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SOLUTION MANUAL FOR RADIO FREQUENCY INTERGRATED CIRCUITS SYSTEMS BY HOOMAN DARABI LATEST UPDATE

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SOLUTION MANUAL FOR RADIO FREQUENCY INTERGRATED CIRCUITS SYSTEMS BY HOOMAN DARABI LATEST UPDATE 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. From Gauss’s law: ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎 2 𝑆 Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏): 𝑎 2 𝐷 = 𝜌𝑆 𝑟 2 𝑎𝑟 Assuming a potential of 𝑉0 between the inner and outer surfaces, we have: 𝑉0 = − � 𝑎 1 𝜌𝑆 𝑎 2 𝑑𝑟 = 𝜌𝑆 𝑎 2 ( 1 − 1 ) Thus: 𝑏 𝜖 𝑄𝑄 𝑟 2 𝜖 𝜌𝑆4𝜋𝑎 2 𝑎 𝑏 4𝜋𝜖 𝐶 = 𝑉 = 𝜌𝑆 1 1 = 1 1 0 𝜖 𝑎 2 ( 𝑎 − 𝑏 ) 𝑎 − 𝑏 In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 = 1 × 36𝜋 10−9 , and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 5 𝑝𝐹 = 0.55𝑝𝐹. 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. 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 ad bottom plates is 𝑉0, based on the line integral we obtain: 𝑑1+𝑑2 𝑉0 = − � 𝐸. 𝑑𝑧 = − � 𝑑2 −𝜌𝑆 𝜖 𝑑𝑧 − � 𝑑1+𝑑2 −𝜌𝑆 𝜖 𝜌𝑆 𝑑𝑧 = 𝜖 𝜌𝑆 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 𝐶 = 𝑉 𝐴 = 𝑑 𝑑 0 𝜖 1 + 𝜖 2 1 2 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? 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.

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