This is the problem I need help on. The answers are: Problem 10, for reference:

Question

This is the problem I need help on. The answers are:

Problem 10, for reference: In a dynamic random access memory computer chip, each memory cell chiefly consists of a capacitor for charge storage. Each of these cells represents a single binary-bit value of 1 when its 35fF (1fF = 1*10^-15F) is charged at 1.5V or 0 when uncharged at 0V.

(a) When it is fully charged, how many excess electrons are on a cell capacitor's negative plate?
(b) After a charge has been placed on a cell capacitor's plate, it slowly "leaks" off at a constant rate of 0.30fC/s. How long does it take for the potential difference across this capacitor to decrease by 1.0% from its fully charged value?

(II) In the DRAM computer chip of Problem 10, the cell capacitor's two conducting parallel plates are separated by a 2.0-nm thick insulating material with dielectric constant K =25. (a) Determine the area A (in m 2) of the cell capacitor's plates. (b) In (older) "planar" designs, the capac- itor was mounted on a silicon-wafer surface with its plates parallel to the plane of the wafer. Assuming the plate area A accounts for half of the area of each cell, estimate how many megabytes of memory can be placed on a 3.0-cm silicon wafer with the planar design? (1 byte = 8 bits.)

Explanation / Answer

part a:

given that capacitance=35 *10^(-15) F

distance between plates=2nm=2*10^(-9) m

dielectric constant=25

as we know, for a parallel plate capacitor, capacitance=dielectric constant*electrical permitivity*area/distance between plates

==>35*10^(-15)=25*8.85*10^(-12)*A/(2*10^(-9))

==>A=3.16384*10^(-13) m^2

as 1 m=10^6 um

1 m^2=10^12 um^2

hence A=0.316384 um^2=0.32 um^2

part b:

as A is half of area of each cell, area of a cell=2*A=0.64 um^2

then number of cells=3 cm^2/0.64 um^2

=(3*10^(-4))/(0.64*10^(-12))=4.6875*10^8

as 1 cell=1 bit

number of bits=4.6875*10^8 bits

=(4.6875*10^8/8) bytes

=58593750 bytes

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