Complete Solutions Manual — Radio Frequency Integrated Circuits and Systems, 2nd Edition — Hooman Darabi, 2020 — ISBN 9781107194755 — All Chapters Covered 1-12 (Latest Update)

Radio Frequency Integrated Circuits
and Systems – 2nd Edition


SOLUTIONS
ST
UV


MANUAL
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PP

Hooman Darabi
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Comprehensive Solutions Manual for
VE

Instructors and Students
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© Hooman Darabi

All rights reserved. Reproduction or distribution without permission is prohibited.




©MedConnoisseur

, 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.
ST
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.
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-
+

+S
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- + a + -

b
+
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-

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

, Area: A
ST
1




d1
2




d2
UV
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
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Accordingly:

𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐
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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
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If the potential between the top ad 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
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Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
𝑄𝑄 𝐴
𝐶= 𝑉 = 𝑑 𝑑
1 2
𝜖1 + 𝜖 2
0
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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
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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?

, Solution: If the conductor is perfect, opposite charges are formed on the surface, but the
capacitance remains the same, that is to say, the electric fields terminate to the conductor, but
are not altered.
If the conductor thickness is greater than zero, but the total distance between the top and
bottom plates is the same (𝑑1 + 𝑑2), we expect the capacitance to increase.
ST
4. Repeat problem 2 if the dielectric boundary were placed normal to the two conducting plates
as shown below.
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d
A1 A2


1 2
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Solution: Similar to 2, the electric flux density is in z direction, and we assume a surface
charge density of +𝜌𝑆1/2 for the top plates, and −𝜌𝑆1/2 for the bottom plates. Assuming a
potential of 𝑉0 between the plates, unlike 2, as 𝐷 is tangent to the surface, in general 𝐷1 ≠
𝐷𝟐𝟐. Thus, we do not assume a uniform charge density on the plates. Furthermore, based on
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the line integral definition, at the boundary the tangent components of the electric field
(which are in z direction) must be equal between the two dielectrics, that is:

𝐸1 = 𝐸𝟐𝟐
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which yields:
𝜌𝑆1 𝜌𝑆2
=
𝜖1 𝜖2

Finally, for the potential the line integral yields:
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𝜌𝑆1 𝜌𝑆2
𝑉0 = 𝑑= 𝑑
𝜖1 𝜖2

The total charge is: 𝑄𝑄 = 𝜌𝑆1𝐴1 + 𝜌𝑆2𝐴2
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Consequently:

𝑄𝑄 𝜖1𝐴1 + 𝜖2𝐴2
𝐶= =
𝑉0 𝑑