EE 42 Homework and Reading No. 7
Homework assigned Thursday 10 March 2005; due at 5:30 pm Thursday 17 March 2005
in the box labeled EE 42 in Room 240 Cory Hall. Note: your graded homeworks will be
returned in your discussion section; your GSI will keep your homework until you pick it
up to maintain privacy.
Reading: Rabaey et al., 3.2.1 to 3.2.2;
Hambley 10.1 to 10.4; Sec. 10-6 pp. 460-462
Problems:
1. Indium-tin oxide (“ITO”) is a very unusual and useful thin-film material: it is both
quite conductive and optically transparent! It is used to coat automobile windshields for
defrosting, liquid-crystal display panels, and so on. The typical composition is 90%
In2O3 and 10% SnO2. Google gives typical properties as follows: For thickness from
120 to 160 nm, it has sheet resistances between 8 and 12 ohms/square; its transparency in
the visible is >83%; its index of refraction 1.95, and the melting point is 1900oC.
a. What is the average resistivity of this ITO? (Hint: you have two values of
thickness and two values of sheet resistance. Which goes with which?)
b. What fraction is the conductivity of this material compared with that of copper?
(You can find from Google that the resistivity of copper is 1.7 x 10-6 -cm.)
c. Suppose you want a 96 transparent resistor made of 12-ohm/square ITO. Show
two different designs indicating the thickness, width and length of each
transparent resistor.
Problems from Hambley:
2. P10.7. (basic diode concepts). Nore: text should read “Assume the turn-on
voltage is 0.6 V for all diodes including the Zener diode in its forward-bias
region”.
3. P10.24 (Zener diode used as a voltage regulator)
4. P10.31 (ideal diodes connected and pointing in opposite directions)
5. P10.32 (analyzing diode circuits using assume states – diodes conducting or not)
6. P10.35 (i-v curves for various diode circuits)
7. P10.43 (rectifier with single diode and a resistor) Hints: Read Hambley Sec.
10.6, pp. 460-462. See p. 462 for definition of the peak inverse voltage – which is
the reverse voltage the diode must be able to withstand in operation. Also recall
the relation between root-mean-square and peak voltage: Vroot-mean-square = Vrms =
V peak / 2