Radio Frequency Integrated Circuits and Systems Hooman Darabi Solution Manual Complete RFIC Design Study Guide 2025/ 2026 with Solution

Radio Frequency Integrated Circuits and Systems Solution Manual Hooman
Radio Frequency
Darabi
1 Integrated Circuits and Systems Solution Manual Hooman Darabi.pdf




Radio Frequency Integrated
Circuits and Systems
Solution Manual

Hooman Darabi




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Radio Frequency
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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.




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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 𝑏𝑏 → ∞, thus, 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎 = 0.55𝑝𝑝𝑝𝑝.

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


-
+

+ρS - + a + -

b
+
-

From Gauss’s law:
�𝑫𝑫 ⋅ 𝑑𝑑𝑺𝑺 = 𝑄𝑄 = +𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2
𝑆𝑆
Thus, inside the sphere (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏):
𝑎𝑎2
𝑫𝑫 = 𝜌𝜌𝑆𝑆
𝒂𝒂
𝑟𝑟 2 𝒓𝒓
Assuming a potential of 𝑉𝑉0 between the inner and outer surfaces, we have:
𝑎𝑎
1 𝑎𝑎2 𝜌𝜌𝑆𝑆 1 1
𝑉𝑉0 = − � 𝜌𝜌𝑆𝑆 2 𝑑𝑑𝑟𝑟 = 𝑎𝑎2 ( − )
𝑏𝑏 𝜖𝜖 𝑟𝑟 𝜖𝜖 𝑎𝑎 𝑏𝑏
Thus:
𝑄𝑄 𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2 4𝜋𝜋𝜖𝜖
𝐶𝐶 = = =
𝑉𝑉0 𝜌𝜌𝑆𝑆 𝑎𝑎2 (1 − 1) 1 − 1
𝜖𝜖 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏
1
In the case of a metallic marble, 𝑏𝑏 → ∞, and hence: 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎. Letting 𝜀𝜀0 = 36𝜋𝜋 ×
5
10−9 , and 𝑎𝑎 = 0.5𝑐𝑐𝑐𝑐, it yields 𝐶𝐶 = 9 𝑝𝑝𝑝𝑝 = 0.55𝑝𝑝𝑝𝑝.



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




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Area: A



ε1




d1
ε2




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:

𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐

Accordingly:

𝜖𝜖1 𝑬𝑬𝟏𝟏 = 𝜖𝜖2 𝑬𝑬𝟐𝟐

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

𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐 = −𝜌𝜌𝑆𝑆 𝒂𝒂𝒛𝒛

If the potential between the top ad bottom plates is 𝑉𝑉0, based on the line integral we obtain:
𝑑𝑑1 +𝑑𝑑2 𝑑𝑑2 𝑑𝑑1 +𝑑𝑑2
−𝜌𝜌𝑆𝑆 −𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆
𝑉𝑉0 = − � 𝑬𝑬. 𝑑𝑑𝒛𝒛 = − � 𝑑𝑑𝑑𝑑 − � 𝑑𝑑𝑑𝑑 = 𝑑𝑑1 + 𝑑𝑑2
0 0 𝜖𝜖2 𝑑𝑑2 𝜖𝜖1 𝜖𝜖1 𝜖𝜖2

Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆 𝐴𝐴, the capacitance is found to be:

𝑄𝑄 𝐴𝐴
𝐶𝐶 = =
𝑉𝑉0 𝑑𝑑1 + 𝑑𝑑2
𝜖𝜖1 𝜖𝜖2

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?




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