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The ground rules and three questions from the midterm 1 exam for the eecs 140 course at uc berkeley, taught by professor r.t. Howe, in the fall of 1990. The exam covers topics such as mosfet characteristics, potential in thermal equilibrium, and pn junction diodes.
Typology: Exams
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U.C Berkeley
EECS 140 Midterm 1: October 8, 1990 Fall 1990 Professor R.T. Howe
Ground Rules:
Closed Book and Notes Do all work on exam pages You have 50 minutes; use your time wisely
QUESTION 1. MOS Inverter [15 points]
(picture 1)
Non-linear iL versus vl characteristics of load device.
iL = kL* squareroot ( vL ) where kL = 800 micro * A * V -1/
(picture 2)
Output characteristics of the MOSFET. The constant mu sub n * Cox (W / L ) = 500 micro * A * V -1/
a.) [5 points] Find an equation relating vo to vI which is valid when the MOSFET is in the triode region.
b.) [5 points] Find an equation relating vo to vI which is valid when the MOSFET is saturated.
c.) [5 points] Using the graphical load line technique, plot the transfer curve vo versus vI on the graph below, using the given current-voltage characteristics of the MOSFET. Label on your plot the points on the transfer curve which mark the boundaries between the cutoff, saturation, and triode regions of operation.
picture 4
QUESTION 2 [17 points]
Potential in Thermal Equilibrium
a.) 6 points Consider an n-type sample with the donor concentration varying as shown in the log-linear plot below. In thermal equilibrium, plot the variation in potential phi (x) for 0 < x < 3 micro metres on the plot below.
c.) [5 points]
Consider a sample which is doped with the superposition of the donor and acceptor concentrations from part a and part b, as shown in the log-linear plot below. In thermal equilibrium, sketch the variation potential phi (x) for 0 < x < 3 micro metres on the plot below. Hint: the width of the deletion region is 1 micro meter
picture 7
picture 5
pn junction diode
Given : pn junction diode with cross sectional area of 10 * 10 -6^ cm 2
p side doping: Na = 2 * 10^16 cm - Nd = 0
n side doping: Na = 1 * 10^16 cm - Nd = 0
minority carrier properties:
Dn = 25 cm 2 s- Taun = 400 ns = .4 micro seconds
Dp = 25 cm 2 s- Taun = 10 microseconds (translators note: Yes the exam redefines tau???)
miscellaneous
kT/q = 26mV ni = 1 * 10^10 cm -
a.) [7 points] Plot the minority carrier concentrations on the linear graphs below for the case of forward bias VD = 0.6 V
picture 8