SOLUTION MANUAL FOR Radio Frequency Integrated Circuits and Systems 2nd Edition by Hooman Darabi ISBN:978-1107194755 COMPLETE GUIDE ALL CHAPTERS COVERED 100% VERIFIED A+ GRADE ASSURED!!!!!NEW LATEST UPDATE!!!!

,1 Chapter One cn




1. Using spherical coordinates, find the capacitance formed by two concentric spherical c
cn cn cn cn cn cn cn cn cn cn cn




onducting shells of radius a, and b. What is the capacitance of a metallic marble with a di
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




ameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
cn cn cn cn cn cn cn cn cn cn cn cn c n cn cn cn




Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge t
cn cn cn cn cn cn cn cn cn cn cn cn cn cn




he same.
cn




-
+

+S - + a + -

b
+
-

From Gauss’s law:
cn cn




ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
cn cn cn cn cn cn


𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏)
cn cn cn cn c n cn cn cn



: 𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟
𝑟
cn cn c n
cn



Assuming a potential of 𝑉0 between the
𝑎 inner
2 and 2 1
outer surfaces, we1have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
cn
1 cn cn cn cn ccnn cn c n cn cn cn cn cn


𝑉 =−
c n
cn
c n
c n cn cn c n c n c n



0 𝑆 ( − ) c n cnc n c n c
n



𝑏 𝑟2 𝜖 𝑎 𝑏 c n c n




Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 c n c n




𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 c n cn cn
c n


−
c n

0
𝜖 𝑎 ( − )
𝑎 𝑏 𝑎 𝑏 cn
cn
cn
cn




1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀 𝑎. Letting 𝜀𝜀0 =
cn cn cn cn cn cn cn cn cn cn cn cn c n cn cn cn
c n c n c n c n ×
36𝜋
0
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
cn
c n cn
cn cn c n cn cn cn c n c n c n


9




2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
cn cn cn cn cn cn cn cn cn cn cn cn




capacitance as a function of the parameters shown in the figure.
cn cn cn cn cn cn cn cn cn cn

, Area: A cn




1




d1
2




d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component of
cn cn cn cn cn cn cn cn cn cn cn cn cn c




the electric flux density has to be equal in each dielectric. That is:
n cn cn cn cn cn cn cn cn cn cn cn cn




𝐷1 = 𝐷𝟐𝟐 c n cn




Accordingly:

𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 c n c n




Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate, the
cn cn cn cn cn cn c n cn cn cn cn cn c n cn cn cn cn c




electric field (or flux has a component only in z direction, and we have:
n cn cn cn cn cn cn cn cn cn cn cn cn cn




𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
cn cn cn cn




If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




𝑑2 𝜌𝑆
𝑑1+𝑑2 cn
−𝜌𝑆 𝑑1+𝑑2 cn
−𝜌 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝜖 𝑆 𝑑𝑧 = 𝜖 𝑑1 + 𝑑2 c n c n

𝜖 𝜖
cn cn cn cn cn cn cn cn cn



𝑑𝑧 cn
c n c n




−
0 0 2 𝑑2 1 1 2

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




= 𝐴 cn



𝑄𝑄 𝑑 𝑑
c n

𝐶 = c n
c n c n



𝑉
0 1 cn
2
+ 𝜖2 cn


𝜖1
which is analogous to two parallel capacitors.
cn cn cn cn cn cn




3. What would be the capacitance of the structure in problem 2 if there were a third conductor
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




with zero thickness at the interface of the dielectrics? How would the electric field lines loo
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




k? How does the capacitance change if the spacing between the top and bottom plates are kept
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




the same, but the conductor thickness is not zero?
cn cn cn cn cn cn cn cn cn

, Solution: If the conductor is perfect, opposite charges are formed on the surface, but the cap
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




acitance remains the same, that is to say, the electric fields terminate to the conductor, but are n
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




ot altered.
cn




If the conductor thickness is greater than zero, but the total distance between the top and botto
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




m plates is the same (𝑑1 + 𝑑2), we expect the capacitance to increase.
cn cn cn cn cn cn cn cn cn cn cn cn cn




4. Repeat problem 2 if the dielectric boundary were placed normal to the two conducting plates a
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




s shown below.
cn cn




d
A1 A2


1 2




Solution: Similar to 2, the electric flux density is in z direction, and we assume a surface c
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




harge density of +𝜌𝑆1/2 for the top plates, and −𝜌𝑆1/2 for the bottom plates. Assuming a pote
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




ntial of 𝑉0 between the plates, unlike 2, as 𝐷 is tangent to the surface, in general 𝐷1 ≠
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn c n




𝐷𝟐𝟐. Thus, we do not assume a uniform charge density on the plates. Furthermore, based on the l
cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn cn




ine integral definition, at the boundary the tangent components of the electric field (which ar
cn cn cn cn cn cn cn cn cn cn cn cn cn cn




e in z direction) must be equal between the two dielectrics, that is:
cn cn cn cn cn cn cn cn cn cn cn cn




𝐸1 = 𝐸𝟐𝟐 c n cn




which yields: cn




𝜌𝑆1 c n
𝜌𝑆2
=
𝜖1 𝜖2

Finally, for the potential the line integral yields:
cn cn cn cn cn cn cn




𝜌𝑆1 𝜌𝑆2
𝑉0 = 𝑑 cn c c n
𝑑
𝜖1 𝜖2
=
n The total charge is: 𝑄𝑄 = 𝜌𝑆1𝐴1 + 𝜌𝑆2𝐴2
cn cn cn cn c n cn cn cn




Consequently:

𝑄𝑄 cn
𝜖1𝐴1 + 𝜖2𝐴2 cn cn


𝐶= cn c n


= 𝑑
𝑉0