem! GIVEN: NdE=2x1018 cm-3 NdB=1017 cm-3 NdC=1016 cm-3 The base and emitter widths are WB=WE=0.5 micrometers. The electron diffusion coefficient in the base is DnB=10 cm2/s and the hole diffusion coefficient in the emitter is DpE=5 cm2/s. a) [3 pts.] Qualitatively sketch the minority carrier concentrations in the emitter, base and collector on the graph below, assuming that the transistor is biased in the forward active region. file:///C|/Documents%20and%20Settings/Jason%20Raft...20-%20Fall%201996%20-%20Howe%20-%20Midterm%202.htm (5 of 6)1/27/2007 4:43:32 PM EE 105, Midterm 2, Fall 1996 Answers b) [3 pts.] For VOUT=0 V what is the numerical value of the minority electron concentration at x=0, npB (0)? You can assume that the transistor is biased in the forward active region. Not available c) [3 pts.] What is the numerical value of the base current IB for the bias condition in part b)? If you couldn't solve b) assume for this part that npB(0) = 1015 cm- 3 -- not the correct answer to b), of course. IB=2.5 micro-amps d) [3 pts.] What is the numerical value of VIN in order that the transistor is biased in the forward active region with VOUT=0 V? Notes: You cannot assume that VBE=0.7 V for this part. If you couldn't solve parts b) and c) you can assume that npB(0)=1015 cm-3 and that IB=4 micro-amps. Neither of these answers are correct, of course. VIN=3.277 V Posted by HKN (Electrical Engineering and Computer Science Honor Society) University of California at Berkeley If you have any questions about these online exams please contact mailto:examfile@hkn.eecs.berkeley.edu file:///C|/Documents%20and%20Settings/Jason%20Raft...20-%20Fall%201996%20-%20Howe%20-%20Midterm%202.htm (6 of 6)1/27/2007 4:43:32 PM