Radio Frequency Integrated Circuits and Systems (2nd Edition) by Hooman Darabi – Complete Solution Manual All Chapters

RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION BY HOOMAN DARABI RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION
BY HOOMAN DARABI




SOLUTION MANUAL FOR RADIO FREQUENCY
INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION BY
HOOMAN DARABI |ALL CHAPTERS |LATEST VERSION

,RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION BY HOOMAN DARABI RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION
BY HOOMAN DARABI




Solutions to Problem Sets
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem
sets depict examples of practical applications of the concepts described in the book, more
detailed analysis of some of the ideas, or in some cases present a new concept.

Note that selected problems have been given answers already in the book.

,RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION BY HOOMAN DARABI RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION
BY HOOMAN DARABI




1 Chapter One
1. Using spherical coordinates, find the capacitance formed by two concentric spherical
conducting shells of radius a, and b. What is the capacitance of a metallic marble with a
diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.

Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge
the same.


-
+

+S - + a + -

b
+
-

From Gauss’s law:
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏):
𝑎2
𝑎𝑟 𝐷 = 𝜌𝑆
𝑟2
Assuming a potential of 𝑉0 between the inner and outer surfaces, we have:
𝑎 2 1 1
𝑉0 = − � 1 𝜌𝑆 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎2( − )
𝑏 𝜖
𝑟2 𝜖 𝑎 𝑏
Thus:
𝑄𝑄 𝜌𝑆4𝜋𝑎2 4𝜋𝜖
𝐶=𝑉 = 1 1 = 1 1
𝜌𝑆 2
0 −
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 𝑏
×
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 = 1
36𝜋
5 𝑝𝐹 = 0.55𝑝𝐹.
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 =
9




2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.

, RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION BY HOOMAN DARABI RADIO FREQUENCY INTEGRATED CIRCUITS AND SYSTEMS 2ND EDITION
BY HOOMAN DARABI




Area: A



1




d1
2




d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component
of the electric flux density has to be equal in each dielectric. That is:

𝐷1 = 𝐷𝟐𝟐

Accordingly:

𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐

Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate, the
electric field (or flux has a component only in z direction, and we have:

𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧

If the potential between the top and bottom plates is 𝑉0, based on the line integral we obtain:
𝑑1+𝑑2 𝑑2
−𝜌𝑆 𝑑1+𝑑2 −𝜌
𝑆 𝜌𝑆 𝜌𝑆
𝑉0 = − � 𝐸. 𝑑𝑧 = − � 𝜖 𝑑𝑧 − � 𝜖 𝑑𝑧 = 𝜖 𝑑1 + 𝜖 𝑑2
0 0 2 𝑑2 1 1 2

Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
𝑄𝑄 𝐴
𝐶=𝑉 = 𝑑 𝑑
0 1 2
𝜖1 + 𝜖2
which is analogous to two parallel capacitors.



3. What would be the capacitance of the structure in problem 2 if there were a third conductor
with zero thickness at the interface of the dielectrics? How would the electric field lines
look? How does the capacitance change if the spacing between the top and bottom plates are
kept the same, but the conductor thickness is not zero?